[ { "images": [ "sample_100_images/kWOTmyoaWJg@561.98_572.0400000000001#1.jpg", null, "sample_100_images/kWOTmyoaWJg@573.2800000000001_585.2800000000001#1.jpg", null, "sample_100_images/kWOTmyoaWJg@587.5400000000001_608.74#1.jpg", null, "sample_100_images/kWOTmyoaWJg@608.9000000000001_632.68#1.jpg", null, null, "sample_100_images/kWOTmyoaWJg@636.14_661.8800000000001#1.jpg", null, null, "sample_100_images/kWOTmyoaWJg@677.0400000000001_683.2#1.jpg", null ], "texts": [ null, " And next we're going to divide both sides by four so twelve divided by four is three therefore x is equal to three", null, " And so, you want to make sure that you know how to perform this cross-multiplication technique. You can only use it if you have two fractions separated by an equal sign.", null, " Now let's move on to our next example. Let's say that we have 3x plus 2 equals 14. What is the value of x? So here we have a multi-step equation and the best way to solve it is to subtract 2 from both sides to begin with. Now the twos will cancel.", null, " On the right side, we have 14 minus 2, which is 12. Then after that, divide by 3. So, 12 divided by 3 is 4. So x is 4. Now, if we check our work: 3 times 4 plus 2, is that equal to 14? 3 times 4 is 12. 12 plus 2 is 14. So the equation is verified.", " X is equal to 4.", null, " So this is going to be the last example for this video. Go ahead and solve the equation. In this example, what we need to do is add three to both sides. Seventeen plus three is twenty. So five x is equal to twenty. Next, let's divide both sides by five. Twenty divided by five is four. And so that's going to be the solution to this equation.", " So now we have x minus 4 plus 2", null, " And now we have x minus 4 plus 2. And that's going to be the solution to this equation." ], "text_ocr_list": [ null, "We can see these text from the image: 12/4 = 4\n\n12/3 = 4.\n And next we're going to divide both sides by four so twelve divided by four is three therefore x is equal to three", null, "We can see these text from the image: 12/4 = 3\n\n12/4 = 4/x\n\n3 = x.\n And so, you want to make sure that you know how to perform this cross-multiplication technique. You can only use it if you have two fractions separated by an equal sign.", null, "We can see these text from the image: \\( 3 \\times \\).\n Now let's move on to our next example. Let's say that we have 3x plus 2 equals 14. What is the value of x? So here we have a multi-step equation and the best way to solve it is to subtract 2 from both sides to begin with. Now the twos will cancel.", null, "We can see these text from the image: 3x + 2 = 14\n-2 -2\n3x = 12.\n On the right side, we have 14 minus 2, which is 12. Then after that, divide by 3. So, 12 divided by 3 is 4. So x is 4. Now, if we check our work: 3 times 4 plus 2, is that equal to 14? 3 times 4 is 12. 12 plus 2 is 14. So the equation is verified.", " X is equal to 4.", null, "We can see these text from the image: 5x - 3 =.\n So this is going to be the last example for this video. Go ahead and solve the equation. In this example, what we need to do is add three to both sides. Seventeen plus three is twenty. So five x is equal to twenty. Next, let's divide both sides by five. Twenty divided by five is four. And so that's going to be the solution to this equation.", " So now we have x minus 4 plus 2", null, " And now we have x minus 4 plus 2. And that's going to be the solution to this equation." ], "metadata": [ { "vid": "kWOTmyoaWJg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/kWOTmyoaWJg/561.98_572.0400000000001.mp4", "refined_asr": " And next we're going to divide both sides by four so twelve divided by four is three therefore x is equal to three", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@561.98_572.0400000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@561.98_572.0400000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@561.98_572.0400000000001#2.jpg" ], "ocr_qwen2_vl_72b": "12/4 = 4\n\n12/3 = 4" }, { "vid": "kWOTmyoaWJg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/kWOTmyoaWJg/573.2800000000001_585.2800000000001.mp4", "refined_asr": " And so, you want to make sure that you know how to perform this cross-multiplication technique. You can only use it if you have two fractions separated by an equal sign.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@573.2800000000001_585.2800000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@573.2800000000001_585.2800000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@573.2800000000001_585.2800000000001#2.jpg" ], "ocr_qwen2_vl_72b": "12/4 = 3\n\n12/4 = 4/x\n\n3 = x" }, { "vid": "kWOTmyoaWJg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/kWOTmyoaWJg/587.5400000000001_608.74.mp4", "refined_asr": " Now let's move on to our next example. Let's say that we have 3x plus 2 equals 14. What is the value of x? So here we have a multi-step equation and the best way to solve it is to subtract 2 from both sides to begin with. Now the twos will cancel.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@587.5400000000001_608.74#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@587.5400000000001_608.74#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@587.5400000000001_608.74#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@587.5400000000001_608.74#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@587.5400000000001_608.74#4.jpg" ], "ocr_qwen2_vl_72b": "\\( 3 \\times \\)" }, { "vid": "kWOTmyoaWJg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/kWOTmyoaWJg/608.9000000000001_632.68.mp4", "refined_asr": " On the right side, we have 14 minus 2, which is 12. Then after that, divide by 3. So, 12 divided by 3 is 4. So x is 4. Now, if we check our work: 3 times 4 plus 2, is that equal to 14? 3 times 4 is 12. 12 plus 2 is 14. So the equation is verified.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@608.9000000000001_632.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@608.9000000000001_632.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@608.9000000000001_632.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@608.9000000000001_632.68#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@608.9000000000001_632.68#4.jpg" ], "ocr_qwen2_vl_72b": "3x + 2 = 14\n-2 -2\n3x = 12" }, { "vid": "kWOTmyoaWJg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/kWOTmyoaWJg/632.68_636.14.mp4", "refined_asr": " X is equal to 4.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "kWOTmyoaWJg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/kWOTmyoaWJg/636.14_661.8800000000001.mp4", "refined_asr": " So this is going to be the last example for this video. Go ahead and solve the equation. In this example, what we need to do is add three to both sides. Seventeen plus three is twenty. So five x is equal to twenty. Next, let's divide both sides by five. Twenty divided by five is four. And so that's going to be the solution to this equation.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@636.14_661.8800000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@636.14_661.8800000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@636.14_661.8800000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@636.14_661.8800000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@636.14_661.8800000000001#4.jpg" ], "ocr_qwen2_vl_72b": "5x - 3 =" }, { "vid": "kWOTmyoaWJg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/kWOTmyoaWJg/674.5600000000001_677.0400000000001.mp4", "refined_asr": " So now we have x minus 4 plus 2", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@674.5600000000001_677.0400000000001#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "kWOTmyoaWJg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/kWOTmyoaWJg/677.0400000000001_683.2.mp4", "refined_asr": " And now we have x minus 4 plus 2. And that's going to be the solution to this equation.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@677.0400000000001_683.2#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kWOTmyoaWJg/kWOTmyoaWJg@677.0400000000001_683.2#1.jpg" ], "ocr_qwen2_vl_72b": null } ], "image_num": 6, "text_num": 370, "token_num": 3826 }, { "images": [ "sample_100_images/rgvysb9emcQ@206.4_210.08#1.jpg", null, "sample_100_images/rgvysb9emcQ@210.08_213.64000000000001#1.jpg", null, "sample_100_images/rgvysb9emcQ@213.64000000000001_217.4#1.jpg", null, null, null, "sample_100_images/rgvysb9emcQ@221.88_231.54000000000002#1.jpg", null ], "texts": [ null, " Negative two, three. Then our next point.", null, " 0.7 Let me do it in that color.", null, " 0, 7. X is 0, Y is 7.", " Right there.", " Zero comma seven.", null, " Then we have this one in green here point two eleven 2 11 would be right about there" ], "text_ocr_list": [ null, "We can see these text from the image: - \\( y = 2x + 7 \\)\n- \\( x \\) | \\( y \\)\n- \\( -2 \\) | \\( 2(-2) + 7 = 3 \\)\n- \\( 0 \\) | \\( 2(0) + 7 = 7 \\)\n- \\( 2 \\) | \\( 2(2) + 7 = 11 \\)\n- \\( 8 \\) | \\( 2(8) + 7 = 23 \\).\n Negative two, three. Then our next point.", null, "We can see these text from the image: - \\( y = 2x + 7 \\)\n- \\( x \\) | \\( y \\)\n- -2 | \\( 2(-2) + 7 = 3 \\)\n- 0 | \\( 2(0) + 7 = 7 \\)\n- 2 | \\( 2(2) + 7 = 11 \\)\n- 8 | \\( 2(8) + 7 = 23 \\)\n- \\( (-2, 3) \\).\n 0.7 Let me do it in that color.", null, "We can see these text from the image: y = 2x + 7\n\n| x | y |\n|---|---|\n|-2| 3 |\n|0 | 7 |\n|2 | 11 |\n|8 | 23 |.\n 0, 7. X is 0, Y is 7.", " Right there.", " Zero comma seven.", null, "We can see these text from the image: y = 2x + 7\n\n| x | y |\n|---|---|\n|-2 | 3 |\n| 0 | 7 |\n| 2 | 11 |\n| 8 | 23 |\n\n(-2, 3)\n(0, 7).\n Then we have this one in green here point two eleven 2 11 would be right about there" ], "metadata": [ { "vid": "rgvysb9emcQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/rgvysb9emcQ/206.4_210.08.mp4", "refined_asr": " Negative two, three. Then our next point.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rgvysb9emcQ/rgvysb9emcQ@206.4_210.08#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rgvysb9emcQ/rgvysb9emcQ@206.4_210.08#1.jpg" ], "ocr_qwen2_vl_72b": "- \\( y = 2x + 7 \\)\n- \\( x \\) | \\( y \\)\n- \\( -2 \\) | \\( 2(-2) + 7 = 3 \\)\n- \\( 0 \\) | \\( 2(0) + 7 = 7 \\)\n- \\( 2 \\) | \\( 2(2) + 7 = 11 \\)\n- \\( 8 \\) | \\( 2(8) + 7 = 23 \\)" }, { "vid": "rgvysb9emcQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/rgvysb9emcQ/210.08_213.64000000000001.mp4", "refined_asr": " 0.7 Let me do it in that color.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rgvysb9emcQ/rgvysb9emcQ@210.08_213.64000000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rgvysb9emcQ/rgvysb9emcQ@210.08_213.64000000000001#1.jpg" ], "ocr_qwen2_vl_72b": "- \\( y = 2x + 7 \\)\n- \\( x \\) | \\( y \\)\n- -2 | \\( 2(-2) + 7 = 3 \\)\n- 0 | \\( 2(0) + 7 = 7 \\)\n- 2 | \\( 2(2) + 7 = 11 \\)\n- 8 | \\( 2(8) + 7 = 23 \\)\n- \\( (-2, 3) \\)" }, { "vid": "rgvysb9emcQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/rgvysb9emcQ/213.64000000000001_217.4.mp4", "refined_asr": " 0, 7. X is 0, Y is 7.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rgvysb9emcQ/rgvysb9emcQ@213.64000000000001_217.4#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rgvysb9emcQ/rgvysb9emcQ@213.64000000000001_217.4#1.jpg" ], "ocr_qwen2_vl_72b": "y = 2x + 7\n\n| x | y |\n|---|---|\n|-2| 3 |\n|0 | 7 |\n|2 | 11 |\n|8 | 23 |" }, { "vid": "rgvysb9emcQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/rgvysb9emcQ/217.4_219.36.mp4", "refined_asr": " Right there.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "rgvysb9emcQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/rgvysb9emcQ/219.36_221.88.mp4", "refined_asr": " Zero comma seven.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "rgvysb9emcQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Solving Linear Equations Tutorial: Basics of Linear Equations in Math\"\n_27.json#####audio#####doingASR#####FinishASR/rgvysb9emcQ/221.88_231.54000000000002.mp4", "refined_asr": " Then we have this one in green here point two eleven 2 11 would be right about there", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rgvysb9emcQ/rgvysb9emcQ@221.88_231.54000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rgvysb9emcQ/rgvysb9emcQ@221.88_231.54000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rgvysb9emcQ/rgvysb9emcQ@221.88_231.54000000000002#2.jpg" ], "ocr_qwen2_vl_72b": "y = 2x + 7\n\n| x | y |\n|---|---|\n|-2 | 3 |\n| 0 | 7 |\n| 2 | 11 |\n| 8 | 23 |\n\n(-2, 3)\n(0, 7)" } ], "image_num": 4, "text_num": 71, "token_num": 2375 }, { "images": [ "sample_100_images/irSq7beNAJ8@474.0_504.0#1.jpg", "sample_100_images/irSq7beNAJ8@474.0_504.0#3.jpg", "sample_100_images/irSq7beNAJ8@474.0_504.0#4.jpg", null, null, null, null, null, null, null, "sample_100_images/irSq7beNAJ8@516.0_520.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@520.0_526.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@526.0_530.0#1.jpg", null, null, "sample_100_images/irSq7beNAJ8@534.0_540.0#1.jpg", null, null, null, "sample_100_images/irSq7beNAJ8@546.0_552.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@552.0_564.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@564.0_594.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@594.0_600.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@602.0_608.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@610.0_616.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@618.0_624.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@624.0_632.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@632.0_638.0#1.jpg", null, "sample_100_images/irSq7beNAJ8@638.0_642.0#1.jpg", null ], "texts": [ null, null, null, " Sometimes it is also called negation. And for note, the symbol we use, one of the two symbols can be used in an exam, either this symbol or this symbol. Both of these symbols are used for negation. For example, if P is a statement, then the negation of P, you will write not P, or you will write not P. Okay, so you will write", " Negation of any statement", " Ok", " For example, we gave a name to P.", " We gave a name to P.", " The Earth is round.", " We gave a name to P.", null, " The Earth is round so what will be its negation?", null, " What will be its reverse? The Earth is not round.", null, " We will give the name 'negation p' to 'not p'.", " Okay, so for noteword", null, " We use negation, which is our first operation, for statements.", " The second operation is the end operation.", " Ok", null, " The second operation is the end operation.", null, " The connective word we use is conjunction, and conjunction is nothing but what you have read in Module 3. A 'meet' is called a conjunction.", null, " And we use 'conjunction' for 'and'. For example, we have two statements, 'p' and 'q'. So if we write 'p meet q', it means 'p' conjunction 'q'. And this word is used for 'and'. For example, if 'and' comes between two statements, then we can use conjunction. OK, we can use conjunction. For example, 'p' statement, 'pi' is greater than three.", null, " The statement is that pi is less than 4", null, " Okay, so we can use 'end', 'word', 'p', and 'q'.", null, " P conjunction Q, P conjunction Q", null, " Pi is greater than 3 and less than 4", null, " So if we use the word 'and' to combine two propositions, then p conjunction q", null, " OK and for conjunction, we use 'meet.' K.", null, " Like you have read in set theory, union, we use 'or.'" ], "text_ocr_list": [ null, null, null, "We can see these text from the image: - Connective Word: Not\n- Name of Connective Word: Negation or Denial\n\n- Connective Word: And\n- Name of Connective Word: Conjunction\n\n- Connective Word: Or\n- Name of Connective Word: Disjunction\n\n- Connective Word: If ... then ...\n- Name of Connective Word: Conditional\n\n- Connective Word: If and only if\n- Name of Connective Word: Biconditional.\n Sometimes it is also called negation. And for note, the symbol we use, one of the two symbols can be used in an exam, either this symbol or this symbol. Both of these symbols are used for negation. For example, if P is a statement, then the negation of P, you will write not P, or you will write not P. Okay, so you will write", " Negation of any statement", " Ok", " For example, we gave a name to P.", " We gave a name to P.", " The Earth is round.", " We gave a name to P.", null, "We can see these text from the image: - Denial\n- Symbol: ~ or \u00ac\n- p = The earth is round\n- ~p = \u00acp.\n The Earth is round so what will be its negation?", null, "We can see these text from the image: - Connective word: Denial\n- Symbol: ~ or \u00ac\n- p = The earth is round\n- ~p\n- \u00acp.\n What will be its reverse? The Earth is not round.", null, "We can see these text from the image: Connective word | Name of connective word | Symbol\nNot | Negation or Denial | ~ or \u00ac\nand | Conjunction |\nor | Disjunction |\nif ... then ... | Conditional |\niff or if and only if | Biconditional |.\n We will give the name 'negation p' to 'not p'.", " Okay, so for noteword", null, "We can see these text from the image: Connective word | Name of connective word | Symbol\nNot | Negation or Denial | ~ or \u00ac\nand | Conjunction |\nor | Disjunction |\nif ... then ... | Conditional |\niff or if and only if | Biconditional |.\n We use negation, which is our first operation, for statements.", " The second operation is the end operation.", " Ok", null, "We can see these text from the image: - Connective Word: Not, And, Or, If ..., Then ..., Iff, If and only if\n- Name of connective word: Negation or Denial, Conjunction, Disjunction, Conditional, Biconditional\n- Symbol: ~ or \u00ac.\n The second operation is the end operation.", null, "We can see these text from the image: Connective word | Name of connective word | Symbol\nNot | Negation or Denial | ~ or \u00ac\nand | Conjunction |\nor | Disjunction |\nif ... then ... | Conditional |\niff or if and only if | Biconditional |.\n The connective word we use is conjunction, and conjunction is nothing but what you have read in Module 3. A 'meet' is called a conjunction.", null, "We can see these text from the image: - Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- Connective word: Disjunction\n- Connective word: Conditional\n- Connective word: Biconditional.\n And we use 'conjunction' for 'and'. For example, we have two statements, 'p' and 'q'. So if we write 'p meet q', it means 'p' conjunction 'q'. And this word is used for 'and'. For example, if 'and' comes between two statements, then we can use conjunction. OK, we can use conjunction. For example, 'p' statement, 'pi' is greater than three.", null, "We can see these text from the image: - Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- p = x > 3\n- q = x < 2.\n The statement is that pi is less than 4", null, "We can see these text from the image: - Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- p = x > 3\n- q = x < 4.\n Okay, so we can use 'end', 'word', 'p', and 'q'.", null, "We can see these text from the image: - Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- p = x > 3\n- q = x < 4\n- p \u2227 q = p \u2228 q\n- =.\n P conjunction Q, P conjunction Q", null, "We can see these text from the image: - Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- p = x > 3\n- q = x < 4\n- p \u2227 q = (x > 3) \u2227 (x < 4).\n Pi is greater than 3 and less than 4", null, "We can see these text from the image: - Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- p \u2261 x > 3\n- q \u2261 x < 4\n- p \u2228 q = p \u2227 q\n- (x > 3) \u2227 (x < 4).\n So if we use the word 'and' to combine two propositions, then p conjunction q", null, "We can see these text from the image: og Connective word\n\nation or Denial\n\nConjunction\n\nDie junction\n\nConditional\n\nBiconditional\n\nSymbol\n\n~ or \u00ac\n\n\u2227.\n OK and for conjunction, we use 'meet.' K.", null, "We can see these text from the image: - Connective word\n- Symbol\n- Negation or Denial: ~\n- Conjunction: \u2227\n- Disjunction: \u2228\n- Conditional: \u2192\n- Biconditional: \u2194.\n Like you have read in set theory, union, we use 'or.'" ], "metadata": [ { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/474.0_504.0.mp4", "refined_asr": " Sometimes it is also called negation. And for note, the symbol we use, one of the two symbols can be used in an exam, either this symbol or this symbol. Both of these symbols are used for negation. For example, if P is a statement, then the negation of P, you will write not P, or you will write not P. Okay, so you will write", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@474.0_504.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@474.0_504.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@474.0_504.0#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@474.0_504.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@474.0_504.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@474.0_504.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@474.0_504.0#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@474.0_504.0#5.jpg" ], "ocr_qwen2_vl_72b": "- Connective Word: Not\n- Name of Connective Word: Negation or Denial\n\n- Connective Word: And\n- Name of Connective Word: Conjunction\n\n- Connective Word: Or\n- Name of Connective Word: Disjunction\n\n- Connective Word: If ... then ...\n- Name of Connective Word: Conditional\n\n- Connective Word: If and only if\n- Name of Connective Word: Biconditional" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/504.0_506.0.mp4", "refined_asr": " Negation of any statement", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/506.0_508.0.mp4", "refined_asr": " Ok", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/508.0_510.0.mp4", "refined_asr": " For example, we gave a name to P.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@508.0_510.0#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/510.0_512.0.mp4", "refined_asr": " We gave a name to P.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/512.0_514.0.mp4", "refined_asr": " The Earth is round.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/514.0_516.0.mp4", "refined_asr": " We gave a name to P.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/516.0_520.0.mp4", "refined_asr": " The Earth is round so what will be its negation?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@516.0_520.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@516.0_520.0#1.jpg" ], "ocr_qwen2_vl_72b": "- Denial\n- Symbol: ~ or \u00ac\n- p = The earth is round\n- ~p = \u00acp" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/520.0_526.0.mp4", "refined_asr": " What will be its reverse? The Earth is not round.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@520.0_526.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@520.0_526.0#1.jpg" ], "ocr_qwen2_vl_72b": "- Connective word: Denial\n- Symbol: ~ or \u00ac\n- p = The earth is round\n- ~p\n- \u00acp" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/526.0_530.0.mp4", "refined_asr": " We will give the name 'negation p' to 'not p'.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@526.0_530.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@526.0_530.0#1.jpg" ], "ocr_qwen2_vl_72b": "Connective word | Name of connective word | Symbol\nNot | Negation or Denial | ~ or \u00ac\nand | Conjunction |\nor | Disjunction |\nif ... then ... | Conditional |\niff or if and only if | Biconditional |" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/530.0_534.0.mp4", "refined_asr": " Okay, so for noteword", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/534.0_540.0.mp4", "refined_asr": " We use negation, which is our first operation, for statements.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@534.0_540.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@534.0_540.0#1.jpg" ], "ocr_qwen2_vl_72b": "Connective word | Name of connective word | Symbol\nNot | Negation or Denial | ~ or \u00ac\nand | Conjunction |\nor | Disjunction |\nif ... then ... | Conditional |\niff or if and only if | Biconditional |" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/540.0_544.0.mp4", "refined_asr": " The second operation is the end operation.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/544.0_546.0.mp4", "refined_asr": " Ok", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/546.0_552.0.mp4", "refined_asr": " The second operation is the end operation.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@546.0_552.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@546.0_552.0#1.jpg" ], "ocr_qwen2_vl_72b": "- Connective Word: Not, And, Or, If ..., Then ..., Iff, If and only if\n- Name of connective word: Negation or Denial, Conjunction, Disjunction, Conditional, Biconditional\n- Symbol: ~ or \u00ac" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/552.0_564.0.mp4", "refined_asr": " The connective word we use is conjunction, and conjunction is nothing but what you have read in Module 3. A 'meet' is called a conjunction.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@552.0_564.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@552.0_564.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@552.0_564.0#2.jpg" ], "ocr_qwen2_vl_72b": "Connective word | Name of connective word | Symbol\nNot | Negation or Denial | ~ or \u00ac\nand | Conjunction |\nor | Disjunction |\nif ... then ... | Conditional |\niff or if and only if | Biconditional |" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/564.0_594.0.mp4", "refined_asr": " And we use 'conjunction' for 'and'. For example, we have two statements, 'p' and 'q'. So if we write 'p meet q', it means 'p' conjunction 'q'. And this word is used for 'and'. For example, if 'and' comes between two statements, then we can use conjunction. OK, we can use conjunction. For example, 'p' statement, 'pi' is greater than three.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@564.0_594.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@564.0_594.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@564.0_594.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@564.0_594.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@564.0_594.0#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@564.0_594.0#5.jpg" ], "ocr_qwen2_vl_72b": "- Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- Connective word: Disjunction\n- Connective word: Conditional\n- Connective word: Biconditional" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/594.0_600.0.mp4", "refined_asr": " The statement is that pi is less than 4", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@594.0_600.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@594.0_600.0#1.jpg" ], "ocr_qwen2_vl_72b": "- Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- p = x > 3\n- q = x < 2" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/602.0_608.0.mp4", "refined_asr": " Okay, so we can use 'end', 'word', 'p', and 'q'.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@602.0_608.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@602.0_608.0#1.jpg" ], "ocr_qwen2_vl_72b": "- Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- p = x > 3\n- q = x < 4" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/610.0_616.0.mp4", "refined_asr": " P conjunction Q, P conjunction Q", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@610.0_616.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@610.0_616.0#1.jpg" ], "ocr_qwen2_vl_72b": "- Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- p = x > 3\n- q = x < 4\n- p \u2227 q = p \u2228 q\n- =" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/618.0_624.0.mp4", "refined_asr": " Pi is greater than 3 and less than 4", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@618.0_624.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@618.0_624.0#1.jpg" ], "ocr_qwen2_vl_72b": "- Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- p = x > 3\n- q = x < 4\n- p \u2227 q = (x > 3) \u2227 (x < 4)" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/624.0_632.0.mp4", "refined_asr": " So if we use the word 'and' to combine two propositions, then p conjunction q", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@624.0_632.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@624.0_632.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@624.0_632.0#2.jpg" ], "ocr_qwen2_vl_72b": "- Connective word: Denial\n- Symbol: ~ or \u00ac\n- Connective word: Conjunction\n- Symbol: \u2227\n- p \u2261 x > 3\n- q \u2261 x < 4\n- p \u2228 q = p \u2227 q\n- (x > 3) \u2227 (x < 4)" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/632.0_638.0.mp4", "refined_asr": " OK and for conjunction, we use 'meet.' K.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@632.0_638.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@632.0_638.0#1.jpg" ], "ocr_qwen2_vl_72b": "og Connective word\n\nation or Denial\n\nConjunction\n\nDie junction\n\nConditional\n\nBiconditional\n\nSymbol\n\n~ or \u00ac\n\n\u2227" }, { "vid": "irSq7beNAJ8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Mathematical Logic tutorial on Well-Formed Formulas in Predicate Logic\"\n_23.json#####audio#####doingASR#####FinishASR/irSq7beNAJ8/638.0_642.0.mp4", "refined_asr": " Like you have read in set theory, union, we use 'or.'", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@638.0_642.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/irSq7beNAJ8/irSq7beNAJ8@638.0_642.0#1.jpg" ], "ocr_qwen2_vl_72b": "- Connective word\n- Symbol\n- Negation or Denial: ~\n- Conjunction: \u2227\n- Disjunction: \u2228\n- Conditional: \u2192\n- Biconditional: \u2194" } ], "image_num": 17, "text_num": 449, "token_num": 10241 }, { "images": [ "sample_100_images/F-G6QNx1Zak@199.0_211.0#1.jpg", "sample_100_images/F-G6QNx1Zak@199.0_211.0#2.jpg", null, null, "sample_100_images/F-G6QNx1Zak@215.0_227.0#1.jpg", "sample_100_images/F-G6QNx1Zak@215.0_227.0#2.jpg", null, "sample_100_images/F-G6QNx1Zak@227.0_238.0#1.jpg", "sample_100_images/F-G6QNx1Zak@227.0_238.0#2.jpg", null, "sample_100_images/F-G6QNx1Zak@238.0_245.0#1.jpg", null, "sample_100_images/F-G6QNx1Zak@245.0_255.0#1.jpg", null, null, null, "sample_100_images/F-G6QNx1Zak@258.0_286.0#1.jpg", "sample_100_images/F-G6QNx1Zak@258.0_286.0#2.jpg", "sample_100_images/F-G6QNx1Zak@258.0_286.0#3.jpg", "sample_100_images/F-G6QNx1Zak@258.0_286.0#4.jpg", null, null, "sample_100_images/F-G6QNx1Zak@287.0_306.0#1.jpg", "sample_100_images/F-G6QNx1Zak@287.0_306.0#2.jpg", "sample_100_images/F-G6QNx1Zak@287.0_306.0#3.jpg", null, "sample_100_images/F-G6QNx1Zak@306.0_312.0#1.jpg", null, null, null, "sample_100_images/F-G6QNx1Zak@316.0_331.22#1.jpg", "sample_100_images/F-G6QNx1Zak@316.0_331.22#2.jpg", null ], "texts": [ null, null, " All the activities in living bodies occur at a particular pH. If there is a change in the pH, then the survival of the organisms becomes difficult.", " Survival of the organisms becomes difficult.", null, null, " Let us suppose here, let us take acid rain. What is that? Acid rain.", null, null, " What is acid rain? Well, when the pH of rain is less than 5.5.", null, " When the pH of the rain is 5.5 then that rain is called acid rain.", null, " So why is there acid rain? Right so in the atmosphere when the pollutants are more like carbon dioxide sulfur dioxide", " Right.", " So next, nitrogen dioxide.", null, null, null, null, " So what happens during the rain? The water comes into contact with these non-metal oxides and they form respective acids. What are these acids? They are carbonic acid, nitric acid, and sulfuric acid. Right. When they reach the Earth's surface, what happens next? They flow over the land and reach the river water.", " Right.", null, null, null, " So, when this acid rain gets mixed up with the water, it will change the pH of the river water.", null, " It will change the pH of the river water.", " So, when acid is getting mixed up with the river water, the pH will decrease.", " When the pH decreases what happens It becomes more acidic due to increased carbon dioxide", null, null, " Acidic character increases. So when the acidic character increases, then what happens? The plants and animals which are living in this river water will be affected. So the change in the pH level impacts the ecosystem." ], "text_ocr_list": [ null, null, "We can see these text from the image: - CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants/animals through chemical warfare\n- pH \u2192 digestion\n- peptidase (inactive)\n- HCl (0.8-1.8)\n- Active peptidase \u2192 protein \u2192 peptones\n- Carbohydrates \u2192 Alkaline medium.\n All the activities in living bodies occur at a particular pH. If there is a change in the pH, then the survival of the organisms becomes difficult.", " Survival of the organisms becomes difficult.", null, null, "We can see these text from the image: - CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants/animals through chemical warfare.\n Let us suppose here, let us take acid rain. What is that? Acid rain.", null, null, "We can see these text from the image: - CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life:\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants/animals through chemical warfare\n- Acid rain.\n What is acid rain? Well, when the pH of rain is less than 5.5.", null, "We can see these text from the image: - CLASS: X\n- Physical science\n- 3. Acids, Bases And Salts\n- Importance of pH in everyday life\n- Are plants and animals are pH sensitive?\n- Is pH change can cause tooth decay?\n- pH in our digestive system\n- pH of soil\n- self defense by plants/animals through chemical warfare\n- Acid rain - pH of rain K 5.5.\n When the pH of the rain is 5.5 then that rain is called acid rain.", null, "We can see these text from the image: - CLASS: X\n- Physical science\n- 3. Acids, Bases And Salts\n- Importance of pH in everyday life\n- Are plants and animals are pH sensitive?\n- Is pH change can cause tooth decay?\n- pH in our digestive system\n- pH of soil\n- self defense by plants/animals through chemical warfare\n- Acid rain\n- pH of rain < 5.5.\n So why is there acid rain? Right so in the atmosphere when the pollutants are more like carbon dioxide sulfur dioxide", " Right.", " So next, nitrogen dioxide.", null, null, null, null, "We can see these text from the image: - CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants & animals through chemical warfare\n- Acid rain\n- pH of rain < 5.5.\n So what happens during the rain? The water comes into contact with these non-metal oxides and they form respective acids. What are these acids? They are carbonic acid, nitric acid, and sulfuric acid. Right. When they reach the Earth's surface, what happens next? They flow over the land and reach the river water.", " Right.", null, null, null, "We can see these text from the image: - CLASS: X.\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n- Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay\n - pH in our digestive system\n - pH of soil\n - Self defense by plants/animals through chemical warfare\n- Acids\n- pH of rain = 5.5\n- HNO3 + H2O -> H3O+ + NO3-.\n So, when this acid rain gets mixed up with the water, it will change the pH of the river water.", null, "We can see these text from the image: - CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants/animals through chemical warfare\n- pH of rain < 5.5\n- pH of river.\n It will change the pH of the river water.", " So, when acid is getting mixed up with the river water, the pH will decrease.", " When the pH decreases what happens It becomes more acidic due to increased carbon dioxide", null, null, "We can see these text from the image: - CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants/animals through chemical warfare\n- Acids\n- pH of rain: K 5.5\n- pH decrease\n- River.\n Acidic character increases. So when the acidic character increases, then what happens? The plants and animals which are living in this river water will be affected. So the change in the pH level impacts the ecosystem." ], "metadata": [ { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/199.0_211.0.mp4", "refined_asr": " All the activities in living bodies occur at a particular pH. If there is a change in the pH, then the survival of the organisms becomes difficult.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@199.0_211.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@199.0_211.0#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@199.0_211.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@199.0_211.0#2.jpg" ], "ocr_qwen2_vl_72b": "- CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants/animals through chemical warfare\n- pH \u2192 digestion\n- peptidase (inactive)\n- HCl (0.8-1.8)\n- Active peptidase \u2192 protein \u2192 peptones\n- Carbohydrates \u2192 Alkaline medium" }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/211.0_215.0.mp4", "refined_asr": " Survival of the organisms becomes difficult.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/215.0_227.0.mp4", "refined_asr": " Let us suppose here, let us take acid rain. What is that? Acid rain.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@215.0_227.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@215.0_227.0#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@215.0_227.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@215.0_227.0#2.jpg" ], "ocr_qwen2_vl_72b": "- CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants/animals through chemical warfare" }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/227.0_238.0.mp4", "refined_asr": " What is acid rain? Well, when the pH of rain is less than 5.5.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@227.0_238.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@227.0_238.0#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@227.0_238.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@227.0_238.0#2.jpg" ], "ocr_qwen2_vl_72b": "- CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life:\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants/animals through chemical warfare\n- Acid rain" }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/238.0_245.0.mp4", "refined_asr": " When the pH of the rain is 5.5 then that rain is called acid rain.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@238.0_245.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@238.0_245.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@238.0_245.0#2.jpg" ], "ocr_qwen2_vl_72b": "- CLASS: X\n- Physical science\n- 3. Acids, Bases And Salts\n- Importance of pH in everyday life\n- Are plants and animals are pH sensitive?\n- Is pH change can cause tooth decay?\n- pH in our digestive system\n- pH of soil\n- self defense by plants/animals through chemical warfare\n- Acid rain - pH of rain K 5.5" }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/245.0_255.0.mp4", "refined_asr": " So why is there acid rain? Right so in the atmosphere when the pollutants are more like carbon dioxide sulfur dioxide", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@245.0_255.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@245.0_255.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@245.0_255.0#2.jpg" ], "ocr_qwen2_vl_72b": "- CLASS: X\n- Physical science\n- 3. Acids, Bases And Salts\n- Importance of pH in everyday life\n- Are plants and animals are pH sensitive?\n- Is pH change can cause tooth decay?\n- pH in our digestive system\n- pH of soil\n- self defense by plants/animals through chemical warfare\n- Acid rain\n- pH of rain < 5.5" }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/255.0_256.0.mp4", "refined_asr": " Right.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/256.0_257.0.mp4", "refined_asr": " So next, nitrogen dioxide.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@256.0_257.0#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/258.0_286.0.mp4", "refined_asr": " So what happens during the rain? The water comes into contact with these non-metal oxides and they form respective acids. What are these acids? They are carbonic acid, nitric acid, and sulfuric acid. Right. When they reach the Earth's surface, what happens next? They flow over the land and reach the river water.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@258.0_286.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@258.0_286.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@258.0_286.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@258.0_286.0#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@258.0_286.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@258.0_286.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@258.0_286.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@258.0_286.0#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@258.0_286.0#5.jpg" ], "ocr_qwen2_vl_72b": "- CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants & animals through chemical warfare\n- Acid rain\n- pH of rain < 5.5" }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/286.0_287.0.mp4", "refined_asr": " Right.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/287.0_306.0.mp4", "refined_asr": " So, when this acid rain gets mixed up with the water, it will change the pH of the river water.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@287.0_306.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@287.0_306.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@287.0_306.0#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@287.0_306.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@287.0_306.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@287.0_306.0#3.jpg" ], "ocr_qwen2_vl_72b": "- CLASS: X.\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n- Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay\n - pH in our digestive system\n - pH of soil\n - Self defense by plants/animals through chemical warfare\n- Acids\n- pH of rain = 5.5\n- HNO3 + H2O -> H3O+ + NO3-" }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/306.0_312.0.mp4", "refined_asr": " It will change the pH of the river water.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@306.0_312.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@306.0_312.0#1.jpg" ], "ocr_qwen2_vl_72b": "- CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants/animals through chemical warfare\n- pH of rain < 5.5\n- pH of river" }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/312.0_314.0.mp4", "refined_asr": " So, when acid is getting mixed up with the river water, the pH will decrease.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@312.0_314.0#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/314.0_316.0.mp4", "refined_asr": " When the pH decreases what happens It becomes more acidic due to increased carbon dioxide", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "F-G6QNx1Zak.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Acids and bases in daily life_100.json#####audio#####doingASR#####FinishASR/F-G6QNx1Zak/316.0_331.22.mp4", "refined_asr": " Acidic character increases. So when the acidic character increases, then what happens? The plants and animals which are living in this river water will be affected. So the change in the pH level impacts the ecosystem.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@316.0_331.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@316.0_331.22#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@316.0_331.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@316.0_331.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/F-G6QNx1Zak/F-G6QNx1Zak@316.0_331.22#3.jpg" ], "ocr_qwen2_vl_72b": "- CLASS: X\n- Physical science\n- Acids, Bases And Salts\n- Importance of pH in everyday life\n - Are plants and animals are pH sensitive?\n - Is pH change can cause tooth decay?\n - pH in our digestive system\n - pH of soil\n - self defense by plants/animals through chemical warfare\n- Acids\n- pH of rain: K 5.5\n- pH decrease\n- River" } ], "image_num": 18, "text_num": 359, "token_num": 10727 }, { "images": [ "sample_100_images/CV6omNpa_u8@1482.8600000000001_1496.8600000000001#1.jpg", null, "sample_100_images/CV6omNpa_u8@1496.8600000000001_1504.8600000000001#1.jpg", null, null, null, "sample_100_images/CV6omNpa_u8@1517.8600000000001_1540.8600000000001#1.jpg", "sample_100_images/CV6omNpa_u8@1517.8600000000001_1540.8600000000001#2.jpg", null, "sample_100_images/CV6omNpa_u8@1540.8600000000001_1546.8600000000001#1.jpg", null, "sample_100_images/CV6omNpa_u8@1546.8600000000001_1554.8600000000001#1.jpg", null, null, "sample_100_images/CV6omNpa_u8@1556.8600000000001_1566.8600000000001#1.jpg", null, "sample_100_images/CV6omNpa_u8@1566.8600000000001_1570.8600000000001#1.jpg", null, "sample_100_images/CV6omNpa_u8@1570.8600000000001_1575.8600000000001#1.jpg", null, null, "sample_100_images/CV6omNpa_u8@1578.8600000000001_1595.8600000000001#1.jpg", null, null, "sample_100_images/CV6omNpa_u8@1597.8600000000001_1602.8600000000001#1.jpg", null, "sample_100_images/CV6omNpa_u8@1602.8600000000001_1611.8600000000001#1.jpg", null, null, "sample_100_images/CV6omNpa_u8@1613.8600000000001_1626.8600000000001#1.jpg", null, null ], "texts": [ null, " And it's the same situation. Add the numbers in the ones place, then the tens place, and the hundreds place, and so on and so forth.", null, " One plus zero plus two is three. One plus two plus one is four. Five plus zero plus zero is five. Blah, blah, blah, blah, blah, blah.", " 543. Rectangle. Boom.", " All right.", null, null, " All right, let's do another one of these. We have 6012 plus 231 plus 113. Add the numbers in the ones place. 2 plus 1 plus 3 is 6.", null, " 1 plus 3 plus 1 is 5. 1 plus 2 plus 0 is 3.", null, " Six plus zero plus zero is six. And that's the final answer.", " 6. (No changes needed)", null, " 6,356. All right.", null, " Now pause the video and try number 19.", null, " And when you come back we'll do it together", " All right, we're back.", null, " Line up the numbers vertically. I'm writing the numbers, the smaller ones on top. But it really doesn't matter. In fact, just to show you that it doesn't matter, I'm going to go ahead and write the smaller ones on top. Just to be thorough here.", " 297.", null, " So I'm reversing the order you're going to get the same answer", null, " 8 plus 7 is 15. Carry the 1. 1 plus 9 plus 1 plus 4 is 15.", " Carry the one.", null, " Add all these numbers up and you get 16 I believe", " 1,000." ], "text_ocr_list": [ null, "We can see these text from the image: 17) 20 + 12 + 511\n\n511\n20\n+ 12\n\n18) 231 + 6,012 + 113\n\n21) 43 + 12 + 31 + 50\n\n22) 171 + 223 + 50.\n And it's the same situation. Add the numbers in the ones place, then the tens place, and the hundreds place, and so on and so forth.", null, "We can see these text from the image: 17 20 + 12 + 511\n\n511\n20\n+ 12\n3\n\n18 231 + 6,012 + 113\n\n21 43 + 12 + 31 + 14\n\n22 171 + 223 + 500.\n One plus zero plus two is three. One plus two plus one is four. Five plus zero plus zero is five. Blah, blah, blah, blah, blah, blah.", " 543. Rectangle. Boom.", " All right.", null, null, "We can see these text from the image: 511\n20\n+ 12\n543\n\n543\n\n18 231 + 6,012 + 113\n22 171 + 223 + 500 + 431.\n All right, let's do another one of these. We have 6012 plus 231 plus 113. Add the numbers in the ones place. 2 plus 1 plus 3 is 6.", null, "We can see these text from the image: + 12 543\n\n231 + 6,012 + 113\n\n6,012\n231\n+ 113\n56\n\n948 + 297 + 410.\n 1 plus 3 plus 1 is 5. 1 plus 2 plus 0 is 3.", null, "We can see these text from the image: 18) 231 + 6,012 + 113\n\n6,012\n+ 231\n+ 113\n_____\n356\n\n19) 948 + 297 + 410\n\n22) 171 + 223 + 500 + 431\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 amu and an atom of oxygen has a mass of 16 amu. What is the total mass of one molecule of carbon monoxide?.\n Six plus zero plus zero is six. And that's the final answer.", " 6. (No changes needed)", null, "We can see these text from the image: 18) 231 + 6,012 + 113\n\n6,012\n+ 231\n+ 113\n_____\n6,356\n\n19) 948 + 297 + 410\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 amu, and an atom of oxygen has a mass of 16 amu. What is the total mass of one molecule of carbon monoxide?.\n 6,356. All right.", null, "We can see these text from the image: 231\n+ 113\n6 356\n\n6,356\n\n19) 948 + 297 + 410\n\n20) 1,321 + 2,137 + 8,153\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 units, and an atom of oxygen has a mass of 16 units. What is the mass of a molecule of carbon monoxide?.\n Now pause the video and try number 19.", null, "We can see these text from the image: 231\n+ 113\n6 356\n\n6,356\n\n19) 948 + 297 + 410\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 units and an atom of oxygen has a mass of 16 units. What is the mass of a molecule of carbon monoxide?\n\n20) 1,321 + 2,137 + 8,153.\n And when you come back we'll do it together", " All right, we're back.", null, "We can see these text from the image: 19) 948 + 297 + 410\n\n948\n410\n2.\n\n20) 1,321 + 2,137 + 8,153\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 units, and an atom of oxygen has a mass of 16 units. What is the mass of one molecule of carbon monoxide?.\n Line up the numbers vertically. I'm writing the numbers, the smaller ones on top. But it really doesn't matter. In fact, just to show you that it doesn't matter, I'm going to go ahead and write the smaller ones on top. Just to be thorough here.", " 297.", null, "We can see these text from the image: 231\n+ 113\n6 356\n\n6,356\n\n19) 948 + 297 + 410\n\n297\n410\n948\n\n20) 1,321 + 2,137 + 8,153\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 units and an atom of oxygen has a mass of 16 units. What is the mass of a molecule of carbon monoxide?.\n So I'm reversing the order you're going to get the same answer", null, "We can see these text from the image: 231\n\n+ 113\n\n6356\n\n6,356\n\n19) 948 + 297 + 410\n\n297\n\n410\n\n948\n\n5\n\n20) 1,321 + 2,137 + 8,153\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 and an atom of oxygen has a mass of 16. What is the mass of a molecule of carbon monoxide?.\n 8 plus 7 is 15. Carry the 1. 1 plus 9 plus 1 plus 4 is 15.", " Carry the one.", null, "We can see these text from the image: + 113\n6 356\n\n6,356\n\n948 + 297 + 410\n\n11\n297\n410\n948\n55\n\n1,321 + 2,137 + 8,153\n\nOne molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 units and an atom of oxygen has a mass of 16 units. What is the mass of one molecule of carbon monoxide?.\n Add all these numbers up and you get 16 I believe", " 1,000." ], "metadata": [ { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1482.8600000000001_1496.8600000000001.mp4", "refined_asr": " And it's the same situation. Add the numbers in the ones place, then the tens place, and the hundreds place, and so on and so forth.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1482.8600000000001_1496.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1482.8600000000001_1496.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1482.8600000000001_1496.8600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1482.8600000000001_1496.8600000000001#3.jpg" ], "ocr_qwen2_vl_72b": "17) 20 + 12 + 511\n\n511\n20\n+ 12\n\n18) 231 + 6,012 + 113\n\n21) 43 + 12 + 31 + 50\n\n22) 171 + 223 + 50" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1496.8600000000001_1504.8600000000001.mp4", "refined_asr": " One plus zero plus two is three. One plus two plus one is four. Five plus zero plus zero is five. Blah, blah, blah, blah, blah, blah.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1496.8600000000001_1504.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1496.8600000000001_1504.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1496.8600000000001_1504.8600000000001#2.jpg" ], "ocr_qwen2_vl_72b": "17 20 + 12 + 511\n\n511\n20\n+ 12\n3\n\n18 231 + 6,012 + 113\n\n21 43 + 12 + 31 + 14\n\n22 171 + 223 + 500" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1504.8600000000001_1507.8600000000001.mp4", "refined_asr": " 543. Rectangle. Boom.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1509.8600000000001_1517.8600000000001.mp4", "refined_asr": " All right.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1517.8600000000001_1540.8600000000001.mp4", "refined_asr": " All right, let's do another one of these. We have 6012 plus 231 plus 113. Add the numbers in the ones place. 2 plus 1 plus 3 is 6.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1517.8600000000001_1540.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1517.8600000000001_1540.8600000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1517.8600000000001_1540.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1517.8600000000001_1540.8600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1517.8600000000001_1540.8600000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1517.8600000000001_1540.8600000000001#4.jpg" ], "ocr_qwen2_vl_72b": "511\n20\n+ 12\n543\n\n543\n\n18 231 + 6,012 + 113\n22 171 + 223 + 500 + 431" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1540.8600000000001_1546.8600000000001.mp4", "refined_asr": " 1 plus 3 plus 1 is 5. 1 plus 2 plus 0 is 3.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1540.8600000000001_1546.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1540.8600000000001_1546.8600000000001#1.jpg" ], "ocr_qwen2_vl_72b": "+ 12 543\n\n231 + 6,012 + 113\n\n6,012\n231\n+ 113\n56\n\n948 + 297 + 410" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1546.8600000000001_1554.8600000000001.mp4", "refined_asr": " Six plus zero plus zero is six. And that's the final answer.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1546.8600000000001_1554.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1546.8600000000001_1554.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1546.8600000000001_1554.8600000000001#2.jpg" ], "ocr_qwen2_vl_72b": "18) 231 + 6,012 + 113\n\n6,012\n+ 231\n+ 113\n_____\n356\n\n19) 948 + 297 + 410\n\n22) 171 + 223 + 500 + 431\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 amu and an atom of oxygen has a mass of 16 amu. What is the total mass of one molecule of carbon monoxide?" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1554.8600000000001_1556.8600000000001.mp4", "refined_asr": " 6. (No changes needed)", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1556.8600000000001_1566.8600000000001.mp4", "refined_asr": " 6,356. All right.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1556.8600000000001_1566.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1556.8600000000001_1566.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1556.8600000000001_1566.8600000000001#2.jpg" ], "ocr_qwen2_vl_72b": "18) 231 + 6,012 + 113\n\n6,012\n+ 231\n+ 113\n_____\n6,356\n\n19) 948 + 297 + 410\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 amu, and an atom of oxygen has a mass of 16 amu. What is the total mass of one molecule of carbon monoxide?" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1566.8600000000001_1570.8600000000001.mp4", "refined_asr": " Now pause the video and try number 19.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1566.8600000000001_1570.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1566.8600000000001_1570.8600000000001#1.jpg" ], "ocr_qwen2_vl_72b": "231\n+ 113\n6 356\n\n6,356\n\n19) 948 + 297 + 410\n\n20) 1,321 + 2,137 + 8,153\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 units, and an atom of oxygen has a mass of 16 units. What is the mass of a molecule of carbon monoxide?" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1570.8600000000001_1575.8600000000001.mp4", "refined_asr": " And when you come back we'll do it together", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1570.8600000000001_1575.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1570.8600000000001_1575.8600000000001#1.jpg" ], "ocr_qwen2_vl_72b": "231\n+ 113\n6 356\n\n6,356\n\n19) 948 + 297 + 410\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 units and an atom of oxygen has a mass of 16 units. What is the mass of a molecule of carbon monoxide?\n\n20) 1,321 + 2,137 + 8,153" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1575.8600000000001_1578.8600000000001.mp4", "refined_asr": " All right, we're back.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1575.8600000000001_1578.8600000000001#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1578.8600000000001_1595.8600000000001.mp4", "refined_asr": " Line up the numbers vertically. I'm writing the numbers, the smaller ones on top. But it really doesn't matter. In fact, just to show you that it doesn't matter, I'm going to go ahead and write the smaller ones on top. Just to be thorough here.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1578.8600000000001_1595.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1578.8600000000001_1595.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1578.8600000000001_1595.8600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1578.8600000000001_1595.8600000000001#3.jpg" ], "ocr_qwen2_vl_72b": "19) 948 + 297 + 410\n\n948\n410\n2.\n\n20) 1,321 + 2,137 + 8,153\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 units, and an atom of oxygen has a mass of 16 units. What is the mass of one molecule of carbon monoxide?" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1595.8600000000001_1597.8600000000001.mp4", "refined_asr": " 297.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1597.8600000000001_1602.8600000000001.mp4", "refined_asr": " So I'm reversing the order you're going to get the same answer", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1597.8600000000001_1602.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1597.8600000000001_1602.8600000000001#1.jpg" ], "ocr_qwen2_vl_72b": "231\n+ 113\n6 356\n\n6,356\n\n19) 948 + 297 + 410\n\n297\n410\n948\n\n20) 1,321 + 2,137 + 8,153\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 units and an atom of oxygen has a mass of 16 units. What is the mass of a molecule of carbon monoxide?" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1602.8600000000001_1611.8600000000001.mp4", "refined_asr": " 8 plus 7 is 15. Carry the 1. 1 plus 9 plus 1 plus 4 is 15.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1602.8600000000001_1611.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1602.8600000000001_1611.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1602.8600000000001_1611.8600000000001#2.jpg" ], "ocr_qwen2_vl_72b": "231\n\n+ 113\n\n6356\n\n6,356\n\n19) 948 + 297 + 410\n\n297\n\n410\n\n948\n\n5\n\n20) 1,321 + 2,137 + 8,153\n\n23) One molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 and an atom of oxygen has a mass of 16. What is the mass of a molecule of carbon monoxide?" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1611.8600000000001_1613.8600000000001.mp4", "refined_asr": " Carry the one.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1613.8600000000001_1626.8600000000001.mp4", "refined_asr": " Add all these numbers up and you get 16 I believe", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1613.8600000000001_1626.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1613.8600000000001_1626.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/CV6omNpa_u8/CV6omNpa_u8@1613.8600000000001_1626.8600000000001#2.jpg" ], "ocr_qwen2_vl_72b": "+ 113\n6 356\n\n6,356\n\n948 + 297 + 410\n\n11\n297\n410\n948\n55\n\n1,321 + 2,137 + 8,153\n\nOne molecule of carbon monoxide contains one atom of carbon and one atom of oxygen. An atom of carbon has a mass of 12 units and an atom of oxygen has a mass of 16 units. What is the mass of one molecule of carbon monoxide?" }, { "vid": "CV6omNpa_u8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Adding Whole Numbers\" in elementary arithmetic math tutorial_30.json#####audio#####doingASR#####FinishASR/CV6omNpa_u8/1626.8600000000001_1628.8600000000001.mp4", "refined_asr": " 1,000.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null } ], "image_num": 13, "text_num": 376, "token_num": 7864 }, { "images": [ "sample_100_images/Rw70zkvqEiE@313.22_339.32000000000005#1.jpg", null, "sample_100_images/Rw70zkvqEiE@339.32000000000005_355.22#1.jpg", null, "sample_100_images/Rw70zkvqEiE@355.22_382.40000000000003#1.jpg", null, "sample_100_images/Rw70zkvqEiE@382.40000000000003_406.1#1.jpg", null, "sample_100_images/Rw70zkvqEiE@406.16_432.32000000000005#1.jpg", null, "sample_100_images/Rw70zkvqEiE@432.32_454.96#1.jpg", null ], "texts": [ null, " Divide the numerator and the denominator by 2. So 26 over 20 becomes 13 over 10. Now 13 times 2 is 26. 4 times 13 is 52. And then we can also reduce those fractions. 26 over 10 reduces to 13 over 5.", null, " 5, if you divide both numbers by 2. And the same is true for 52 over 10. If you divide the top and bottom by 2, you get 26 over 5. So this vector here is the vector that is parallel to, or rather, it is.", null, " This is the component of vector u that is parallel to vector v. So, that is the projection of u onto v. Now, the last thing we need to do is find w2, which is the vector component of u, orthogonal or perpendicular to vector v.", null, " And w2 is going to be vector u minus w1. So it's 3 comma 5 minus 13 over 5 comma 26 over 5. Now let's get common denominators. So 3 is the same as 15 over 5.", null, " Over 5 and 5 is the same as 25 over 5. So 15 minus 13 is 2 and 25 minus 20 is 5. So this is 2/5. So this is the component of vector u that is orthogonal.", null, " Or perpendicular to vector v. Now let's work on another example. So let's say that vector u is 6i minus 3j plus 9k. And let's say that vector v is 4i minus j plus 8k." ], "text_ocr_list": [ null, "We can see these text from the image: 1. (a) Find W1 - the projection of u onto v. (b) Find W2 - the vector component of u orthogonal to v.\n\nV = <3, 5> V = <2, 4>\n\nW1 = (6 + 20) / (4 + 16) <2, 4> = 26 / 20 <2, 4>\n\nW1 -.\n Divide the numerator and the denominator by 2. So 26 over 20 becomes 13 over 10. Now 13 times 2 is 26. 4 times 13 is 52. And then we can also reduce those fractions. 26 over 10 reduces to 13 over 5.", null, "We can see these text from the image: 1. (a) Find W1 - the projection of u onto v. (b) Find W2 - the vector component of u orthogonal to v.\n\nv = <3, 5> V = <2, 4>\n\nW1 = (6 + 20) / (4 + 16) <2, 4> = 26 / 20 <2, 4>\n\nW1 = 13 / 10 <2, 4> = <26 / 10, 52 / 10>\n\nW1 = <13 / 5, 1>.\n 5, if you divide both numbers by 2. And the same is true for 52 over 10. If you divide the top and bottom by 2, you get 26 over 5. So this vector here is the vector that is parallel to, or rather, it is.", null, "We can see these text from the image: 1. (a) Find W1 - the projection of u onto v. (b) Find W2 - the vector component of u orthogonal to v.\n\nV = <3, 5> V = <2, 4>\n\nW1 = (6 + 20) <2, 4> = 26 <2, 4>\n4 + 16 20\n\nW1 = 13 <2, 4> = <26, 52>\n10 10\n\nW1 = <13/5, 26/5>.\n This is the component of vector u that is parallel to vector v. So, that is the projection of u onto v. Now, the last thing we need to do is find w2, which is the vector component of u, orthogonal or perpendicular to vector v.", null, "We can see these text from the image: 1. (a) Find W1 - the projection of u onto v. (b) Find W2 - the vector component of u orthogonal to v.\n\nv = <3, 5> V = <2, 4>\n\nW2 = U\n\nW1 = <13/5, 26/5> W1 || V.\n And w2 is going to be vector u minus w1. So it's 3 comma 5 minus 13 over 5 comma 26 over 5. Now let's get common denominators. So 3 is the same as 15 over 5.", null, "We can see these text from the image: 1. (a) Find W1 - the projection of u onto v. (b) Find W2 - the vector component of u orthogonal to v.\n\nu = <3, 5> V = <2, 4>\n\nW2 = U - W1\n\n= <3, 5> - <13/5, 26/5>\n\n= <15/5, 25>.\n Over 5 and 5 is the same as 25 over 5. So 15 minus 13 is 2 and 25 minus 20 is 5. So this is 2/5. So this is the component of vector u that is orthogonal.", null, "We can see these text from the image: 1. (a) Find W1 - the projection of u onto v.\n(b) Find W2 - the vector component of u orthogonal to v..\n Or perpendicular to vector v. Now let's work on another example. So let's say that vector u is 6i minus 3j plus 9k. And let's say that vector v is 4i minus j plus 8k." ], "metadata": [ { "vid": "Rw70zkvqEiE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dot Product in Vectors_30.json#####audio#####doingASR#####FinishASR/Rw70zkvqEiE/313.22_339.32000000000005.mp4", "refined_asr": " Divide the numerator and the denominator by 2. So 26 over 20 becomes 13 over 10. Now 13 times 2 is 26. 4 times 13 is 52. And then we can also reduce those fractions. 26 over 10 reduces to 13 over 5.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@313.22_339.32000000000005#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@313.22_339.32000000000005#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@313.22_339.32000000000005#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@313.22_339.32000000000005#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@313.22_339.32000000000005#4.jpg" ], "ocr_qwen2_vl_72b": "1. (a) Find W1 - the projection of u onto v. (b) Find W2 - the vector component of u orthogonal to v.\n\nV = <3, 5> V = <2, 4>\n\nW1 = (6 + 20) / (4 + 16) <2, 4> = 26 / 20 <2, 4>\n\nW1 -" }, { "vid": "Rw70zkvqEiE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dot Product in Vectors_30.json#####audio#####doingASR#####FinishASR/Rw70zkvqEiE/339.32000000000005_355.22.mp4", "refined_asr": " 5, if you divide both numbers by 2. And the same is true for 52 over 10. If you divide the top and bottom by 2, you get 26 over 5. So this vector here is the vector that is parallel to, or rather, it is.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@339.32000000000005_355.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@339.32000000000005_355.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@339.32000000000005_355.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@339.32000000000005_355.22#3.jpg" ], "ocr_qwen2_vl_72b": "1. (a) Find W1 - the projection of u onto v. (b) Find W2 - the vector component of u orthogonal to v.\n\nv = <3, 5> V = <2, 4>\n\nW1 = (6 + 20) / (4 + 16) <2, 4> = 26 / 20 <2, 4>\n\nW1 = 13 / 10 <2, 4> = <26 / 10, 52 / 10>\n\nW1 = <13 / 5, 1>" }, { "vid": "Rw70zkvqEiE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dot Product in Vectors_30.json#####audio#####doingASR#####FinishASR/Rw70zkvqEiE/355.22_382.40000000000003.mp4", "refined_asr": " This is the component of vector u that is parallel to vector v. So, that is the projection of u onto v. Now, the last thing we need to do is find w2, which is the vector component of u, orthogonal or perpendicular to vector v.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@355.22_382.40000000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@355.22_382.40000000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@355.22_382.40000000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@355.22_382.40000000000003#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@355.22_382.40000000000003#4.jpg" ], "ocr_qwen2_vl_72b": "1. (a) Find W1 - the projection of u onto v. (b) Find W2 - the vector component of u orthogonal to v.\n\nV = <3, 5> V = <2, 4>\n\nW1 = (6 + 20) <2, 4> = 26 <2, 4>\n4 + 16 20\n\nW1 = 13 <2, 4> = <26, 52>\n10 10\n\nW1 = <13/5, 26/5>" }, { "vid": "Rw70zkvqEiE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dot Product in Vectors_30.json#####audio#####doingASR#####FinishASR/Rw70zkvqEiE/382.40000000000003_406.1.mp4", "refined_asr": " And w2 is going to be vector u minus w1. So it's 3 comma 5 minus 13 over 5 comma 26 over 5. Now let's get common denominators. So 3 is the same as 15 over 5.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@382.40000000000003_406.1#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@382.40000000000003_406.1#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@382.40000000000003_406.1#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@382.40000000000003_406.1#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@382.40000000000003_406.1#4.jpg" ], "ocr_qwen2_vl_72b": "1. (a) Find W1 - the projection of u onto v. (b) Find W2 - the vector component of u orthogonal to v.\n\nv = <3, 5> V = <2, 4>\n\nW2 = U\n\nW1 = <13/5, 26/5> W1 || V" }, { "vid": "Rw70zkvqEiE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dot Product in Vectors_30.json#####audio#####doingASR#####FinishASR/Rw70zkvqEiE/406.16_432.32000000000005.mp4", "refined_asr": " Over 5 and 5 is the same as 25 over 5. So 15 minus 13 is 2 and 25 minus 20 is 5. So this is 2/5. So this is the component of vector u that is orthogonal.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@406.16_432.32000000000005#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@406.16_432.32000000000005#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@406.16_432.32000000000005#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@406.16_432.32000000000005#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@406.16_432.32000000000005#4.jpg" ], "ocr_qwen2_vl_72b": "1. (a) Find W1 - the projection of u onto v. (b) Find W2 - the vector component of u orthogonal to v.\n\nu = <3, 5> V = <2, 4>\n\nW2 = U - W1\n\n= <3, 5> - <13/5, 26/5>\n\n= <15/5, 25>" }, { "vid": "Rw70zkvqEiE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dot Product in Vectors_30.json#####audio#####doingASR#####FinishASR/Rw70zkvqEiE/432.32_454.96.mp4", "refined_asr": " Or perpendicular to vector v. Now let's work on another example. So let's say that vector u is 6i minus 3j plus 9k. And let's say that vector v is 4i minus j plus 8k.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@432.32_454.96#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@432.32_454.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@432.32_454.96#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@432.32_454.96#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/Rw70zkvqEiE/Rw70zkvqEiE@432.32_454.96#4.jpg" ], "ocr_qwen2_vl_72b": "1. (a) Find W1 - the projection of u onto v.\n(b) Find W2 - the vector component of u orthogonal to v." } ], "image_num": 6, "text_num": 381, "token_num": 3837 }, { "images": [ "sample_100_images/wQhFQFeYdJU@326.8_356.16#1.jpg", "sample_100_images/wQhFQFeYdJU@326.8_356.16#2.jpg", "sample_100_images/wQhFQFeYdJU@326.8_356.16#4.jpg", null, "sample_100_images/wQhFQFeYdJU@356.8_368.08#1.jpg", null, "sample_100_images/wQhFQFeYdJU@368.08_383.68#1.jpg", "sample_100_images/wQhFQFeYdJU@368.08_383.68#2.jpg", null, null, "sample_100_images/wQhFQFeYdJU@386.32_391.44#1.jpg", null, "sample_100_images/wQhFQFeYdJU@391.44_402.48#1.jpg", "sample_100_images/wQhFQFeYdJU@391.44_402.48#2.jpg", null, "sample_100_images/wQhFQFeYdJU@402.48_410.4#1.jpg", null, "sample_100_images/wQhFQFeYdJU@410.4_414.08#1.jpg", null, "sample_100_images/wQhFQFeYdJU@414.08_450.32#1.jpg", "sample_100_images/wQhFQFeYdJU@414.08_450.32#2.jpg", "sample_100_images/wQhFQFeYdJU@414.08_450.32#3.jpg", "sample_100_images/wQhFQFeYdJU@414.08_450.32#4.jpg", "sample_100_images/wQhFQFeYdJU@414.08_450.32#5.jpg", null ], "texts": [ null, null, null, " The DRC, as a resource-rich place, was highly coveted by colonizers. And this jump-started political unrest and suffering that's still ongoing. In 1960, when the Democratic Republic of the Congo gained independence from Belgium, this resource-rich country had a population that had been denied human rights like education for generations, leaving the population with a mostly primary economy. By 1996, war erupted in the DRC, and it's considered by the International Rescue Committee as the deadliest war since World War II.", null, " Even in 2022, regional conflicts are ongoing. And the war there has layers of tensions, like ethnic conflicts and distrust between villages and the central government, that make it hard to broker a lasting peace.", null, null, " Some of the fighting is also tied to access to land, especially land containing minerals. Because the process to legally mine in DRC is time-consuming and burdensome, many mines are illegal. And that means there's often not a lot of oversight over how the land is used or how the workers are being treated.", " These mines become sites of abuse of those working there.", null, " And mine profits can end up funding the militias that defend the territory from outsiders and the government.", null, null, " The miners themselves tend to do hard work with hand tools. And miners, their families, and their communities are exposed to a range of toxic materials, adding poor health to the list of troubles the people of DRC face.", null, " Yet, Congolese people still come to work in these mines because there are few ways to earn a living. Though those who work in the small, unregulated mines still don't make a lot of money.", null, " By not paying much for the labor, it keeps the price of cobalt low. As the world comes to understand the cost of mining on the land, it's important to realize that the cost of mining on the land is not necessarily lower than the cost of mining elsewhere.", null, null, null, null, null, " As the world comes to understand the cost of mining on the mining communities themselves, there is mounting pressure put on governments and corporations to stop using unregulated minerals. This might help to ensure some of the mining wealth stays in the mining communities, and to stop the use of child labor in DRC mines. There's also a debate between international human rights advocates, mining communities, and international corporations over whether to classify cobalt from the DRC as a conflict mineral, which are minerals that create and perpetuate conflict and war. And if cobalt was a conflict mineral, suppliers would have to prove that cobalt came from mines not funding militant groups and perpetuating the conflict in the DRC." ], "text_ocr_list": [ null, null, null, "We can see these text from the image: - France\n- Britain\n- Germany\n- Belgium\n- Portugal\n- Spain\n- Independent.\n The DRC, as a resource-rich place, was highly coveted by colonizers. And this jump-started political unrest and suffering that's still ongoing. In 1960, when the Democratic Republic of the Congo gained independence from Belgium, this resource-rich country had a population that had been denied human rights like education for generations, leaving the population with a mostly primary economy. By 1996, war erupted in the DRC, and it's considered by the International Rescue Committee as the deadliest war since World War II.", null, "We can see these text from the image: 2022\n\nDEMOCRATIC REPUBLIC OF THE CONGO.\n Even in 2022, regional conflicts are ongoing. And the war there has layers of tensions, like ethnic conflicts and distrust between villages and the central government, that make it hard to broker a lasting peace.", null, null, "We can see these text from the image: IGA BARRIERE, DRC.\n Some of the fighting is also tied to access to land, especially land containing minerals. Because the process to legally mine in DRC is time-consuming and burdensome, many mines are illegal. And that means there's often not a lot of oversight over how the land is used or how the workers are being treated.", " These mines become sites of abuse of those working there.", null, "We can see these text from the image: There is no text or formula present in the image..\n And mine profits can end up funding the militias that defend the territory from outsiders and the government.", null, null, "We can see these text from the image: RUBAYA, DRC.\n The miners themselves tend to do hard work with hand tools. And miners, their families, and their communities are exposed to a range of toxic materials, adding poor health to the list of troubles the people of DRC face.", null, "We can see these text from the image: NORTH KIVU, DRC.\n Yet, Congolese people still come to work in these mines because there are few ways to earn a living. Though those who work in the small, unregulated mines still don't make a lot of money.", null, "We can see these text from the image: No text or formulas present in the image..\n By not paying much for the labor, it keeps the price of cobalt low. As the world comes to understand the cost of mining on the land, it's important to realize that the cost of mining on the land is not necessarily lower than the cost of mining elsewhere.", null, null, null, null, null, "We can see these text from the image: There is no text or formula present in the image..\n As the world comes to understand the cost of mining on the mining communities themselves, there is mounting pressure put on governments and corporations to stop using unregulated minerals. This might help to ensure some of the mining wealth stays in the mining communities, and to stop the use of child labor in DRC mines. There's also a debate between international human rights advocates, mining communities, and international corporations over whether to classify cobalt from the DRC as a conflict mineral, which are minerals that create and perpetuate conflict and war. And if cobalt was a conflict mineral, suppliers would have to prove that cobalt came from mines not funding militant groups and perpetuating the conflict in the DRC." ], "metadata": [ { "vid": "wQhFQFeYdJU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Geology and Society: Environmental Regulations in Geology Policy\" tutorial videos._22.json#####audio#####doingASR#####FinishASR/wQhFQFeYdJU/326.8_356.16.mp4", "refined_asr": " The DRC, as a resource-rich place, was highly coveted by colonizers. And this jump-started political unrest and suffering that's still ongoing. In 1960, when the Democratic Republic of the Congo gained independence from Belgium, this resource-rich country had a population that had been denied human rights like education for generations, leaving the population with a mostly primary economy. By 1996, war erupted in the DRC, and it's considered by the International Rescue Committee as the deadliest war since World War II.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@326.8_356.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@326.8_356.16#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@326.8_356.16#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@326.8_356.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@326.8_356.16#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@326.8_356.16#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@326.8_356.16#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@326.8_356.16#5.jpg" ], "ocr_qwen2_vl_72b": "- France\n- Britain\n- Germany\n- Belgium\n- Portugal\n- Spain\n- Independent" }, { "vid": "wQhFQFeYdJU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Geology and Society: Environmental Regulations in Geology Policy\" tutorial videos._22.json#####audio#####doingASR#####FinishASR/wQhFQFeYdJU/356.8_368.08.mp4", "refined_asr": " Even in 2022, regional conflicts are ongoing. And the war there has layers of tensions, like ethnic conflicts and distrust between villages and the central government, that make it hard to broker a lasting peace.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@356.8_368.08#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@356.8_368.08#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@356.8_368.08#2.jpg" ], "ocr_qwen2_vl_72b": "2022\n\nDEMOCRATIC REPUBLIC OF THE CONGO" }, { "vid": "wQhFQFeYdJU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Geology and Society: Environmental Regulations in Geology Policy\" tutorial videos._22.json#####audio#####doingASR#####FinishASR/wQhFQFeYdJU/368.08_383.68.mp4", "refined_asr": " Some of the fighting is also tied to access to land, especially land containing minerals. Because the process to legally mine in DRC is time-consuming and burdensome, many mines are illegal. And that means there's often not a lot of oversight over how the land is used or how the workers are being treated.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@368.08_383.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@368.08_383.68#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@368.08_383.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@368.08_383.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@368.08_383.68#3.jpg" ], "ocr_qwen2_vl_72b": "IGA BARRIERE, DRC" }, { "vid": "wQhFQFeYdJU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Geology and Society: Environmental Regulations in Geology Policy\" tutorial videos._22.json#####audio#####doingASR#####FinishASR/wQhFQFeYdJU/383.68_386.32.mp4", "refined_asr": " These mines become sites of abuse of those working there.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@383.68_386.32#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "wQhFQFeYdJU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Geology and Society: Environmental Regulations in Geology Policy\" tutorial videos._22.json#####audio#####doingASR#####FinishASR/wQhFQFeYdJU/386.32_391.44.mp4", "refined_asr": " And mine profits can end up funding the militias that defend the territory from outsiders and the government.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@386.32_391.44#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@386.32_391.44#1.jpg" ], "ocr_qwen2_vl_72b": "There is no text or formula present in the image." }, { "vid": "wQhFQFeYdJU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Geology and Society: Environmental Regulations in Geology Policy\" tutorial videos._22.json#####audio#####doingASR#####FinishASR/wQhFQFeYdJU/391.44_402.48.mp4", "refined_asr": " The miners themselves tend to do hard work with hand tools. And miners, their families, and their communities are exposed to a range of toxic materials, adding poor health to the list of troubles the people of DRC face.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@391.44_402.48#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@391.44_402.48#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@391.44_402.48#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@391.44_402.48#2.jpg" ], "ocr_qwen2_vl_72b": "RUBAYA, DRC" }, { "vid": "wQhFQFeYdJU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Geology and Society: Environmental Regulations in Geology Policy\" tutorial videos._22.json#####audio#####doingASR#####FinishASR/wQhFQFeYdJU/402.48_410.4.mp4", "refined_asr": " Yet, Congolese people still come to work in these mines because there are few ways to earn a living. Though those who work in the small, unregulated mines still don't make a lot of money.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@402.48_410.4#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@402.48_410.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@402.48_410.4#2.jpg" ], "ocr_qwen2_vl_72b": "NORTH KIVU, DRC" }, { "vid": "wQhFQFeYdJU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Geology and Society: Environmental Regulations in Geology Policy\" tutorial videos._22.json#####audio#####doingASR#####FinishASR/wQhFQFeYdJU/410.4_414.08.mp4", "refined_asr": " By not paying much for the labor, it keeps the price of cobalt low. As the world comes to understand the cost of mining on the land, it's important to realize that the cost of mining on the land is not necessarily lower than the cost of mining elsewhere.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@410.4_414.08#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@410.4_414.08#1.jpg" ], "ocr_qwen2_vl_72b": "No text or formulas present in the image." }, { "vid": "wQhFQFeYdJU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Geology and Society: Environmental Regulations in Geology Policy\" tutorial videos._22.json#####audio#####doingASR#####FinishASR/wQhFQFeYdJU/414.08_450.32.mp4", "refined_asr": " As the world comes to understand the cost of mining on the mining communities themselves, there is mounting pressure put on governments and corporations to stop using unregulated minerals. This might help to ensure some of the mining wealth stays in the mining communities, and to stop the use of child labor in DRC mines. There's also a debate between international human rights advocates, mining communities, and international corporations over whether to classify cobalt from the DRC as a conflict mineral, which are minerals that create and perpetuate conflict and war. And if cobalt was a conflict mineral, suppliers would have to prove that cobalt came from mines not funding militant groups and perpetuating the conflict in the DRC.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#5.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/wQhFQFeYdJU/wQhFQFeYdJU@414.08_450.32#6.jpg" ], "ocr_qwen2_vl_72b": "There is no text or formula present in the image." } ], "image_num": 16, "text_num": 611, "token_num": 9827 }, { "images": [ "sample_100_images/kyjlxsLW1Is@61.440000000000005_83.2#1.jpg", "sample_100_images/kyjlxsLW1Is@61.440000000000005_83.2#3.jpg", null, "sample_100_images/kyjlxsLW1Is@83.2_105.5#1.jpg", null, "sample_100_images/kyjlxsLW1Is@106.10000000000001_131.0#1.jpg", null, "sample_100_images/kyjlxsLW1Is@132.3_153.9#1.jpg", null, "sample_100_images/kyjlxsLW1Is@155.1_173.0#1.jpg", null, "sample_100_images/kyjlxsLW1Is@173.6_188.9#1.jpg", null ], "texts": [ null, null, " And to keep it light and interesting I've themed all of the examples in this video on my latest obsession which is the NBA Despite proudly following Australian sports my brother's getting me hopelessly addicted to American basketball So here we go The first thing we're going to delve into is what types of data", null, " We're going to encounter different types of data when dealing with statistics. Broadly, we can categorize data into two main classes: categorical data and numerical data. These terms are quite self-explanatory; numerical refers to data involving numbers, while categorical pertains to data grouped into categories.", null, " So I'll give you some examples of those in a second. But categorical data can be further split into nominal categorical data and ordinal categorical data. Nominal meaning there is no order to the various categories of a particular variable, and ordinal means that there is some kind of order to the categories. We'll see some examples in a sec.", null, " Numerical data can be further split into discrete numerical data or continuous numerical data. And again we'll have a look at some examples right now actually. So if I was to ask you what team does Stephen Curry play for you can see that clearly the answer to that question is not going to be numerical.", null, " So it's a categorical piece of data. And here in these brackets, I've put what's called the sample space for this particular question. Now, Steph Curry could either play for the Atlanta Hawks, the Boston Celtics, et cetera, et cetera.", null, " Turns out he plays for the Golden State Warriors. But all these potential values for what team Steph Curry could play for when we combine them we call that the sample space. And you can see that there's no order to the teams." ], "text_ocr_list": [ null, null, "We can see these text from the image: An introduction to statistics\n\nzstatistics.com\n\nData types\nDistributions\nSampling and Estimation\nHypothesis testing\np-values.\n And to keep it light and interesting I've themed all of the examples in this video on my latest obsession which is the NBA Despite proudly following Australian sports my brother's getting me hopelessly addicted to American basketball So here we go The first thing we're going to delve into is what types of data", null, "We can see these text from the image: Data types\n\nCategorical Numerical.\n We're going to encounter different types of data when dealing with statistics. Broadly, we can categorize data into two main classes: categorical data and numerical data. These terms are quite self-explanatory; numerical refers to data involving numbers, while categorical pertains to data grouped into categories.", null, "We can see these text from the image: Data types\n\nCategorical\n- Nominal\n- Ordinal\n\nNumerical.\n So I'll give you some examples of those in a second. But categorical data can be further split into nominal categorical data and ordinal categorical data. Nominal meaning there is no order to the various categories of a particular variable, and ordinal means that there is some kind of order to the categories. We'll see some examples in a sec.", null, "We can see these text from the image: Data types\n\nCategorical\nNominal Ordinal\n\nNumerical\nDiscrete Continuous.\n Numerical data can be further split into discrete numerical data or continuous numerical data. And again we'll have a look at some examples right now actually. So if I was to ask you what team does Stephen Curry play for you can see that clearly the answer to that question is not going to be numerical.", null, "We can see these text from the image: Data types\n\nCategorical Numerical\n\nNominal Ordinal Discrete Continuous\n\n\"What team does Steph Curry play for?\"\n\n{Atlanta Hawks, Boston Celtics...}.\n So it's a categorical piece of data. And here in these brackets, I've put what's called the sample space for this particular question. Now, Steph Curry could either play for the Atlanta Hawks, the Boston Celtics, et cetera, et cetera.", null, "We can see these text from the image: Data types\n\nCategorical Numerical\n\nNominal Ordinal Discrete Continuous\n\n\"What team does Steph Curry play for?\"\n{Atlanta Hawks, Boston Celtics...}.\n Turns out he plays for the Golden State Warriors. But all these potential values for what team Steph Curry could play for when we combine them we call that the sample space. And you can see that there's no order to the teams." ], "metadata": [ { "vid": "kyjlxsLW1Is.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Standard Deviation in Probability and Statistics_30.json#####audio#####doingASR#####FinishASR/kyjlxsLW1Is/61.440000000000005_83.2.mp4", "refined_asr": " And to keep it light and interesting I've themed all of the examples in this video on my latest obsession which is the NBA Despite proudly following Australian sports my brother's getting me hopelessly addicted to American basketball So here we go The first thing we're going to delve into is what types of data", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@61.440000000000005_83.2#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@61.440000000000005_83.2#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@61.440000000000005_83.2#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@61.440000000000005_83.2#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@61.440000000000005_83.2#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@61.440000000000005_83.2#4.jpg" ], "ocr_qwen2_vl_72b": "An introduction to statistics\n\nzstatistics.com\n\nData types\nDistributions\nSampling and Estimation\nHypothesis testing\np-values" }, { "vid": "kyjlxsLW1Is.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Standard Deviation in Probability and Statistics_30.json#####audio#####doingASR#####FinishASR/kyjlxsLW1Is/83.2_105.5.mp4", "refined_asr": " We're going to encounter different types of data when dealing with statistics. Broadly, we can categorize data into two main classes: categorical data and numerical data. These terms are quite self-explanatory; numerical refers to data involving numbers, while categorical pertains to data grouped into categories.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@83.2_105.5#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@83.2_105.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@83.2_105.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@83.2_105.5#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@83.2_105.5#4.jpg" ], "ocr_qwen2_vl_72b": "Data types\n\nCategorical Numerical" }, { "vid": "kyjlxsLW1Is.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Standard Deviation in Probability and Statistics_30.json#####audio#####doingASR#####FinishASR/kyjlxsLW1Is/106.10000000000001_131.0.mp4", "refined_asr": " So I'll give you some examples of those in a second. But categorical data can be further split into nominal categorical data and ordinal categorical data. Nominal meaning there is no order to the various categories of a particular variable, and ordinal means that there is some kind of order to the categories. We'll see some examples in a sec.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@106.10000000000001_131.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@106.10000000000001_131.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@106.10000000000001_131.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@106.10000000000001_131.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@106.10000000000001_131.0#4.jpg" ], "ocr_qwen2_vl_72b": "Data types\n\nCategorical\n- Nominal\n- Ordinal\n\nNumerical" }, { "vid": "kyjlxsLW1Is.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Standard Deviation in Probability and Statistics_30.json#####audio#####doingASR#####FinishASR/kyjlxsLW1Is/132.3_153.9.mp4", "refined_asr": " Numerical data can be further split into discrete numerical data or continuous numerical data. And again we'll have a look at some examples right now actually. So if I was to ask you what team does Stephen Curry play for you can see that clearly the answer to that question is not going to be numerical.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@132.3_153.9#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@132.3_153.9#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@132.3_153.9#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@132.3_153.9#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@132.3_153.9#4.jpg" ], "ocr_qwen2_vl_72b": "Data types\n\nCategorical\nNominal Ordinal\n\nNumerical\nDiscrete Continuous" }, { "vid": "kyjlxsLW1Is.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Standard Deviation in Probability and Statistics_30.json#####audio#####doingASR#####FinishASR/kyjlxsLW1Is/155.1_173.0.mp4", "refined_asr": " So it's a categorical piece of data. And here in these brackets, I've put what's called the sample space for this particular question. Now, Steph Curry could either play for the Atlanta Hawks, the Boston Celtics, et cetera, et cetera.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@155.1_173.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@155.1_173.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@155.1_173.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@155.1_173.0#3.jpg" ], "ocr_qwen2_vl_72b": "Data types\n\nCategorical Numerical\n\nNominal Ordinal Discrete Continuous\n\n\"What team does Steph Curry play for?\"\n\n{Atlanta Hawks, Boston Celtics...}" }, { "vid": "kyjlxsLW1Is.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Standard Deviation in Probability and Statistics_30.json#####audio#####doingASR#####FinishASR/kyjlxsLW1Is/173.6_188.9.mp4", "refined_asr": " Turns out he plays for the Golden State Warriors. But all these potential values for what team Steph Curry could play for when we combine them we call that the sample space. And you can see that there's no order to the teams.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@173.6_188.9#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@173.6_188.9#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@173.6_188.9#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/kyjlxsLW1Is/kyjlxsLW1Is@173.6_188.9#3.jpg" ], "ocr_qwen2_vl_72b": "Data types\n\nCategorical Numerical\n\nNominal Ordinal Discrete Continuous\n\n\"What team does Steph Curry play for?\"\n{Atlanta Hawks, Boston Celtics...}" } ], "image_num": 7, "text_num": 392, "token_num": 4424 }, { "images": [ "sample_100_images/ZOZTX2xDnSM@387.14_408.4#1.jpg", "sample_100_images/ZOZTX2xDnSM@387.14_408.4#2.jpg", "sample_100_images/ZOZTX2xDnSM@387.14_408.4#3.jpg", null, "sample_100_images/ZOZTX2xDnSM@408.4_420.7#1.jpg", "sample_100_images/ZOZTX2xDnSM@408.4_420.7#2.jpg", null, null, "sample_100_images/ZOZTX2xDnSM@422.7_444.0#1.jpg", "sample_100_images/ZOZTX2xDnSM@422.7_444.0#2.jpg", "sample_100_images/ZOZTX2xDnSM@422.7_444.0#3.jpg", null, "sample_100_images/ZOZTX2xDnSM@444.0_482.5#1.jpg", "sample_100_images/ZOZTX2xDnSM@444.0_482.5#3.jpg", "sample_100_images/ZOZTX2xDnSM@444.0_482.5#4.jpg", "sample_100_images/ZOZTX2xDnSM@444.0_482.5#5.jpg", "sample_100_images/ZOZTX2xDnSM@444.0_482.5#6.jpg", null, "sample_100_images/ZOZTX2xDnSM@482.5_486.0#1.jpg", null, null, "sample_100_images/ZOZTX2xDnSM@489.0_496.0#1.jpg", null, "sample_100_images/ZOZTX2xDnSM@496.0_507.5#1.jpg", "sample_100_images/ZOZTX2xDnSM@496.0_507.5#2.jpg", null, "sample_100_images/ZOZTX2xDnSM@507.5_513.5#1.jpg", null ], "texts": [ null, null, null, " This is a very similar result, so I'm not going to go through the details here. You would just need this result I mentioned earlier that if an element has order infinity then all of its powers are distinct. That way you would be able to draw a bijection from the infinite cyclic group it generates to the set of integers.", null, null, " This, of course, also means that every infinite cyclic group is, in fact, isomorphic to the set of integers. Therefore, any two infinite cyclic groups are isomorphic to each other.", " One more thing before we go.", null, null, null, " Certainly, the integers mod 15 is a cyclic group. For example, it's entirely generated by 1. But it also contains some subgroups that are cyclic. For example, consider the powers of 3. The 0th power, or in this case, we're talking about addition. So let's say 0.", null, null, null, null, null, " Let's say multiple. The 0th multiple of 3 would be 0. Then 3, then 6, then 9, then 12, and then 15. Which is 0. So we'd actually just come right back around. If it wasn't already clear, that's why these are called cyclic groups. Because, at least in the case of finite order, a generator will eventually complete an entire cycle. And then just keep looping through the group repeatedly. A couple things you might notice here is that if we combine any of these multiples of 3, we get another multiple of 3.", null, " Three plus six, for example, brings us to nine.", " Nine plus twelve brings us to twenty-one.", null, " Mod 15 is 6. If we combine multiples of 3, we get another multiple of 3.", null, null, " The second thing you might notice is that if we take the inverse of any of these multiples of three like negative sixteen for example this is also a multiple of three", null, " Negative 6 mod 15 is congruent to 9, the third multiple of 3." ], "text_ocr_list": [ null, null, null, "We can see these text from the image: - \\(\\mathbb{Z}_n = \\{0, 1, 2, ..., n-1\\}\\)\n- \\(f: \\mathbb{Z}_n \\rightarrow \\langle a \\rangle\\)\n- \\(f(i) = a^i\\)\n- \\(f(i + j) = a^{i+j} = a^i a^j = f(i)f(j)\\)\n- \\(\\mathbb{Z}_n \\cong \\langle a \\rangle\\)\n\nFor every \\(n \\in \\mathbb{Z}^+\\), every cyclic group of order \\(n\\) is isomorphic to \\(\\mathbb{Z}_n\\).\n\nEvery infinite cyclic group is isomorphic to \\(\\mathbb{Z}\\)..\n This is a very similar result, so I'm not going to go through the details here. You would just need this result I mentioned earlier that if an element has order infinity then all of its powers are distinct. That way you would be able to draw a bijection from the infinite cyclic group it generates to the set of integers.", null, null, "We can see these text from the image: - \\( \\mathbb{Z}_n = \\{0, 1, 2, \\ldots, n-1\\} \\)\n- \\( f: \\mathbb{Z}_n \\rightarrow \\langle a \\rangle \\)\n- \\( f(i) = a^i \\)\n- \\( f(i + j) = a^{i+j} = a^i a^j = f(i)f(j) \\)\n- \\( \\mathbb{Z}_n \\cong \\langle a \\rangle \\)\n\nFor every \\( n \\in \\mathbb{Z}^+ \\), every cyclic group of order \\( n \\) is isomorphic to \\( \\mathbb{Z}_n \\).\n\nEvery infinite cyclic group is isomorphic to \\( \\mathbb{Z} \\)..\n This, of course, also means that every infinite cyclic group is, in fact, isomorphic to the set of integers. Therefore, any two infinite cyclic groups are isomorphic to each other.", " One more thing before we go.", null, null, null, "We can see these text from the image: Z15.\n Certainly, the integers mod 15 is a cyclic group. For example, it's entirely generated by 1. But it also contains some subgroups that are cyclic. For example, consider the powers of 3. The 0th power, or in this case, we're talking about addition. So let's say 0.", null, null, null, null, null, "We can see these text from the image: Z\u2081\u2085 = <1>.\n Let's say multiple. The 0th multiple of 3 would be 0. Then 3, then 6, then 9, then 12, and then 15. Which is 0. So we'd actually just come right back around. If it wasn't already clear, that's why these are called cyclic groups. Because, at least in the case of finite order, a generator will eventually complete an entire cycle. And then just keep looping through the group repeatedly. A couple things you might notice here is that if we combine any of these multiples of 3, we get another multiple of 3.", null, "We can see these text from the image: Z\u2081\u2085 = <3>\n\n0, 3, 6, 9, 12.\n Three plus six, for example, brings us to nine.", " Nine plus twelve brings us to twenty-one.", null, "We can see these text from the image: Z\u2081\u2085 = <1>\n\n0, 3, 6, 9, 12.\n Mod 15 is 6. If we combine multiples of 3, we get another multiple of 3.", null, null, "We can see these text from the image: Z\u2081\u2085 = <1>\n\n{0, 3, 6, 9, 12}.\n The second thing you might notice is that if we take the inverse of any of these multiples of three like negative sixteen for example this is also a multiple of three", null, "We can see these text from the image: Z15 = <3>\n\n0, 3, 6, 9, 12\n\n-6 \u2261 9.\n Negative 6 mod 15 is congruent to 9, the third multiple of 3." ], "metadata": [ { "vid": "ZOZTX2xDnSM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra explained_100.json#####audio#####doingASR#####FinishASR/ZOZTX2xDnSM/387.14_408.4.mp4", "refined_asr": " This is a very similar result, so I'm not going to go through the details here. You would just need this result I mentioned earlier that if an element has order infinity then all of its powers are distinct. That way you would be able to draw a bijection from the infinite cyclic group it generates to the set of integers.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@387.14_408.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@387.14_408.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@387.14_408.4#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@387.14_408.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@387.14_408.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@387.14_408.4#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@387.14_408.4#4.jpg" ], "ocr_qwen2_vl_72b": "- \\(\\mathbb{Z}_n = \\{0, 1, 2, ..., n-1\\}\\)\n- \\(f: \\mathbb{Z}_n \\rightarrow \\langle a \\rangle\\)\n- \\(f(i) = a^i\\)\n- \\(f(i + j) = a^{i+j} = a^i a^j = f(i)f(j)\\)\n- \\(\\mathbb{Z}_n \\cong \\langle a \\rangle\\)\n\nFor every \\(n \\in \\mathbb{Z}^+\\), every cyclic group of order \\(n\\) is isomorphic to \\(\\mathbb{Z}_n\\).\n\nEvery infinite cyclic group is isomorphic to \\(\\mathbb{Z}\\)." }, { "vid": "ZOZTX2xDnSM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra explained_100.json#####audio#####doingASR#####FinishASR/ZOZTX2xDnSM/408.4_420.7.mp4", "refined_asr": " This, of course, also means that every infinite cyclic group is, in fact, isomorphic to the set of integers. Therefore, any two infinite cyclic groups are isomorphic to each other.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@408.4_420.7#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@408.4_420.7#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@408.4_420.7#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@408.4_420.7#2.jpg" ], "ocr_qwen2_vl_72b": "- \\( \\mathbb{Z}_n = \\{0, 1, 2, \\ldots, n-1\\} \\)\n- \\( f: \\mathbb{Z}_n \\rightarrow \\langle a \\rangle \\)\n- \\( f(i) = a^i \\)\n- \\( f(i + j) = a^{i+j} = a^i a^j = f(i)f(j) \\)\n- \\( \\mathbb{Z}_n \\cong \\langle a \\rangle \\)\n\nFor every \\( n \\in \\mathbb{Z}^+ \\), every cyclic group of order \\( n \\) is isomorphic to \\( \\mathbb{Z}_n \\).\n\nEvery infinite cyclic group is isomorphic to \\( \\mathbb{Z} \\)." }, { "vid": "ZOZTX2xDnSM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra explained_100.json#####audio#####doingASR#####FinishASR/ZOZTX2xDnSM/420.7_422.7.mp4", "refined_asr": " One more thing before we go.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@420.7_422.7#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "ZOZTX2xDnSM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra explained_100.json#####audio#####doingASR#####FinishASR/ZOZTX2xDnSM/422.7_444.0.mp4", "refined_asr": " Certainly, the integers mod 15 is a cyclic group. For example, it's entirely generated by 1. But it also contains some subgroups that are cyclic. For example, consider the powers of 3. The 0th power, or in this case, we're talking about addition. So let's say 0.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@422.7_444.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@422.7_444.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@422.7_444.0#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@422.7_444.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@422.7_444.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@422.7_444.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@422.7_444.0#4.jpg" ], "ocr_qwen2_vl_72b": "Z15" }, { "vid": "ZOZTX2xDnSM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra explained_100.json#####audio#####doingASR#####FinishASR/ZOZTX2xDnSM/444.0_482.5.mp4", "refined_asr": " Let's say multiple. The 0th multiple of 3 would be 0. Then 3, then 6, then 9, then 12, and then 15. Which is 0. So we'd actually just come right back around. If it wasn't already clear, that's why these are called cyclic groups. Because, at least in the case of finite order, a generator will eventually complete an entire cycle. And then just keep looping through the group repeatedly. A couple things you might notice here is that if we combine any of these multiples of 3, we get another multiple of 3.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#6.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@444.0_482.5#6.jpg" ], "ocr_qwen2_vl_72b": "Z\u2081\u2085 = <1>" }, { "vid": "ZOZTX2xDnSM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra explained_100.json#####audio#####doingASR#####FinishASR/ZOZTX2xDnSM/482.5_486.0.mp4", "refined_asr": " Three plus six, for example, brings us to nine.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@482.5_486.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@482.5_486.0#1.jpg" ], "ocr_qwen2_vl_72b": "Z\u2081\u2085 = <3>\n\n0, 3, 6, 9, 12" }, { "vid": "ZOZTX2xDnSM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra explained_100.json#####audio#####doingASR#####FinishASR/ZOZTX2xDnSM/486.0_489.0.mp4", "refined_asr": " Nine plus twelve brings us to twenty-one.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@486.0_489.0#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "ZOZTX2xDnSM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra explained_100.json#####audio#####doingASR#####FinishASR/ZOZTX2xDnSM/489.0_496.0.mp4", "refined_asr": " Mod 15 is 6. If we combine multiples of 3, we get another multiple of 3.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@489.0_496.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@489.0_496.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@489.0_496.0#2.jpg" ], "ocr_qwen2_vl_72b": "Z\u2081\u2085 = <1>\n\n0, 3, 6, 9, 12" }, { "vid": "ZOZTX2xDnSM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra explained_100.json#####audio#####doingASR#####FinishASR/ZOZTX2xDnSM/496.0_507.5.mp4", "refined_asr": " The second thing you might notice is that if we take the inverse of any of these multiples of three like negative sixteen for example this is also a multiple of three", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@496.0_507.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@496.0_507.5#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@496.0_507.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@496.0_507.5#2.jpg" ], "ocr_qwen2_vl_72b": "Z\u2081\u2085 = <1>\n\n{0, 3, 6, 9, 12}" }, { "vid": "ZOZTX2xDnSM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra explained_100.json#####audio#####doingASR#####FinishASR/ZOZTX2xDnSM/507.5_513.5.mp4", "refined_asr": " Negative 6 mod 15 is congruent to 9, the third multiple of 3.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@507.5_513.5#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZOZTX2xDnSM/ZOZTX2xDnSM@507.5_513.5#1.jpg" ], "ocr_qwen2_vl_72b": "Z15 = <3>\n\n0, 3, 6, 9, 12\n\n-6 \u2261 9" } ], "image_num": 18, "text_num": 444, "token_num": 10812 }, { "images": [ "sample_100_images/XOIFmvozB8s@2075.86_2084.64#1.jpg", null, "sample_100_images/XOIFmvozB8s@2084.64_2098.18#1.jpg", null, "sample_100_images/XOIFmvozB8s@2098.18_2112.36#1.jpg", null, "sample_100_images/XOIFmvozB8s@2112.36_2128.84#1.jpg", null, "sample_100_images/XOIFmvozB8s@2128.84_2139.08#1.jpg", null, "sample_100_images/XOIFmvozB8s@2139.08_2149.44#1.jpg", null ], "texts": [ null, " X over 4 squared, which is 1 to the power of 9, equals 220. So, 220 lots of, and we get x cubed over 4 cubed. And 4 cubed is 64.", null, " So it's x cubed over 64. There we go. And then obviously just simplifying this down again. 1 at the start stays as 1. 12 divided by 4. There, obviously, we're doing a quarter of this, so that's 3. So 3x.", null, " Then we have 66 divided by 16 for this next piece, which simplifies to 33 over 8. Not very nice. So we've got 6 divided by 16, which also simplifies to 3 over 8. Lots of x squared. And then we've got the last piece there.", null, " Which is 220 divided by 64? Which gives us 55 over 16 as a fraction. So 55 over 16, x cubed. There we go. As I said, again, we're getting these fractional coefficients of x squared and x cubed. But again, that shouldn't put you off.", null, " Obviously we're just typing it into a calculator. So 66 over 16 just gave us that, 33 over 8, and that's 55 over 16. Okay, but there we go. There's our final answer. There is the first four terms in ascending powers of X.", null, " And obviously, just showing you there not to be put off by these fractional pieces. But there we go. We're going to have a couple of questions for you to have a go at. So let's have a look at those now. Okay, so here are your two questions." ], "text_ocr_list": [ null, "We can see these text from the image: Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + x/4)^12 giving each term in its simplest form.\n\n12C0 = 1\n12C1 = 12\n12C2 = 66\n12C3 = 220\n\n1 + 12(x/4) + 66(x^2/16) + 220(x^3/64)\n\n1 + 12(x/4) + 66(x^2/16) + 220(x^3/64).\n X over 4 squared, which is 1 to the power of 9, equals 220. So, 220 lots of, and we get x cubed over 4 cubed. And 4 cubed is 64.", null, "We can see these text from the image: A LEVEL MATHS TUTOR\n\nFind the first 4 terms, in ascending powers of \\( x \\), of the binomial expansion of \\( \\left(1 + \\frac{x}{4}\\right)^{12} \\) giving each term in its simplest form.\n\n\\( 12C0 = 1 \\)\n\\( 12C1 = 12 \\)\n\\( 12C2 = 66 \\)\n\\( 12C3 = 220 \\)\n\n\\( 1 + 12\\left(\\frac{x}{4}\\right) + 66\\left(\\frac{x^2}{16}\\right) + 220\\left(\\frac{x^3}{64}\\right) \\)\n\nGCSE Maths Tutor.\n So it's x cubed over 64. There we go. And then obviously just simplifying this down again. 1 at the start stays as 1. 12 divided by 4. There, obviously, we're doing a quarter of this, so that's 3. So 3x.", null, "We can see these text from the image: A LEVEL MATHS TUTOR\n\nFind the first 4 terms, in ascending powers of \\( x \\), of the binomial expansion of \\( \\left(1 + \\frac{x}{4}\\right)^{12} \\) giving each term in its simplest form.\n\n\\( 12C0 = 1 \\)\n\n\\( 12C1 = 12 \\)\n\n\\( 12C2 = 66 \\)\n\n\\( 12C3 = 220 \\)\n\n\\( 1 + 12\\left(\\frac{x}{4}\\right) + 66\\left(\\frac{x^2}{16}\\right) + 220\\left(\\frac{x^3}{64}\\right) \\)\n\n\\( 1 + 3x + \\).\n Then we have 66 divided by 16 for this next piece, which simplifies to 33 over 8. Not very nice. So we've got 6 divided by 16, which also simplifies to 3 over 8. Lots of x squared. And then we've got the last piece there.", null, "We can see these text from the image: Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + x/4)^12 giving each term in its simplest form.\n\n12C0 = 1\n12C1 = 12\n12C2 = 66\n12C3 = 220\n\n1 + 12(x/4) + 66(x^2/16) + 220(x^3/64)\n\n1 + 3x + 33/8 x^2 +.\n Which is 220 divided by 64? Which gives us 55 over 16 as a fraction. So 55 over 16, x cubed. There we go. As I said, again, we're getting these fractional coefficients of x squared and x cubed. But again, that shouldn't put you off.", null, "We can see these text from the image: A LEVEL MATHS TUTOR\n\nFind the first 4 terms, in ascending powers of \\( x \\), of the binomial expansion of \\( (1 + \\frac{x}{4})^{12} \\) giving each term in its simplest form.\n\n\\( 12C0 = 1 \\)\n\\( 12C1 = 12 \\)\n\\( 12C2 = 66 \\)\n\\( 12C3 = 220 \\)\n\n\\( 1 + 12(\\frac{x}{4}) + 66(\\frac{x^2}{16}) + 220(\\frac{x^3}{64}) \\)\n\n\\( 1 + 3x + \\frac{33}{8}x^2 + \\frac{55}{16}x^3 \\)\n\nGCSE Maths Tutor\n\nSubscribe.\n Obviously we're just typing it into a calculator. So 66 over 16 just gave us that, 33 over 8, and that's 55 over 16. Okay, but there we go. There's our final answer. There is the first four terms in ascending powers of X.", null, "We can see these text from the image: A LEVEL MATHS TUTOR\n\nFind the first 4 terms, in ascending powers of \\( x \\), of the binomial expansion of \\( \\left(1 + \\frac{x}{4}\\right)^{12} \\) giving each term in its simplest form.\n\n\\( 12C0 = 1 \\)\n\\( 12C1 = 12 \\)\n\\( 12C2 = 66 \\)\n\\( 12C3 = 220 \\)\n\n\\( 1 + 12\\left(\\frac{x}{4}\\right) + 66\\left(\\frac{x^2}{16}\\right) + 220\\left(\\frac{x^3}{64}\\right) \\)\n\n\\( 1 + 3x + \\frac{33}{8}x^2 + \\frac{55}{16}x^3 \\).\n And obviously, just showing you there not to be put off by these fractional pieces. But there we go. We're going to have a couple of questions for you to have a go at. So let's have a look at those now. Okay, so here are your two questions." ], "metadata": [ { "vid": "XOIFmvozB8s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Sequences and Series with focus on Binomial Theorem_30.json#####audio#####doingASR#####FinishASR/XOIFmvozB8s/2075.86_2084.64.mp4", "refined_asr": " X over 4 squared, which is 1 to the power of 9, equals 220. So, 220 lots of, and we get x cubed over 4 cubed. And 4 cubed is 64.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2075.86_2084.64#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2075.86_2084.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2075.86_2084.64#2.jpg" ], "ocr_qwen2_vl_72b": "Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + x/4)^12 giving each term in its simplest form.\n\n12C0 = 1\n12C1 = 12\n12C2 = 66\n12C3 = 220\n\n1 + 12(x/4) + 66(x^2/16) + 220(x^3/64)\n\n1 + 12(x/4) + 66(x^2/16) + 220(x^3/64)" }, { "vid": "XOIFmvozB8s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Sequences and Series with focus on Binomial Theorem_30.json#####audio#####doingASR#####FinishASR/XOIFmvozB8s/2084.64_2098.18.mp4", "refined_asr": " So it's x cubed over 64. There we go. And then obviously just simplifying this down again. 1 at the start stays as 1. 12 divided by 4. There, obviously, we're doing a quarter of this, so that's 3. So 3x.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2084.64_2098.18#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2084.64_2098.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2084.64_2098.18#2.jpg" ], "ocr_qwen2_vl_72b": "A LEVEL MATHS TUTOR\n\nFind the first 4 terms, in ascending powers of \\( x \\), of the binomial expansion of \\( \\left(1 + \\frac{x}{4}\\right)^{12} \\) giving each term in its simplest form.\n\n\\( 12C0 = 1 \\)\n\\( 12C1 = 12 \\)\n\\( 12C2 = 66 \\)\n\\( 12C3 = 220 \\)\n\n\\( 1 + 12\\left(\\frac{x}{4}\\right) + 66\\left(\\frac{x^2}{16}\\right) + 220\\left(\\frac{x^3}{64}\\right) \\)\n\nGCSE Maths Tutor" }, { "vid": "XOIFmvozB8s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Sequences and Series with focus on Binomial Theorem_30.json#####audio#####doingASR#####FinishASR/XOIFmvozB8s/2098.18_2112.36.mp4", "refined_asr": " Then we have 66 divided by 16 for this next piece, which simplifies to 33 over 8. Not very nice. So we've got 6 divided by 16, which also simplifies to 3 over 8. Lots of x squared. And then we've got the last piece there.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2098.18_2112.36#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2098.18_2112.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2098.18_2112.36#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2098.18_2112.36#3.jpg" ], "ocr_qwen2_vl_72b": "A LEVEL MATHS TUTOR\n\nFind the first 4 terms, in ascending powers of \\( x \\), of the binomial expansion of \\( \\left(1 + \\frac{x}{4}\\right)^{12} \\) giving each term in its simplest form.\n\n\\( 12C0 = 1 \\)\n\n\\( 12C1 = 12 \\)\n\n\\( 12C2 = 66 \\)\n\n\\( 12C3 = 220 \\)\n\n\\( 1 + 12\\left(\\frac{x}{4}\\right) + 66\\left(\\frac{x^2}{16}\\right) + 220\\left(\\frac{x^3}{64}\\right) \\)\n\n\\( 1 + 3x + \\)" }, { "vid": "XOIFmvozB8s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Sequences and Series with focus on Binomial Theorem_30.json#####audio#####doingASR#####FinishASR/XOIFmvozB8s/2112.36_2128.84.mp4", "refined_asr": " Which is 220 divided by 64? Which gives us 55 over 16 as a fraction. So 55 over 16, x cubed. There we go. As I said, again, we're getting these fractional coefficients of x squared and x cubed. But again, that shouldn't put you off.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2112.36_2128.84#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2112.36_2128.84#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2112.36_2128.84#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2112.36_2128.84#3.jpg" ], "ocr_qwen2_vl_72b": "Find the first 4 terms, in ascending powers of x, of the binomial expansion of (1 + x/4)^12 giving each term in its simplest form.\n\n12C0 = 1\n12C1 = 12\n12C2 = 66\n12C3 = 220\n\n1 + 12(x/4) + 66(x^2/16) + 220(x^3/64)\n\n1 + 3x + 33/8 x^2 +" }, { "vid": "XOIFmvozB8s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Sequences and Series with focus on Binomial Theorem_30.json#####audio#####doingASR#####FinishASR/XOIFmvozB8s/2128.84_2139.08.mp4", "refined_asr": " Obviously we're just typing it into a calculator. So 66 over 16 just gave us that, 33 over 8, and that's 55 over 16. Okay, but there we go. There's our final answer. There is the first four terms in ascending powers of X.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2128.84_2139.08#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2128.84_2139.08#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2128.84_2139.08#2.jpg" ], "ocr_qwen2_vl_72b": "A LEVEL MATHS TUTOR\n\nFind the first 4 terms, in ascending powers of \\( x \\), of the binomial expansion of \\( (1 + \\frac{x}{4})^{12} \\) giving each term in its simplest form.\n\n\\( 12C0 = 1 \\)\n\\( 12C1 = 12 \\)\n\\( 12C2 = 66 \\)\n\\( 12C3 = 220 \\)\n\n\\( 1 + 12(\\frac{x}{4}) + 66(\\frac{x^2}{16}) + 220(\\frac{x^3}{64}) \\)\n\n\\( 1 + 3x + \\frac{33}{8}x^2 + \\frac{55}{16}x^3 \\)\n\nGCSE Maths Tutor\n\nSubscribe" }, { "vid": "XOIFmvozB8s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Sequences and Series with focus on Binomial Theorem_30.json#####audio#####doingASR#####FinishASR/XOIFmvozB8s/2139.08_2149.44.mp4", "refined_asr": " And obviously, just showing you there not to be put off by these fractional pieces. But there we go. We're going to have a couple of questions for you to have a go at. So let's have a look at those now. Okay, so here are your two questions.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2139.08_2149.44#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2139.08_2149.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XOIFmvozB8s/XOIFmvozB8s@2139.08_2149.44#2.jpg" ], "ocr_qwen2_vl_72b": "A LEVEL MATHS TUTOR\n\nFind the first 4 terms, in ascending powers of \\( x \\), of the binomial expansion of \\( \\left(1 + \\frac{x}{4}\\right)^{12} \\) giving each term in its simplest form.\n\n\\( 12C0 = 1 \\)\n\\( 12C1 = 12 \\)\n\\( 12C2 = 66 \\)\n\\( 12C3 = 220 \\)\n\n\\( 1 + 12\\left(\\frac{x}{4}\\right) + 66\\left(\\frac{x^2}{16}\\right) + 220\\left(\\frac{x^3}{64}\\right) \\)\n\n\\( 1 + 3x + \\frac{33}{8}x^2 + \\frac{55}{16}x^3 \\)" } ], "image_num": 6, "text_num": 412, "token_num": 3868 }, { "images": [ "sample_100_images/6BiqCTqc67Q@1640.02_1648.98#1.jpg", null, "sample_100_images/6BiqCTqc67Q@1650.82_1660.18#1.jpg", null, "sample_100_images/6BiqCTqc67Q@1663.1399999999999_1673.22#1.jpg", null, "sample_100_images/6BiqCTqc67Q@1674.5_1701.1399999999999#1.jpg", null, "sample_100_images/6BiqCTqc67Q@1701.3799999999999_1709.22#1.jpg", null, "sample_100_images/6BiqCTqc67Q@1711.1399999999999_1720.58#1.jpg", null, "sample_100_images/6BiqCTqc67Q@1723.94_1728.82#1.jpg", null, "sample_100_images/6BiqCTqc67Q@1728.82_1738.22#1.jpg", null, null, "sample_100_images/6BiqCTqc67Q@1745.54_1749.1399999999999#1.jpg", null, "sample_100_images/6BiqCTqc67Q@1753.86_1763.2#1.jpg", null ], "texts": [ null, " Okay, so if I'm spinning around y equals negative two, I know there's going to be a hole in the middle of this, or a cylinder drilled out most likely.", null, " So I'm going to put a lot of my graph in the negative Y direction Oh you know what it doesn't want to draw over there", null, " Good. Okay so here's the y-axis and then I'm only going to go up to a little past one right because for sine of x it's not going to go up very high.", null, " Here's the x-axis. We're only interested in the area from zero to pi. So here's one, and then we're spinning around y equals negative two. Spinning around this, perpendicular to y equals negative two, the slices are going to be stacked in the x-axis. Oh!", null, " Alright, so I think that's the x direction, so I'm going to integrate with respect to x.", null, " Okay, so now let's graph sine x. Starting at zero, zero, then pi over two, one; and at pi, zero. We also bound below by the x-axis.", null, " So what are we spinning? We're just spinning this region here.", null, " Okay and I'm spinning it around y equals negative two. So look I have this gap of two units so I'm gonna mirror it on the other side.", " Okay, so this is the result from revolving.", null, " And then I'll go down to negative five there.", null, " And let's change this up so it's gray as well since this was not part of the original region, right? So this is just the result from revolving." ], "text_ocr_list": [ null, "We can see these text from the image: 5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\)..\n Okay, so if I'm spinning around y equals negative two, I know there's going to be a hole in the middle of this, or a cylinder drilled out most likely.", null, "We can see these text from the image: 5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\)..\n So I'm going to put a lot of my graph in the negative Y direction Oh you know what it doesn't want to draw over there", null, "We can see these text from the image: 5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\)..\n Good. Okay so here's the y-axis and then I'm only going to go up to a little past one right because for sine of x it's not going to go up very high.", null, "We can see these text from the image: 5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by y = 0, y = sin x, 0 \u2264 x \u2264 \u03c0 about y = -2..\n Here's the x-axis. We're only interested in the area from zero to pi. So here's one, and then we're spinning around y equals negative two. Spinning around this, perpendicular to y equals negative two, the slices are going to be stacked in the x-axis. Oh!", null, "We can see these text from the image: 5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\)..\n Alright, so I think that's the x direction, so I'm going to integrate with respect to x.", null, "We can see these text from the image: 5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\).\n\nIntegrate w.r.t \\( x \\)\n\n\\( \\int_{\\frac{\\pi}{2}}^{\\pi} \\).\n Okay, so now let's graph sine x. Starting at zero, zero, then pi over two, one; and at pi, zero. We also bound below by the x-axis.", null, "We can see these text from the image: 5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\).\n\nIntegrate w.r.t \\( x \\)\n\n\\( \\int_{0}^{\\pi} \\pi (2 + \\sin x)^2 dx \\).\n So what are we spinning? We're just spinning this region here.", null, "We can see these text from the image: 5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\).\n\nIntegrate w.r.t \\( x \\).\n Okay and I'm spinning it around y equals negative two. So look I have this gap of two units so I'm gonna mirror it on the other side.", " Okay, so this is the result from revolving.", null, "We can see these text from the image: 5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\).\n\nIntegrate w.r.t \\( x \\).\n And then I'll go down to negative five there.", null, "We can see these text from the image: 5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\).\n\nIntegrate w.r.t \\( x \\).\n And let's change this up so it's gray as well since this was not part of the original region, right? So this is just the result from revolving." ], "metadata": [ { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1640.02_1648.98.mp4", "refined_asr": " Okay, so if I'm spinning around y equals negative two, I know there's going to be a hole in the middle of this, or a cylinder drilled out most likely.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1640.02_1648.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1640.02_1648.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1640.02_1648.98#2.jpg" ], "ocr_qwen2_vl_72b": "5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\)." }, { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1650.82_1660.18.mp4", "refined_asr": " So I'm going to put a lot of my graph in the negative Y direction Oh you know what it doesn't want to draw over there", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1650.82_1660.18#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1650.82_1660.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1650.82_1660.18#2.jpg" ], "ocr_qwen2_vl_72b": "5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\)." }, { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1663.1399999999999_1673.22.mp4", "refined_asr": " Good. Okay so here's the y-axis and then I'm only going to go up to a little past one right because for sine of x it's not going to go up very high.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1663.1399999999999_1673.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1663.1399999999999_1673.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1663.1399999999999_1673.22#2.jpg" ], "ocr_qwen2_vl_72b": "5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\)." }, { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1674.5_1701.1399999999999.mp4", "refined_asr": " Here's the x-axis. We're only interested in the area from zero to pi. So here's one, and then we're spinning around y equals negative two. Spinning around this, perpendicular to y equals negative two, the slices are going to be stacked in the x-axis. Oh!", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1674.5_1701.1399999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1674.5_1701.1399999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1674.5_1701.1399999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1674.5_1701.1399999999999#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1674.5_1701.1399999999999#4.jpg" ], "ocr_qwen2_vl_72b": "5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by y = 0, y = sin x, 0 \u2264 x \u2264 \u03c0 about y = -2." }, { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1701.3799999999999_1709.22.mp4", "refined_asr": " Alright, so I think that's the x direction, so I'm going to integrate with respect to x.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1701.3799999999999_1709.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1701.3799999999999_1709.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1701.3799999999999_1709.22#2.jpg" ], "ocr_qwen2_vl_72b": "5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\)." }, { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1711.1399999999999_1720.58.mp4", "refined_asr": " Okay, so now let's graph sine x. Starting at zero, zero, then pi over two, one; and at pi, zero. We also bound below by the x-axis.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1711.1399999999999_1720.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1711.1399999999999_1720.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1711.1399999999999_1720.58#2.jpg" ], "ocr_qwen2_vl_72b": "5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\).\n\nIntegrate w.r.t \\( x \\)\n\n\\( \\int_{\\frac{\\pi}{2}}^{\\pi} \\)" }, { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1723.94_1728.82.mp4", "refined_asr": " So what are we spinning? We're just spinning this region here.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1723.94_1728.82#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1723.94_1728.82#1.jpg" ], "ocr_qwen2_vl_72b": "5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\).\n\nIntegrate w.r.t \\( x \\)\n\n\\( \\int_{0}^{\\pi} \\pi (2 + \\sin x)^2 dx \\)" }, { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1728.82_1738.22.mp4", "refined_asr": " Okay and I'm spinning it around y equals negative two. So look I have this gap of two units so I'm gonna mirror it on the other side.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1728.82_1738.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1728.82_1738.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1728.82_1738.22#2.jpg" ], "ocr_qwen2_vl_72b": "5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\).\n\nIntegrate w.r.t \\( x \\)" }, { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1740.12_1742.9399999999998.mp4", "refined_asr": " Okay, so this is the result from revolving.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1740.12_1742.9399999999998#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1745.54_1749.1399999999999.mp4", "refined_asr": " And then I'll go down to negative five there.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1745.54_1749.1399999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1745.54_1749.1399999999999#1.jpg" ], "ocr_qwen2_vl_72b": "5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\).\n\nIntegrate w.r.t \\( x \\)" }, { "vid": "6BiqCTqc67Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/6BiqCTqc67Q/1753.86_1763.2.mp4", "refined_asr": " And let's change this up so it's gray as well since this was not part of the original region, right? So this is just the result from revolving.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1753.86_1763.2#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1753.86_1763.2#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/6BiqCTqc67Q/6BiqCTqc67Q@1753.86_1763.2#2.jpg" ], "ocr_qwen2_vl_72b": "5) Set up only an integral for the volume of the solid obtained by rotating the region bounded by \\( y = 0 \\), \\( y = \\sin x \\), \\( 0 \\leq x \\leq \\pi \\) about \\( y = -2 \\).\n\nIntegrate w.r.t \\( x \\)" } ], "image_num": 10, "text_num": 377, "token_num": 6137 }, { "images": [ null, null, null, null, "sample_100_images/GJOJl47l2_4@565.72_581.72#1.jpg", "sample_100_images/GJOJl47l2_4@565.72_581.72#2.jpg", null, null, "sample_100_images/GJOJl47l2_4@583.72_588.72#1.jpg", null, "sample_100_images/GJOJl47l2_4@588.72_594.72#1.jpg", null, "sample_100_images/GJOJl47l2_4@594.72_599.72#1.jpg", null, null, "sample_100_images/GJOJl47l2_4@600.72_607.72#1.jpg", null, "sample_100_images/GJOJl47l2_4@607.72_611.72#1.jpg", null, "sample_100_images/GJOJl47l2_4@611.72_616.72#1.jpg", null, null, "sample_100_images/GJOJl47l2_4@618.72_645.72#1.jpg", "sample_100_images/GJOJl47l2_4@618.72_645.72#2.jpg", "sample_100_images/GJOJl47l2_4@618.72_645.72#4.jpg", null, null, null, "sample_100_images/GJOJl47l2_4@671.72_700.72#1.jpg", "sample_100_images/GJOJl47l2_4@671.72_700.72#2.jpg", "sample_100_images/GJOJl47l2_4@671.72_700.72#4.jpg", null, null, null ], "texts": [ " And we'd slice it like this.", " There we go.", " We do some sort of a slice.", " Now tell me something", null, null, " If this distance is x it's on the x-axis and this distance is let's say oh what did I call it? I think I called it y for that. And this distance is y.", " And this is Z.", null, " Could you find the surface area of the cross-section?", null, " Would it have anything to do with X?", null, " What would it have to do with just y and z for the cross-section?", " So check this out.", null, " If I asked you to find the volume of this whole thing here's how you'd do it. You'd probably say oh well I know the volume of a rectangular prism.", null, " The volume of a rectangular prism is just the base times the height times the width.", null, " So basically it's y times z times x.", " Do you follow me on that?", null, null, null, " Now I'm going to kind of prove to you that this is the cross-sectional area times the length. Do you agree that this is the volume of this figure, no matter what? What's y times z? Y times z is the surface area of the cross-section. That's what that is. So basically, this says, oh, that's the surface area of the cross-section times x, the length.", " So this is surface area of cross-section.", " And that's the length.", null, null, null, " Well, now here's the idea. If that's the case, then what we're basically doing here is making a whole bunch of these things. Do you see it? Making a whole bunch of those things and then adding up all the volumes. Very similar to making rectangles for areas, we're making rectangular prisms for volumes. And we're going to add them all up. So the idea I needed to get across to you was that a volume is basically a surface area times the length.", " How many of you feel okay with that?", " Surface area times the length gives you the volume Good deal" ], "text_ocr_list": [ " And we'd slice it like this.", " There we go.", " We do some sort of a slice.", " Now tell me something", null, null, "We can see these text from the image: - Lesson 5.2 - Volume: Disks/Washers\n- Volume of Solids by Slicing:\n - Cut into thin slabs, then use summations to set up an integral.\n - Do this, find area of cross-section.\n If this distance is x it's on the x-axis and this distance is let's say oh what did I call it? I think I called it y for that. And this distance is y.", " And this is Z.", null, "We can see these text from the image: - Lesson 5.2 - Volume: Disks/Washers\n- Volume of Solids by Slicing\n- Cut into thin slabs, then use summations to set up an integral.\n- To do this, find area of cross-section.\n Could you find the surface area of the cross-section?", null, "We can see these text from the image: - LESSON 5.2 - VOLUME: DISKS/WASHERS\n- VOLUME OF SOLIDS BY SLICING:\n - CUT INTO THIN SLABS, THEN USE SUMMATIONS TO SET UP AN INTEGRAL.\n - TO DO THIS, FIND AREA OF CROSS-SECTION.\n Would it have anything to do with X?", null, "We can see these text from the image: - LESSON 5.2 - VOLUME: DISKS/WASHERS\n- VOLUME OF SOLIDS BY SLICING\n- CUT INTO THIN SLABS, THEN USE SUMMATIONS TO SET UP AN INTEGRAL.\n- TO DO THIS, FIND AREA OF CROSS-SECTION.\n What would it have to do with just y and z for the cross-section?", " So check this out.", null, "We can see these text from the image: - Lesson 5.2 - Volume: Disks/Washers\n- Volume of Solids by Slicing:\n - Cut into thin slabs. Then use summations to set up an integral.\n - To do this, find the area of cross-section..\n If I asked you to find the volume of this whole thing here's how you'd do it. You'd probably say oh well I know the volume of a rectangular prism.", null, "We can see these text from the image: - LESSON 5.2 - VOLUME: DISKS/WASHERS\n- VOLUME OF SOLIDS BY SLICING:\n - CUT INTO THIN SLABS, THEN USE SUMMATIONS TO SET UP AN INTEGRAL.\n - FIND AREA OF CROSS-SECTION.\n The volume of a rectangular prism is just the base times the height times the width.", null, "We can see these text from the image: - LESSON 5.2 - VOLUME: DISKS/WASHERS\n- VOLUME OF SOLIDS BY SLICING:\n - CUT INTO THIN SLABS. THEN USE SUMMATIONS TO SET UP AN INTEGRAL.\n - TO DO THIS, AREA OF CROSS-SECTION\n - \\( V = \\pi r^2 \\).\n So basically it's y times z times x.", " Do you follow me on that?", null, null, null, "We can see these text from the image: - Lesson 5.2 - Volume: Disks/Washers\n- Volume of Solids by Slicing\n- Cut into thin slabs, then use summations to set up an integral.\n- To do this, find area of cross-section\n- V = \u222b A(x) dx.\n Now I'm going to kind of prove to you that this is the cross-sectional area times the length. Do you agree that this is the volume of this figure, no matter what? What's y times z? Y times z is the surface area of the cross-section. That's what that is. So basically, this says, oh, that's the surface area of the cross-section times x, the length.", " So this is surface area of cross-section.", " And that's the length.", null, null, null, "We can see these text from the image: - Lesson 5.2: Volume: Disks/Washers\n- Volume of Solids by Slicing\n- Cut into thin slabs, then use summations to set up an integral.\n- To do this, find area of cross-section\n- \\( V = (y \\cdot z) \\cdot x \\)\n- Shape area of cross-section\n- Length.\n Well, now here's the idea. If that's the case, then what we're basically doing here is making a whole bunch of these things. Do you see it? Making a whole bunch of those things and then adding up all the volumes. Very similar to making rectangles for areas, we're making rectangular prisms for volumes. And we're going to add them all up. So the idea I needed to get across to you was that a volume is basically a surface area times the length.", " How many of you feel okay with that?", " Surface area times the length gives you the volume Good deal" ], "metadata": [ { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/540.72_542.72.mp4", "refined_asr": " And we'd slice it like this.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@540.72_542.72#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/560.72_561.72.mp4", "refined_asr": " There we go.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/561.72_563.72.mp4", "refined_asr": " We do some sort of a slice.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@561.72_563.72#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/563.72_565.72.mp4", "refined_asr": " Now tell me something", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/565.72_581.72.mp4", "refined_asr": " If this distance is x it's on the x-axis and this distance is let's say oh what did I call it? I think I called it y for that. And this distance is y.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@565.72_581.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@565.72_581.72#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@565.72_581.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@565.72_581.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@565.72_581.72#3.jpg" ], "ocr_qwen2_vl_72b": "- Lesson 5.2 - Volume: Disks/Washers\n- Volume of Solids by Slicing:\n - Cut into thin slabs, then use summations to set up an integral.\n - Do this, find area of cross-section" }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/581.72_583.72.mp4", "refined_asr": " And this is Z.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/583.72_588.72.mp4", "refined_asr": " Could you find the surface area of the cross-section?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@583.72_588.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@583.72_588.72#1.jpg" ], "ocr_qwen2_vl_72b": "- Lesson 5.2 - Volume: Disks/Washers\n- Volume of Solids by Slicing\n- Cut into thin slabs, then use summations to set up an integral.\n- To do this, find area of cross-section" }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/588.72_594.72.mp4", "refined_asr": " Would it have anything to do with X?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@588.72_594.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@588.72_594.72#1.jpg" ], "ocr_qwen2_vl_72b": "- LESSON 5.2 - VOLUME: DISKS/WASHERS\n- VOLUME OF SOLIDS BY SLICING:\n - CUT INTO THIN SLABS, THEN USE SUMMATIONS TO SET UP AN INTEGRAL.\n - TO DO THIS, FIND AREA OF CROSS-SECTION" }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/594.72_599.72.mp4", "refined_asr": " What would it have to do with just y and z for the cross-section?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@594.72_599.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@594.72_599.72#1.jpg" ], "ocr_qwen2_vl_72b": "- LESSON 5.2 - VOLUME: DISKS/WASHERS\n- VOLUME OF SOLIDS BY SLICING\n- CUT INTO THIN SLABS, THEN USE SUMMATIONS TO SET UP AN INTEGRAL.\n- TO DO THIS, FIND AREA OF CROSS-SECTION" }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/599.72_600.72.mp4", "refined_asr": " So check this out.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/600.72_607.72.mp4", "refined_asr": " If I asked you to find the volume of this whole thing here's how you'd do it. You'd probably say oh well I know the volume of a rectangular prism.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@600.72_607.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@600.72_607.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@600.72_607.72#2.jpg" ], "ocr_qwen2_vl_72b": "- Lesson 5.2 - Volume: Disks/Washers\n- Volume of Solids by Slicing:\n - Cut into thin slabs. Then use summations to set up an integral.\n - To do this, find the area of cross-section." }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/607.72_611.72.mp4", "refined_asr": " The volume of a rectangular prism is just the base times the height times the width.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@607.72_611.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@607.72_611.72#1.jpg" ], "ocr_qwen2_vl_72b": "- LESSON 5.2 - VOLUME: DISKS/WASHERS\n- VOLUME OF SOLIDS BY SLICING:\n - CUT INTO THIN SLABS, THEN USE SUMMATIONS TO SET UP AN INTEGRAL.\n - FIND AREA OF CROSS-SECTION" }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/611.72_616.72.mp4", "refined_asr": " So basically it's y times z times x.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@611.72_616.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@611.72_616.72#1.jpg" ], "ocr_qwen2_vl_72b": "- LESSON 5.2 - VOLUME: DISKS/WASHERS\n- VOLUME OF SOLIDS BY SLICING:\n - CUT INTO THIN SLABS. THEN USE SUMMATIONS TO SET UP AN INTEGRAL.\n - TO DO THIS, AREA OF CROSS-SECTION\n - \\( V = \\pi r^2 \\)" }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/616.72_618.72.mp4", "refined_asr": " Do you follow me on that?", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@616.72_618.72#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/618.72_645.72.mp4", "refined_asr": " Now I'm going to kind of prove to you that this is the cross-sectional area times the length. Do you agree that this is the volume of this figure, no matter what? What's y times z? Y times z is the surface area of the cross-section. That's what that is. So basically, this says, oh, that's the surface area of the cross-section times x, the length.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@618.72_645.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@618.72_645.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@618.72_645.72#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@618.72_645.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@618.72_645.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@618.72_645.72#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@618.72_645.72#4.jpg" ], "ocr_qwen2_vl_72b": "- Lesson 5.2 - Volume: Disks/Washers\n- Volume of Solids by Slicing\n- Cut into thin slabs, then use summations to set up an integral.\n- To do this, find area of cross-section\n- V = \u222b A(x) dx" }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/645.72_647.72.mp4", "refined_asr": " So this is surface area of cross-section.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@645.72_647.72#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/662.72_671.72.mp4", "refined_asr": " And that's the length.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/671.72_700.72.mp4", "refined_asr": " Well, now here's the idea. If that's the case, then what we're basically doing here is making a whole bunch of these things. Do you see it? Making a whole bunch of those things and then adding up all the volumes. Very similar to making rectangles for areas, we're making rectangular prisms for volumes. And we're going to add them all up. So the idea I needed to get across to you was that a volume is basically a surface area times the length.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@671.72_700.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@671.72_700.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@671.72_700.72#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@671.72_700.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@671.72_700.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@671.72_700.72#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@671.72_700.72#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@671.72_700.72#5.jpg" ], "ocr_qwen2_vl_72b": "- Lesson 5.2: Volume: Disks/Washers\n- Volume of Solids by Slicing\n- Cut into thin slabs, then use summations to set up an integral.\n- To do this, find area of cross-section\n- \\( V = (y \\cdot z) \\cdot x \\)\n- Shape area of cross-section\n- Length" }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/700.72_702.72.mp4", "refined_asr": " How many of you feel okay with that?", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@700.72_702.72#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "GJOJl47l2_4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Calculus tutorial on Applications of Integration with a focus on Volumes of Revolution - Disk and Washer Methods\"\n_28.json#####audio#####doingASR#####FinishASR/GJOJl47l2_4/702.72_705.72.mp4", "refined_asr": " Surface area times the length gives you the volume Good deal", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GJOJl47l2_4/GJOJl47l2_4@702.72_705.72#1.jpg" ], "ocr_qwen2_vl_72b": null } ], "image_num": 14, "text_num": 451, "token_num": 8515 }, { "images": [ "sample_100_images/pFSkS6HwXD8@2339.42_2356.7000000000003#1.jpg", null, "sample_100_images/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#1.jpg", "sample_100_images/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#3.jpg", "sample_100_images/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#4.jpg", null, "sample_100_images/pFSkS6HwXD8@2385.5600000000004_2392.6200000000003#1.jpg", null, "sample_100_images/pFSkS6HwXD8@2396.38_2402.2200000000003#1.jpg", null, "sample_100_images/pFSkS6HwXD8@2402.2200000000003_2416.38#1.jpg", null, "sample_100_images/pFSkS6HwXD8@2416.38_2427.98#1.jpg", "sample_100_images/pFSkS6HwXD8@2416.38_2427.98#2.jpg", null, "sample_100_images/pFSkS6HwXD8@2427.98_2434.54#1.jpg", null, "sample_100_images/pFSkS6HwXD8@2435.1800000000003_2445.98#1.jpg", null, "sample_100_images/pFSkS6HwXD8@2446.38_2458.86#1.jpg", null, "sample_100_images/pFSkS6HwXD8@2460.46_2475.9#1.jpg", "sample_100_images/pFSkS6HwXD8@2460.46_2475.9#2.jpg", null, "sample_100_images/pFSkS6HwXD8@2476.38_2488.54#1.jpg", null ], "texts": [ null, " Right there we have it, so we're good to go. So therefore, angle KOL is in fact not 48, but it is 42 degrees. And again, the reason it doesn't change is because triangle KOL is isosceles. So therefore,", null, null, null, " We see that these angles are equal. All right, so next we're looking at angle MLK. Let's scroll up here. Angle MLK, where is that? So under MLK, that is all of this, inclusive of that, say here, 48. So it's", null, " Under MLK key all of this inclusive of the 48 here Well in this case here it is in fact 42", null, " All right so it was supposed to be in fact 42 so the angle we're looking for is inclusive of", null, " This is 42 degrees that we have here. So now all you need to know is how to find this angle over here. We're looking for the entire angle here, inclusive of the 42 degrees. So, I need to", null, null, " Find this angle right in here. Now, it's a little challenging to find that angle. So what I'm going to be doing is using my tangent theorem here.", null, " States that if I have an angle that is created between a chord and a tangent", null, " Then that angle is going to be exactly equal to the angle that is created by the chord in the alternate segment. So this 48 degrees here, this", null, " Chord here KL created an angle all the way up here. So this angle right in here is in fact also 48. Let me just go ahead and write a statement for that.", null, null, " So we can name this angle here an inadvertent angle that we have to calculate. So it is KML, that's our starting statement. So angle KML is in fact equal to", null, " 188 degrees and the reason is simply because the angle between a tangent" ], "text_ocr_list": [ null, "We can see these text from the image: - AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- Angle \\( KLO \\)\n- \\( \\angle KLO = 48^\\circ \\) (\\( \\triangle KOL \\) is an isosceles).\n Right there we have it, so we're good to go. So therefore, angle KOL is in fact not 48, but it is 42 degrees. And again, the reason it doesn't change is because triangle KOL is isosceles. So therefore,", null, null, null, "We can see these text from the image: - AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- Angle \\( KLO \\)\n- \\( \\angle KLO = 42^\\circ \\) (\u0394KOL is an isosceles base angles.).\n We see that these angles are equal. All right, so next we're looking at angle MLK. Let's scroll up here. Angle MLK, where is that? So under MLK, that is all of this, inclusive of that, say here, 48. So it's", null, "We can see these text from the image: - AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKM = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- \u2234 KOL = 180 - 2 x 42\n- 160 - 84 (The sum of the angles)\n- = 96\u00b0 (In a \u0394 are subtended by).\n Under MLK key all of this inclusive of the 48 here Well in this case here it is in fact 42", null, "We can see these text from the image: - AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKH = 90 - 48 (radius bisects tangent)\n- = 42\u00b0\n- \u2234 KOL = 180 - 2 x 42\n- 160 - 84\u00b0 (The sum of the angles in a triangle are subtended by)\n- = 96\u00b0.\n All right so it was supposed to be in fact 42 so the angle we're looking for is inclusive of", null, "We can see these text from the image: - AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKM = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- \u2234 KOL = 180 - 2 x 42\n- 160 - 84\u00b0 (The sum of the angles in a triangle are supplementary)\n- = 96\u00b0.\n This is 42 degrees that we have here. So now all you need to know is how to find this angle over here. We're looking for the entire angle here, inclusive of the 42 degrees. So, I need to", null, null, "We can see these text from the image: - AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKH = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- \u2234 KOL = 180 - 2 x 42\n- 160 - 84 (The sum of the angles in a triangle are supplementary)\n- = 96\u00b0.\n Find this angle right in here. Now, it's a little challenging to find that angle. So what I'm going to be doing is using my tangent theorem here.", null, "We can see these text from the image: - AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKH = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- KOL = 180 - 2 x 42\n- 160 - 84 (The sum of the angles in a triangle are subtended).\n States that if I have an angle that is created between a chord and a tangent", null, "We can see these text from the image: - AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKL = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- KOL = 180 - 2 x 42\n- 160 - 84 (The sum of the angles in a triangle is 180\u00b0).\n Then that angle is going to be exactly equal to the angle that is created by the chord in the alternate segment. So this 48 degrees here, this", null, "We can see these text from the image: - AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKM = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- \u2234 KOL = 180 - 2 x 42\n- 160 - 84\u00b0 (The sum of the angles in a triangle are supplementary)\n- 76\u00b0.\n Chord here KL created an angle all the way up here. So this angle right in here is in fact also 48. Let me just go ahead and write a statement for that.", null, null, "We can see these text from the image: AP3X INSTITUTE OF MATHEMATICS EDUCATION\n\nCSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n\nOKH = 90 - 48 (radius b to tangent)\n= 42\u00b0\n\nKOL = 180 - 2 x 42\n180 - 84\nThe sum of the angles.\n So we can name this angle here an inadvertent angle that we have to calculate. So it is KML, that's our starting statement. So angle KML is in fact equal to", null, "We can see these text from the image: - AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- Angle MLK\n- Angle KML = 48\u00b0\n- Angle KNM.\n 188 degrees and the reason is simply because the angle between a tangent" ], "metadata": [ { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2339.42_2356.7000000000003.mp4", "refined_asr": " Right there we have it, so we're good to go. So therefore, angle KOL is in fact not 48, but it is 42 degrees. And again, the reason it doesn't change is because triangle KOL is isosceles. So therefore,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2339.42_2356.7000000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2339.42_2356.7000000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2339.42_2356.7000000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2339.42_2356.7000000000003#3.jpg" ], "ocr_qwen2_vl_72b": "- AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- Angle \\( KLO \\)\n- \\( \\angle KLO = 48^\\circ \\) (\\( \\triangle KOL \\) is an isosceles)" }, { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2356.7000000000003_2385.5600000000004.mp4", "refined_asr": " We see that these angles are equal. All right, so next we're looking at angle MLK. Let's scroll up here. Angle MLK, where is that? So under MLK, that is all of this, inclusive of that, say here, 48. So it's", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2356.7000000000003_2385.5600000000004#5.jpg" ], "ocr_qwen2_vl_72b": "- AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- Angle \\( KLO \\)\n- \\( \\angle KLO = 42^\\circ \\) (\u0394KOL is an isosceles base angles.)" }, { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2385.5600000000004_2392.6200000000003.mp4", "refined_asr": " Under MLK key all of this inclusive of the 48 here Well in this case here it is in fact 42", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2385.5600000000004_2392.6200000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2385.5600000000004_2392.6200000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2385.5600000000004_2392.6200000000003#2.jpg" ], "ocr_qwen2_vl_72b": "- AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKM = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- \u2234 KOL = 180 - 2 x 42\n- 160 - 84 (The sum of the angles)\n- = 96\u00b0 (In a \u0394 are subtended by)" }, { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2396.38_2402.2200000000003.mp4", "refined_asr": " All right so it was supposed to be in fact 42 so the angle we're looking for is inclusive of", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2396.38_2402.2200000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2396.38_2402.2200000000003#1.jpg" ], "ocr_qwen2_vl_72b": "- AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKH = 90 - 48 (radius bisects tangent)\n- = 42\u00b0\n- \u2234 KOL = 180 - 2 x 42\n- 160 - 84\u00b0 (The sum of the angles in a triangle are subtended by)\n- = 96\u00b0" }, { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2402.2200000000003_2416.38.mp4", "refined_asr": " This is 42 degrees that we have here. So now all you need to know is how to find this angle over here. We're looking for the entire angle here, inclusive of the 42 degrees. So, I need to", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2402.2200000000003_2416.38#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2402.2200000000003_2416.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2402.2200000000003_2416.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2402.2200000000003_2416.38#3.jpg" ], "ocr_qwen2_vl_72b": "- AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKM = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- \u2234 KOL = 180 - 2 x 42\n- 160 - 84\u00b0 (The sum of the angles in a triangle are supplementary)\n- = 96\u00b0" }, { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2416.38_2427.98.mp4", "refined_asr": " Find this angle right in here. Now, it's a little challenging to find that angle. So what I'm going to be doing is using my tangent theorem here.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2416.38_2427.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2416.38_2427.98#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2416.38_2427.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2416.38_2427.98#2.jpg" ], "ocr_qwen2_vl_72b": "- AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKH = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- \u2234 KOL = 180 - 2 x 42\n- 160 - 84 (The sum of the angles in a triangle are supplementary)\n- = 96\u00b0" }, { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2427.98_2434.54.mp4", "refined_asr": " States that if I have an angle that is created between a chord and a tangent", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2427.98_2434.54#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2427.98_2434.54#1.jpg" ], "ocr_qwen2_vl_72b": "- AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKH = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- KOL = 180 - 2 x 42\n- 160 - 84 (The sum of the angles in a triangle are subtended)" }, { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2435.1800000000003_2445.98.mp4", "refined_asr": " Then that angle is going to be exactly equal to the angle that is created by the chord in the alternate segment. So this 48 degrees here, this", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2435.1800000000003_2445.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2435.1800000000003_2445.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2435.1800000000003_2445.98#2.jpg" ], "ocr_qwen2_vl_72b": "- AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKL = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- KOL = 180 - 2 x 42\n- 160 - 84 (The sum of the angles in a triangle is 180\u00b0)" }, { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2446.38_2458.86.mp4", "refined_asr": " Chord here KL created an angle all the way up here. So this angle right in here is in fact also 48. Let me just go ahead and write a statement for that.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2446.38_2458.86#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2446.38_2458.86#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2446.38_2458.86#2.jpg" ], "ocr_qwen2_vl_72b": "- AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- OKM = 90 - 48 (radius to tangent)\n- = 42\u00b0\n- \u2234 KOL = 180 - 2 x 42\n- 160 - 84\u00b0 (The sum of the angles in a triangle are supplementary)\n- 76\u00b0" }, { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2460.46_2475.9.mp4", "refined_asr": " So we can name this angle here an inadvertent angle that we have to calculate. So it is KML, that's our starting statement. So angle KML is in fact equal to", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2460.46_2475.9#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2460.46_2475.9#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2460.46_2475.9#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2460.46_2475.9#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2460.46_2475.9#3.jpg" ], "ocr_qwen2_vl_72b": "AP3X INSTITUTE OF MATHEMATICS EDUCATION\n\nCSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n\nOKH = 90 - 48 (radius b to tangent)\n= 42\u00b0\n\nKOL = 180 - 2 x 42\n180 - 84\nThe sum of the angles" }, { "vid": "pFSkS6HwXD8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/\"Intermediate Algebra tutorials on Quadratic Equations using Quadratic Formula\"\n_30.json#####audio#####doingASR#####FinishASR/pFSkS6HwXD8/2476.38_2488.54.mp4", "refined_asr": " 188 degrees and the reason is simply because the angle between a tangent", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2476.38_2488.54#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2476.38_2488.54#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pFSkS6HwXD8/pFSkS6HwXD8@2476.38_2488.54#2.jpg" ], "ocr_qwen2_vl_72b": "- AP3X INSTITUTE OF MATHEMATICS EDUCATION\n- CSEC, CAPE & COLLEGE MATH | 2024 PAST PAPER MATHEMATICS\n- Angle MLK\n- Angle KML = 48\u00b0\n- Angle KNM" } ], "image_num": 15, "text_num": 425, "token_num": 9065 }, { "images": [ "sample_100_images/PqKyvTWYjPQ@13429.16_13437.619999999999#1.jpg", null, "sample_100_images/PqKyvTWYjPQ@13437.64_13443.16#1.jpg", null, "sample_100_images/PqKyvTWYjPQ@13443.76_13456.859999999999#1.jpg", null, "sample_100_images/PqKyvTWYjPQ@13457.34_13467.64#1.jpg", null, "sample_100_images/PqKyvTWYjPQ@13467.64_13483.779999999999#1.jpg", null, "sample_100_images/PqKyvTWYjPQ@13483.779999999999_13497.4#1.jpg", null, "sample_100_images/PqKyvTWYjPQ@13497.64_13508.42#1.jpg", null, "sample_100_images/PqKyvTWYjPQ@13508.42_13538.060000000001#1.jpg", null, "sample_100_images/PqKyvTWYjPQ@13538.320000000002_13550.24#1.jpg", null ], "texts": [ null, " It's 68. Okay so the first part of this formula is 68 and then for the second part we're going to multiply now.", null, " So starting at the bottom left, I'm going to go up. Seven times two is 14. Times one is still 14.", null, " And then I'm going to multiply up. Nine times one times one is nine. Then plus, I'm going to multiply one times five times three which is 15. So 14 plus nine is 23, plus 15 is 38. Therefore, you have 68 minus 38,", null, " Which is 30. Okay so the first part of this is that the determinant became 30. So the area, okay the area,", null, " It is equal to, in this case, because we have a positive determinant, I'm simply going to use plus one half. So, I'm just going to calculate one half times 30, which gives us 15. Now, when working with triangles, remember the formula from basic geometry: one half times the base times the", null, " Height. Remember you have units involved with the base and the height. So you've got to think about the fact that it's going to be square units. So if you're working with inches, it would be you know square inches or inches squared or whatever that is. So you would really want to give a precise answer and say it's 15.", null, " units. So I'll just say the area, and let me erase this, is going to be 15 square units. Okay?", null, " Units. Now really quickly just to prove this to you let's say I swap these two rows. So let's say I put five two and one on the top. I'm just going to erase that and I'll drag this up here. So really this would correspond to what it would correspond to me choosing this point to be X sub one Y sub one okay and this point to be X sub two Y sub two okay and because I swapped two rows", null, " The only effect this is going to have is it's going to change the sign of the determinant. Right? So now the determinant is going to be negative 30. And so to get the area, I'm going to multiply by negative one half. And so I'm still going to get 15. And it's going to be 15 square." ], "text_ocr_list": [ null, "We can see these text from the image: Area = 1/2 | 1 3 1 |\n| 5 2 1 |\n| 7 9 1 |\n\nGreeneMath.com.\n It's 68. Okay so the first part of this formula is 68 and then for the second part we're going to multiply now.", null, "We can see these text from the image: Area = \u00b1 1/2 | 1 3 1 |\n| 5 2 1 |\n| 7 9 1 |\n\n68 - 3\n\nGreeneMath.com.\n So starting at the bottom left, I'm going to go up. Seven times two is 14. Times one is still 14.", null, "We can see these text from the image: Area = 1/2 | 1 3 1 |\n| 5 2 1 |\n| 7 9 1 |\n\n68 - 3(14 + 9).\n And then I'm going to multiply up. Nine times one times one is nine. Then plus, I'm going to multiply one times five times three which is 15. So 14 plus nine is 23, plus 15 is 38. Therefore, you have 68 minus 38,", null, "We can see these text from the image: Area = \u00b1 1/2 | 1 3 1 |\n| 5 2 1 |\n| 7 9 1 |\n\n68 - 38 = 30\n\nGreeneMath.com.\n Which is 30. Okay so the first part of this is that the determinant became 30. So the area, okay the area,", null, "We can see these text from the image: Area = 1/2 | 1 3 1 |\n| 5 2 1 |\n| 7 9 1 |\n\nArea =.\n It is equal to, in this case, because we have a positive determinant, I'm simply going to use plus one half. So, I'm just going to calculate one half times 30, which gives us 15. Now, when working with triangles, remember the formula from basic geometry: one half times the base times the", null, "We can see these text from the image: Area = 1/2 | 1 3 1 |\n | 5 2 1 |\n | 7 9 1 |\n\nArea = 1/2 * 30 = 15\n\nGreeneMath.com.\n Height. Remember you have units involved with the base and the height. So you've got to think about the fact that it's going to be square units. So if you're working with inches, it would be you know square inches or inches squared or whatever that is. So you would really want to give a precise answer and say it's 15.", null, "We can see these text from the image: Vertices:\n(1,3)\n(5,2)\n(7,9).\n units. So I'll just say the area, and let me erase this, is going to be 15 square units. Okay?", null, "We can see these text from the image: Area = \u00b1 1/2 | x\u2081 y\u2081 1 |\n | x\u2082 y\u2082 1 |\n | x\u2083 y\u2083 1 |\n\n(x\u2081,y\u2081), (x\u2082,y\u2082), (x\u2083,y\u2083)\n\n(1,3), (5,2), (7,9)\n\nArea = \u00b1 1/2 | 1 3 1 |\n | 5 2 1 |\n | 7 9 1 |.\n Units. Now really quickly just to prove this to you let's say I swap these two rows. So let's say I put five two and one on the top. I'm just going to erase that and I'll drag this up here. So really this would correspond to what it would correspond to me choosing this point to be X sub one Y sub one okay and this point to be X sub two Y sub two okay and because I swapped two rows", null, "We can see these text from the image: Area = \u00b1 1/2\n\n| x\u2081 y\u2081 1 |\n| x\u2082 y\u2082 1 |\n| x\u2083 y\u2083 1 |\n\n(x\u2081, y\u2081), (x\u2082, y\u2082), (x\u2083, y\u2083)\n\n(1,3), (5,8), (7,9)\n\nArea = \u00b1 1/2\n\n| 5 8 1 |\n| 1 3 1 |\n| 7 9 1 |.\n The only effect this is going to have is it's going to change the sign of the determinant. Right? So now the determinant is going to be negative 30. And so to get the area, I'm going to multiply by negative one half. And so I'm still going to get 15. And it's going to be 15 square." ], "metadata": [ { "vid": "PqKyvTWYjPQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Matrices tutorial focusing on Determinants_30.json#####audio#####doingASR#####FinishASR/PqKyvTWYjPQ/13429.16_13437.619999999999.mp4", "refined_asr": " It's 68. Okay so the first part of this formula is 68 and then for the second part we're going to multiply now.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13429.16_13437.619999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13429.16_13437.619999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13429.16_13437.619999999999#2.jpg" ], "ocr_qwen2_vl_72b": "Area = 1/2 | 1 3 1 |\n| 5 2 1 |\n| 7 9 1 |\n\nGreeneMath.com" }, { "vid": "PqKyvTWYjPQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Matrices tutorial focusing on Determinants_30.json#####audio#####doingASR#####FinishASR/PqKyvTWYjPQ/13437.64_13443.16.mp4", "refined_asr": " So starting at the bottom left, I'm going to go up. Seven times two is 14. Times one is still 14.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13437.64_13443.16#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13437.64_13443.16#1.jpg" ], "ocr_qwen2_vl_72b": "Area = \u00b1 1/2 | 1 3 1 |\n| 5 2 1 |\n| 7 9 1 |\n\n68 - 3\n\nGreeneMath.com" }, { "vid": "PqKyvTWYjPQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Matrices tutorial focusing on Determinants_30.json#####audio#####doingASR#####FinishASR/PqKyvTWYjPQ/13443.76_13456.859999999999.mp4", "refined_asr": " And then I'm going to multiply up. Nine times one times one is nine. Then plus, I'm going to multiply one times five times three which is 15. So 14 plus nine is 23, plus 15 is 38. Therefore, you have 68 minus 38,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13443.76_13456.859999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13443.76_13456.859999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13443.76_13456.859999999999#2.jpg" ], "ocr_qwen2_vl_72b": "Area = 1/2 | 1 3 1 |\n| 5 2 1 |\n| 7 9 1 |\n\n68 - 3(14 + 9)" }, { "vid": "PqKyvTWYjPQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Matrices tutorial focusing on Determinants_30.json#####audio#####doingASR#####FinishASR/PqKyvTWYjPQ/13457.34_13467.64.mp4", "refined_asr": " Which is 30. Okay so the first part of this is that the determinant became 30. So the area, okay the area,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13457.34_13467.64#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13457.34_13467.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13457.34_13467.64#2.jpg" ], "ocr_qwen2_vl_72b": "Area = \u00b1 1/2 | 1 3 1 |\n| 5 2 1 |\n| 7 9 1 |\n\n68 - 38 = 30\n\nGreeneMath.com" }, { "vid": "PqKyvTWYjPQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Matrices tutorial focusing on Determinants_30.json#####audio#####doingASR#####FinishASR/PqKyvTWYjPQ/13467.64_13483.779999999999.mp4", "refined_asr": " It is equal to, in this case, because we have a positive determinant, I'm simply going to use plus one half. So, I'm just going to calculate one half times 30, which gives us 15. Now, when working with triangles, remember the formula from basic geometry: one half times the base times the", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13467.64_13483.779999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13467.64_13483.779999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13467.64_13483.779999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13467.64_13483.779999999999#3.jpg" ], "ocr_qwen2_vl_72b": "Area = 1/2 | 1 3 1 |\n| 5 2 1 |\n| 7 9 1 |\n\nArea =" }, { "vid": "PqKyvTWYjPQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Matrices tutorial focusing on Determinants_30.json#####audio#####doingASR#####FinishASR/PqKyvTWYjPQ/13483.779999999999_13497.4.mp4", "refined_asr": " Height. Remember you have units involved with the base and the height. So you've got to think about the fact that it's going to be square units. So if you're working with inches, it would be you know square inches or inches squared or whatever that is. So you would really want to give a precise answer and say it's 15.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13483.779999999999_13497.4#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13483.779999999999_13497.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13483.779999999999_13497.4#2.jpg" ], "ocr_qwen2_vl_72b": "Area = 1/2 | 1 3 1 |\n | 5 2 1 |\n | 7 9 1 |\n\nArea = 1/2 * 30 = 15\n\nGreeneMath.com" }, { "vid": "PqKyvTWYjPQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Matrices tutorial focusing on Determinants_30.json#####audio#####doingASR#####FinishASR/PqKyvTWYjPQ/13497.64_13508.42.mp4", "refined_asr": " units. So I'll just say the area, and let me erase this, is going to be 15 square units. Okay?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13497.64_13508.42#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13497.64_13508.42#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13497.64_13508.42#2.jpg" ], "ocr_qwen2_vl_72b": "Vertices:\n(1,3)\n(5,2)\n(7,9)" }, { "vid": "PqKyvTWYjPQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Matrices tutorial focusing on Determinants_30.json#####audio#####doingASR#####FinishASR/PqKyvTWYjPQ/13508.42_13538.060000000001.mp4", "refined_asr": " Units. Now really quickly just to prove this to you let's say I swap these two rows. So let's say I put five two and one on the top. I'm just going to erase that and I'll drag this up here. So really this would correspond to what it would correspond to me choosing this point to be X sub one Y sub one okay and this point to be X sub two Y sub two okay and because I swapped two rows", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13508.42_13538.060000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13508.42_13538.060000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13508.42_13538.060000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13508.42_13538.060000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13508.42_13538.060000000001#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13508.42_13538.060000000001#5.jpg" ], "ocr_qwen2_vl_72b": "Area = \u00b1 1/2 | x\u2081 y\u2081 1 |\n | x\u2082 y\u2082 1 |\n | x\u2083 y\u2083 1 |\n\n(x\u2081,y\u2081), (x\u2082,y\u2082), (x\u2083,y\u2083)\n\n(1,3), (5,2), (7,9)\n\nArea = \u00b1 1/2 | 1 3 1 |\n | 5 2 1 |\n | 7 9 1 |" }, { "vid": "PqKyvTWYjPQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Matrices tutorial focusing on Determinants_30.json#####audio#####doingASR#####FinishASR/PqKyvTWYjPQ/13538.320000000002_13550.24.mp4", "refined_asr": " The only effect this is going to have is it's going to change the sign of the determinant. Right? So now the determinant is going to be negative 30. And so to get the area, I'm going to multiply by negative one half. And so I'm still going to get 15. And it's going to be 15 square.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13538.320000000002_13550.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13538.320000000002_13550.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PqKyvTWYjPQ/PqKyvTWYjPQ@13538.320000000002_13550.24#2.jpg" ], "ocr_qwen2_vl_72b": "Area = \u00b1 1/2\n\n| x\u2081 y\u2081 1 |\n| x\u2082 y\u2082 1 |\n| x\u2083 y\u2083 1 |\n\n(x\u2081, y\u2081), (x\u2082, y\u2082), (x\u2083, y\u2083)\n\n(1,3), (5,8), (7,9)\n\nArea = \u00b1 1/2\n\n| 5 8 1 |\n| 1 3 1 |\n| 7 9 1 |" } ], "image_num": 9, "text_num": 534, "token_num": 5718 }, { "images": [ "sample_100_images/-YcQu_rZYSE@267.96000000000004_284.76000000000005#1.jpg", null, "sample_100_images/-YcQu_rZYSE@284.76000000000005_305.94#1.jpg", null, "sample_100_images/-YcQu_rZYSE@305.94_320.52#1.jpg", null, "sample_100_images/-YcQu_rZYSE@320.52_333.94#1.jpg", null, "sample_100_images/-YcQu_rZYSE@333.94_350.97999999999996#1.jpg", null, "sample_100_images/-YcQu_rZYSE@351.38_372.09999999999997#1.jpg", "sample_100_images/-YcQu_rZYSE@351.38_372.09999999999997#3.jpg", null ], "texts": [ null, " How can we make sense of this? Well for starters we could write this in some more compact notation using sigma notation. That's what the sum would look like in sigma notation. I'll leave a link in the description to my lesson introducing this notation. It's very important that you're familiar with it.", null, " Get comfortable with it because we'll be using it a lot. The crux of this problem of course is the idea of adding infinitely many positive numbers. It's not at all obvious that adding infinitely many positive numbers could produce a finite number since adding a positive number just makes the sum bigger.", null, " And we're adding infinitely many. It seems like this could diverge to infinity. And it's not like we can just add up the infinitely many terms to see if that's true or not. So what could we do to try to analyze this series?", null, " Well I can't add up infinitely many numbers but I can add up one number I could also add up two numbers I could probably add up three numbers or I might even be able to handle four", null, " Just the first number, that would of course be one half. If we were to add up the first two numbers, that would be a half plus a fourth, which happens to equal three fourths. Adding up the first three numbers gives us a half plus a fourth plus an eighth.", null, null, " Plus an eighth which is equal to seven eighths and so on. We could continue calculating these partial sums as long as we wanted. And by calculating these partial sums we've created a sequence. We've got the sum of the first number then the sum of the first two numbers then the sum of the three numbers and so on." ], "text_ocr_list": [ null, "We can see these text from the image: 1\n\n1/2\n\n1/4\n\n1/8\n\n1/16\n\n1/2 + 1/4 + 1/8 + ....\n How can we make sense of this? Well for starters we could write this in some more compact notation using sigma notation. That's what the sum would look like in sigma notation. I'll leave a link in the description to my lesson introducing this notation. It's very important that you're familiar with it.", null, "We can see these text from the image: \\[\n\\sum_{k=1}^{\\infty} \\frac{1}{2^k} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots\n\\].\n Get comfortable with it because we'll be using it a lot. The crux of this problem of course is the idea of adding infinitely many positive numbers. It's not at all obvious that adding infinitely many positive numbers could produce a finite number since adding a positive number just makes the sum bigger.", null, "We can see these text from the image: \\[\n\\sum_{k=1}^{\\infty} \\frac{1}{2^k} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots = 1\n\\].\n And we're adding infinitely many. It seems like this could diverge to infinity. And it's not like we can just add up the infinitely many terms to see if that's true or not. So what could we do to try to analyze this series?", null, "We can see these text from the image: \\[\n\\sum_{k=1}^{\\infty} \\frac{1}{2^k} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots = 1\n\\].\n Well I can't add up infinitely many numbers but I can add up one number I could also add up two numbers I could probably add up three numbers or I might even be able to handle four", null, "We can see these text from the image: \\[\n\\sum_{k=1}^{\\infty} \\frac{1}{2^k} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots = \\infty\n\\].\n Just the first number, that would of course be one half. If we were to add up the first two numbers, that would be a half plus a fourth, which happens to equal three fourths. Adding up the first three numbers gives us a half plus a fourth plus an eighth.", null, null, "We can see these text from the image: 1\n\n1/2\n\n1/4\n\n1/8\n\n1/16\n\n\u2211 1 = 1 + 1 + 1 + ... = \u221e\nk=1 2^k 2 4 8\n\n1/2, 3/4, 1/2 + 1/4 + 1/8 =.\n Plus an eighth which is equal to seven eighths and so on. We could continue calculating these partial sums as long as we wanted. And by calculating these partial sums we've created a sequence. We've got the sum of the first number then the sum of the first two numbers then the sum of the three numbers and so on." ], "metadata": [ { "vid": "-YcQu_rZYSE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Infinite Series_30.json#####audio#####doingASR#####FinishASR/-YcQu_rZYSE/267.96000000000004_284.76000000000005.mp4", "refined_asr": " How can we make sense of this? Well for starters we could write this in some more compact notation using sigma notation. That's what the sum would look like in sigma notation. I'll leave a link in the description to my lesson introducing this notation. It's very important that you're familiar with it.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@267.96000000000004_284.76000000000005#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@267.96000000000004_284.76000000000005#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@267.96000000000004_284.76000000000005#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@267.96000000000004_284.76000000000005#3.jpg" ], "ocr_qwen2_vl_72b": "1\n\n1/2\n\n1/4\n\n1/8\n\n1/16\n\n1/2 + 1/4 + 1/8 + ..." }, { "vid": "-YcQu_rZYSE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Infinite Series_30.json#####audio#####doingASR#####FinishASR/-YcQu_rZYSE/284.76000000000005_305.94.mp4", "refined_asr": " Get comfortable with it because we'll be using it a lot. The crux of this problem of course is the idea of adding infinitely many positive numbers. It's not at all obvious that adding infinitely many positive numbers could produce a finite number since adding a positive number just makes the sum bigger.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@284.76000000000005_305.94#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@284.76000000000005_305.94#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@284.76000000000005_305.94#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@284.76000000000005_305.94#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@284.76000000000005_305.94#4.jpg" ], "ocr_qwen2_vl_72b": "\\[\n\\sum_{k=1}^{\\infty} \\frac{1}{2^k} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots\n\\]" }, { "vid": "-YcQu_rZYSE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Infinite Series_30.json#####audio#####doingASR#####FinishASR/-YcQu_rZYSE/305.94_320.52.mp4", "refined_asr": " And we're adding infinitely many. It seems like this could diverge to infinity. And it's not like we can just add up the infinitely many terms to see if that's true or not. So what could we do to try to analyze this series?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@305.94_320.52#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@305.94_320.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@305.94_320.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@305.94_320.52#3.jpg" ], "ocr_qwen2_vl_72b": "\\[\n\\sum_{k=1}^{\\infty} \\frac{1}{2^k} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots = 1\n\\]" }, { "vid": "-YcQu_rZYSE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Infinite Series_30.json#####audio#####doingASR#####FinishASR/-YcQu_rZYSE/320.52_333.94.mp4", "refined_asr": " Well I can't add up infinitely many numbers but I can add up one number I could also add up two numbers I could probably add up three numbers or I might even be able to handle four", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@320.52_333.94#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@320.52_333.94#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@320.52_333.94#2.jpg" ], "ocr_qwen2_vl_72b": "\\[\n\\sum_{k=1}^{\\infty} \\frac{1}{2^k} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots = 1\n\\]" }, { "vid": "-YcQu_rZYSE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Infinite Series_30.json#####audio#####doingASR#####FinishASR/-YcQu_rZYSE/333.94_350.97999999999996.mp4", "refined_asr": " Just the first number, that would of course be one half. If we were to add up the first two numbers, that would be a half plus a fourth, which happens to equal three fourths. Adding up the first three numbers gives us a half plus a fourth plus an eighth.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@333.94_350.97999999999996#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@333.94_350.97999999999996#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@333.94_350.97999999999996#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@333.94_350.97999999999996#3.jpg" ], "ocr_qwen2_vl_72b": "\\[\n\\sum_{k=1}^{\\infty} \\frac{1}{2^k} = \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{8} + \\ldots = \\infty\n\\]" }, { "vid": "-YcQu_rZYSE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Infinite Series_30.json#####audio#####doingASR#####FinishASR/-YcQu_rZYSE/351.38_372.09999999999997.mp4", "refined_asr": " Plus an eighth which is equal to seven eighths and so on. We could continue calculating these partial sums as long as we wanted. And by calculating these partial sums we've created a sequence. We've got the sum of the first number then the sum of the first two numbers then the sum of the three numbers and so on.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@351.38_372.09999999999997#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@351.38_372.09999999999997#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@351.38_372.09999999999997#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@351.38_372.09999999999997#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-YcQu_rZYSE/-YcQu_rZYSE@351.38_372.09999999999997#3.jpg" ], "ocr_qwen2_vl_72b": "1\n\n1/2\n\n1/4\n\n1/8\n\n1/16\n\n\u2211 1 = 1 + 1 + 1 + ... = \u221e\nk=1 2^k 2 4 8\n\n1/2, 3/4, 1/2 + 1/4 + 1/8 =" } ], "image_num": 7, "text_num": 363, "token_num": 4395 }, { "images": [ "sample_100_images/9WoRv2d6hw4@283.12_334.32#1.jpg", null, "sample_100_images/9WoRv2d6hw4@334.7_361.98#1.jpg", null, "sample_100_images/9WoRv2d6hw4@363.74_381.84#1.jpg", null, "sample_100_images/9WoRv2d6hw4@385.84_389.92#1.jpg", null ], "texts": [ null, " Recall that a rational number is a number of the form a over b where a and b are integers and b is non-zero. We identify rational numbers a over b and c over d whenever ad is equal to bc. For example 1 half and 3 sixths represent the same rational number because 1 times 6 equals 2 times 3. We denote the set of rational numbers by the symbol Q. So we have Q equal to the set of quotients a over b such that a and b are integers and b is not zero. We identify the rational number a over 1 with the integer a.", null, " In this way, we have that the set of integers is a subset of the set of rational numbers. We add two rational numbers using the rule: a over b plus c over d is equal to a times d plus b times c, all over b times d. Note that zero, which equals zero over one, is an identity for the set of rational numbers together with addition.", null, " Because a over b is a set of integers and b is a subset of the set of rational numbers. a over b plus zero over one is equal to a times one plus b times zero over b times one, which is a over b. And zero over one plus a over b is equal to zero times b plus one times a over one times b, which is also a over b.", null, " The set of rational numbers together with addition is a commutative group." ], "text_ocr_list": [ null, "We can see these text from the image: Lesson 3\nAbstract Algebra\n\nInverses and Groups\n\nRecall that a rational number is a number of the form \\(\\frac{a}{b}\\), where \\(a\\) and \\(b\\) are integers and \\(b \\neq 0\\)..\n Recall that a rational number is a number of the form a over b where a and b are integers and b is non-zero. We identify rational numbers a over b and c over d whenever ad is equal to bc. For example 1 half and 3 sixths represent the same rational number because 1 times 6 equals 2 times 3. We denote the set of rational numbers by the symbol Q. So we have Q equal to the set of quotients a over b such that a and b are integers and b is not zero. We identify the rational number a over 1 with the integer a.", null, "We can see these text from the image: Lesson 3\n\nAbstract Algebra\n\nInverses and Groups\n\nRecall that a rational number is a number of the form \\(\\frac{a}{b}\\), where \\(a\\) and \\(b\\) are integers and \\(b \\neq 0\\).\n\nWe identify rational numbers \\(\\frac{a}{b}\\) and \\(\\frac{c}{d}\\) whenever \\(ad = bc\\).\n\nFor example, \\(\\frac{1}{2}\\) and \\(\\frac{3}{6}\\) represent the same rational number because \\(1 \\cdot 6 = 6\\) and \\(2 \\cdot 3 = 6\\).\n\nWe denote the set of rational numbers by \\(\\mathbb{Q}\\). So, we have \\(\\mathbb{Q} = \\left\\{\\frac{a}{b} \\mid a, b \\in \\mathbb{Z}, b \\neq 0\\right\\}\\).\n\nIn words, \\(\\mathbb{Q}\\) is \u201cthe set of quotients \\(a\\) over \\(b\\) such that \\(a\\) and \\(b\\) are integers and \\(b\\) is not zero.\u201d\n\nWe identify the rational number \\(\\frac{a}{1}\\) with the integer \\(a\\). In this way, we have \\(\\mathbb{Z} \\subseteq \\mathbb{Q}\\)..\n In this way, we have that the set of integers is a subset of the set of rational numbers. We add two rational numbers using the rule: a over b plus c over d is equal to a times d plus b times c, all over b times d. Note that zero, which equals zero over one, is an identity for the set of rational numbers together with addition.", null, "We can see these text from the image: Lesson 3\nAbstract Algebra\n\nInverses and Groups\n\nRecall that a rational number is a number of the form \\(\\frac{a}{b}\\), where \\(a\\) and \\(b\\) are integers and \\(b \\neq 0\\).\n\nWe identify rational numbers \\(\\frac{a}{b}\\) and \\(\\frac{c}{d}\\) whenever \\(ad = bc\\).\n\nFor example, \\(\\frac{1}{2}\\) and \\(\\frac{3}{6}\\) represent the same rational number because \\(1 \\cdot 6 = 6\\) and \\(2 \\cdot 3 = 6\\).\n\nWe denote the set of rational numbers by \\(\\mathbb{Q}\\). So, we have \\(\\mathbb{Q} = \\left\\{\\frac{a}{b} \\mid a, b \\in \\mathbb{Z}, b \\neq 0\\right\\}\\).\n\nIn words, \\(\\mathbb{Q}\\) is \u201cthe set of quotients \\(a\\) over \\(b\\) such that \\(a\\) and \\(b\\) are integers and \\(b\\) is not zero.\u201d\n\nWe identify the rational number \\(\\frac{a}{1}\\) with the integer \\(a\\). In this way, we have \\(\\mathbb{Z} \\subseteq \\mathbb{Q}\\).\n\nWe add two rational numbers using the rule \\(\\frac{a}{b} + \\frac{c}{d} = \\frac{a \\cdot d + b \\cdot c}{b \\cdot d}\\).\n\nNote that \\(0 = \\frac{0}{1}\\) is an identity for \\((\\mathbb{Q}, +)\\) because \\(\\frac{a}{b} + \\frac{0}{1} = \\frac{a \\cdot 1 + b \\cdot 0}{b \\cdot 1}\\)..\n Because a over b is a set of integers and b is a subset of the set of rational numbers. a over b plus zero over one is equal to a times one plus b times zero over b times one, which is a over b. And zero over one plus a over b is equal to zero times b plus one times a over one times b, which is also a over b.", null, "We can see these text from the image: Lesson 3\n\nAbstract Algebra\n\nInverses and Groups\n\nRecall that a rational number is a number of the form \\(\\frac{a}{b}\\), where \\(a\\) and \\(b\\) are integers and \\(b \\neq 0\\).\n\nWe identify rational numbers \\(\\frac{a}{b}\\) and \\(\\frac{c}{d}\\) whenever \\(ad = bc\\).\n\nFor example, \\(\\frac{1}{2}\\) and \\(\\frac{3}{6}\\) represent the same rational number because \\(1 \\cdot 6 = 6\\) and \\(2 \\cdot 3 = 6\\).\n\nWe denote the set of rational numbers by \\(\\mathbb{Q}\\). So, we have \\(\\mathbb{Q} = \\left\\{\\frac{a}{b} \\mid a, b \\in \\mathbb{Z}, b \\neq 0\\right\\}\\).\n\nIn words, \\(\\mathbb{Q}\\) is \u201cthe set of quotients \\(a\\) over \\(b\\) such that \\(a\\) and \\(b\\) are integers and \\(b\\) is not zero.\u201d\n\nWe identify the rational number \\(\\frac{a}{1}\\) with the integer \\(a\\). In this way, we have \\(\\mathbb{Z} \\subseteq \\mathbb{Q}\\).\n\nWe add two rational numbers using the rule \\(\\frac{a}{b} + \\frac{c}{d} = \\frac{a \\cdot d + b \\cdot c}{b \\cdot d}\\).\n\nNote that \\(0 = \\frac{0}{1}\\) is an identity for \\((\\mathbb{Q}, +)\\) because \\(\\frac{a}{b} + \\frac{0}{1} = \\frac{a \\cdot 1 + b \\cdot 0}{b \\cdot 1} = \\frac{a}{b}\\) and \\(\\frac{0}{1} + \\frac{a}{b} = \\frac{0 \\cdot b + 1 \\cdot a}{1 \\cdot b} = \\frac{a}{b}\\).\n\n\\((\\mathbb{Q}, +)\\) is a commutative group..\n The set of rational numbers together with addition is a commutative group." ], "metadata": [ { "vid": "9WoRv2d6hw4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/9WoRv2d6hw4/283.12_334.32.mp4", "refined_asr": " Recall that a rational number is a number of the form a over b where a and b are integers and b is non-zero. We identify rational numbers a over b and c over d whenever ad is equal to bc. For example 1 half and 3 sixths represent the same rational number because 1 times 6 equals 2 times 3. We denote the set of rational numbers by the symbol Q. So we have Q equal to the set of quotients a over b such that a and b are integers and b is not zero. We identify the rational number a over 1 with the integer a.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@283.12_334.32#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@283.12_334.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@283.12_334.32#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@283.12_334.32#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@283.12_334.32#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@283.12_334.32#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@283.12_334.32#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@283.12_334.32#7.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@283.12_334.32#8.jpg" ], "ocr_qwen2_vl_72b": "Lesson 3\nAbstract Algebra\n\nInverses and Groups\n\nRecall that a rational number is a number of the form \\(\\frac{a}{b}\\), where \\(a\\) and \\(b\\) are integers and \\(b \\neq 0\\)." }, { "vid": "9WoRv2d6hw4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/9WoRv2d6hw4/334.7_361.98.mp4", "refined_asr": " In this way, we have that the set of integers is a subset of the set of rational numbers. We add two rational numbers using the rule: a over b plus c over d is equal to a times d plus b times c, all over b times d. Note that zero, which equals zero over one, is an identity for the set of rational numbers together with addition.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@334.7_361.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@334.7_361.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@334.7_361.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@334.7_361.98#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@334.7_361.98#4.jpg" ], "ocr_qwen2_vl_72b": "Lesson 3\n\nAbstract Algebra\n\nInverses and Groups\n\nRecall that a rational number is a number of the form \\(\\frac{a}{b}\\), where \\(a\\) and \\(b\\) are integers and \\(b \\neq 0\\).\n\nWe identify rational numbers \\(\\frac{a}{b}\\) and \\(\\frac{c}{d}\\) whenever \\(ad = bc\\).\n\nFor example, \\(\\frac{1}{2}\\) and \\(\\frac{3}{6}\\) represent the same rational number because \\(1 \\cdot 6 = 6\\) and \\(2 \\cdot 3 = 6\\).\n\nWe denote the set of rational numbers by \\(\\mathbb{Q}\\). So, we have \\(\\mathbb{Q} = \\left\\{\\frac{a}{b} \\mid a, b \\in \\mathbb{Z}, b \\neq 0\\right\\}\\).\n\nIn words, \\(\\mathbb{Q}\\) is \u201cthe set of quotients \\(a\\) over \\(b\\) such that \\(a\\) and \\(b\\) are integers and \\(b\\) is not zero.\u201d\n\nWe identify the rational number \\(\\frac{a}{1}\\) with the integer \\(a\\). In this way, we have \\(\\mathbb{Z} \\subseteq \\mathbb{Q}\\)." }, { "vid": "9WoRv2d6hw4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/9WoRv2d6hw4/363.74_381.84.mp4", "refined_asr": " Because a over b is a set of integers and b is a subset of the set of rational numbers. a over b plus zero over one is equal to a times one plus b times zero over b times one, which is a over b. And zero over one plus a over b is equal to zero times b plus one times a over one times b, which is also a over b.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@363.74_381.84#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@363.74_381.84#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@363.74_381.84#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@363.74_381.84#3.jpg" ], "ocr_qwen2_vl_72b": "Lesson 3\nAbstract Algebra\n\nInverses and Groups\n\nRecall that a rational number is a number of the form \\(\\frac{a}{b}\\), where \\(a\\) and \\(b\\) are integers and \\(b \\neq 0\\).\n\nWe identify rational numbers \\(\\frac{a}{b}\\) and \\(\\frac{c}{d}\\) whenever \\(ad = bc\\).\n\nFor example, \\(\\frac{1}{2}\\) and \\(\\frac{3}{6}\\) represent the same rational number because \\(1 \\cdot 6 = 6\\) and \\(2 \\cdot 3 = 6\\).\n\nWe denote the set of rational numbers by \\(\\mathbb{Q}\\). So, we have \\(\\mathbb{Q} = \\left\\{\\frac{a}{b} \\mid a, b \\in \\mathbb{Z}, b \\neq 0\\right\\}\\).\n\nIn words, \\(\\mathbb{Q}\\) is \u201cthe set of quotients \\(a\\) over \\(b\\) such that \\(a\\) and \\(b\\) are integers and \\(b\\) is not zero.\u201d\n\nWe identify the rational number \\(\\frac{a}{1}\\) with the integer \\(a\\). In this way, we have \\(\\mathbb{Z} \\subseteq \\mathbb{Q}\\).\n\nWe add two rational numbers using the rule \\(\\frac{a}{b} + \\frac{c}{d} = \\frac{a \\cdot d + b \\cdot c}{b \\cdot d}\\).\n\nNote that \\(0 = \\frac{0}{1}\\) is an identity for \\((\\mathbb{Q}, +)\\) because \\(\\frac{a}{b} + \\frac{0}{1} = \\frac{a \\cdot 1 + b \\cdot 0}{b \\cdot 1}\\)." }, { "vid": "9WoRv2d6hw4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/9WoRv2d6hw4/385.84_389.92.mp4", "refined_asr": " The set of rational numbers together with addition is a commutative group.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@385.84_389.92#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/9WoRv2d6hw4/9WoRv2d6hw4@385.84_389.92#1.jpg" ], "ocr_qwen2_vl_72b": "Lesson 3\n\nAbstract Algebra\n\nInverses and Groups\n\nRecall that a rational number is a number of the form \\(\\frac{a}{b}\\), where \\(a\\) and \\(b\\) are integers and \\(b \\neq 0\\).\n\nWe identify rational numbers \\(\\frac{a}{b}\\) and \\(\\frac{c}{d}\\) whenever \\(ad = bc\\).\n\nFor example, \\(\\frac{1}{2}\\) and \\(\\frac{3}{6}\\) represent the same rational number because \\(1 \\cdot 6 = 6\\) and \\(2 \\cdot 3 = 6\\).\n\nWe denote the set of rational numbers by \\(\\mathbb{Q}\\). So, we have \\(\\mathbb{Q} = \\left\\{\\frac{a}{b} \\mid a, b \\in \\mathbb{Z}, b \\neq 0\\right\\}\\).\n\nIn words, \\(\\mathbb{Q}\\) is \u201cthe set of quotients \\(a\\) over \\(b\\) such that \\(a\\) and \\(b\\) are integers and \\(b\\) is not zero.\u201d\n\nWe identify the rational number \\(\\frac{a}{1}\\) with the integer \\(a\\). In this way, we have \\(\\mathbb{Z} \\subseteq \\mathbb{Q}\\).\n\nWe add two rational numbers using the rule \\(\\frac{a}{b} + \\frac{c}{d} = \\frac{a \\cdot d + b \\cdot c}{b \\cdot d}\\).\n\nNote that \\(0 = \\frac{0}{1}\\) is an identity for \\((\\mathbb{Q}, +)\\) because \\(\\frac{a}{b} + \\frac{0}{1} = \\frac{a \\cdot 1 + b \\cdot 0}{b \\cdot 1} = \\frac{a}{b}\\) and \\(\\frac{0}{1} + \\frac{a}{b} = \\frac{0 \\cdot b + 1 \\cdot a}{1 \\cdot b} = \\frac{a}{b}\\).\n\n\\((\\mathbb{Q}, +)\\) is a commutative group." } ], "image_num": 4, "text_num": 302, "token_num": 2606 }, { "images": [ "sample_100_images/sbbYntt5CJk@27310.62_27330.62#1.jpg", null, "sample_100_images/sbbYntt5CJk@27330.62_27349.62#1.jpg", "sample_100_images/sbbYntt5CJk@27330.62_27349.62#3.jpg", null, "sample_100_images/sbbYntt5CJk@27349.62_27372.62#1.jpg", "sample_100_images/sbbYntt5CJk@27349.62_27372.62#2.jpg", null, "sample_100_images/sbbYntt5CJk@27372.62_27391.62#1.jpg", null, "sample_100_images/sbbYntt5CJk@27391.62_27412.62#1.jpg", null, "sample_100_images/sbbYntt5CJk@27412.62_27443.62#1.jpg", "sample_100_images/sbbYntt5CJk@27412.62_27443.62#3.jpg", null ], "texts": [ null, " No more than four questions so the probability of missing will be 0.1. With 50 questions, this is the exact distribution of X. This is the exact probability to calculate. We can approximate this binomial with a normal. Y will follow a normal distribution.", null, null, " The expected value is 50 times 0.1 and the variance is 50 times 0.1 times 0.9. We need to calculate the probability that Y is less than or equal to 4.5. Oops, that's too far. It should be 40%.", null, null, " The real answer is 43%. So I'm off by 2.5%. Many toothpaste commercials advertise that 304 dentists recommend their brand of toothpaste. Using a normal distribution to estimate the probability that in a random survey of 400 dentists,", null, " So we'll be using random sampling, simple random sampling. Exactly 300 will recommend Brand X toothpaste. We're going to assume the commercials are correct and therefore there's a 75% chance that any given dentist will recommend Brand X. So if X is the number,", null, " In that sample of 400, X follows exactly a binomial distribution with N equals 400 and P equals 0.75. We need to calculate the probability of X equals 300. The approximation Y will follow a normal distribution with an expected value of 400 times 0.75.", null, null, " Variance is 400 times 0.75 times 0.25. And we need to calculate the probability that Y is between 299.5 and 300.5. Now we get 0.0460403, and the real answer is 0.04602432." ], "text_ocr_list": [ null, "We can see these text from the image: Example 3: The Stats Examination\n\nExample\n\nAfter many hours of studying for your statistics examination, you believe that you have a 90% probability of answering any given question correctly. Your test includes 50 true/false questions. Assuming that your estimate is the true probability that you will answer a question correctly, use a Normal distribution to estimate the probability that you will miss no more than 4 questions.\n\nIn other words, if \\( X \\) is the number of problems you get wrong on this examination, then\n\n\\[ X \\sim Bin(50, 0.100) \\]\n\nand we are to calculate \\( P[X \\leq 4] \\)..\n No more than four questions so the probability of missing will be 0.1. With 50 questions, this is the exact distribution of X. This is the exact probability to calculate. We can approximate this binomial with a normal. Y will follow a normal distribution.", null, null, "We can see these text from the image: Example 3: The Stats Examination\n\nExample\n\nAfter many hours of studying for your statistics examination, you believe that you have a 90% probability of answering any given question correctly. Your test includes 50 true/false questions. Assuming that your estimate is the true probability that you will answer a question correctly, use a Normal distribution to estimate the probability that you will miss no more than 4 questions.\n\nIn other words, if \\(X\\) is the number of problems you get wrong on this examination, then\n\n\\[ X \\sim Bin(50, 0.100) \\]\n\nand we are to calculate \\(P[X \\leq 4]\\)..\n The expected value is 50 times 0.1 and the variance is 50 times 0.1 times 0.9. We need to calculate the probability that Y is less than or equal to 4.5. Oops, that's too far. It should be 40%.", null, null, "We can see these text from the image: Example 3: The Stats Examination\n\nSolution\n\nX ~ Bin(50, 0.100)\n\nY ~ N(\u03bc = 5, \u03c3\u00b2 = 4.5)\n\nP[X \u2264 4] \u2248 P[Y \u2264 4.5]\n\u2248 0.4068319\n\nThis last calculation came from R's command\npnorm(4.5, m=5, s=sqrt(4.5))\n\nThe real answer is\npbinom(4, size=50, prob=0.100) = 0.4311984..\n The real answer is 43%. So I'm off by 2.5%. Many toothpaste commercials advertise that 304 dentists recommend their brand of toothpaste. Using a normal distribution to estimate the probability that in a random survey of 400 dentists,", null, "We can see these text from the image: Example 4: Toothpaste\n\nExample\nMany toothpaste commercials advertise that 3 out of 4 dentists recommend their brand of toothpaste. Use a normal distribution to estimate the probability that in a random survey of 400 dentists, exactly 300 will recommend Brand X toothpaste. Assume that the commercials are correct, and therefore, there is a 75% chance that any given dentist will recommend Brand X toothpaste.\n\nIn other words, if \\(X\\) is the number of dentists recommending Brand X toothpaste, then\n\\[X \\sim Bin(400, 0.750)\\]\nand we are to calculate \\(P[X = 300]\\)..\n So we'll be using random sampling, simple random sampling. Exactly 300 will recommend Brand X toothpaste. We're going to assume the commercials are correct and therefore there's a 75% chance that any given dentist will recommend Brand X. So if X is the number,", null, "We can see these text from the image: Example 4: Toothpaste\n\nExample\nMany toothpaste commercials advertise that 3 out of 4 dentists recommend their brand of toothpaste. Use a normal distribution to estimate the probability that in a random survey of 400 dentists, exactly 300 will recommend Brand X toothpaste. Assume that the commercials are correct, and therefore, there is a 75% chance that any given dentist will recommend Brand X toothpaste.\n\nIn other words, if \\(X\\) is the number of dentists recommending Brand X toothpaste, then\n\\[X \\sim Bin(400, 0.750)\\]\nand we are to calculate \\(P[X = 300]\\)..\n In that sample of 400, X follows exactly a binomial distribution with N equals 400 and P equals 0.75. We need to calculate the probability of X equals 300. The approximation Y will follow a normal distribution with an expected value of 400 times 0.75.", null, null, "We can see these text from the image: Example 4: Toothpaste\n\nExample\n\nMany toothpaste commercials advertise that 3 out of 4 dentists recommend their brand of toothpaste. Use a normal distribution to estimate the probability that in a random survey of 400 dentists, exactly 300 will recommend Brand X toothpaste. Assume that the commercials are correct, and therefore, there is a 75% chance that any given dentist will recommend Brand X toothpaste.\n\nIn other words, if \\(X\\) is the number of dentists recommending Brand X toothpaste, then\n\n\\[X \\sim Bin(400, 0.750)\\]\n\nand we are to calculate \\(P[X = 300]\\)..\n Variance is 400 times 0.75 times 0.25. And we need to calculate the probability that Y is between 299.5 and 300.5. Now we get 0.0460403, and the real answer is 0.04602432." ], "metadata": [ { "vid": "sbbYntt5CJk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/sbbYntt5CJk/27310.62_27330.62.mp4", "refined_asr": " No more than four questions so the probability of missing will be 0.1. With 50 questions, this is the exact distribution of X. This is the exact probability to calculate. We can approximate this binomial with a normal. Y will follow a normal distribution.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27310.62_27330.62#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27310.62_27330.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27310.62_27330.62#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27310.62_27330.62#3.jpg" ], "ocr_qwen2_vl_72b": "Example 3: The Stats Examination\n\nExample\n\nAfter many hours of studying for your statistics examination, you believe that you have a 90% probability of answering any given question correctly. Your test includes 50 true/false questions. Assuming that your estimate is the true probability that you will answer a question correctly, use a Normal distribution to estimate the probability that you will miss no more than 4 questions.\n\nIn other words, if \\( X \\) is the number of problems you get wrong on this examination, then\n\n\\[ X \\sim Bin(50, 0.100) \\]\n\nand we are to calculate \\( P[X \\leq 4] \\)." }, { "vid": "sbbYntt5CJk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/sbbYntt5CJk/27330.62_27349.62.mp4", "refined_asr": " The expected value is 50 times 0.1 and the variance is 50 times 0.1 times 0.9. We need to calculate the probability that Y is less than or equal to 4.5. Oops, that's too far. It should be 40%.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27330.62_27349.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27330.62_27349.62#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27330.62_27349.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27330.62_27349.62#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27330.62_27349.62#3.jpg" ], "ocr_qwen2_vl_72b": "Example 3: The Stats Examination\n\nExample\n\nAfter many hours of studying for your statistics examination, you believe that you have a 90% probability of answering any given question correctly. Your test includes 50 true/false questions. Assuming that your estimate is the true probability that you will answer a question correctly, use a Normal distribution to estimate the probability that you will miss no more than 4 questions.\n\nIn other words, if \\(X\\) is the number of problems you get wrong on this examination, then\n\n\\[ X \\sim Bin(50, 0.100) \\]\n\nand we are to calculate \\(P[X \\leq 4]\\)." }, { "vid": "sbbYntt5CJk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/sbbYntt5CJk/27349.62_27372.62.mp4", "refined_asr": " The real answer is 43%. So I'm off by 2.5%. Many toothpaste commercials advertise that 304 dentists recommend their brand of toothpaste. Using a normal distribution to estimate the probability that in a random survey of 400 dentists,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27349.62_27372.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27349.62_27372.62#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27349.62_27372.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27349.62_27372.62#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27349.62_27372.62#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27349.62_27372.62#4.jpg" ], "ocr_qwen2_vl_72b": "Example 3: The Stats Examination\n\nSolution\n\nX ~ Bin(50, 0.100)\n\nY ~ N(\u03bc = 5, \u03c3\u00b2 = 4.5)\n\nP[X \u2264 4] \u2248 P[Y \u2264 4.5]\n\u2248 0.4068319\n\nThis last calculation came from R's command\npnorm(4.5, m=5, s=sqrt(4.5))\n\nThe real answer is\npbinom(4, size=50, prob=0.100) = 0.4311984." }, { "vid": "sbbYntt5CJk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/sbbYntt5CJk/27372.62_27391.62.mp4", "refined_asr": " So we'll be using random sampling, simple random sampling. Exactly 300 will recommend Brand X toothpaste. We're going to assume the commercials are correct and therefore there's a 75% chance that any given dentist will recommend Brand X. So if X is the number,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27372.62_27391.62#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27372.62_27391.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27372.62_27391.62#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27372.62_27391.62#3.jpg" ], "ocr_qwen2_vl_72b": "Example 4: Toothpaste\n\nExample\nMany toothpaste commercials advertise that 3 out of 4 dentists recommend their brand of toothpaste. Use a normal distribution to estimate the probability that in a random survey of 400 dentists, exactly 300 will recommend Brand X toothpaste. Assume that the commercials are correct, and therefore, there is a 75% chance that any given dentist will recommend Brand X toothpaste.\n\nIn other words, if \\(X\\) is the number of dentists recommending Brand X toothpaste, then\n\\[X \\sim Bin(400, 0.750)\\]\nand we are to calculate \\(P[X = 300]\\)." }, { "vid": "sbbYntt5CJk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/sbbYntt5CJk/27391.62_27412.62.mp4", "refined_asr": " In that sample of 400, X follows exactly a binomial distribution with N equals 400 and P equals 0.75. We need to calculate the probability of X equals 300. The approximation Y will follow a normal distribution with an expected value of 400 times 0.75.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27391.62_27412.62#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27391.62_27412.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27391.62_27412.62#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27391.62_27412.62#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27391.62_27412.62#4.jpg" ], "ocr_qwen2_vl_72b": "Example 4: Toothpaste\n\nExample\nMany toothpaste commercials advertise that 3 out of 4 dentists recommend their brand of toothpaste. Use a normal distribution to estimate the probability that in a random survey of 400 dentists, exactly 300 will recommend Brand X toothpaste. Assume that the commercials are correct, and therefore, there is a 75% chance that any given dentist will recommend Brand X toothpaste.\n\nIn other words, if \\(X\\) is the number of dentists recommending Brand X toothpaste, then\n\\[X \\sim Bin(400, 0.750)\\]\nand we are to calculate \\(P[X = 300]\\)." }, { "vid": "sbbYntt5CJk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/sbbYntt5CJk/27412.62_27443.62.mp4", "refined_asr": " Variance is 400 times 0.75 times 0.25. And we need to calculate the probability that Y is between 299.5 and 300.5. Now we get 0.0460403, and the real answer is 0.04602432.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27412.62_27443.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27412.62_27443.62#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27412.62_27443.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27412.62_27443.62#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27412.62_27443.62#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27412.62_27443.62#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sbbYntt5CJk/sbbYntt5CJk@27412.62_27443.62#5.jpg" ], "ocr_qwen2_vl_72b": "Example 4: Toothpaste\n\nExample\n\nMany toothpaste commercials advertise that 3 out of 4 dentists recommend their brand of toothpaste. Use a normal distribution to estimate the probability that in a random survey of 400 dentists, exactly 300 will recommend Brand X toothpaste. Assume that the commercials are correct, and therefore, there is a 75% chance that any given dentist will recommend Brand X toothpaste.\n\nIn other words, if \\(X\\) is the number of dentists recommending Brand X toothpaste, then\n\n\\[X \\sim Bin(400, 0.750)\\]\n\nand we are to calculate \\(P[X = 300]\\)." } ], "image_num": 9, "text_num": 404, "token_num": 5588 }, { "images": [ "sample_100_images/5TRFpFBccQM@0.0_24.16#1.jpg", "sample_100_images/5TRFpFBccQM@0.0_24.16#2.jpg", null, "sample_100_images/5TRFpFBccQM@24.16_48.4#1.jpg", "sample_100_images/5TRFpFBccQM@24.16_48.4#2.jpg", "sample_100_images/5TRFpFBccQM@24.16_48.4#3.jpg", null, "sample_100_images/5TRFpFBccQM@48.4_70.8#1.jpg", "sample_100_images/5TRFpFBccQM@48.4_70.8#2.jpg", null, "sample_100_images/5TRFpFBccQM@72.0_94.32000000000001#1.jpg", null, "sample_100_images/5TRFpFBccQM@94.88000000000001_113.28#1.jpg", null, "sample_100_images/5TRFpFBccQM@113.28_131.28#1.jpg", "sample_100_images/5TRFpFBccQM@113.28_131.28#2.jpg", "sample_100_images/5TRFpFBccQM@113.28_131.28#3.jpg", null ], "texts": [ null, null, " Why is Redis so fast? What fundamental design decisions did the developers make more than a decade ago that stood the test of time? Let's take a look. Redis is a very popular in-memory database. It's rock solid, easy to use, and fast. These", null, null, null, " Attributes explain why it is one of the most loved databases according to Stack Overflow's annual developer survey. The first reason Redis is fast is because it is an in-memory database. Memory access is several orders of magnitude faster than random disk I/O. Pure memory access provides high read and write throughput and low latency. The trade-off is", null, null, " The data set cannot be larger than memory. Code-wise, in-memory data structures are also much easier to implement than their on-disk counterparts. This keeps the code simple and contributes to Redis' rock-solid stability. Another reason Redis is fast, which might seem counterintuitive, is that it is primarily single-threaded.", null, " Why would a single-threaded design lead to high performance? Wouldn't it be faster if it used threads to leverage all the CPU cores? Multi-threaded applications require locks or other synchronization mechanisms. They are notoriously hard to reason about. In many applications, the added complexity is bug-prone and sacrifices stability.", null, " Making it difficult to justify the performance gain. In the case of Redis, the single-threaded code path is easy to understand. How does a single-threaded code base handle many thousands of incoming requests and outgoing responses at the same time? Won't the thread get blocked?", null, null, null, " Waiting for the completion of each request individually is where I/O might be helpful. I/O multiplexing comes into the picture. With I/O multiplexing, the operating system allows a single thread to wait on many socket connections simultaneously. Traditionally, this is done with the 'select' function." ], "text_ocr_list": [ null, null, "We can see these text from the image: ByteByteGo.com\n\nSystem Design Video Series.\n Why is Redis so fast? What fundamental design decisions did the developers make more than a decade ago that stood the test of time? Let's take a look. Redis is a very popular in-memory database. It's rock solid, easy to use, and fast. These", null, null, null, "We can see these text from the image: Why is Redis so fast\n\n- Rock solid\n- Easy to use\n- Fast.\n Attributes explain why it is one of the most loved databases according to Stack Overflow's annual developer survey. The first reason Redis is fast is because it is an in-memory database. Memory access is several orders of magnitude faster than random disk I/O. Pure memory access provides high read and write throughput and low latency. The trade-off is", null, null, "We can see these text from the image: Why is Redis so fast\n\n50-150us\n\nHigh read/write throughput\nLow latency\nDataset cannot be larger than memory\n\n120ns\n\nMemory access Random disk I/O.\n The data set cannot be larger than memory. Code-wise, in-memory data structures are also much easier to implement than their on-disk counterparts. This keeps the code simple and contributes to Redis' rock-solid stability. Another reason Redis is fast, which might seem counterintuitive, is that it is primarily single-threaded.", null, "We can see these text from the image: Why is Redis so fast\n\nSingle-threaded process\n\ncode data files\ncode files\n\nthread.\n Why would a single-threaded design lead to high performance? Wouldn't it be faster if it used threads to leverage all the CPU cores? Multi-threaded applications require locks or other synchronization mechanisms. They are notoriously hard to reason about. In many applications, the added complexity is bug-prone and sacrifices stability.", null, "We can see these text from the image: Why is Redis so fast\n\nThread 1\nLPUSH\nLength + 1\n\nThread 2\nLPOP\nLength - 1\n\nList.length\nList.\n Making it difficult to justify the performance gain. In the case of Redis, the single-threaded code path is easy to understand. How does a single-threaded code base handle many thousands of incoming requests and outgoing responses at the same time? Won't the thread get blocked?", null, null, null, "We can see these text from the image: Why is Redis so fast\n\nredis.\n Waiting for the completion of each request individually is where I/O might be helpful. I/O multiplexing comes into the picture. With I/O multiplexing, the operating system allows a single thread to wait on many socket connections simultaneously. Traditionally, this is done with the 'select' function." ], "metadata": [ { "vid": "5TRFpFBccQM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/5TRFpFBccQM/0.0_24.16.mp4", "refined_asr": " Why is Redis so fast? What fundamental design decisions did the developers make more than a decade ago that stood the test of time? Let's take a look. Redis is a very popular in-memory database. It's rock solid, easy to use, and fast. These", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@0.0_24.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@0.0_24.16#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@0.0_24.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@0.0_24.16#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@0.0_24.16#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@0.0_24.16#4.jpg" ], "ocr_qwen2_vl_72b": "ByteByteGo.com\n\nSystem Design Video Series" }, { "vid": "5TRFpFBccQM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/5TRFpFBccQM/24.16_48.4.mp4", "refined_asr": " Attributes explain why it is one of the most loved databases according to Stack Overflow's annual developer survey. The first reason Redis is fast is because it is an in-memory database. Memory access is several orders of magnitude faster than random disk I/O. Pure memory access provides high read and write throughput and low latency. The trade-off is", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@24.16_48.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@24.16_48.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@24.16_48.4#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@24.16_48.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@24.16_48.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@24.16_48.4#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@24.16_48.4#4.jpg" ], "ocr_qwen2_vl_72b": "Why is Redis so fast\n\n- Rock solid\n- Easy to use\n- Fast" }, { "vid": "5TRFpFBccQM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/5TRFpFBccQM/48.4_70.8.mp4", "refined_asr": " The data set cannot be larger than memory. Code-wise, in-memory data structures are also much easier to implement than their on-disk counterparts. This keeps the code simple and contributes to Redis' rock-solid stability. Another reason Redis is fast, which might seem counterintuitive, is that it is primarily single-threaded.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@48.4_70.8#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@48.4_70.8#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@48.4_70.8#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@48.4_70.8#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@48.4_70.8#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@48.4_70.8#4.jpg" ], "ocr_qwen2_vl_72b": "Why is Redis so fast\n\n50-150us\n\nHigh read/write throughput\nLow latency\nDataset cannot be larger than memory\n\n120ns\n\nMemory access Random disk I/O" }, { "vid": "5TRFpFBccQM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/5TRFpFBccQM/72.0_94.32000000000001.mp4", "refined_asr": " Why would a single-threaded design lead to high performance? Wouldn't it be faster if it used threads to leverage all the CPU cores? Multi-threaded applications require locks or other synchronization mechanisms. They are notoriously hard to reason about. In many applications, the added complexity is bug-prone and sacrifices stability.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@72.0_94.32000000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@72.0_94.32000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@72.0_94.32000000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@72.0_94.32000000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@72.0_94.32000000000001#4.jpg" ], "ocr_qwen2_vl_72b": "Why is Redis so fast\n\nSingle-threaded process\n\ncode data files\ncode files\n\nthread" }, { "vid": "5TRFpFBccQM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/5TRFpFBccQM/94.88000000000001_113.28.mp4", "refined_asr": " Making it difficult to justify the performance gain. In the case of Redis, the single-threaded code path is easy to understand. How does a single-threaded code base handle many thousands of incoming requests and outgoing responses at the same time? Won't the thread get blocked?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@94.88000000000001_113.28#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@94.88000000000001_113.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@94.88000000000001_113.28#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@94.88000000000001_113.28#3.jpg" ], "ocr_qwen2_vl_72b": "Why is Redis so fast\n\nThread 1\nLPUSH\nLength + 1\n\nThread 2\nLPOP\nLength - 1\n\nList.length\nList" }, { "vid": "5TRFpFBccQM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/5TRFpFBccQM/113.28_131.28.mp4", "refined_asr": " Waiting for the completion of each request individually is where I/O might be helpful. I/O multiplexing comes into the picture. With I/O multiplexing, the operating system allows a single thread to wait on many socket connections simultaneously. Traditionally, this is done with the 'select' function.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@113.28_131.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@113.28_131.28#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@113.28_131.28#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@113.28_131.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@113.28_131.28#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5TRFpFBccQM/5TRFpFBccQM@113.28_131.28#3.jpg" ], "ocr_qwen2_vl_72b": "Why is Redis so fast\n\nredis" } ], "image_num": 12, "text_num": 418, "token_num": 7330 }, { "images": [ "sample_100_images/87f_TWmhkzI@503.0_517.0#1.jpg", null, "sample_100_images/87f_TWmhkzI@517.0_531.0#1.jpg", null, "sample_100_images/87f_TWmhkzI@531.0_548.0#1.jpg", null, "sample_100_images/87f_TWmhkzI@548.0_562.0#1.jpg", null, "sample_100_images/87f_TWmhkzI@562.0_580.0#1.jpg", "sample_100_images/87f_TWmhkzI@562.0_580.0#3.jpg", null, "sample_100_images/87f_TWmhkzI@580.0_592.0#1.jpg", "sample_100_images/87f_TWmhkzI@580.0_592.0#2.jpg", null, "sample_100_images/87f_TWmhkzI@592.0_610.0#1.jpg", null, "sample_100_images/87f_TWmhkzI@610.0_623.0#1.jpg", null ], "texts": [ null, " When you're thinking about applying the Diels-Alder reaction synthetically, you should look for this specific structure within the target. Wherever you see a six-membered ring within a target, the Diels-Alder reaction is a good candidate reaction to use.", null, " But you want to be careful to identify a cyclohexene like this specifically. For example, in the substrate that we see down here, there are two six-membered rings, but only one of them can be made using a Diels-Alder reaction.", null, " The one that can be made using a Diels-Alder reaction is the ring on the left because this is a cyclohexene. The ring on the right is not, strictly speaking, a cyclohexene since these carbons involved in the CO double bonds are sp2 hybridized.", null, " We need sp3 hybridization at these carbons in the middle of the ring here in order to use the Diels-Alder reaction to construct that ring. And so to deepen our understanding of this reaction, let's just take a minute and work backwards from this product or target.", null, null, " To the substrates that could have been combined to form it. In other words, what diene and dienophile can this be made from? Well if we look at the example above we see that the reaction establishes two new sigma bonds that are two bonds away from the double bond within the cyclohexene.", null, null, " And those bonds within the cyclohexene this target at the bottom are here and here. One of the nice things about pericyclic reactions and a theme you'll notice as we study the Diels-Alder reaction in more detail.", null, " To work backwards, you need only flow electrons in the opposite direction. So we can use cyclic electron flow in a retrosynthetic or backwards sense to generate the diene and dienophile that can be used to construct this cyclohexene using a Diels-Alder reaction.", null, " So I can just follow my cyclic electron flow follow the arrows backwards to generate the diene and dienophile here. So here's the diene and notice that I've got two new double bonds one from cleavage quote-unquote." ], "text_ocr_list": [ null, "We can see these text from the image: The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\n1,4-Cyclohexadienes can also be synthesized from alkyne dienophiles. :).\n When you're thinking about applying the Diels-Alder reaction synthetically, you should look for this specific structure within the target. Wherever you see a six-membered ring within a target, the Diels-Alder reaction is a good candidate reaction to use.", null, "We can see these text from the image: The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nCN\ndiene\ndienophile\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nH3CO\nO\n\n1,4-Cyclohexadienes can also be synthesized from alkyne dienophiles. :).\n But you want to be careful to identify a cyclohexene like this specifically. For example, in the substrate that we see down here, there are two six-membered rings, but only one of them can be made using a Diels-Alder reaction.", null, "We can see these text from the image: The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nCN\ndiene\ndienophile\n\nCN\ncyclohexene\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nH\u2083CO.\n The one that can be made using a Diels-Alder reaction is the ring on the left because this is a cyclohexene. The ring on the right is not, strictly speaking, a cyclohexene since these carbons involved in the CO double bonds are sp2 hybridized.", null, "We can see these text from the image: The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nCN\nCN\ndiene\ndienophile\nH3CO.\n We need sp3 hybridization at these carbons in the middle of the ring here in order to use the Diels-Alder reaction to construct that ring. And so to deepen our understanding of this reaction, let's just take a minute and work backwards from this product or target.", null, null, "We can see these text from the image: The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nCN\nCN\ncyclohexene\n\nH3CO.\n To the substrates that could have been combined to form it. In other words, what diene and dienophile can this be made from? Well if we look at the example above we see that the reaction establishes two new sigma bonds that are two bonds away from the double bond within the cyclohexene.", null, null, "We can see these text from the image: The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nCN\n[4+2]\ncyclohexene\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nH\u2083CO\nO\nO.\n And those bonds within the cyclohexene this target at the bottom are here and here. One of the nice things about pericyclic reactions and a theme you'll notice as we study the Diels-Alder reaction in more detail.", null, "We can see these text from the image: The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nCN\ndiene\ndienophile\nH3CO\nCN\ncyclohexene\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0).\n To work backwards, you need only flow electrons in the opposite direction. So we can use cyclic electron flow in a retrosynthetic or backwards sense to generate the diene and dienophile that can be used to construct this cyclohexene using a Diels-Alder reaction.", null, "We can see these text from the image: The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nCN\nCN\ncyclohexene\n\ndiene\ndienophile\n\nH3CO.\n So I can just follow my cyclic electron flow follow the arrows backwards to generate the diene and dienophile here. So here's the diene and notice that I've got two new double bonds one from cleavage quote-unquote." ], "metadata": [ { "vid": "87f_TWmhkzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/87f_TWmhkzI/503.0_517.0.mp4", "refined_asr": " When you're thinking about applying the Diels-Alder reaction synthetically, you should look for this specific structure within the target. Wherever you see a six-membered ring within a target, the Diels-Alder reaction is a good candidate reaction to use.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@503.0_517.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@503.0_517.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@503.0_517.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@503.0_517.0#3.jpg" ], "ocr_qwen2_vl_72b": "The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\n1,4-Cyclohexadienes can also be synthesized from alkyne dienophiles. :)" }, { "vid": "87f_TWmhkzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/87f_TWmhkzI/517.0_531.0.mp4", "refined_asr": " But you want to be careful to identify a cyclohexene like this specifically. For example, in the substrate that we see down here, there are two six-membered rings, but only one of them can be made using a Diels-Alder reaction.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@517.0_531.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@517.0_531.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@517.0_531.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@517.0_531.0#3.jpg" ], "ocr_qwen2_vl_72b": "The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nCN\ndiene\ndienophile\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nH3CO\nO\n\n1,4-Cyclohexadienes can also be synthesized from alkyne dienophiles. :)" }, { "vid": "87f_TWmhkzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/87f_TWmhkzI/531.0_548.0.mp4", "refined_asr": " The one that can be made using a Diels-Alder reaction is the ring on the left because this is a cyclohexene. The ring on the right is not, strictly speaking, a cyclohexene since these carbons involved in the CO double bonds are sp2 hybridized.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@531.0_548.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@531.0_548.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@531.0_548.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@531.0_548.0#3.jpg" ], "ocr_qwen2_vl_72b": "The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nCN\ndiene\ndienophile\n\nCN\ncyclohexene\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nH\u2083CO" }, { "vid": "87f_TWmhkzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/87f_TWmhkzI/548.0_562.0.mp4", "refined_asr": " We need sp3 hybridization at these carbons in the middle of the ring here in order to use the Diels-Alder reaction to construct that ring. And so to deepen our understanding of this reaction, let's just take a minute and work backwards from this product or target.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@548.0_562.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@548.0_562.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@548.0_562.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@548.0_562.0#3.jpg" ], "ocr_qwen2_vl_72b": "The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nCN\nCN\ndiene\ndienophile\nH3CO" }, { "vid": "87f_TWmhkzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/87f_TWmhkzI/562.0_580.0.mp4", "refined_asr": " To the substrates that could have been combined to form it. In other words, what diene and dienophile can this be made from? Well if we look at the example above we see that the reaction establishes two new sigma bonds that are two bonds away from the double bond within the cyclohexene.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@562.0_580.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@562.0_580.0#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@562.0_580.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@562.0_580.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@562.0_580.0#3.jpg" ], "ocr_qwen2_vl_72b": "The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nCN\nCN\ncyclohexene\n\nH3CO" }, { "vid": "87f_TWmhkzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/87f_TWmhkzI/580.0_592.0.mp4", "refined_asr": " And those bonds within the cyclohexene this target at the bottom are here and here. One of the nice things about pericyclic reactions and a theme you'll notice as we study the Diels-Alder reaction in more detail.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@580.0_592.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@580.0_592.0#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@580.0_592.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@580.0_592.0#2.jpg" ], "ocr_qwen2_vl_72b": "The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nCN\n[4+2]\ncyclohexene\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nH\u2083CO\nO\nO" }, { "vid": "87f_TWmhkzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/87f_TWmhkzI/592.0_610.0.mp4", "refined_asr": " To work backwards, you need only flow electrons in the opposite direction. So we can use cyclic electron flow in a retrosynthetic or backwards sense to generate the diene and dienophile that can be used to construct this cyclohexene using a Diels-Alder reaction.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@592.0_610.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@592.0_610.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@592.0_610.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@592.0_610.0#3.jpg" ], "ocr_qwen2_vl_72b": "The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nCN\ndiene\ndienophile\nH3CO\nCN\ncyclohexene\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)" }, { "vid": "87f_TWmhkzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/87f_TWmhkzI/610.0_623.0.mp4", "refined_asr": " So I can just follow my cyclic electron flow follow the arrows backwards to generate the diene and dienophile here. So here's the diene and notice that I've got two new double bonds one from cleavage quote-unquote.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@610.0_623.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@610.0_623.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/87f_TWmhkzI/87f_TWmhkzI@610.0_623.0#2.jpg" ], "ocr_qwen2_vl_72b": "The Diels-Alder Reaction: [4+2] Cycloaddition\n\nConjugated dienes (1,3-dienes) react with alkenes and alkynes in a [4+2] cycloaddition process called the Diels-Alder reaction.\n\nComplex molecule synthesis!\nAtom economy!\n\nImportant terminology:\nDiene (4\u03c0) and dienophile (2\u03c0)\n\nCN\nCN\ncyclohexene\n\ndiene\ndienophile\n\nH3CO" } ], "image_num": 10, "text_num": 487, "token_num": 6247 }, { "images": [ "sample_100_images/vEM4YnRFGoA@4973.58_4994.58#1.jpg", "sample_100_images/vEM4YnRFGoA@4973.58_4994.58#3.jpg", null, "sample_100_images/vEM4YnRFGoA@4994.58_5010.58#1.jpg", "sample_100_images/vEM4YnRFGoA@4994.58_5010.58#2.jpg", null, "sample_100_images/vEM4YnRFGoA@5010.58_5028.58#1.jpg", "sample_100_images/vEM4YnRFGoA@5010.58_5028.58#2.jpg", null, "sample_100_images/vEM4YnRFGoA@5028.58_5037.58#1.jpg", null, "sample_100_images/vEM4YnRFGoA@5037.58_5056.58#1.jpg", "sample_100_images/vEM4YnRFGoA@5037.58_5056.58#2.jpg", "sample_100_images/vEM4YnRFGoA@5037.58_5056.58#3.jpg", null, "sample_100_images/vEM4YnRFGoA@5056.58_5081.58#1.jpg", "sample_100_images/vEM4YnRFGoA@5056.58_5081.58#3.jpg", null, "sample_100_images/vEM4YnRFGoA@5081.58_5095.58#1.jpg", null ], "texts": [ null, null, " Whatever they are. I lost count already. I'm just adding 1.8. Where are we? We're adding 1.8 to each of those. Notice that it may make sense to do a class width of 2 instead of 1.8.", null, null, " One, it's a lot easier to do in your head. Two, it makes a lot more sense in terms of presenting the data. So instead of 1.8, let's round this up to 2. So the boundaries will be 1, 3, 5, 7, 9, 11, 13.", null, null, " Much easier. Now, the lower class limit is the limit of our first class. One is the lowest. So if we just want to follow this, we could start with a one. I would say let's start with zero. That would kind of make sense.", null, " So it would be 0, 2, 4, 6, 8, 10, 12. That would include all the data, instead of 1, 3, 5, 7, 9, 11, 13.", null, null, null, " Which would also include all the data. But maybe 0, 2, 4, 6, 8, etc., looks better to the client. However, we're going with 1. So 1, 3, 5, 7, 9, 11.", null, null, " 3, 5, 7, 9, 11, 13. We're backing it off by 0.1. Now we do the frequencies. There are our midpoints. And again, don't know any statistician that would say the midpoint is 1.95.", null, " Every statistician I know would say the midpoint is 2, 4, 6, 8, 10, 12. And the class boundaries for the class would be 1 to 3, 3 to 5, 5 to 7, and so on." ], "text_ocr_list": [ null, null, "We can see these text from the image: Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nSolution\n\nSince this example calls for six classes, a good starting point for the class width is calculated as follows.\n\nNumbers of Miles Professors Drive to Work Each Day\n\n| 3.8 | 2.7 | 9.3 | 6.5 | 5.8 | 7 |\n|-----|-----|-----|-----|-----|---|\n| 10.2 | 1 | 3.7 | 9.1 | 6.2 | 11 |\n| 11.9 | 5.5 | 4.8 | 7.3 | 9.1 | 1.4 |\n\n\\[\n\\frac{11.9 - 1}{6} = 1.816 \\approx 1.8\n\\].\n Whatever they are. I lost count already. I'm just adding 1.8. Where are we? We're adding 1.8 to each of those. Notice that it may make sense to do a class width of 2 instead of 1.8.", null, null, "We can see these text from the image: Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nBecause our data are in miles, a more sensible class width to use is 2.\n\nNext, to choose the lower class limit of the first class, begin by considering the smallest data value, which is 1 mile. In this case, 1.0 is a reasonable place to begin our classes. Adding the class width of 2 gives us the following table.\n\nCopyright \u00a9 by Hawkes Learning\nAll rights reserved..\n One, it's a lot easier to do in your head. Two, it makes a lot more sense in terms of presenting the data. So instead of 1.8, let's round this up to 2. So the boundaries will be 1, 3, 5, 7, 9, 11, 13.", null, null, "We can see these text from the image: Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nBecause our data are in miles, a more sensible class width to use is 2.\n\nNext, to choose the lower class limit of the first class, begin by considering the smallest data value, which is 1 mile. In this case, 1.0 is a reasonable place to begin our classes. Adding the class width of 2 gives us the following table.\n\nCopyright \u00a9 by Hawkes Learning\nAll rights reserved..\n Much easier. Now, the lower class limit is the limit of our first class. One is the lowest. So if we just want to follow this, we could start with a one. I would say let's start with zero. That would kind of make sense.", null, "We can see these text from the image: Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nSolution\n\nSince this example calls for six classes, a good starting point for the class width is calculated as follows.\n\nNumbers of Miles Professors Drive to Work Each Day\n\n| 3.8 | 2.7 | 9.3 | 6.5 | 5.8 | 7 |\n| 10.2 | 1 | 3.7 | 9.1 | 6.2 | 11 |\n| 11.9 | 5.5 | 4.8 | 7.3 | 9.1 | 1.4 |\n\n\\(\\frac{11.9 - 1}{6} = 1.816 \\approx 1.8\\).\n So it would be 0, 2, 4, 6, 8, 10, 12. That would include all the data, instead of 1, 3, 5, 7, 9, 11, 13.", null, null, null, "We can see these text from the image: Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nSolution\n\nSince this example calls for six classes, a good starting point for the class width is calculated as follows.\n\nNumbers of Miles Professors Drive to Work Each Day\n\n| 3.8 | 2.7 | 9.3 | 6.5 | 5.8 | 7 |\n| 10.2 | 1 | 3.7 | 9.1 | 6.2 | 11 |\n| 11.9 | 5.5 | 4.8 | 7.3 | 9.1 | 1.4 |\n\n\\[\n\\frac{11.9 - 1}{6} = 1.816 \\approx 1.8\n\\].\n Which would also include all the data. But maybe 0, 2, 4, 6, 8, etc., looks better to the client. However, we're going with 1. So 1, 3, 5, 7, 9, 11.", null, null, "We can see these text from the image: Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nNumbers of Miles Professors Drive to Work Each Day\n\n| Class | Frequency |\n|---------------|-----------|\n| 1.0 - 2.9 | |\n| 3.0 - 4.9 | |\n| 5.0 - 6.9 | |\n| 7.0 - 8.9 | |\n| 9.0 - 10.9 | |\n| 11.0 - 12.9 | |\n\nOnce again, note that all of the data values fall within the range of the class limits. So, no adjustments in the classes are necessary.\n\nCopyright \u00a9 by Hawkes Learning\nAll rights reserved..\n 3, 5, 7, 9, 11, 13. We're backing it off by 0.1. Now we do the frequencies. There are our midpoints. And again, don't know any statistician that would say the midpoint is 1.95.", null, "We can see these text from the image: Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nNumbers of Miles Professors Drive to Work Each Day\n\n| Class | Frequency | Class Boundaries | Midpoint | Relative Frequency | Cumulative Frequency |\n|-------------|-----------|------------------|----------|--------------------|----------------------|\n| 1.0\u20132.9 | 3 | 0.95\u20132.95 | 1.95 | \\(\\frac{3}{18} = \\frac{1}{6} = 0.16 \\approx 17\\%\\) | 3 |\n| 3.0\u20134.9 | 3 | 2.95\u20134.95 | 3.95 | \\(\\frac{3}{18} = \\frac{1}{6} = 0.16 \\approx 17\\%\\) | 6 |\n| 5.0\u20136.9 | 4 | 4.95\u20136.95 | 5.95 | \\(\\frac{4}{18} = \\frac{2}{9} = 0.2 \\approx 22\\%\\) | 10 |\n| 7.0\u20138.9 | 2 | 6.95\u20138.95 | 7.95 | \\(\\frac{2}{18} = \\frac{1}{9} = 0.1 \\approx 11\\%\\) | 12 |\n| 9.0\u201310.9 | 4 | 8.95\u201310.95 | 9.95 | \\(\\frac{4}{18} = \\frac{2}{9} = 0.2 \\approx 22\\%\\) | 16 |\n| 11.0\u201312.9 | 2 | 10.95\u201312.95 | 11.95 | \\(\\frac{2}{18} = \\frac{1}{9} = 0.1 \\approx 11\\%\\) | 18 |\n\nCopyright by Hawkes Learning\nAll rights reserved..\n Every statistician I know would say the midpoint is 2, 4, 6, 8, 10, 12. And the class boundaries for the class would be 1 to 3, 3 to 5, 5 to 7, and so on." ], "metadata": [ { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/4973.58_4994.58.mp4", "refined_asr": " Whatever they are. I lost count already. I'm just adding 1.8. Where are we? We're adding 1.8 to each of those. Notice that it may make sense to do a class width of 2 instead of 1.8.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4973.58_4994.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4973.58_4994.58#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4973.58_4994.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4973.58_4994.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4973.58_4994.58#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4973.58_4994.58#4.jpg" ], "ocr_qwen2_vl_72b": "Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nSolution\n\nSince this example calls for six classes, a good starting point for the class width is calculated as follows.\n\nNumbers of Miles Professors Drive to Work Each Day\n\n| 3.8 | 2.7 | 9.3 | 6.5 | 5.8 | 7 |\n|-----|-----|-----|-----|-----|---|\n| 10.2 | 1 | 3.7 | 9.1 | 6.2 | 11 |\n| 11.9 | 5.5 | 4.8 | 7.3 | 9.1 | 1.4 |\n\n\\[\n\\frac{11.9 - 1}{6} = 1.816 \\approx 1.8\n\\]" }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/4994.58_5010.58.mp4", "refined_asr": " One, it's a lot easier to do in your head. Two, it makes a lot more sense in terms of presenting the data. So instead of 1.8, let's round this up to 2. So the boundaries will be 1, 3, 5, 7, 9, 11, 13.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4994.58_5010.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4994.58_5010.58#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4994.58_5010.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4994.58_5010.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@4994.58_5010.58#3.jpg" ], "ocr_qwen2_vl_72b": "Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nBecause our data are in miles, a more sensible class width to use is 2.\n\nNext, to choose the lower class limit of the first class, begin by considering the smallest data value, which is 1 mile. In this case, 1.0 is a reasonable place to begin our classes. Adding the class width of 2 gives us the following table.\n\nCopyright \u00a9 by Hawkes Learning\nAll rights reserved." }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/5010.58_5028.58.mp4", "refined_asr": " Much easier. Now, the lower class limit is the limit of our first class. One is the lowest. So if we just want to follow this, we could start with a one. I would say let's start with zero. That would kind of make sense.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5010.58_5028.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5010.58_5028.58#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5010.58_5028.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5010.58_5028.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5010.58_5028.58#3.jpg" ], "ocr_qwen2_vl_72b": "Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nBecause our data are in miles, a more sensible class width to use is 2.\n\nNext, to choose the lower class limit of the first class, begin by considering the smallest data value, which is 1 mile. In this case, 1.0 is a reasonable place to begin our classes. Adding the class width of 2 gives us the following table.\n\nCopyright \u00a9 by Hawkes Learning\nAll rights reserved." }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/5028.58_5037.58.mp4", "refined_asr": " So it would be 0, 2, 4, 6, 8, 10, 12. That would include all the data, instead of 1, 3, 5, 7, 9, 11, 13.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5028.58_5037.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5028.58_5037.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5028.58_5037.58#2.jpg" ], "ocr_qwen2_vl_72b": "Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nSolution\n\nSince this example calls for six classes, a good starting point for the class width is calculated as follows.\n\nNumbers of Miles Professors Drive to Work Each Day\n\n| 3.8 | 2.7 | 9.3 | 6.5 | 5.8 | 7 |\n| 10.2 | 1 | 3.7 | 9.1 | 6.2 | 11 |\n| 11.9 | 5.5 | 4.8 | 7.3 | 9.1 | 1.4 |\n\n\\(\\frac{11.9 - 1}{6} = 1.816 \\approx 1.8\\)" }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/5037.58_5056.58.mp4", "refined_asr": " Which would also include all the data. But maybe 0, 2, 4, 6, 8, etc., looks better to the client. However, we're going with 1. So 1, 3, 5, 7, 9, 11.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5037.58_5056.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5037.58_5056.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5037.58_5056.58#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5037.58_5056.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5037.58_5056.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5037.58_5056.58#3.jpg" ], "ocr_qwen2_vl_72b": "Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nSolution\n\nSince this example calls for six classes, a good starting point for the class width is calculated as follows.\n\nNumbers of Miles Professors Drive to Work Each Day\n\n| 3.8 | 2.7 | 9.3 | 6.5 | 5.8 | 7 |\n| 10.2 | 1 | 3.7 | 9.1 | 6.2 | 11 |\n| 11.9 | 5.5 | 4.8 | 7.3 | 9.1 | 1.4 |\n\n\\[\n\\frac{11.9 - 1}{6} = 1.816 \\approx 1.8\n\\]" }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/5056.58_5081.58.mp4", "refined_asr": " 3, 5, 7, 9, 11, 13. We're backing it off by 0.1. Now we do the frequencies. There are our midpoints. And again, don't know any statistician that would say the midpoint is 1.95.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5056.58_5081.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5056.58_5081.58#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5056.58_5081.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5056.58_5081.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5056.58_5081.58#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5056.58_5081.58#4.jpg" ], "ocr_qwen2_vl_72b": "Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nNumbers of Miles Professors Drive to Work Each Day\n\n| Class | Frequency |\n|---------------|-----------|\n| 1.0 - 2.9 | |\n| 3.0 - 4.9 | |\n| 5.0 - 6.9 | |\n| 7.0 - 8.9 | |\n| 9.0 - 10.9 | |\n| 11.0 - 12.9 | |\n\nOnce again, note that all of the data values fall within the range of the class limits. So, no adjustments in the classes are necessary.\n\nCopyright \u00a9 by Hawkes Learning\nAll rights reserved." }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/5081.58_5095.58.mp4", "refined_asr": " Every statistician I know would say the midpoint is 2, 4, 6, 8, 10, 12. And the class boundaries for the class would be 1 to 3, 3 to 5, 5 to 7, and so on.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5081.58_5095.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5081.58_5095.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5081.58_5095.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@5081.58_5095.58#3.jpg" ], "ocr_qwen2_vl_72b": "Example 2.6: Characteristics of a Frequency Distribution (cont.)\n\nNumbers of Miles Professors Drive to Work Each Day\n\n| Class | Frequency | Class Boundaries | Midpoint | Relative Frequency | Cumulative Frequency |\n|-------------|-----------|------------------|----------|--------------------|----------------------|\n| 1.0\u20132.9 | 3 | 0.95\u20132.95 | 1.95 | \\(\\frac{3}{18} = \\frac{1}{6} = 0.16 \\approx 17\\%\\) | 3 |\n| 3.0\u20134.9 | 3 | 2.95\u20134.95 | 3.95 | \\(\\frac{3}{18} = \\frac{1}{6} = 0.16 \\approx 17\\%\\) | 6 |\n| 5.0\u20136.9 | 4 | 4.95\u20136.95 | 5.95 | \\(\\frac{4}{18} = \\frac{2}{9} = 0.2 \\approx 22\\%\\) | 10 |\n| 7.0\u20138.9 | 2 | 6.95\u20138.95 | 7.95 | \\(\\frac{2}{18} = \\frac{1}{9} = 0.1 \\approx 11\\%\\) | 12 |\n| 9.0\u201310.9 | 4 | 8.95\u201310.95 | 9.95 | \\(\\frac{4}{18} = \\frac{2}{9} = 0.2 \\approx 22\\%\\) | 16 |\n| 11.0\u201312.9 | 2 | 10.95\u201312.95 | 11.95 | \\(\\frac{2}{18} = \\frac{1}{9} = 0.1 \\approx 11\\%\\) | 18 |\n\nCopyright by Hawkes Learning\nAll rights reserved." } ], "image_num": 13, "text_num": 448, "token_num": 7936 }, { "images": [ "sample_100_images/vEM4YnRFGoA@9386.58_9408.58#1.jpg", "sample_100_images/vEM4YnRFGoA@9386.58_9408.58#2.jpg", null, "sample_100_images/vEM4YnRFGoA@9408.58_9426.58#1.jpg", null, "sample_100_images/vEM4YnRFGoA@9426.58_9442.58#1.jpg", null, "sample_100_images/vEM4YnRFGoA@9442.58_9462.58#1.jpg", "sample_100_images/vEM4YnRFGoA@9442.58_9462.58#3.jpg", null, "sample_100_images/vEM4YnRFGoA@9462.58_9482.58#1.jpg", null, "sample_100_images/vEM4YnRFGoA@9482.58_9508.58#1.jpg", "sample_100_images/vEM4YnRFGoA@9482.58_9508.58#3.jpg", "sample_100_images/vEM4YnRFGoA@9482.58_9508.58#4.jpg", null, "sample_100_images/vEM4YnRFGoA@9508.58_9526.58#1.jpg", null ], "texts": [ null, null, " So that later you can transpose it into Moodle. Pause. And back we are. Third one is mode - the most observed value. Whether you're dealing with data or a population, it's the most observed value.", null, " Here's how we calculate it by hand. You just do a frequency distribution. Remember that from Chapter Two? A frequency distribution of the data and find the value that occurs most frequently. Some terminology: if there is only one value that occurs most often.", null, " It's called unimodal distribution. If exactly two values occur equally often, it's bimodal data or the variable is termed bimodal. If it's more than two values, it doesn't fall under these classifications.", null, null, " If values occur equally often it's multimodal however if the data values only occur once or an equal number of times each we say there is no mode so here's some examples finding the mode for the first one I see two sixes but I see three sevens", null, " So seven is the mode of A and A is unimodal because there's only one mode. For B I don't see anything occurring more than once so there is no mode. For C I see seven and two both occurring twice.", null, null, null, " So the modes are seven and two and it is bimodal. For D, everything occurs twice so there is no mode, which is what these solutions say. Doing this in R, note that the mode is not really a helpful measure of center for most.", null, " Data that we deal with can often lack a clear mode. Therefore, most statistical programs don't have a built-in function for finding the mode. I've created one for you. It's called 'modal'. That's the function: modal. And it just takes your data. However, you need to source it first." ], "text_ocr_list": [ null, null, "We can see these text from the image: Intra-Lecture Question\n\nTo help ensure that you are making the most of the lecture presentations, I am including these \"Intra-Lecture Questions\" for you to answer in the Moodle quiz for the lecture. Not all lectures have these; most will. I encourage you to write the questions and answers in your notes. That will allow you to take the quiz easier.\n\nQuestion 2:\n\nWhat is the largest number of medians that a variable can have?.\n So that later you can transpose it into Moodle. Pause. And back we are. Third one is mode - the most observed value. Whether you're dealing with data or a population, it's the most observed value.", null, "We can see these text from the image: Mode\n\nCalculation by Hand\n\nThe mode is a value in the data set that occurs most frequently.\n- Unimodal: One value occurs most often.\n- Bimodal: Exactly two values occur equally often.\n- Multimodal: More than two values occur equally often.\n\nCopyright \u00a9 by Hawkes Learning,\nmodified by Ole J. Forsberg,\nAll rights reserved..\n Here's how we calculate it by hand. You just do a frequency distribution. Remember that from Chapter Two? A frequency distribution of the data and find the value that occurs most frequently. Some terminology: if there is only one value that occurs most often.", null, "We can see these text from the image: Mode\n\nCalculation by Hand\n\nThe mode is a value in the data set that occurs most frequently.\n- Unimodal: One value occurs most often.\n- Bimodal: Exactly two values occur equally often.\n- Multimodal: More than two values occur equally often.\n\nCopyright \u00a9 by Hawkes Learning,\nmodified by Ole J. Forsberg,\nAll rights reserved..\n It's called unimodal distribution. If exactly two values occur equally often, it's bimodal data or the variable is termed bimodal. If it's more than two values, it doesn't fall under these classifications.", null, null, "We can see these text from the image: Mode\n\nCalculation by Hand\n\nThe mode is a value in the data set that occurs most frequently.\n- Unimodal: One value occurs most often.\n- Bimodal: Exactly two values occur equally often.\n- Multimodal: More than two values occur equally often.\n\nNote: If the data values only occur once or an equal number of times, we say there is no mode..\n If values occur equally often it's multimodal however if the data values only occur once or an equal number of times each we say there is no mode so here's some examples finding the mode for the first one I see two sixes but I see three sevens", null, "We can see these text from the image: Ex 3.6: Finding the Mode\n\nGiven the number of phone calls received each hour during business hours for four different companies, find the mode of each, and state if the data set is unimodal, bimodal, or neither.\n\na. 6, 4, 6, 1, 7, 8, 7, 2, 5, 7\nb. 3, 4, 7, 8, 1, 6, 9\nc. 2, 5, 7, 2, 8, 7, 9, 3\nd. 2, 2, 3, 3, 4, 4, 5, 5.\n So seven is the mode of A and A is unimodal because there's only one mode. For B I don't see anything occurring more than once so there is no mode. For C I see seven and two both occurring twice.", null, null, null, "We can see these text from the image: Ex 3.6: Finding the Mode\n\nGiven the number of phone calls received each hour during business hours for four different companies, find the mode of each, and state if the data set is unimodal, bimodal, or neither.\n\na. 6, 4, 6, 1, 7, 8, 7, 2, 5, 7\nb. 3, 4, 7, 8, 1, 6, 9\nc. 2, 5, 7, 2, 8, 7, 9, 3\nd. 2, 2, 3, 3, 4, 4, 5, 5.\n So the modes are seven and two and it is bimodal. For D, everything occurs twice so there is no mode, which is what these solutions say. Doing this in R, note that the mode is not really a helpful measure of center for most.", null, "We can see these text from the image: Ex 3.6: Finding the Mode\n\nUsing the R Statistical Environment\n\nNote that, although a few statistical software packages will identify the mode of a data set, the mode is not one of the descriptive statistics natively listed by R. This should indicate that the mode is not as useful to the statistician as are other statistics.\n\nHowever, the stat200 file provides the modal function that calculates the mode(s) of a vector:\n\nsource(\"https://rfs.kvasaheim.com/stat200.R\")\nmodal(c(6, 4, 6, 1, 7, 8, 7, 2, 5, 7)).\n Data that we deal with can often lack a clear mode. Therefore, most statistical programs don't have a built-in function for finding the mode. I've created one for you. It's called 'modal'. That's the function: modal. And it just takes your data. However, you need to source it first." ], "metadata": [ { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/9386.58_9408.58.mp4", "refined_asr": " So that later you can transpose it into Moodle. Pause. And back we are. Third one is mode - the most observed value. Whether you're dealing with data or a population, it's the most observed value.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9386.58_9408.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9386.58_9408.58#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9386.58_9408.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9386.58_9408.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9386.58_9408.58#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9386.58_9408.58#4.jpg" ], "ocr_qwen2_vl_72b": "Intra-Lecture Question\n\nTo help ensure that you are making the most of the lecture presentations, I am including these \"Intra-Lecture Questions\" for you to answer in the Moodle quiz for the lecture. Not all lectures have these; most will. I encourage you to write the questions and answers in your notes. That will allow you to take the quiz easier.\n\nQuestion 2:\n\nWhat is the largest number of medians that a variable can have?" }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/9408.58_9426.58.mp4", "refined_asr": " Here's how we calculate it by hand. You just do a frequency distribution. Remember that from Chapter Two? A frequency distribution of the data and find the value that occurs most frequently. Some terminology: if there is only one value that occurs most often.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9408.58_9426.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9408.58_9426.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9408.58_9426.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9408.58_9426.58#3.jpg" ], "ocr_qwen2_vl_72b": "Mode\n\nCalculation by Hand\n\nThe mode is a value in the data set that occurs most frequently.\n- Unimodal: One value occurs most often.\n- Bimodal: Exactly two values occur equally often.\n- Multimodal: More than two values occur equally often.\n\nCopyright \u00a9 by Hawkes Learning,\nmodified by Ole J. Forsberg,\nAll rights reserved." }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/9426.58_9442.58.mp4", "refined_asr": " It's called unimodal distribution. If exactly two values occur equally often, it's bimodal data or the variable is termed bimodal. If it's more than two values, it doesn't fall under these classifications.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9426.58_9442.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9426.58_9442.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9426.58_9442.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9426.58_9442.58#3.jpg" ], "ocr_qwen2_vl_72b": "Mode\n\nCalculation by Hand\n\nThe mode is a value in the data set that occurs most frequently.\n- Unimodal: One value occurs most often.\n- Bimodal: Exactly two values occur equally often.\n- Multimodal: More than two values occur equally often.\n\nCopyright \u00a9 by Hawkes Learning,\nmodified by Ole J. Forsberg,\nAll rights reserved." }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/9442.58_9462.58.mp4", "refined_asr": " If values occur equally often it's multimodal however if the data values only occur once or an equal number of times each we say there is no mode so here's some examples finding the mode for the first one I see two sixes but I see three sevens", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9442.58_9462.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9442.58_9462.58#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9442.58_9462.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9442.58_9462.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9442.58_9462.58#3.jpg" ], "ocr_qwen2_vl_72b": "Mode\n\nCalculation by Hand\n\nThe mode is a value in the data set that occurs most frequently.\n- Unimodal: One value occurs most often.\n- Bimodal: Exactly two values occur equally often.\n- Multimodal: More than two values occur equally often.\n\nNote: If the data values only occur once or an equal number of times, we say there is no mode." }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/9462.58_9482.58.mp4", "refined_asr": " So seven is the mode of A and A is unimodal because there's only one mode. For B I don't see anything occurring more than once so there is no mode. For C I see seven and two both occurring twice.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9462.58_9482.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9462.58_9482.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9462.58_9482.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9462.58_9482.58#3.jpg" ], "ocr_qwen2_vl_72b": "Ex 3.6: Finding the Mode\n\nGiven the number of phone calls received each hour during business hours for four different companies, find the mode of each, and state if the data set is unimodal, bimodal, or neither.\n\na. 6, 4, 6, 1, 7, 8, 7, 2, 5, 7\nb. 3, 4, 7, 8, 1, 6, 9\nc. 2, 5, 7, 2, 8, 7, 9, 3\nd. 2, 2, 3, 3, 4, 4, 5, 5" }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/9482.58_9508.58.mp4", "refined_asr": " So the modes are seven and two and it is bimodal. For D, everything occurs twice so there is no mode, which is what these solutions say. Doing this in R, note that the mode is not really a helpful measure of center for most.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9482.58_9508.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9482.58_9508.58#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9482.58_9508.58#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9482.58_9508.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9482.58_9508.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9482.58_9508.58#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9482.58_9508.58#4.jpg" ], "ocr_qwen2_vl_72b": "Ex 3.6: Finding the Mode\n\nGiven the number of phone calls received each hour during business hours for four different companies, find the mode of each, and state if the data set is unimodal, bimodal, or neither.\n\na. 6, 4, 6, 1, 7, 8, 7, 2, 5, 7\nb. 3, 4, 7, 8, 1, 6, 9\nc. 2, 5, 7, 2, 8, 7, 9, 3\nd. 2, 2, 3, 3, 4, 4, 5, 5" }, { "vid": "vEM4YnRFGoA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Statistics Tutorial on Bayesian Probability in Statistics and Probability Course_30.json#####audio#####doingASR#####FinishASR/vEM4YnRFGoA/9508.58_9526.58.mp4", "refined_asr": " Data that we deal with can often lack a clear mode. Therefore, most statistical programs don't have a built-in function for finding the mode. I've created one for you. It's called 'modal'. That's the function: modal. And it just takes your data. However, you need to source it first.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9508.58_9526.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9508.58_9526.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9508.58_9526.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vEM4YnRFGoA/vEM4YnRFGoA@9508.58_9526.58#3.jpg" ], "ocr_qwen2_vl_72b": "Ex 3.6: Finding the Mode\n\nUsing the R Statistical Environment\n\nNote that, although a few statistical software packages will identify the mode of a data set, the mode is not one of the descriptive statistics natively listed by R. This should indicate that the mode is not as useful to the statistician as are other statistics.\n\nHowever, the stat200 file provides the modal function that calculates the mode(s) of a vector:\n\nsource(\"https://rfs.kvasaheim.com/stat200.R\")\nmodal(c(6, 4, 6, 1, 7, 8, 7, 2, 5, 7))" } ], "image_num": 11, "text_num": 401, "token_num": 6737 }, { "images": [ "sample_100_images/dz7Ntp7KQGA@35506.42_35525.520000000004#1.jpg", "sample_100_images/dz7Ntp7KQGA@35506.42_35525.520000000004#2.jpg", null, "sample_100_images/dz7Ntp7KQGA@35525.520000000004_35542.420000000006#1.jpg", "sample_100_images/dz7Ntp7KQGA@35525.520000000004_35542.420000000006#2.jpg", null, "sample_100_images/dz7Ntp7KQGA@35542.420000000006_35560.12#1.jpg", null, "sample_100_images/dz7Ntp7KQGA@35560.12_35578.82000000001#1.jpg", "sample_100_images/dz7Ntp7KQGA@35560.12_35578.82000000001#2.jpg", null, "sample_100_images/dz7Ntp7KQGA@35578.82000000001_35596.82000000001#1.jpg", "sample_100_images/dz7Ntp7KQGA@35578.82000000001_35596.82000000001#3.jpg", null, "sample_100_images/dz7Ntp7KQGA@35596.82000000001_35612.82000000001#1.jpg", "sample_100_images/dz7Ntp7KQGA@35596.82000000001_35612.82000000001#2.jpg", null, "sample_100_images/dz7Ntp7KQGA@35612.82000000001_35629.62#1.jpg", null, "sample_100_images/dz7Ntp7KQGA@35629.62_35644.12#1.jpg", "sample_100_images/dz7Ntp7KQGA@35629.62_35644.12#2.jpg", null ], "texts": [ null, null, " With the bank that everything has indeed been straightened out. So what is exactly wrong here? Well account and deposit information is sensitive data that could be used for identity theft. Sending this or any kind of sensitive information by email is very very risky because email is typically not private or secure. Anyone who knows", null, null, " How can access be granted anywhere along its route? As an alternative, the two offices could have called each other or worked with the ITS to send the information in a more secure fashion. Okay, moving on to the next scenario: The mouse on your computer screen starts to move around on its own and click.", null, " Imagine you have several things on your desktop. What do you do in such a situation? Let's consider these options: A colleague or co-worker leans over so they can see. B: Unplug your mouse. C: Disconnect your computer from the network. D: Tell your supervisor. E: Turn the computer off. F: Run an antivirus. Or G: All of the above.", null, null, " So we have to select all the options that apply in the situation. The options that apply are B and D: which is basically, disconnect your computer from the network. And tell your supervisor - so this is definitely suspicious - immediately report the problem to your supervisor and the ITS support center also since it", null, null, " It seems possible that someone is controlling the computer remotely. It's best if you can disconnect the computer from the network and turn it off until help arrives. If possible, don't turn off the computer. Okay, time for scenario number eight. Below is a list of passwords pulled out of a database. Now,", null, null, " Which of the following passwords meets the UCSC password requirement? Okay, so the third password, which is option number C, is the only one that meets all the UCSC requirements: it has at least eight characters in length and it contains at least three of the following four types of characters:", null, " Characters which are lowercase characters uppercase characters numbers and special characters and not a word is preceded or followed by a digit so it's the third option which is correct in this situation Moving on to the second last scenario we have for today is you receive an email from your bank telling you there is a", null, null, " Problem with your account. The email provides instructions and links so you can log in to fix your account and fix the problem in doing so. So what should you do? Well, we have to delete the email and, better yet, use the web client that is, Gmail, Yahoo Mail." ], "text_ocr_list": [ null, null, "We can see these text from the image: Scenario Based Cybersecurity Questions\n\nTwo different offices on campus are working to straighten out an error in an employee's bank account due to a direct deposit mistake.\n\nOffice #1 emails the correct account and deposit information to office #2, which promptly fixes the problem.\n\nThe employee confirms with the bank that everything has, indeed, been straightened out.\n\n6 What is wrong here?.\n With the bank that everything has indeed been straightened out. So what is exactly wrong here? Well account and deposit information is sensitive data that could be used for identity theft. Sending this or any kind of sensitive information by email is very very risky because email is typically not private or secure. Anyone who knows", null, null, "We can see these text from the image: Scenario Based Cybersecurity Questions: Answer\n\n- Account and deposit information is sensitive data that could be used for identity theft. Sending this or any kind of sensitive information by email is very risky because email is typically not private or secure. Anyone who knows how can access it anywhere along its route.\n- As an alternative, the two offices could have called each other or worked with ITS to send the information a more secure way.\n\nCYBERSECURITY CERTIFICATION COURSE\nwww.edureka.co/cybersecurity-certification-training.\n How can access be granted anywhere along its route? As an alternative, the two offices could have called each other or worked with the ITS to send the information in a more secure fashion. Okay, moving on to the next scenario: The mouse on your computer screen starts to move around on its own and click.", null, "We can see these text from the image: Scenario Based Cybersecurity Questions\n\nThe mouse on your computer screen starts to move around on its own and click on things on your desktop. What do you do?\na) Call your co-workers over so they can see\nb) Disconnect your computer from the network\nc) Unplug your mouse\nd) Tell your supervisor\ne) Turn your computer off\nf) Run anti-virus\ng) All of the above\n\n7 Select all the options that apply.\n\nedureka!\n\nCYBERSECURITY CERTIFICATION COURSE\n\nwww.edureka.co/cybersecurity-certification-training.\n Imagine you have several things on your desktop. What do you do in such a situation? Let's consider these options: A colleague or co-worker leans over so they can see. B: Unplug your mouse. C: Disconnect your computer from the network. D: Tell your supervisor. E: Turn the computer off. F: Run an antivirus. Or G: All of the above.", null, null, "We can see these text from the image: Scenario Based Cybersecurity Questions\n\nThe mouse on your computer screen starts to move around on its own and click on things on your desktop. What do you do?\na) Call your co-workers over so they can see\nb) Disconnect your computer from the network\nc) Unplug your mouse\nd) Tell your supervisor\ne) Turn your computer off\nf) Run anti-virus\ng) All of the above\n\nSelect all the options that apply.\n\nCYBERSECURITY CERTIFICATION COURSE\nwww.edureka.co/cybersecurity-certification-training.\n So we have to select all the options that apply in the situation. The options that apply are B and D: which is basically, disconnect your computer from the network. And tell your supervisor - so this is definitely suspicious - immediately report the problem to your supervisor and the ITS support center also since it", null, null, "We can see these text from the image: Right answer is B & D.\n\nThis is definitely suspicious. Immediately report the problem to your supervisor and the ITS Support Center: itrequest.ucsc.edu, 459-HELP (4357), help@ucsc.edu or Kerr Hall room 54, M-F 8AM-5PM\n\nAlso, since it seems possible that someone is controlling the computer remotely, it is best if you can disconnect the computer from the network (and turn off wireless if you have it) until help arrives. If possible, don\u2019t turn off the computer..\n It seems possible that someone is controlling the computer remotely. It's best if you can disconnect the computer from the network and turn it off until help arrives. If possible, don't turn off the computer. Okay, time for scenario number eight. Below is a list of passwords pulled out of a database. Now,", null, null, "We can see these text from the image: Scenario Based Cybersecurity Questions\n\nBelow are a list of passwords pulled out a database.\nA. @#$)*&^%\nB. akHGksmLN\nC. UcSc4Evr!\nD. Password1\n\nWhich of the following passwords meets UCSC's password requirements?.\n Which of the following passwords meets the UCSC password requirement? Okay, so the third password, which is option number C, is the only one that meets all the UCSC requirements: it has at least eight characters in length and it contains at least three of the following four types of characters:", null, "We can see these text from the image: Scenario Based Cybersecurity Questions: Answer\n\nAnswer is UcSc4Evr!\n\n- This is the only choice that meets all of the following UCSC requirements:\n - At least 8 characters in length\n - Contains at least 3 of the following 4 types of characters: lower case letters, upper case letters, numbers, special characters\n - Not a word preceded or followed by a digit\n\nCYBERSECURITY CERTIFICATION COURSE\nwww.edureka.co/cybersecurity-certification-training.\n Characters which are lowercase characters uppercase characters numbers and special characters and not a word is preceded or followed by a digit so it's the third option which is correct in this situation Moving on to the second last scenario we have for today is you receive an email from your bank telling you there is a", null, null, "We can see these text from the image: Scenario Based Cybersecurity Questions\n\nYou receive an email from your bank telling you there is a problem with your account. The email provides instructions and a link so you can log in to your account and fix the problem.\n\nScenario 9\n\nWhat should you do?\n\nedureka!\n\nCYBERSECURITY CERTIFICATION COURSE\n\nwww.edureka.co/cybersecurity-certification-training.\n Problem with your account. The email provides instructions and links so you can log in to fix your account and fix the problem in doing so. So what should you do? Well, we have to delete the email and, better yet, use the web client that is, Gmail, Yahoo Mail." ], "metadata": [ { "vid": "dz7Ntp7KQGA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Cybersecurity: Ethical Hacking Tutorial_29.json#####audio#####doingASR#####FinishASR/dz7Ntp7KQGA/35506.42_35525.520000000004.mp4", "refined_asr": " With the bank that everything has indeed been straightened out. So what is exactly wrong here? Well account and deposit information is sensitive data that could be used for identity theft. Sending this or any kind of sensitive information by email is very very risky because email is typically not private or secure. Anyone who knows", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35506.42_35525.520000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35506.42_35525.520000000004#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35506.42_35525.520000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35506.42_35525.520000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35506.42_35525.520000000004#3.jpg" ], "ocr_qwen2_vl_72b": "Scenario Based Cybersecurity Questions\n\nTwo different offices on campus are working to straighten out an error in an employee's bank account due to a direct deposit mistake.\n\nOffice #1 emails the correct account and deposit information to office #2, which promptly fixes the problem.\n\nThe employee confirms with the bank that everything has, indeed, been straightened out.\n\n6 What is wrong here?" }, { "vid": "dz7Ntp7KQGA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Cybersecurity: Ethical Hacking Tutorial_29.json#####audio#####doingASR#####FinishASR/dz7Ntp7KQGA/35525.520000000004_35542.420000000006.mp4", "refined_asr": " How can access be granted anywhere along its route? As an alternative, the two offices could have called each other or worked with the ITS to send the information in a more secure fashion. Okay, moving on to the next scenario: The mouse on your computer screen starts to move around on its own and click.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35525.520000000004_35542.420000000006#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35525.520000000004_35542.420000000006#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35525.520000000004_35542.420000000006#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35525.520000000004_35542.420000000006#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35525.520000000004_35542.420000000006#3.jpg" ], "ocr_qwen2_vl_72b": "Scenario Based Cybersecurity Questions: Answer\n\n- Account and deposit information is sensitive data that could be used for identity theft. Sending this or any kind of sensitive information by email is very risky because email is typically not private or secure. Anyone who knows how can access it anywhere along its route.\n- As an alternative, the two offices could have called each other or worked with ITS to send the information a more secure way.\n\nCYBERSECURITY CERTIFICATION COURSE\nwww.edureka.co/cybersecurity-certification-training" }, { "vid": "dz7Ntp7KQGA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Cybersecurity: Ethical Hacking Tutorial_29.json#####audio#####doingASR#####FinishASR/dz7Ntp7KQGA/35542.420000000006_35560.12.mp4", "refined_asr": " Imagine you have several things on your desktop. What do you do in such a situation? Let's consider these options: A colleague or co-worker leans over so they can see. B: Unplug your mouse. C: Disconnect your computer from the network. D: Tell your supervisor. E: Turn the computer off. F: Run an antivirus. Or G: All of the above.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35542.420000000006_35560.12#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35542.420000000006_35560.12#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35542.420000000006_35560.12#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35542.420000000006_35560.12#3.jpg" ], "ocr_qwen2_vl_72b": "Scenario Based Cybersecurity Questions\n\nThe mouse on your computer screen starts to move around on its own and click on things on your desktop. What do you do?\na) Call your co-workers over so they can see\nb) Disconnect your computer from the network\nc) Unplug your mouse\nd) Tell your supervisor\ne) Turn your computer off\nf) Run anti-virus\ng) All of the above\n\n7 Select all the options that apply.\n\nedureka!\n\nCYBERSECURITY CERTIFICATION COURSE\n\nwww.edureka.co/cybersecurity-certification-training" }, { "vid": "dz7Ntp7KQGA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Cybersecurity: Ethical Hacking Tutorial_29.json#####audio#####doingASR#####FinishASR/dz7Ntp7KQGA/35560.12_35578.82000000001.mp4", "refined_asr": " So we have to select all the options that apply in the situation. The options that apply are B and D: which is basically, disconnect your computer from the network. And tell your supervisor - so this is definitely suspicious - immediately report the problem to your supervisor and the ITS support center also since it", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35560.12_35578.82000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35560.12_35578.82000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35560.12_35578.82000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35560.12_35578.82000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35560.12_35578.82000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Scenario Based Cybersecurity Questions\n\nThe mouse on your computer screen starts to move around on its own and click on things on your desktop. What do you do?\na) Call your co-workers over so they can see\nb) Disconnect your computer from the network\nc) Unplug your mouse\nd) Tell your supervisor\ne) Turn your computer off\nf) Run anti-virus\ng) All of the above\n\nSelect all the options that apply.\n\nCYBERSECURITY CERTIFICATION COURSE\nwww.edureka.co/cybersecurity-certification-training" }, { "vid": "dz7Ntp7KQGA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Cybersecurity: Ethical Hacking Tutorial_29.json#####audio#####doingASR#####FinishASR/dz7Ntp7KQGA/35578.82000000001_35596.82000000001.mp4", "refined_asr": " It seems possible that someone is controlling the computer remotely. It's best if you can disconnect the computer from the network and turn it off until help arrives. If possible, don't turn off the computer. Okay, time for scenario number eight. Below is a list of passwords pulled out of a database. Now,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35578.82000000001_35596.82000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35578.82000000001_35596.82000000001#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35578.82000000001_35596.82000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35578.82000000001_35596.82000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35578.82000000001_35596.82000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Right answer is B & D.\n\nThis is definitely suspicious. Immediately report the problem to your supervisor and the ITS Support Center: itrequest.ucsc.edu, 459-HELP (4357), help@ucsc.edu or Kerr Hall room 54, M-F 8AM-5PM\n\nAlso, since it seems possible that someone is controlling the computer remotely, it is best if you can disconnect the computer from the network (and turn off wireless if you have it) until help arrives. If possible, don\u2019t turn off the computer." }, { "vid": "dz7Ntp7KQGA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Cybersecurity: Ethical Hacking Tutorial_29.json#####audio#####doingASR#####FinishASR/dz7Ntp7KQGA/35596.82000000001_35612.82000000001.mp4", "refined_asr": " Which of the following passwords meets the UCSC password requirement? Okay, so the third password, which is option number C, is the only one that meets all the UCSC requirements: it has at least eight characters in length and it contains at least three of the following four types of characters:", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35596.82000000001_35612.82000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35596.82000000001_35612.82000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35596.82000000001_35612.82000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35596.82000000001_35612.82000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35596.82000000001_35612.82000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Scenario Based Cybersecurity Questions\n\nBelow are a list of passwords pulled out a database.\nA. @#$)*&^%\nB. akHGksmLN\nC. UcSc4Evr!\nD. Password1\n\nWhich of the following passwords meets UCSC's password requirements?" }, { "vid": "dz7Ntp7KQGA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Cybersecurity: Ethical Hacking Tutorial_29.json#####audio#####doingASR#####FinishASR/dz7Ntp7KQGA/35612.82000000001_35629.62.mp4", "refined_asr": " Characters which are lowercase characters uppercase characters numbers and special characters and not a word is preceded or followed by a digit so it's the third option which is correct in this situation Moving on to the second last scenario we have for today is you receive an email from your bank telling you there is a", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35612.82000000001_35629.62#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35612.82000000001_35629.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35612.82000000001_35629.62#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35612.82000000001_35629.62#3.jpg" ], "ocr_qwen2_vl_72b": "Scenario Based Cybersecurity Questions: Answer\n\nAnswer is UcSc4Evr!\n\n- This is the only choice that meets all of the following UCSC requirements:\n - At least 8 characters in length\n - Contains at least 3 of the following 4 types of characters: lower case letters, upper case letters, numbers, special characters\n - Not a word preceded or followed by a digit\n\nCYBERSECURITY CERTIFICATION COURSE\nwww.edureka.co/cybersecurity-certification-training" }, { "vid": "dz7Ntp7KQGA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Cybersecurity: Ethical Hacking Tutorial_29.json#####audio#####doingASR#####FinishASR/dz7Ntp7KQGA/35629.62_35644.12.mp4", "refined_asr": " Problem with your account. The email provides instructions and links so you can log in to fix your account and fix the problem in doing so. So what should you do? Well, we have to delete the email and, better yet, use the web client that is, Gmail, Yahoo Mail.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35629.62_35644.12#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35629.62_35644.12#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35629.62_35644.12#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35629.62_35644.12#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dz7Ntp7KQGA/dz7Ntp7KQGA@35629.62_35644.12#3.jpg" ], "ocr_qwen2_vl_72b": "Scenario Based Cybersecurity Questions\n\nYou receive an email from your bank telling you there is a problem with your account. The email provides instructions and a link so you can log in to your account and fix the problem.\n\nScenario 9\n\nWhat should you do?\n\nedureka!\n\nCYBERSECURITY CERTIFICATION COURSE\n\nwww.edureka.co/cybersecurity-certification-training" } ], "image_num": 14, "text_num": 546, "token_num": 8610 }, { "images": [ "sample_100_images/BdvOjT7bh2c@1028.58_1043.2#1.jpg", null, "sample_100_images/BdvOjT7bh2c@1043.88_1060.1200000000001#1.jpg", null, "sample_100_images/BdvOjT7bh2c@1060.66_1087.64#1.jpg", "sample_100_images/BdvOjT7bh2c@1060.66_1087.64#3.jpg", null, "sample_100_images/BdvOjT7bh2c@1088.4_1111.8600000000001#1.jpg", "sample_100_images/BdvOjT7bh2c@1088.4_1111.8600000000001#2.jpg", "sample_100_images/BdvOjT7bh2c@1088.4_1111.8600000000001#3.jpg", null, "sample_100_images/BdvOjT7bh2c@1112.0400000000002_1125.68#1.jpg", "sample_100_images/BdvOjT7bh2c@1112.0400000000002_1125.68#2.jpg", null, "sample_100_images/BdvOjT7bh2c@1125.7800000000002_1147.6200000000001#1.jpg", "sample_100_images/BdvOjT7bh2c@1125.7800000000002_1147.6200000000001#2.jpg", "sample_100_images/BdvOjT7bh2c@1125.7800000000002_1147.6200000000001#3.jpg", null, "sample_100_images/BdvOjT7bh2c@1148.6200000000001_1173.38#1.jpg", "sample_100_images/BdvOjT7bh2c@1148.6200000000001_1173.38#2.jpg", "sample_100_images/BdvOjT7bh2c@1148.6200000000001_1173.38#4.jpg", null ], "texts": [ null, " So if we work on this now, this is equal to 9 times 11 which is 99. To solve further, take the LCM here. The LCM of 3 and 2 is 6.", null, " So 6 over 3 is 2 times pi which is 2 pi. Plus 6 over 2 is 3 times pi which is 3 pi. So I have this plus i sine. Take the LCM of 3 and 2.", null, null, " And that again is 6. 6 divided by 3 is 2 times pi which is 2 pi. Plus 6 divided by 2 is 3 times pi which is 3 pi. So we have this. Alright, so it means that z1, z2 is simply equal to 99 into cos.", null, null, null, " Um, into cosine 2 pi plus 3 pi gives you 5 pi all over 6. So I have this plus i sine. I have 2 pi plus 3 pi that's 5 pi all over 6.", null, null, " So I have this. So this becomes the answer in radians. So this is the value of Z1, Z2, and so on, in radians. If you're asked to provide this value in terms of degrees, alright, you'd say, leave your answer in terms of degrees.", null, null, null, " What do you do? Very simple. We said 1 pi is equal to 180 degrees. So simply replace pi with 180 degrees. Then we'll have that z1, z2 is equal to 99 into cos.", null, null, null, " 5 times 180. That's pi 180 all over 6. Plus i sine 5 times pi, 5 times 180 over 6. So we have this value. Now let me clarify something for you. The value of pi in angular measurement is 180 degrees." ], "text_ocr_list": [ null, "We can see these text from the image: - WhatsApp: +2349032558166\n- Mathematics\n- Complex Numbers (Polar Form)\n- \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n- \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n- \\( z_1 z_2 = 9 \\times 11 \\left[ \\cos \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) \\right] \\)\n- \\( z_1 z_2 = r_1 r_2 \\left[ \\cos (\\theta_1 + \\theta_2) + i \\sin (\\theta_1 + \\theta_2) \\right] \\)\n- Jonah Emmanuel.\n So if we work on this now, this is equal to 9 times 11 which is 99. To solve further, take the LCM here. The LCM of 3 and 2 is 6.", null, "We can see these text from the image: - WhatsApp: +2349032558166\n- Mathematics\n- Complex Numbers (Polar Form)\n- \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n- \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n- \\( z_1 \\cdot z_2 = r_1 r_2 \\left[ \\cos (\\theta_1 + \\theta_2) + i \\sin (\\theta_1 + \\theta_2) \\right] \\)\n- Jonah Emmanuel.\n So 6 over 3 is 2 times pi which is 2 pi. Plus 6 over 2 is 3 times pi which is 3 pi. So I have this plus i sine. Take the LCM of 3 and 2.", null, null, "We can see these text from the image: - WhatsApp: +2349032558166\n- Jonah Emmanuel\n\nMathematics:\n- Complex Numbers (Polar Form)\n - \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n - \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n - \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) \\right] \\)\n - \\( z_1 z_2 = r_1 r_2 \\left[ \\cos (\\theta_1 + \\theta_2) + i \\sin (\\theta_1 + \\theta_2) \\right] \\).\n And that again is 6. 6 divided by 3 is 2 times pi which is 2 pi. Plus 6 divided by 2 is 3 times pi which is 3 pi. So we have this. Alright, so it means that z1, z2 is simply equal to 99 into cos.", null, null, null, "We can see these text from the image: - WhatsApp: +2349032558166\n- Jonah Emmanuel\n\nMathematics:\n- Complex Numbers (Polar Form)\n - \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n - \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n - \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{2\\pi + 3\\pi}{6} \\right) + i \\sin \\left( \\frac{2\\pi + 3\\pi}{6} \\right) \\right] \\)\n - \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{5\\pi}{6} \\right) + i \\sin \\left( \\frac{5\\pi}{6} \\right) \\right] \\).\n Um, into cosine 2 pi plus 3 pi gives you 5 pi all over 6. So I have this plus i sine. I have 2 pi plus 3 pi that's 5 pi all over 6.", null, null, "We can see these text from the image: - WhatsApp: +2349032558166\n- Jonah Emmanuel\n\nMathematics:\n- Complex Numbers (Polar Form)\n - \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n - \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n - \\( z_1 z_2 = 9 \\times 11 \\left[ \\cos \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) \\right] \\)\n - \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{2\\pi + 3\\pi}{6} \\right) + i \\sin \\left( \\frac{2\\pi + 3\\pi}{6} \\right) \\right] \\)\n\n- \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{5\\pi}{6} \\right) + i \\sin \\left( \\frac{5\\pi}{6} \\right) \\right] \\)\n- \\( z_1 z_2 = r_1 r_2 \\left[ \\cos (\\theta_1 + \\theta_2) + i \\sin (\\theta_1 + \\theta_2) \\right] \\).\n So I have this. So this becomes the answer in radians. So this is the value of Z1, Z2, and so on, in radians. If you're asked to provide this value in terms of degrees, alright, you'd say, leave your answer in terms of degrees.", null, null, null, "We can see these text from the image: - WhatsApp: +2349032558166\n- Mathematics\n- Complex Numbers (Polar Form)\n- \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n- \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n- \\( z_1 \\cdot z_2 = 99 \\left[ \\cos \\left( \\frac{2\\pi + 3\\pi}{6} \\right) + i \\sin \\left( \\frac{2\\pi + 3\\pi}{6} \\right) \\right] \\)\n- \\( z_1 \\cdot z_2 = 99 \\left[ \\cos \\left( \\frac{5\\pi}{6} \\right) + i \\sin \\left( \\frac{5\\pi}{6} \\right) \\right] \\)\n- Jonah Emmanuel.\n What do you do? Very simple. We said 1 pi is equal to 180 degrees. So simply replace pi with 180 degrees. Then we'll have that z1, z2 is equal to 99 into cos.", null, null, null, "We can see these text from the image: - WhatsApp: +2349032558166\n- Mathematics\n- Complex Numbers (Polar Form)\n- \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n- \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n- \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{2\\pi + 3\\pi}{6} \\right) + i \\sin \\left( \\frac{2\\pi + 3\\pi}{6} \\right) \\right] \\)\n- \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{5\\pi}{6} \\right) + i \\sin \\left( \\frac{5\\pi}{6} \\right) \\right] \\)\n- \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{5 \\times 180}{6} \\right) + i \\sin \\left( \\frac{5 \\times 180}{6} \\right) \\right] \\)\n- Jonah Emmanuel.\n 5 times 180. That's pi 180 all over 6. Plus i sine 5 times pi, 5 times 180 over 6. So we have this value. Now let me clarify something for you. The value of pi in angular measurement is 180 degrees." ], "metadata": [ { "vid": "BdvOjT7bh2c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Multiplication and Division of Complex Numbers_30.json#####audio#####doingASR#####FinishASR/BdvOjT7bh2c/1028.58_1043.2.mp4", "refined_asr": " So if we work on this now, this is equal to 9 times 11 which is 99. To solve further, take the LCM here. The LCM of 3 and 2 is 6.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1028.58_1043.2#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1028.58_1043.2#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1028.58_1043.2#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1028.58_1043.2#3.jpg" ], "ocr_qwen2_vl_72b": "- WhatsApp: +2349032558166\n- Mathematics\n- Complex Numbers (Polar Form)\n- \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n- \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n- \\( z_1 z_2 = 9 \\times 11 \\left[ \\cos \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) \\right] \\)\n- \\( z_1 z_2 = r_1 r_2 \\left[ \\cos (\\theta_1 + \\theta_2) + i \\sin (\\theta_1 + \\theta_2) \\right] \\)\n- Jonah Emmanuel" }, { "vid": "BdvOjT7bh2c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Multiplication and Division of Complex Numbers_30.json#####audio#####doingASR#####FinishASR/BdvOjT7bh2c/1043.88_1060.1200000000001.mp4", "refined_asr": " So 6 over 3 is 2 times pi which is 2 pi. Plus 6 over 2 is 3 times pi which is 3 pi. So I have this plus i sine. Take the LCM of 3 and 2.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1043.88_1060.1200000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1043.88_1060.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1043.88_1060.1200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1043.88_1060.1200000000001#3.jpg" ], "ocr_qwen2_vl_72b": "- WhatsApp: +2349032558166\n- Mathematics\n- Complex Numbers (Polar Form)\n- \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n- \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n- \\( z_1 \\cdot z_2 = r_1 r_2 \\left[ \\cos (\\theta_1 + \\theta_2) + i \\sin (\\theta_1 + \\theta_2) \\right] \\)\n- Jonah Emmanuel" }, { "vid": "BdvOjT7bh2c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Multiplication and Division of Complex Numbers_30.json#####audio#####doingASR#####FinishASR/BdvOjT7bh2c/1060.66_1087.64.mp4", "refined_asr": " And that again is 6. 6 divided by 3 is 2 times pi which is 2 pi. Plus 6 divided by 2 is 3 times pi which is 3 pi. So we have this. Alright, so it means that z1, z2 is simply equal to 99 into cos.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1060.66_1087.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1060.66_1087.64#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1060.66_1087.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1060.66_1087.64#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1060.66_1087.64#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1060.66_1087.64#4.jpg" ], "ocr_qwen2_vl_72b": "- WhatsApp: +2349032558166\n- Jonah Emmanuel\n\nMathematics:\n- Complex Numbers (Polar Form)\n - \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n - \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n - \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) \\right] \\)\n - \\( z_1 z_2 = r_1 r_2 \\left[ \\cos (\\theta_1 + \\theta_2) + i \\sin (\\theta_1 + \\theta_2) \\right] \\)" }, { "vid": "BdvOjT7bh2c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Multiplication and Division of Complex Numbers_30.json#####audio#####doingASR#####FinishASR/BdvOjT7bh2c/1088.4_1111.8600000000001.mp4", "refined_asr": " Um, into cosine 2 pi plus 3 pi gives you 5 pi all over 6. So I have this plus i sine. I have 2 pi plus 3 pi that's 5 pi all over 6.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1088.4_1111.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1088.4_1111.8600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1088.4_1111.8600000000001#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1088.4_1111.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1088.4_1111.8600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1088.4_1111.8600000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1088.4_1111.8600000000001#4.jpg" ], "ocr_qwen2_vl_72b": "- WhatsApp: +2349032558166\n- Jonah Emmanuel\n\nMathematics:\n- Complex Numbers (Polar Form)\n - \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n - \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n - \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{2\\pi + 3\\pi}{6} \\right) + i \\sin \\left( \\frac{2\\pi + 3\\pi}{6} \\right) \\right] \\)\n - \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{5\\pi}{6} \\right) + i \\sin \\left( \\frac{5\\pi}{6} \\right) \\right] \\)" }, { "vid": "BdvOjT7bh2c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Multiplication and Division of Complex Numbers_30.json#####audio#####doingASR#####FinishASR/BdvOjT7bh2c/1112.0400000000002_1125.68.mp4", "refined_asr": " So I have this. So this becomes the answer in radians. So this is the value of Z1, Z2, and so on, in radians. If you're asked to provide this value in terms of degrees, alright, you'd say, leave your answer in terms of degrees.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1112.0400000000002_1125.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1112.0400000000002_1125.68#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1112.0400000000002_1125.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1112.0400000000002_1125.68#2.jpg" ], "ocr_qwen2_vl_72b": "- WhatsApp: +2349032558166\n- Jonah Emmanuel\n\nMathematics:\n- Complex Numbers (Polar Form)\n - \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n - \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n - \\( z_1 z_2 = 9 \\times 11 \\left[ \\cos \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{3} + \\frac{\\pi}{2} \\right) \\right] \\)\n - \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{2\\pi + 3\\pi}{6} \\right) + i \\sin \\left( \\frac{2\\pi + 3\\pi}{6} \\right) \\right] \\)\n\n- \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{5\\pi}{6} \\right) + i \\sin \\left( \\frac{5\\pi}{6} \\right) \\right] \\)\n- \\( z_1 z_2 = r_1 r_2 \\left[ \\cos (\\theta_1 + \\theta_2) + i \\sin (\\theta_1 + \\theta_2) \\right] \\)" }, { "vid": "BdvOjT7bh2c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Multiplication and Division of Complex Numbers_30.json#####audio#####doingASR#####FinishASR/BdvOjT7bh2c/1125.7800000000002_1147.6200000000001.mp4", "refined_asr": " What do you do? Very simple. We said 1 pi is equal to 180 degrees. So simply replace pi with 180 degrees. Then we'll have that z1, z2 is equal to 99 into cos.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1125.7800000000002_1147.6200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1125.7800000000002_1147.6200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1125.7800000000002_1147.6200000000001#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1125.7800000000002_1147.6200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1125.7800000000002_1147.6200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1125.7800000000002_1147.6200000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1125.7800000000002_1147.6200000000001#4.jpg" ], "ocr_qwen2_vl_72b": "- WhatsApp: +2349032558166\n- Mathematics\n- Complex Numbers (Polar Form)\n- \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n- \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n- \\( z_1 \\cdot z_2 = 99 \\left[ \\cos \\left( \\frac{2\\pi + 3\\pi}{6} \\right) + i \\sin \\left( \\frac{2\\pi + 3\\pi}{6} \\right) \\right] \\)\n- \\( z_1 \\cdot z_2 = 99 \\left[ \\cos \\left( \\frac{5\\pi}{6} \\right) + i \\sin \\left( \\frac{5\\pi}{6} \\right) \\right] \\)\n- Jonah Emmanuel" }, { "vid": "BdvOjT7bh2c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Multiplication and Division of Complex Numbers_30.json#####audio#####doingASR#####FinishASR/BdvOjT7bh2c/1148.6200000000001_1173.38.mp4", "refined_asr": " 5 times 180. That's pi 180 all over 6. Plus i sine 5 times pi, 5 times 180 over 6. So we have this value. Now let me clarify something for you. The value of pi in angular measurement is 180 degrees.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1148.6200000000001_1173.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1148.6200000000001_1173.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1148.6200000000001_1173.38#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1148.6200000000001_1173.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1148.6200000000001_1173.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1148.6200000000001_1173.38#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/BdvOjT7bh2c/BdvOjT7bh2c@1148.6200000000001_1173.38#4.jpg" ], "ocr_qwen2_vl_72b": "- WhatsApp: +2349032558166\n- Mathematics\n- Complex Numbers (Polar Form)\n- \\( z_1 = 9 \\left[ \\cos \\left( \\frac{\\pi}{3} \\right) + i \\sin \\left( \\frac{\\pi}{3} \\right) \\right] \\)\n- \\( z_2 = 11 \\left[ \\cos \\left( \\frac{\\pi}{2} \\right) + i \\sin \\left( \\frac{\\pi}{2} \\right) \\right] \\)\n- \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{2\\pi + 3\\pi}{6} \\right) + i \\sin \\left( \\frac{2\\pi + 3\\pi}{6} \\right) \\right] \\)\n- \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{5\\pi}{6} \\right) + i \\sin \\left( \\frac{5\\pi}{6} \\right) \\right] \\)\n- \\( z_1 z_2 = 99 \\left[ \\cos \\left( \\frac{5 \\times 180}{6} \\right) + i \\sin \\left( \\frac{5 \\times 180}{6} \\right) \\right] \\)\n- Jonah Emmanuel" } ], "image_num": 15, "text_num": 413, "token_num": 9053 }, { "images": [ "sample_100_images/4jr7UWP9JtM@1021.82_1042.38#1.jpg", "sample_100_images/4jr7UWP9JtM@1021.82_1042.38#2.jpg", null, "sample_100_images/4jr7UWP9JtM@1043.0_1052.76#1.jpg", null, "sample_100_images/4jr7UWP9JtM@1053.3_1072.48#1.jpg", null, "sample_100_images/4jr7UWP9JtM@1073.52_1098.7#1.jpg", null, "sample_100_images/4jr7UWP9JtM@1098.7_1122.42#1.jpg", "sample_100_images/4jr7UWP9JtM@1098.7_1122.42#3.jpg", null, "sample_100_images/4jr7UWP9JtM@1122.42_1146.2600000000002#1.jpg", null, "sample_100_images/4jr7UWP9JtM@1146.2600000000002_1164.0200000000002#1.jpg", null ], "texts": [ null, null, " And then you take half of that vector, and then you take half of it for a triangle. Okay. So let's talk about the triple scalar product. A triple scalar product is very simple. You're just taking the dot product of a cross product.", null, " But the easiest way to set it up is actually in a 3x3 matrix and you can use a calculator for that. So we've got a dot. We've got a dot. We've got a dot. We've got a dot.", null, " And again, the simplest way is just what we call a triple scalar product. So we're going to take the coordinates for each vector and just set them up in each row. And then we're going to find the determinant of that 3x3 matrix.", null, " So an application, a geometric property of a triple scalar is very simple. If we have three vectors and they are in component form, we can set them up into this triple scalar. What it does is it gives us the volume of this three-dimensional shape.", null, null, " This shape is called a parallelepiped. Okay. And that is the volume. So once we do this triple scalar product, it's the volume of that three-dimensional figure, which is very cool. So let's try this. We've got three vectors here. All of them are written well. We're going to write them in.", null, " Component form. So this position or this point is 3, negative 5, 1. So since it's being generated from the origin, I can write that as the vector: 3, negative 5, 1. My second vector is at the point or the", null, " The vertex of this three-dimensional figure is at the point 3, 1, 1. So I can write that as a vector, because that side is a vector. And this third one is a vector at 0, 2, negative 2." ], "text_ocr_list": [ null, null, "We can see these text from the image: i j k\n1 1 6 = (5-12)i - (5-0)j + (-2-0)k\n0 -2 5 -> 17i - 5j - 2k\n\n3. Find 1/2 the magnitude of the resultant vector\n\n1/2 ||AB x AC||\n\n\u221a(17\u00b2 + (-5)\u00b2 + (-2)\u00b2\n\u221a289 + 25 + 4 = 17.83.\n And then you take half of that vector, and then you take half of it for a triangle. Okay. So let's talk about the triple scalar product. A triple scalar product is very simple. You're just taking the dot product of a cross product.", null, "We can see these text from the image: The Triple Scalar Product\n\nFor \\( \\mathbf{u} = u_1 \\mathbf{i} + u_2 \\mathbf{j} + u_3 \\mathbf{k} \\), \\( \\mathbf{v} = v_1 \\mathbf{i} + v_2 \\mathbf{j} + v_3 \\mathbf{k} \\), and \\( \\mathbf{w} = w_1 \\mathbf{i} + w_2 \\mathbf{j} + w_3 \\mathbf{k} \\), the triple scalar product is given by\n\n\\[ \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\begin{vmatrix} u_1 & u_2 & u_3 \\\\ v_1 & v_2 & v_3 \\\\ w_1 & w_2 & w_3 \\end{vmatrix}. \\]\n\nDot product.\n But the easiest way to set it up is actually in a 3x3 matrix and you can use a calculator for that. So we've got a dot. We've got a dot. We've got a dot. We've got a dot.", null, "We can see these text from the image: For \\( \\mathbf{u} = u_1 \\mathbf{i} + u_2 \\mathbf{j} + u_3 \\mathbf{k} \\), \\( \\mathbf{v} = v_1 \\mathbf{i} + v_2 \\mathbf{j} + v_3 \\mathbf{k} \\), and \\( \\mathbf{w} = w_1 \\mathbf{i} + w_2 \\mathbf{j} + w_3 \\mathbf{k} \\), the triple scalar product is given by\n\n\\[ \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\begin{vmatrix} u_1 & u_2 & u_3 \\\\ v_1 & v_2 & v_3 \\\\ w_1 & w_2 & w_3 \\end{vmatrix}. \\].\n And again, the simplest way is just what we call a triple scalar product. So we're going to take the coordinates for each vector and just set them up in each row. And then we're going to find the determinant of that 3x3 matrix.", null, "We can see these text from the image: - \\( \\mathbf{v} \\times \\mathbf{w} \\)\n- \\( \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) \\)\n\nArea of base = \\( \\| \\mathbf{v} \\times \\mathbf{w} \\| \\)\n\nVolume of parallelepiped = \\( | \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})| \\)\n\nFigure 11.17\n\nGeometric Property of the Triple Scalar Product\n\nThe volume \\( V \\) of a parallelepiped with vectors \\( \\mathbf{u} \\), \\( \\mathbf{v} \\), and \\( \\mathbf{w} \\) as adjacent edges is\n\n\\( V = | \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})| \\).\n\nEXAMPLE 6 Volume by the Triple Scalar Product.\n So an application, a geometric property of a triple scalar is very simple. If we have three vectors and they are in component form, we can set them up into this triple scalar. What it does is it gives us the volume of this three-dimensional shape.", null, null, "We can see these text from the image: - \\( \\mathbf{v} \\times \\mathbf{w} \\)\n- Area of base = \\( \\| \\mathbf{v} \\times \\mathbf{w} \\| \\)\n- Volume of parallelepiped = \\( | \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) | \\)\n\n- Figure 11.17\n\n- Geometric Property of the Triple Scalar Product\n - The volume \\( V \\) of a parallelepiped with vectors \\( \\mathbf{u} \\), \\( \\mathbf{v} \\), and \\( \\mathbf{w} \\) as adjacent edges is\n - \\( V = | \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) | \\).\n\n- Example: Volume by the Triple Scalar Product.\n This shape is called a parallelepiped. Okay. And that is the volume. So once we do this triple scalar product, it's the volume of that three-dimensional figure, which is very cool. So let's try this. We've got three vectors here. All of them are written well. We're going to write them in.", null, "We can see these text from the image: Find the volume of using the triple scalar product\n\nFigure 11.18.\n Component form. So this position or this point is 3, negative 5, 1. So since it's being generated from the origin, I can write that as the vector: 3, negative 5, 1. My second vector is at the point or the", null, "We can see these text from the image: Find the volume of using the triple scalar product\n\n<3, -5, 1>\n\nFigure 11.18.\n The vertex of this three-dimensional figure is at the point 3, 1, 1. So I can write that as a vector, because that side is a vector. And this third one is a vector at 0, 2, negative 2." ], "metadata": [ { "vid": "4jr7UWP9JtM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Tutorial on Vector Applications_30.json#####audio#####doingASR#####FinishASR/4jr7UWP9JtM/1021.82_1042.38.mp4", "refined_asr": " And then you take half of that vector, and then you take half of it for a triangle. Okay. So let's talk about the triple scalar product. A triple scalar product is very simple. You're just taking the dot product of a cross product.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1021.82_1042.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1021.82_1042.38#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1021.82_1042.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1021.82_1042.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1021.82_1042.38#3.jpg" ], "ocr_qwen2_vl_72b": "i j k\n1 1 6 = (5-12)i - (5-0)j + (-2-0)k\n0 -2 5 -> 17i - 5j - 2k\n\n3. Find 1/2 the magnitude of the resultant vector\n\n1/2 ||AB x AC||\n\n\u221a(17\u00b2 + (-5)\u00b2 + (-2)\u00b2\n\u221a289 + 25 + 4 = 17.83" }, { "vid": "4jr7UWP9JtM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Tutorial on Vector Applications_30.json#####audio#####doingASR#####FinishASR/4jr7UWP9JtM/1043.0_1052.76.mp4", "refined_asr": " But the easiest way to set it up is actually in a 3x3 matrix and you can use a calculator for that. So we've got a dot. We've got a dot. We've got a dot. We've got a dot.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1043.0_1052.76#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1043.0_1052.76#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1043.0_1052.76#2.jpg" ], "ocr_qwen2_vl_72b": "The Triple Scalar Product\n\nFor \\( \\mathbf{u} = u_1 \\mathbf{i} + u_2 \\mathbf{j} + u_3 \\mathbf{k} \\), \\( \\mathbf{v} = v_1 \\mathbf{i} + v_2 \\mathbf{j} + v_3 \\mathbf{k} \\), and \\( \\mathbf{w} = w_1 \\mathbf{i} + w_2 \\mathbf{j} + w_3 \\mathbf{k} \\), the triple scalar product is given by\n\n\\[ \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\begin{vmatrix} u_1 & u_2 & u_3 \\\\ v_1 & v_2 & v_3 \\\\ w_1 & w_2 & w_3 \\end{vmatrix}. \\]\n\nDot product" }, { "vid": "4jr7UWP9JtM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Tutorial on Vector Applications_30.json#####audio#####doingASR#####FinishASR/4jr7UWP9JtM/1053.3_1072.48.mp4", "refined_asr": " And again, the simplest way is just what we call a triple scalar product. So we're going to take the coordinates for each vector and just set them up in each row. And then we're going to find the determinant of that 3x3 matrix.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1053.3_1072.48#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1053.3_1072.48#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1053.3_1072.48#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1053.3_1072.48#3.jpg" ], "ocr_qwen2_vl_72b": "For \\( \\mathbf{u} = u_1 \\mathbf{i} + u_2 \\mathbf{j} + u_3 \\mathbf{k} \\), \\( \\mathbf{v} = v_1 \\mathbf{i} + v_2 \\mathbf{j} + v_3 \\mathbf{k} \\), and \\( \\mathbf{w} = w_1 \\mathbf{i} + w_2 \\mathbf{j} + w_3 \\mathbf{k} \\), the triple scalar product is given by\n\n\\[ \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\begin{vmatrix} u_1 & u_2 & u_3 \\\\ v_1 & v_2 & v_3 \\\\ w_1 & w_2 & w_3 \\end{vmatrix}. \\]" }, { "vid": "4jr7UWP9JtM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Tutorial on Vector Applications_30.json#####audio#####doingASR#####FinishASR/4jr7UWP9JtM/1073.52_1098.7.mp4", "refined_asr": " So an application, a geometric property of a triple scalar is very simple. If we have three vectors and they are in component form, we can set them up into this triple scalar. What it does is it gives us the volume of this three-dimensional shape.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1073.52_1098.7#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1073.52_1098.7#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1073.52_1098.7#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1073.52_1098.7#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1073.52_1098.7#4.jpg" ], "ocr_qwen2_vl_72b": "- \\( \\mathbf{v} \\times \\mathbf{w} \\)\n- \\( \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) \\)\n\nArea of base = \\( \\| \\mathbf{v} \\times \\mathbf{w} \\| \\)\n\nVolume of parallelepiped = \\( | \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})| \\)\n\nFigure 11.17\n\nGeometric Property of the Triple Scalar Product\n\nThe volume \\( V \\) of a parallelepiped with vectors \\( \\mathbf{u} \\), \\( \\mathbf{v} \\), and \\( \\mathbf{w} \\) as adjacent edges is\n\n\\( V = | \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})| \\).\n\nEXAMPLE 6 Volume by the Triple Scalar Product" }, { "vid": "4jr7UWP9JtM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Tutorial on Vector Applications_30.json#####audio#####doingASR#####FinishASR/4jr7UWP9JtM/1098.7_1122.42.mp4", "refined_asr": " This shape is called a parallelepiped. Okay. And that is the volume. So once we do this triple scalar product, it's the volume of that three-dimensional figure, which is very cool. So let's try this. We've got three vectors here. All of them are written well. We're going to write them in.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1098.7_1122.42#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1098.7_1122.42#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1098.7_1122.42#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1098.7_1122.42#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1098.7_1122.42#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1098.7_1122.42#4.jpg" ], "ocr_qwen2_vl_72b": "- \\( \\mathbf{v} \\times \\mathbf{w} \\)\n- Area of base = \\( \\| \\mathbf{v} \\times \\mathbf{w} \\| \\)\n- Volume of parallelepiped = \\( | \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) | \\)\n\n- Figure 11.17\n\n- Geometric Property of the Triple Scalar Product\n - The volume \\( V \\) of a parallelepiped with vectors \\( \\mathbf{u} \\), \\( \\mathbf{v} \\), and \\( \\mathbf{w} \\) as adjacent edges is\n - \\( V = | \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) | \\).\n\n- Example: Volume by the Triple Scalar Product" }, { "vid": "4jr7UWP9JtM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Tutorial on Vector Applications_30.json#####audio#####doingASR#####FinishASR/4jr7UWP9JtM/1122.42_1146.2600000000002.mp4", "refined_asr": " Component form. So this position or this point is 3, negative 5, 1. So since it's being generated from the origin, I can write that as the vector: 3, negative 5, 1. My second vector is at the point or the", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1122.42_1146.2600000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1122.42_1146.2600000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1122.42_1146.2600000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1122.42_1146.2600000000002#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1122.42_1146.2600000000002#4.jpg" ], "ocr_qwen2_vl_72b": "Find the volume of using the triple scalar product\n\nFigure 11.18" }, { "vid": "4jr7UWP9JtM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Tutorial on Vector Applications_30.json#####audio#####doingASR#####FinishASR/4jr7UWP9JtM/1146.2600000000002_1164.0200000000002.mp4", "refined_asr": " The vertex of this three-dimensional figure is at the point 3, 1, 1. So I can write that as a vector, because that side is a vector. And this third one is a vector at 0, 2, negative 2.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1146.2600000000002_1164.0200000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1146.2600000000002_1164.0200000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1146.2600000000002_1164.0200000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4jr7UWP9JtM/4jr7UWP9JtM@1146.2600000000002_1164.0200000000002#3.jpg" ], "ocr_qwen2_vl_72b": "Find the volume of using the triple scalar product\n\n<3, -5, 1>\n\nFigure 11.18" } ], "image_num": 9, "text_num": 415, "token_num": 5599 }, { "images": [ "sample_100_images/WG6k74VSOOU@467.96_485.84000000000003#1.jpg", null, "sample_100_images/WG6k74VSOOU@485.84000000000003_503.82000000000005#1.jpg", "sample_100_images/WG6k74VSOOU@485.84000000000003_503.82000000000005#2.jpg", null, "sample_100_images/WG6k74VSOOU@503.82000000000005_524.38#1.jpg", "sample_100_images/WG6k74VSOOU@503.82000000000005_524.38#2.jpg", null, "sample_100_images/WG6k74VSOOU@524.38_541.54#1.jpg", null, "sample_100_images/WG6k74VSOOU@541.54_559.74#1.jpg", "sample_100_images/WG6k74VSOOU@541.54_559.74#3.jpg", null, "sample_100_images/WG6k74VSOOU@559.9_578.26#1.jpg", null, "sample_100_images/WG6k74VSOOU@578.26_599.4599999999999#1.jpg", "sample_100_images/WG6k74VSOOU@578.26_599.4599999999999#3.jpg", null ], "texts": [ null, " Reads requests which is great. But what happens if you don't scale reads? It's not great as most modern applications are very read-intensive. They have a high proportion of reads compared to writes. Yet the system loses all abilities to serve write requests. And that's", null, null, " Really bad because it affects the user experience. The clock is ticking and the system urgently needs to regain its capabilities through write requests. If you lose your main node in the master-replica pattern, you can use the replicas to act as the new master and start accepting writes.", null, null, " Requests. The big question is how to decide which replica should be the new main. Due to the asynchronous nature of replication in the master-replica pattern, all replica nodes are at a different stage, replaying the latest changes. Once the master node is down, all of those replays stop. So the most", null, " A reasonable choice for promotion is the replica that is most up-to-date. Once promoted, all other replicas need to start replaying the changes. This newly promoted main is ahead of them. As stated, the most up-to-date replica is promoted.", null, null, " In most cases no replica is actually identical to the main. That means all data that isn't replicated in the moment the main goes down is lost. Last but not least to also regain its full read capacity the system needs to spin up a new replica.", null, " Replaces the just-promoted one. That's how a replicated database reacts to failure scenarios. It's very straightforward to see how it significantly increases the system's availability. But we're not done yet with exploring the main replica pattern. Availability is just one of the desired properties.", null, null, " System needs to serve large-scale applications. Another crucial one is scalability. Let's investigate which methods you can leverage to tweak your main replica architecture in order to keep up with an ever-growing user base. To give you an idea, Netflix just grew their user base to 200 million users." ], "text_ocr_list": [ null, "We can see these text from the image: Main\n\nReplica\nread\n\nReplica\nread.\n Reads requests which is great. But what happens if you don't scale reads? It's not great as most modern applications are very read-intensive. They have a high proportion of reads compared to writes. Yet the system loses all abilities to serve write requests. And that's", null, null, "We can see these text from the image: - Main\n- write\n- Replica\n- read\n- Replica\n- read.\n Really bad because it affects the user experience. The clock is ticking and the system urgently needs to regain its capabilities through write requests. If you lose your main node in the master-replica pattern, you can use the replicas to act as the new master and start accepting writes.", null, null, "We can see these text from the image: write\n\nMain\n\nReplica\nReplica\nReplica.\n Requests. The big question is how to decide which replica should be the new main. Due to the asynchronous nature of replication in the master-replica pattern, all replica nodes are at a different stage, replaying the latest changes. Once the master node is down, all of those replays stop. So the most", null, "We can see these text from the image: Replica\n\nReplica\n\nReplica.\n A reasonable choice for promotion is the replica that is most up-to-date. Once promoted, all other replicas need to start replaying the changes. This newly promoted main is ahead of them. As stated, the most up-to-date replica is promoted.", null, null, "We can see these text from the image: No replica is identical with the main node.\n In most cases no replica is actually identical to the main. That means all data that isn't replicated in the moment the main goes down is lost. Last but not least to also regain its full read capacity the system needs to spin up a new replica.", null, "We can see these text from the image: API.\n Replaces the just-promoted one. That's how a replicated database reacts to failure scenarios. It's very straightforward to see how it significantly increases the system's availability. But we're not done yet with exploring the main replica pattern. Availability is just one of the desired properties.", null, null, "We can see these text from the image: Availability is just one of the desired properties..\n System needs to serve large-scale applications. Another crucial one is scalability. Let's investigate which methods you can leverage to tweak your main replica architecture in order to keep up with an ever-growing user base. To give you an idea, Netflix just grew their user base to 200 million users." ], "metadata": [ { "vid": "WG6k74VSOOU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/WG6k74VSOOU/467.96_485.84000000000003.mp4", "refined_asr": " Reads requests which is great. But what happens if you don't scale reads? It's not great as most modern applications are very read-intensive. They have a high proportion of reads compared to writes. Yet the system loses all abilities to serve write requests. And that's", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@467.96_485.84000000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@467.96_485.84000000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@467.96_485.84000000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@467.96_485.84000000000003#3.jpg" ], "ocr_qwen2_vl_72b": "Main\n\nReplica\nread\n\nReplica\nread" }, { "vid": "WG6k74VSOOU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/WG6k74VSOOU/485.84000000000003_503.82000000000005.mp4", "refined_asr": " Really bad because it affects the user experience. The clock is ticking and the system urgently needs to regain its capabilities through write requests. If you lose your main node in the master-replica pattern, you can use the replicas to act as the new master and start accepting writes.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@485.84000000000003_503.82000000000005#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@485.84000000000003_503.82000000000005#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@485.84000000000003_503.82000000000005#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@485.84000000000003_503.82000000000005#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@485.84000000000003_503.82000000000005#3.jpg" ], "ocr_qwen2_vl_72b": "- Main\n- write\n- Replica\n- read\n- Replica\n- read" }, { "vid": "WG6k74VSOOU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/WG6k74VSOOU/503.82000000000005_524.38.mp4", "refined_asr": " Requests. The big question is how to decide which replica should be the new main. Due to the asynchronous nature of replication in the master-replica pattern, all replica nodes are at a different stage, replaying the latest changes. Once the master node is down, all of those replays stop. So the most", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@503.82000000000005_524.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@503.82000000000005_524.38#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@503.82000000000005_524.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@503.82000000000005_524.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@503.82000000000005_524.38#3.jpg" ], "ocr_qwen2_vl_72b": "write\n\nMain\n\nReplica\nReplica\nReplica" }, { "vid": "WG6k74VSOOU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/WG6k74VSOOU/524.38_541.54.mp4", "refined_asr": " A reasonable choice for promotion is the replica that is most up-to-date. Once promoted, all other replicas need to start replaying the changes. This newly promoted main is ahead of them. As stated, the most up-to-date replica is promoted.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@524.38_541.54#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@524.38_541.54#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@524.38_541.54#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@524.38_541.54#3.jpg" ], "ocr_qwen2_vl_72b": "Replica\n\nReplica\n\nReplica" }, { "vid": "WG6k74VSOOU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/WG6k74VSOOU/541.54_559.74.mp4", "refined_asr": " In most cases no replica is actually identical to the main. That means all data that isn't replicated in the moment the main goes down is lost. Last but not least to also regain its full read capacity the system needs to spin up a new replica.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@541.54_559.74#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@541.54_559.74#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@541.54_559.74#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@541.54_559.74#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@541.54_559.74#3.jpg" ], "ocr_qwen2_vl_72b": "No replica is identical with the main node" }, { "vid": "WG6k74VSOOU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/WG6k74VSOOU/559.9_578.26.mp4", "refined_asr": " Replaces the just-promoted one. That's how a replicated database reacts to failure scenarios. It's very straightforward to see how it significantly increases the system's availability. But we're not done yet with exploring the main replica pattern. Availability is just one of the desired properties.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@559.9_578.26#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@559.9_578.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@559.9_578.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@559.9_578.26#3.jpg" ], "ocr_qwen2_vl_72b": "API" }, { "vid": "WG6k74VSOOU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Database Topics tutorial on Data Replication in Distributed Databases_30.json#####audio#####doingASR#####FinishASR/WG6k74VSOOU/578.26_599.4599999999999.mp4", "refined_asr": " System needs to serve large-scale applications. Another crucial one is scalability. Let's investigate which methods you can leverage to tweak your main replica architecture in order to keep up with an ever-growing user base. To give you an idea, Netflix just grew their user base to 200 million users.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@578.26_599.4599999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@578.26_599.4599999999999#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@578.26_599.4599999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@578.26_599.4599999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@578.26_599.4599999999999#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/WG6k74VSOOU/WG6k74VSOOU@578.26_599.4599999999999#4.jpg" ], "ocr_qwen2_vl_72b": "Availability is just one of the desired properties." } ], "image_num": 11, "text_num": 457, "token_num": 6793 }, { "images": [ "sample_100_images/XphDeHpIdOI@140.62_173.68#1.jpg", "sample_100_images/XphDeHpIdOI@140.62_173.68#5.jpg", null, "sample_100_images/XphDeHpIdOI@174.72_197.84#1.jpg", "sample_100_images/XphDeHpIdOI@174.72_197.84#2.jpg", "sample_100_images/XphDeHpIdOI@174.72_197.84#3.jpg", null, "sample_100_images/XphDeHpIdOI@197.84_226.86#1.jpg", "sample_100_images/XphDeHpIdOI@197.84_226.86#2.jpg", "sample_100_images/XphDeHpIdOI@197.84_226.86#3.jpg", "sample_100_images/XphDeHpIdOI@197.84_226.86#4.jpg", null, "sample_100_images/XphDeHpIdOI@226.86_258.82#1.jpg", "sample_100_images/XphDeHpIdOI@226.86_258.82#2.jpg", "sample_100_images/XphDeHpIdOI@226.86_258.82#5.jpg", null, "sample_100_images/XphDeHpIdOI@258.82_287.79999999999995#1.jpg", "sample_100_images/XphDeHpIdOI@258.82_287.79999999999995#4.jpg", null ], "texts": [ null, null, " The types of reactions we discuss in electrocyclic reactions include one called the ring-closing reaction and another known as the ring-opening reaction. In the case of ring-opening reactions, there are two types that we consider. So, in a ring-closing reaction, the FMO method is involved. FMO stands for Frontal Molecular Orbital.", null, null, null, " FOMO is considered, or we can say it's considered during the analysis. Whereas in a ring-opening reaction, I will consider both FOMO and LUMO. So, FOMO and LUMO are usually considered. Otherwise, we can also consider the FOMO only. But here, the FOMO is considered. And here, the LUMO is considered. But here, the FOMO is considered. And here, the T assimilation reaction is considered, involving the loan of LUMO to FOMO. Tomorrow, I will discuss FOMO for the consideration of the crack reaction, sliding reaction. I' mani\u00e8re dalla dietera vedetah. (Note: The last sentence seems to be in a different language and may require a native speaker for accurate refinement.)", null, null, null, null, " Usually, otherwise we can also consider the FOMOs, but here I will consider HOMO as well as LUMO for ring-opening reactions. Okay, now let's first see the ring-closing reaction. So first of all, I will discuss the ring-closing reaction. As I told you, two types of systems we usually take into account.", null, null, null, " Pericyclic reactions involve 4n electron systems and 4n plus 2 electron systems. So if I consider a 4n electron system for example if I take butadiene upon heating or exposure to light it would be converted into", null, null, " Cyclic building that would be converted into cyclic building now. So in this reaction, this is a ring closing reaction. This is the ring closing reaction. Okay, so what I've told you, in a ring closing reaction, we take the HOMO of the reactant. The HOMO of the reactant is taken. So in this case, this is deutlich. And for the term, hmm, which contains carta, theology is shown. Falta volta electricity is written. Note: The refined transcription maintains the original phrasing and unusual terms as they may be specific to a certain context or subject matter, such as chemistry. However, it adds necessary punctuation for clarity." ], "text_ocr_list": [ null, null, "We can see these text from the image: - HOMO LUMO\n- PERICYCLIC REACTIONS\n- Analysis of Electrocyclic Reactions by Frontier Molecular Orbital Method\n- FMO methods\n- Ring\n- Table:\n - \u0394\n - 4n\u03c0: Con\n - (4n+2)\u03c0: Dis\n - h\u03bd\n - 4n\u03c0: Dis\n - (4n+2)\u03c0: Con.\n The types of reactions we discuss in electrocyclic reactions include one called the ring-closing reaction and another known as the ring-opening reaction. In the case of ring-opening reactions, there are two types that we consider. So, in a ring-closing reaction, the FMO method is involved. FMO stands for Frontal Molecular Orbital.", null, null, null, "We can see these text from the image: - PERICYCLIC REACTIONS\n- Analysis of Electrocyclic Reactions by Frontier Molecular Orbital Method\n- HOMO LUMO\n- Ring Reaction: HOMO\n- Ring Reaction: LUMO\n- Table:\n - \u0394\n - 4n\u03c0: Con\n - (4n+2)\u03c0: Dis\n - hu\n - 4n\u03c0: Dis\n - (4n+2)\u03c0: Con.\n FOMO is considered, or we can say it's considered during the analysis. Whereas in a ring-opening reaction, I will consider both FOMO and LUMO. So, FOMO and LUMO are usually considered. Otherwise, we can also consider the FOMO only. But here, the FOMO is considered. And here, the LUMO is considered. But here, the FOMO is considered. And here, the T assimilation reaction is considered, involving the loan of LUMO to FOMO. Tomorrow, I will discuss FOMO for the consideration of the crack reaction, sliding reaction. I' mani\u00e8re dalla dietera vedetah. (Note: The last sentence seems to be in a different language and may require a native speaker for accurate refinement.)", null, null, null, null, "We can see these text from the image: - PERICYCLIC REACTIONS\n- Analysis of Electrocyclic Reactions by Frontier Molecular Orbital Method\n- FMO methods:\n - Ring Closing Reaction: HOMO is considered\n - Ring Opening Reaction: LUMO is considered\n- Table:\n - \u0394: 4n\u03c0 Con Dis\n - hu: (4n+2)\u03c0 Dis Con.\n Usually, otherwise we can also consider the FOMOs, but here I will consider HOMO as well as LUMO for ring-opening reactions. Okay, now let's first see the ring-closing reaction. So first of all, I will discuss the ring-closing reaction. As I told you, two types of systems we usually take into account.", null, null, null, "We can see these text from the image: - PERICYCLIC REACTIONS\n- Analysis of Electrocyclic Reactions by Frontier Molecular Orbital Method\n- HOMO LUMO\n- 2 methods:\n - Ring Opening Reaction: HOMO is considered\n- Table:\n - \\( \\Delta \\)\n - \\( 4n\\pi \\): Con\n - \\( (4m+2)\\pi \\): Dis\n - \\( h\\nu \\)\n - \\( 4n\\pi \\): Dis\n - \\( (4m+2)\\pi \\): Con.\n Pericyclic reactions involve 4n electron systems and 4n plus 2 electron systems. So if I consider a 4n electron system for example if I take butadiene upon heating or exposure to light it would be converted into", null, null, "We can see these text from the image: - PERICYCLIC REACTIONS\n- Analysis of Electrocyclic Reactions by Frontier Molecular Orbital Method\n- FMO methods\n - Ring Closing Reaction: HOMO is considered\n - \\(4n\\pi\\) Systems\n - \\(\\Delta\\) or h\\nu\n - \\(4n + 2\\pi\\) Systems\n - \\(\\Delta\\) or h\\nu\n- Table:\n - \\(\\Delta\\)\n - \\(4n\\pi\\): Con\n - \\(4n + 2\\pi\\): Dis\n - h\\nu\n - \\(4n\\pi\\): Dis\n - \\(4n + 2\\pi\\): Con.\n Cyclic building that would be converted into cyclic building now. So in this reaction, this is a ring closing reaction. This is the ring closing reaction. Okay, so what I've told you, in a ring closing reaction, we take the HOMO of the reactant. The HOMO of the reactant is taken. So in this case, this is deutlich. And for the term, hmm, which contains carta, theology is shown. Falta volta electricity is written. Note: The refined transcription maintains the original phrasing and unusual terms as they may be specific to a certain context or subject matter, such as chemistry. However, it adds necessary punctuation for clarity." ], "metadata": [ { "vid": "XphDeHpIdOI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/XphDeHpIdOI/140.62_173.68.mp4", "refined_asr": " The types of reactions we discuss in electrocyclic reactions include one called the ring-closing reaction and another known as the ring-opening reaction. In the case of ring-opening reactions, there are two types that we consider. So, in a ring-closing reaction, the FMO method is involved. FMO stands for Frontal Molecular Orbital.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@140.62_173.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@140.62_173.68#5.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@140.62_173.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@140.62_173.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@140.62_173.68#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@140.62_173.68#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@140.62_173.68#5.jpg" ], "ocr_qwen2_vl_72b": "- HOMO LUMO\n- PERICYCLIC REACTIONS\n- Analysis of Electrocyclic Reactions by Frontier Molecular Orbital Method\n- FMO methods\n- Ring\n- Table:\n - \u0394\n - 4n\u03c0: Con\n - (4n+2)\u03c0: Dis\n - h\u03bd\n - 4n\u03c0: Dis\n - (4n+2)\u03c0: Con" }, { "vid": "XphDeHpIdOI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/XphDeHpIdOI/174.72_197.84.mp4", "refined_asr": " FOMO is considered, or we can say it's considered during the analysis. Whereas in a ring-opening reaction, I will consider both FOMO and LUMO. So, FOMO and LUMO are usually considered. Otherwise, we can also consider the FOMO only. But here, the FOMO is considered. And here, the LUMO is considered. But here, the FOMO is considered. And here, the T assimilation reaction is considered, involving the loan of LUMO to FOMO. Tomorrow, I will discuss FOMO for the consideration of the crack reaction, sliding reaction. I' mani\u00e8re dalla dietera vedetah. (Note: The last sentence seems to be in a different language and may require a native speaker for accurate refinement.)", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@174.72_197.84#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@174.72_197.84#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@174.72_197.84#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@174.72_197.84#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@174.72_197.84#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@174.72_197.84#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@174.72_197.84#4.jpg" ], "ocr_qwen2_vl_72b": "- PERICYCLIC REACTIONS\n- Analysis of Electrocyclic Reactions by Frontier Molecular Orbital Method\n- HOMO LUMO\n- Ring Reaction: HOMO\n- Ring Reaction: LUMO\n- Table:\n - \u0394\n - 4n\u03c0: Con\n - (4n+2)\u03c0: Dis\n - hu\n - 4n\u03c0: Dis\n - (4n+2)\u03c0: Con" }, { "vid": "XphDeHpIdOI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/XphDeHpIdOI/197.84_226.86.mp4", "refined_asr": " Usually, otherwise we can also consider the FOMOs, but here I will consider HOMO as well as LUMO for ring-opening reactions. Okay, now let's first see the ring-closing reaction. So first of all, I will discuss the ring-closing reaction. As I told you, two types of systems we usually take into account.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@197.84_226.86#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@197.84_226.86#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@197.84_226.86#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@197.84_226.86#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@197.84_226.86#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@197.84_226.86#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@197.84_226.86#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@197.84_226.86#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@197.84_226.86#5.jpg" ], "ocr_qwen2_vl_72b": "- PERICYCLIC REACTIONS\n- Analysis of Electrocyclic Reactions by Frontier Molecular Orbital Method\n- FMO methods:\n - Ring Closing Reaction: HOMO is considered\n - Ring Opening Reaction: LUMO is considered\n- Table:\n - \u0394: 4n\u03c0 Con Dis\n - hu: (4n+2)\u03c0 Dis Con" }, { "vid": "XphDeHpIdOI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/XphDeHpIdOI/226.86_258.82.mp4", "refined_asr": " Pericyclic reactions involve 4n electron systems and 4n plus 2 electron systems. So if I consider a 4n electron system for example if I take butadiene upon heating or exposure to light it would be converted into", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@226.86_258.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@226.86_258.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@226.86_258.82#5.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@226.86_258.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@226.86_258.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@226.86_258.82#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@226.86_258.82#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@226.86_258.82#5.jpg" ], "ocr_qwen2_vl_72b": "- PERICYCLIC REACTIONS\n- Analysis of Electrocyclic Reactions by Frontier Molecular Orbital Method\n- HOMO LUMO\n- 2 methods:\n - Ring Opening Reaction: HOMO is considered\n- Table:\n - \\( \\Delta \\)\n - \\( 4n\\pi \\): Con\n - \\( (4m+2)\\pi \\): Dis\n - \\( h\\nu \\)\n - \\( 4n\\pi \\): Dis\n - \\( (4m+2)\\pi \\): Con" }, { "vid": "XphDeHpIdOI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Organic Chemistry tutorial on Electrocyclic Reactions in Pericyclic Reactions_30.json#####audio#####doingASR#####FinishASR/XphDeHpIdOI/258.82_287.79999999999995.mp4", "refined_asr": " Cyclic building that would be converted into cyclic building now. So in this reaction, this is a ring closing reaction. This is the ring closing reaction. Okay, so what I've told you, in a ring closing reaction, we take the HOMO of the reactant. The HOMO of the reactant is taken. So in this case, this is deutlich. And for the term, hmm, which contains carta, theology is shown. Falta volta electricity is written. Note: The refined transcription maintains the original phrasing and unusual terms as they may be specific to a certain context or subject matter, such as chemistry. However, it adds necessary punctuation for clarity.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@258.82_287.79999999999995#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@258.82_287.79999999999995#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@258.82_287.79999999999995#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@258.82_287.79999999999995#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@258.82_287.79999999999995#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@258.82_287.79999999999995#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/XphDeHpIdOI/XphDeHpIdOI@258.82_287.79999999999995#5.jpg" ], "ocr_qwen2_vl_72b": "- PERICYCLIC REACTIONS\n- Analysis of Electrocyclic Reactions by Frontier Molecular Orbital Method\n- FMO methods\n - Ring Closing Reaction: HOMO is considered\n - \\(4n\\pi\\) Systems\n - \\(\\Delta\\) or h\\nu\n - \\(4n + 2\\pi\\) Systems\n - \\(\\Delta\\) or h\\nu\n- Table:\n - \\(\\Delta\\)\n - \\(4n\\pi\\): Con\n - \\(4n + 2\\pi\\): Dis\n - h\\nu\n - \\(4n\\pi\\): Dis\n - \\(4n + 2\\pi\\): Con" } ], "image_num": 14, "text_num": 545, "token_num": 8609 }, { "images": [ "sample_100_images/GAN-jgzYsIo@45921.700000000004_45948.700000000004#1.jpg", null, null, "sample_100_images/GAN-jgzYsIo@45950.700000000004_45972.700000000004#1.jpg", null, null, null, "sample_100_images/GAN-jgzYsIo@45978.700000000004_45992.700000000004#1.jpg", null, "sample_100_images/GAN-jgzYsIo@45992.700000000004_46004.700000000004#1.jpg", null, "sample_100_images/GAN-jgzYsIo@46004.700000000004_46017.700000000004#1.jpg", null, null, "sample_100_images/GAN-jgzYsIo@46018.700000000004_46022.700000000004#1.jpg", null, "sample_100_images/GAN-jgzYsIo@46022.700000000004_46049.700000000004#1.jpg", null, "sample_100_images/GAN-jgzYsIo@46049.700000000004_46053.700000000004#1.jpg", null ], "texts": [ null, " So if I combine like terms, you would see that you get the exact same thing as you got here. You would end up with 4r squared. Negative 6r plus 14r again would be plus 8r. And then you'd have minus 21. So this FOIL method is not producing a different result. It's just a convenient way to remember how to get the product of two binomials. And again, once you start using this, you're going to love it. It's something you're going to use for the rest of your life.", " All right, let's take a look at this one.", null, " We have 5n plus 5 times 6n minus 2. So let's use FOIL. I want to start out with the F, the first terms. The first terms we have 5n times 6n. 5 times 6 is 30. n times n is n squared. Now I want to use my outer terms. Here's the outer term. Here's the outer term.", " So, my outside terms.", " 5n times negative 2 would be negative 10n.", null, " Now I want to do my inner terms. So I want to do 5 times 6n. These are on the inside. 5 times 6 is 30. Then times n. So 30n. Then I want to do my last or final terms.", null, " So I got 5 here as the final term. Negative 2 here is the final term. 5 times negative 2 is negative 10. And so I've got 30n squared minus 10n plus 30n minus 10.", null, " So all I need to do now is just combine like terms here. Negative 10n plus 30n is 20n. So this gives me a final answer of 30n squared plus 20n minus 10.", " All right.", null, " For the next one, I have 4x minus 5 times 5x plus 7.", null, " So again, I'm going to use FOIL. First terms, I have 4x and I have 5x. 4x times 5x is 20x squared. Then I'm going to move to the outer terms, the O. So that's 4x times 7. 4x times 7 is 28x. So plus 28x. Then I'm going to move to my inside terms. So I have negative 5 and I have 5x.", null, " So negative 5 times 5x is negative 25x." ], "text_ocr_list": [ null, "We can see these text from the image: (2r + 7)(2r - 3) = 4r^2 + 8r - 21\n\nF -> 2r * 2r = 4r^2\nO -> 2r * -3 = -6r\nI -> 7 * 2r = 14r\nL -> 7 * -3 = -21.\n So if I combine like terms, you would see that you get the exact same thing as you got here. You would end up with 4r squared. Negative 6r plus 14r again would be plus 8r. And then you'd have minus 21. So this FOIL method is not producing a different result. It's just a convenient way to remember how to get the product of two binomials. And again, once you start using this, you're going to love it. It's something you're going to use for the rest of your life.", " All right, let's take a look at this one.", null, "We can see these text from the image: (5n + 5)(6n - 2).\n We have 5n plus 5 times 6n minus 2. So let's use FOIL. I want to start out with the F, the first terms. The first terms we have 5n times 6n. 5 times 6 is 30. n times n is n squared. Now I want to use my outer terms. Here's the outer term. Here's the outer term.", " So, my outside terms.", " 5n times negative 2 would be negative 10n.", null, "We can see these text from the image: (5n + 5)(6n - 2) =\n30n^2 - 10n.\n Now I want to do my inner terms. So I want to do 5 times 6n. These are on the inside. 5 times 6 is 30. Then times n. So 30n. Then I want to do my last or final terms.", null, "We can see these text from the image: (5n + 5)(6n - 2) = 30n^2 - 10n + 30n.\n So I got 5 here as the final term. Negative 2 here is the final term. 5 times negative 2 is negative 10. And so I've got 30n squared minus 10n plus 30n minus 10.", null, "We can see these text from the image: (5n + 5)(6n - 2) =\n30n^2 - 10n + 30n - 10 =.\n So all I need to do now is just combine like terms here. Negative 10n plus 30n is 20n. So this gives me a final answer of 30n squared plus 20n minus 10.", " All right.", null, "We can see these text from the image: (4x - 5)(5x + 7).\n For the next one, I have 4x minus 5 times 5x plus 7.", null, "We can see these text from the image: (4x - 5)(5x + 7) =.\n So again, I'm going to use FOIL. First terms, I have 4x and I have 5x. 4x times 5x is 20x squared. Then I'm going to move to the outer terms, the O. So that's 4x times 7. 4x times 7 is 28x. So plus 28x. Then I'm going to move to my inside terms. So I have negative 5 and I have 5x.", null, "We can see these text from the image: (4x - 5)(5x + 7) =\n20x^2 + 28x - 25x.\n So negative 5 times 5x is negative 25x." ], "metadata": [ { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/45921.700000000004_45948.700000000004.mp4", "refined_asr": " So if I combine like terms, you would see that you get the exact same thing as you got here. You would end up with 4r squared. Negative 6r plus 14r again would be plus 8r. And then you'd have minus 21. So this FOIL method is not producing a different result. It's just a convenient way to remember how to get the product of two binomials. And again, once you start using this, you're going to love it. It's something you're going to use for the rest of your life.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45921.700000000004_45948.700000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45921.700000000004_45948.700000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45921.700000000004_45948.700000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45921.700000000004_45948.700000000004#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45921.700000000004_45948.700000000004#4.jpg" ], "ocr_qwen2_vl_72b": "(2r + 7)(2r - 3) = 4r^2 + 8r - 21\n\nF -> 2r * 2r = 4r^2\nO -> 2r * -3 = -6r\nI -> 7 * 2r = 14r\nL -> 7 * -3 = -21" }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/45948.700000000004_45950.700000000004.mp4", "refined_asr": " All right, let's take a look at this one.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45948.700000000004_45950.700000000004#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/45950.700000000004_45972.700000000004.mp4", "refined_asr": " We have 5n plus 5 times 6n minus 2. So let's use FOIL. I want to start out with the F, the first terms. The first terms we have 5n times 6n. 5 times 6 is 30. n times n is n squared. Now I want to use my outer terms. Here's the outer term. Here's the outer term.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45950.700000000004_45972.700000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45950.700000000004_45972.700000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45950.700000000004_45972.700000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45950.700000000004_45972.700000000004#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45950.700000000004_45972.700000000004#4.jpg" ], "ocr_qwen2_vl_72b": "(5n + 5)(6n - 2)" }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/45972.700000000004_45974.700000000004.mp4", "refined_asr": " So, my outside terms.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/45974.700000000004_45977.700000000004.mp4", "refined_asr": " 5n times negative 2 would be negative 10n.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45974.700000000004_45977.700000000004#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/45978.700000000004_45992.700000000004.mp4", "refined_asr": " Now I want to do my inner terms. So I want to do 5 times 6n. These are on the inside. 5 times 6 is 30. Then times n. So 30n. Then I want to do my last or final terms.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45978.700000000004_45992.700000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45978.700000000004_45992.700000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45978.700000000004_45992.700000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45978.700000000004_45992.700000000004#3.jpg" ], "ocr_qwen2_vl_72b": "(5n + 5)(6n - 2) =\n30n^2 - 10n" }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/45992.700000000004_46004.700000000004.mp4", "refined_asr": " So I got 5 here as the final term. Negative 2 here is the final term. 5 times negative 2 is negative 10. And so I've got 30n squared minus 10n plus 30n minus 10.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45992.700000000004_46004.700000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45992.700000000004_46004.700000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@45992.700000000004_46004.700000000004#2.jpg" ], "ocr_qwen2_vl_72b": "(5n + 5)(6n - 2) = 30n^2 - 10n + 30n" }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/46004.700000000004_46017.700000000004.mp4", "refined_asr": " So all I need to do now is just combine like terms here. Negative 10n plus 30n is 20n. So this gives me a final answer of 30n squared plus 20n minus 10.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46004.700000000004_46017.700000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46004.700000000004_46017.700000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46004.700000000004_46017.700000000004#2.jpg" ], "ocr_qwen2_vl_72b": "(5n + 5)(6n - 2) =\n30n^2 - 10n + 30n - 10 =" }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/46017.700000000004_46018.700000000004.mp4", "refined_asr": " All right.", "keyframe_ssim": null, "extracted_frames": null, "ocr_qwen2_vl_72b": null }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/46018.700000000004_46022.700000000004.mp4", "refined_asr": " For the next one, I have 4x minus 5 times 5x plus 7.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46018.700000000004_46022.700000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46018.700000000004_46022.700000000004#1.jpg" ], "ocr_qwen2_vl_72b": "(4x - 5)(5x + 7)" }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/46022.700000000004_46049.700000000004.mp4", "refined_asr": " So again, I'm going to use FOIL. First terms, I have 4x and I have 5x. 4x times 5x is 20x squared. Then I'm going to move to the outer terms, the O. So that's 4x times 7. 4x times 7 is 28x. So plus 28x. Then I'm going to move to my inside terms. So I have negative 5 and I have 5x.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46022.700000000004_46049.700000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46022.700000000004_46049.700000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46022.700000000004_46049.700000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46022.700000000004_46049.700000000004#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46022.700000000004_46049.700000000004#4.jpg" ], "ocr_qwen2_vl_72b": "(4x - 5)(5x + 7) =" }, { "vid": "GAN-jgzYsIo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Abstract algebra video series for beginners_100.json#####audio#####doingASR#####FinishASR/GAN-jgzYsIo/46049.700000000004_46053.700000000004.mp4", "refined_asr": " So negative 5 times 5x is negative 25x.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46049.700000000004_46053.700000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/GAN-jgzYsIo/GAN-jgzYsIo@46049.700000000004_46053.700000000004#1.jpg" ], "ocr_qwen2_vl_72b": "(4x - 5)(5x + 7) =\n20x^2 + 28x - 25x" } ], "image_num": 8, "text_num": 588, "token_num": 5196 }, { "images": [ "sample_100_images/PF8jlg4ZVHc@297.44_343.56#1.jpg", "sample_100_images/PF8jlg4ZVHc@297.44_343.56#2.jpg", "sample_100_images/PF8jlg4ZVHc@297.44_343.56#3.jpg", "sample_100_images/PF8jlg4ZVHc@297.44_343.56#4.jpg", "sample_100_images/PF8jlg4ZVHc@297.44_343.56#5.jpg", "sample_100_images/PF8jlg4ZVHc@297.44_343.56#6.jpg", "sample_100_images/PF8jlg4ZVHc@297.44_343.56#7.jpg", null, "sample_100_images/PF8jlg4ZVHc@343.56_400.08#1.jpg", "sample_100_images/PF8jlg4ZVHc@343.56_400.08#2.jpg", "sample_100_images/PF8jlg4ZVHc@343.56_400.08#3.jpg", "sample_100_images/PF8jlg4ZVHc@343.56_400.08#4.jpg", "sample_100_images/PF8jlg4ZVHc@343.56_400.08#5.jpg", "sample_100_images/PF8jlg4ZVHc@343.56_400.08#6.jpg", "sample_100_images/PF8jlg4ZVHc@343.56_400.08#7.jpg", "sample_100_images/PF8jlg4ZVHc@343.56_400.08#8.jpg", null ], "texts": [ null, null, null, null, null, null, null, "\nYou're exerting force on the end of your wrench, for instance. So if your wrench is, let's say, eight inches long, that's the radius vector. And the force that you're applying at the end of the wrench is, hypothetically, two newtons. Then the cross product of these two vectors is going to give you the torque. The torque is going to be about the axis of rotation of that wrench, right? That's basically what's happening here.\n\nSo for a picture, okay, I'm going to do my best. This is going to look a little weird, probably, but here's a nut or a bolt, however you want to think of it. I'll try to draw it kind of big like this. And it's drilled into the board. I'm trying to loosen it, all right? So what I would do is I would stick a wrench on that and try to, you know, try to loosen it or tighten it. And so my wrench, and I'm not a great artist, but my wrench is basically going to look... Let me switch colors. I like my wrenches to look different in color.", null, null, null, null, null, null, null, null, " So, my wrench might come out here and attach to here. Maybe I've got some jaws that kind of come out here and meet here. And then, of course, my wrench is about this long. I know it looks a little bit weird, but you can get the idea. \n\nYou're going to want to make sure that you can do this. Stick it on there, and you want to either tighten it or loosen it or whatever, and you're exerting a force way out here at the end. And you already know from experience that you get more of an effect, more of a mechanical advantage, more torque, if your force is at the very end. \n\nIf you have a wrench that's really long, you're going to be able to loosen that nut much, much easier than if you have a really little tiny wrench. It's not going to be so helpful, alright? And that's because you can see the F is right here, or the R is right here. If the radius or the length of your wrench is a little bit longer, you are going to get a bigger torque, right? If your force is bigger, you're going to get a bigger torque. \n\nSo, anyhow, what we're doing when we do this, typically, is we're applying a force straight up and down. You also know that you really want to do straight up and" ], "text_ocr_list": [ null, null, null, null, null, null, null, "We can see these text from the image: Torque\n\\( \\tau = r \\times F \\).\n\nYou're exerting force on the end of your wrench, for instance. So if your wrench is, let's say, eight inches long, that's the radius vector. And the force that you're applying at the end of the wrench is, hypothetically, two newtons. Then the cross product of these two vectors is going to give you the torque. The torque is going to be about the axis of rotation of that wrench, right? That's basically what's happening here.\n\nSo for a picture, okay, I'm going to do my best. This is going to look a little weird, probably, but here's a nut or a bolt, however you want to think of it. I'll try to draw it kind of big like this. And it's drilled into the board. I'm trying to loosen it, all right? So what I would do is I would stick a wrench on that and try to, you know, try to loosen it or tighten it. And so my wrench, and I'm not a great artist, but my wrench is basically going to look... Let me switch colors. I like my wrenches to look different in color.", null, null, null, null, null, null, null, null, "We can see these text from the image: - B\nF.\n So, my wrench might come out here and attach to here. Maybe I've got some jaws that kind of come out here and meet here. And then, of course, my wrench is about this long. I know it looks a little bit weird, but you can get the idea. \n\nYou're going to want to make sure that you can do this. Stick it on there, and you want to either tighten it or loosen it or whatever, and you're exerting a force way out here at the end. And you already know from experience that you get more of an effect, more of a mechanical advantage, more torque, if your force is at the very end. \n\nIf you have a wrench that's really long, you're going to be able to loosen that nut much, much easier than if you have a really little tiny wrench. It's not going to be so helpful, alright? And that's because you can see the F is right here, or the R is right here. If the radius or the length of your wrench is a little bit longer, you are going to get a bigger torque, right? If your force is bigger, you're going to get a bigger torque. \n\nSo, anyhow, what we're doing when we do this, typically, is we're applying a force straight up and down. You also know that you really want to do straight up and" ], "metadata": [ { "vid": "PF8jlg4ZVHc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Cross Product of Vectors_30.json#####audio#####doingASR#####FinishASR/PF8jlg4ZVHc/297.44_343.56.mp4", "refined_asr": "\nYou're exerting force on the end of your wrench, for instance. So if your wrench is, let's say, eight inches long, that's the radius vector. And the force that you're applying at the end of the wrench is, hypothetically, two newtons. Then the cross product of these two vectors is going to give you the torque. The torque is going to be about the axis of rotation of that wrench, right? That's basically what's happening here.\n\nSo for a picture, okay, I'm going to do my best. This is going to look a little weird, probably, but here's a nut or a bolt, however you want to think of it. I'll try to draw it kind of big like this. And it's drilled into the board. I'm trying to loosen it, all right? So what I would do is I would stick a wrench on that and try to, you know, try to loosen it or tighten it. And so my wrench, and I'm not a great artist, but my wrench is basically going to look... Let me switch colors. I like my wrenches to look different in color.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#7.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@297.44_343.56#7.jpg" ], "ocr_qwen2_vl_72b": "Torque\n\\( \\tau = r \\times F \\)" }, { "vid": "PF8jlg4ZVHc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Cross Product of Vectors_30.json#####audio#####doingASR#####FinishASR/PF8jlg4ZVHc/343.56_400.08.mp4", "refined_asr": " So, my wrench might come out here and attach to here. Maybe I've got some jaws that kind of come out here and meet here. And then, of course, my wrench is about this long. I know it looks a little bit weird, but you can get the idea. \n\nYou're going to want to make sure that you can do this. Stick it on there, and you want to either tighten it or loosen it or whatever, and you're exerting a force way out here at the end. And you already know from experience that you get more of an effect, more of a mechanical advantage, more torque, if your force is at the very end. \n\nIf you have a wrench that's really long, you're going to be able to loosen that nut much, much easier than if you have a really little tiny wrench. It's not going to be so helpful, alright? And that's because you can see the F is right here, or the R is right here. If the radius or the length of your wrench is a little bit longer, you are going to get a bigger torque, right? If your force is bigger, you're going to get a bigger torque. \n\nSo, anyhow, what we're doing when we do this, typically, is we're applying a force straight up and down. You also know that you really want to do straight up and", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#7.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#8.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#7.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#8.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@343.56_400.08#9.jpg" ], "ocr_qwen2_vl_72b": "- B\nF" } ], "image_num": 15, "text_num": 588, "token_num": 9228 }, { "images": [ "sample_100_images/PF8jlg4ZVHc@674.9599999999999_680.8#1.jpg", null, "sample_100_images/PF8jlg4ZVHc@680.8_734.02#1.jpg", "sample_100_images/PF8jlg4ZVHc@680.8_734.02#2.jpg", "sample_100_images/PF8jlg4ZVHc@680.8_734.02#3.jpg", "sample_100_images/PF8jlg4ZVHc@680.8_734.02#4.jpg", "sample_100_images/PF8jlg4ZVHc@680.8_734.02#6.jpg", "sample_100_images/PF8jlg4ZVHc@680.8_734.02#8.jpg", null, "sample_100_images/PF8jlg4ZVHc@734.02_773.16#1.jpg", "sample_100_images/PF8jlg4ZVHc@734.02_773.16#3.jpg", "sample_100_images/PF8jlg4ZVHc@734.02_773.16#6.jpg", null ], "texts": [ null, " But for typical mechanics and things like that we're always talking about three dimensions", null, null, null, null, null, null, " So let's just cut to the chase. And let's say you're given on a test that your vector A is three in the I direction, plus two in the J direction, plus four in the K direction. These are the X, Y, and Z components. And your vector B is simply one in the I direction, plus two in the J direction, minus two in the K direction. These are the two vectors you have, with their X, Y, and Z components. And I ask you, find A cross B. Now, when I find A cross B, I don't want to get just the magnitude; I don't want to get just the direction. I want to know the actual vector, the X component, the Y component, and the Z component of that vector. Because what you typically learn in Physics 1 is the magnitude of the cross product is the magnitude of A, times the magnitude of B, times the sine of the angle between them.", null, null, null, " In order to calculate that, I would need to find the magnitude of A and B, which I know how to do with the Pythagorean theorem. But I would also need to find the angle between these two vectors. Can you look at this and tell me what the angle between those two vectors is? Well, no. It's something you can calculate, something you can find from this, but it's not something that just jumps off the page. You don't know the angle between the two vectors readily, okay? So typically, when you find the cross product of two vectors, you dispense with all that and just calculate the actual vector resulting from the cross product. This is the way you set it up: your first row is just I, J, and K." ], "text_ocr_list": [ null, "We can see these text from the image: a \u00d7 b = det | i j k |\n| ax ay az |\n| bx by bz |.\n But for typical mechanics and things like that we're always talking about three dimensions", null, null, null, null, null, null, "We can see these text from the image: a x b = det | i j k |\n| ax ay az |\n| bx by bz |.\n So let's just cut to the chase. And let's say you're given on a test that your vector A is three in the I direction, plus two in the J direction, plus four in the K direction. These are the X, Y, and Z components. And your vector B is simply one in the I direction, plus two in the J direction, minus two in the K direction. These are the two vectors you have, with their X, Y, and Z components. And I ask you, find A cross B. Now, when I find A cross B, I don't want to get just the magnitude; I don't want to get just the direction. I want to know the actual vector, the X component, the Y component, and the Z component of that vector. Because what you typically learn in Physics 1 is the magnitude of the cross product is the magnitude of A, times the magnitude of B, times the sine of the angle between them.", null, null, null, "We can see these text from the image: - \\(\\overline{a} \\times \\overline{b} = \\text{det}\\begin{vmatrix} i & j & k \\\\ a_x & a_y & a_z \\\\ b_x & b_y & b_z \\end{vmatrix}\\)\n- \\(\\overline{a} = 3i + 2j + 4k\\)\n- \\(\\overline{b} = i + 2j - 2k\\).\n In order to calculate that, I would need to find the magnitude of A and B, which I know how to do with the Pythagorean theorem. But I would also need to find the angle between these two vectors. Can you look at this and tell me what the angle between those two vectors is? Well, no. It's something you can calculate, something you can find from this, but it's not something that just jumps off the page. You don't know the angle between the two vectors readily, okay? So typically, when you find the cross product of two vectors, you dispense with all that and just calculate the actual vector resulting from the cross product. This is the way you set it up: your first row is just I, J, and K." ], "metadata": [ { "vid": "PF8jlg4ZVHc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Cross Product of Vectors_30.json#####audio#####doingASR#####FinishASR/PF8jlg4ZVHc/674.9599999999999_680.8.mp4", "refined_asr": " But for typical mechanics and things like that we're always talking about three dimensions", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@674.9599999999999_680.8#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@674.9599999999999_680.8#1.jpg" ], "ocr_qwen2_vl_72b": "a \u00d7 b = det | i j k |\n| ax ay az |\n| bx by bz |" }, { "vid": "PF8jlg4ZVHc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Cross Product of Vectors_30.json#####audio#####doingASR#####FinishASR/PF8jlg4ZVHc/680.8_734.02.mp4", "refined_asr": " So let's just cut to the chase. And let's say you're given on a test that your vector A is three in the I direction, plus two in the J direction, plus four in the K direction. These are the X, Y, and Z components. And your vector B is simply one in the I direction, plus two in the J direction, minus two in the K direction. These are the two vectors you have, with their X, Y, and Z components. And I ask you, find A cross B. Now, when I find A cross B, I don't want to get just the magnitude; I don't want to get just the direction. I want to know the actual vector, the X component, the Y component, and the Z component of that vector. Because what you typically learn in Physics 1 is the magnitude of the cross product is the magnitude of A, times the magnitude of B, times the sine of the angle between them.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#8.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#7.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@680.8_734.02#8.jpg" ], "ocr_qwen2_vl_72b": "a x b = det | i j k |\n| ax ay az |\n| bx by bz |" }, { "vid": "PF8jlg4ZVHc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra Tutorial on Cross Product of Vectors_30.json#####audio#####doingASR#####FinishASR/PF8jlg4ZVHc/734.02_773.16.mp4", "refined_asr": " In order to calculate that, I would need to find the magnitude of A and B, which I know how to do with the Pythagorean theorem. But I would also need to find the angle between these two vectors. Can you look at this and tell me what the angle between those two vectors is? Well, no. It's something you can calculate, something you can find from this, but it's not something that just jumps off the page. You don't know the angle between the two vectors readily, okay? So typically, when you find the cross product of two vectors, you dispense with all that and just calculate the actual vector resulting from the cross product. This is the way you set it up: your first row is just I, J, and K.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@734.02_773.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@734.02_773.16#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@734.02_773.16#6.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@734.02_773.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@734.02_773.16#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@734.02_773.16#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@734.02_773.16#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@734.02_773.16#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PF8jlg4ZVHc/PF8jlg4ZVHc@734.02_773.16#6.jpg" ], "ocr_qwen2_vl_72b": "- \\(\\overline{a} \\times \\overline{b} = \\text{det}\\begin{vmatrix} i & j & k \\\\ a_x & a_y & a_z \\\\ b_x & b_y & b_z \\end{vmatrix}\\)\n- \\(\\overline{a} = 3i + 2j + 4k\\)\n- \\(\\overline{b} = i + 2j - 2k\\)" } ], "image_num": 10, "text_num": 390, "token_num": 6150 }, { "images": [ "sample_100_images/T7D1W1oD8wo@149.9_171.70000000000002#1.jpg", null, "sample_100_images/T7D1W1oD8wo@171.70000000000002_190.50000000000003#1.jpg", "sample_100_images/T7D1W1oD8wo@171.70000000000002_190.50000000000003#2.jpg", null, "sample_100_images/T7D1W1oD8wo@190.70000000000002_211.3#1.jpg", null, "sample_100_images/T7D1W1oD8wo@211.5_234.3#1.jpg", null, "sample_100_images/T7D1W1oD8wo@234.5_254.3#1.jpg", "sample_100_images/T7D1W1oD8wo@234.5_254.3#3.jpg", null, "sample_100_images/T7D1W1oD8wo@254.5_277.1#1.jpg", null, "sample_100_images/T7D1W1oD8wo@277.3_300.90000000000003#1.jpg", null ], "texts": [ null, " Tan is equal to sine by cos. Cot, opposite to tan. Cosec. Here, since all C's are on the right and what remains is sec. Are you ready? You're going to learn a whole set of trigonometric formulae. This video is fun, but also very long, as we will be covering.", null, null, " Most of the trigonometric formulae we have studied before will help us understand the first set of formulae. Let's try going around the hexagon in a clockwise fashion. We proceed as follows: This tells us that tan theta equals sine theta over cos theta.", null, " Well that's how we started making our hexagon. Now let's see the next three functions in clockwise direction. These three tell us that sine theta equals cosine theta over cotangent theta. The first one equals the second one divided by the third one.", null, " Take the next three functions clockwise. First equals second by third. Hence cos theta equals cot theta by cosec theta. And if we take the next three functions we can say that cot theta equals cosec theta by sec theta. And we will get two more formulae if we go clockwise.", null, null, " Cosec is equal to sec over tan. And sec is equal to tan over sine. Wasn't that simple? One hexagon and six formulae without any effort. Great. But what if you forget that you had to go clockwise and you go anticlockwise instead?", null, " Let's say you started like this, in the anticlockwise direction. Don't worry, the pattern still holds true. Tan theta equals Sec theta over Cosec theta. If we take these three functions anticlockwise, first will equal second over third.", null, " We can write: sine theta equals tan theta over sec theta. If these three, then cos theta equals sine theta over tan theta. And the list goes on. Whether you go anticlockwise or clockwise, first will always equal second over third. We can easily get three more here." ], "text_ocr_list": [ null, "We can see these text from the image: sin cos tan 1.\n Tan is equal to sine by cos. Cot, opposite to tan. Cosec. Here, since all C's are on the right and what remains is sec. Are you ready? You're going to learn a whole set of trigonometric formulae. This video is fun, but also very long, as we will be covering.", null, null, "We can see these text from the image: sin cos tan cot sec cosec 1.\n Most of the trigonometric formulae we have studied before will help us understand the first set of formulae. Let's try going around the hexagon in a clockwise fashion. We proceed as follows: This tells us that tan theta equals sine theta over cos theta.", null, "We can see these text from the image: Clockwise\n\ntan\u03b8 = sin\u03b8 / cos\u03b8\n\nsin cos tan cot sec cosec.\n Well that's how we started making our hexagon. Now let's see the next three functions in clockwise direction. These three tell us that sine theta equals cosine theta over cotangent theta. The first one equals the second one divided by the third one.", null, "We can see these text from the image: Clockwise\n\ntan\u03b8 = sin\u03b8 / cos\u03b8\n\nsin\u03b8 = cos\u03b8 / cot\u03b8.\n Take the next three functions clockwise. First equals second by third. Hence cos theta equals cot theta by cosec theta. And if we take the next three functions we can say that cot theta equals cosec theta by sec theta. And we will get two more formulae if we go clockwise.", null, null, "We can see these text from the image: Clockwise\n\ntan\u03b8 = sin\u03b8 / cos\u03b8\nsin\u03b8 = cos\u03b8 / cot\u03b8\ncos\u03b8 = cot\u03b8 / cosec\u03b8\ncot\u03b8 = cosec\u03b8 / sec\u03b8\n\ncosec\u03b8 = sec\u03b8 / tan\u03b8.\n Cosec is equal to sec over tan. And sec is equal to tan over sine. Wasn't that simple? One hexagon and six formulae without any effort. Great. But what if you forget that you had to go clockwise and you go anticlockwise instead?", null, "We can see these text from the image: Clockwise\n\ntan\u03b8 = sin\u03b8/cos\u03b8\nsin\u03b8 = cos\u03b8/cot\u03b8\ncos\u03b8 = cot\u03b8/cosec\u03b8\ncot\u03b8 = cosec\u03b8/sec\u03b8.\n Let's say you started like this, in the anticlockwise direction. Don't worry, the pattern still holds true. Tan theta equals Sec theta over Cosec theta. If we take these three functions anticlockwise, first will equal second over third.", null, "We can see these text from the image: Clockwise\n\ntan\u03b8 = sin\u03b8 / cos\u03b8\nsin\u03b8 = cos\u03b8 / cot\u03b8\ncos\u03b8 = cot\u03b8 / cosec\u03b8\ncot\u03b8 = cosec\u03b8 / sec\u03b8\n\nAnti-Clockwise\n\ntan\u03b8 = sec\u03b8 / cosec\u03b8\nsin\u03b8 = tan\u03b8.\n We can write: sine theta equals tan theta over sec theta. If these three, then cos theta equals sine theta over tan theta. And the list goes on. Whether you go anticlockwise or clockwise, first will always equal second over third. We can easily get three more here." ], "metadata": [ { "vid": "T7D1W1oD8wo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Trigonometry tutorial on Trigonometric Identities_30.json#####audio#####doingASR#####FinishASR/T7D1W1oD8wo/149.9_171.70000000000002.mp4", "refined_asr": " Tan is equal to sine by cos. Cot, opposite to tan. Cosec. Here, since all C's are on the right and what remains is sec. Are you ready? You're going to learn a whole set of trigonometric formulae. This video is fun, but also very long, as we will be covering.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@149.9_171.70000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@149.9_171.70000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@149.9_171.70000000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@149.9_171.70000000000002#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@149.9_171.70000000000002#4.jpg" ], "ocr_qwen2_vl_72b": "sin cos tan 1" }, { "vid": "T7D1W1oD8wo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Trigonometry tutorial on Trigonometric Identities_30.json#####audio#####doingASR#####FinishASR/T7D1W1oD8wo/171.70000000000002_190.50000000000003.mp4", "refined_asr": " Most of the trigonometric formulae we have studied before will help us understand the first set of formulae. Let's try going around the hexagon in a clockwise fashion. We proceed as follows: This tells us that tan theta equals sine theta over cos theta.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@171.70000000000002_190.50000000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@171.70000000000002_190.50000000000003#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@171.70000000000002_190.50000000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@171.70000000000002_190.50000000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@171.70000000000002_190.50000000000003#3.jpg" ], "ocr_qwen2_vl_72b": "sin cos tan cot sec cosec 1" }, { "vid": "T7D1W1oD8wo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Trigonometry tutorial on Trigonometric Identities_30.json#####audio#####doingASR#####FinishASR/T7D1W1oD8wo/190.70000000000002_211.3.mp4", "refined_asr": " Well that's how we started making our hexagon. Now let's see the next three functions in clockwise direction. These three tell us that sine theta equals cosine theta over cotangent theta. The first one equals the second one divided by the third one.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@190.70000000000002_211.3#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@190.70000000000002_211.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@190.70000000000002_211.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@190.70000000000002_211.3#3.jpg" ], "ocr_qwen2_vl_72b": "Clockwise\n\ntan\u03b8 = sin\u03b8 / cos\u03b8\n\nsin cos tan cot sec cosec" }, { "vid": "T7D1W1oD8wo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Trigonometry tutorial on Trigonometric Identities_30.json#####audio#####doingASR#####FinishASR/T7D1W1oD8wo/211.5_234.3.mp4", "refined_asr": " Take the next three functions clockwise. First equals second by third. Hence cos theta equals cot theta by cosec theta. And if we take the next three functions we can say that cot theta equals cosec theta by sec theta. And we will get two more formulae if we go clockwise.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@211.5_234.3#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@211.5_234.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@211.5_234.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@211.5_234.3#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@211.5_234.3#4.jpg" ], "ocr_qwen2_vl_72b": "Clockwise\n\ntan\u03b8 = sin\u03b8 / cos\u03b8\n\nsin\u03b8 = cos\u03b8 / cot\u03b8" }, { "vid": "T7D1W1oD8wo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Trigonometry tutorial on Trigonometric Identities_30.json#####audio#####doingASR#####FinishASR/T7D1W1oD8wo/234.5_254.3.mp4", "refined_asr": " Cosec is equal to sec over tan. And sec is equal to tan over sine. Wasn't that simple? One hexagon and six formulae without any effort. Great. But what if you forget that you had to go clockwise and you go anticlockwise instead?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@234.5_254.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@234.5_254.3#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@234.5_254.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@234.5_254.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@234.5_254.3#3.jpg" ], "ocr_qwen2_vl_72b": "Clockwise\n\ntan\u03b8 = sin\u03b8 / cos\u03b8\nsin\u03b8 = cos\u03b8 / cot\u03b8\ncos\u03b8 = cot\u03b8 / cosec\u03b8\ncot\u03b8 = cosec\u03b8 / sec\u03b8\n\ncosec\u03b8 = sec\u03b8 / tan\u03b8" }, { "vid": "T7D1W1oD8wo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Trigonometry tutorial on Trigonometric Identities_30.json#####audio#####doingASR#####FinishASR/T7D1W1oD8wo/254.5_277.1.mp4", "refined_asr": " Let's say you started like this, in the anticlockwise direction. Don't worry, the pattern still holds true. Tan theta equals Sec theta over Cosec theta. If we take these three functions anticlockwise, first will equal second over third.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@254.5_277.1#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@254.5_277.1#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@254.5_277.1#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@254.5_277.1#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@254.5_277.1#4.jpg" ], "ocr_qwen2_vl_72b": "Clockwise\n\ntan\u03b8 = sin\u03b8/cos\u03b8\nsin\u03b8 = cos\u03b8/cot\u03b8\ncos\u03b8 = cot\u03b8/cosec\u03b8\ncot\u03b8 = cosec\u03b8/sec\u03b8" }, { "vid": "T7D1W1oD8wo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Geometry Trigonometry tutorial on Trigonometric Identities_30.json#####audio#####doingASR#####FinishASR/T7D1W1oD8wo/277.3_300.90000000000003.mp4", "refined_asr": " We can write: sine theta equals tan theta over sec theta. If these three, then cos theta equals sine theta over tan theta. And the list goes on. Whether you go anticlockwise or clockwise, first will always equal second over third. We can easily get three more here.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@277.3_300.90000000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@277.3_300.90000000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@277.3_300.90000000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@277.3_300.90000000000003#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/T7D1W1oD8wo/T7D1W1oD8wo@277.3_300.90000000000003#4.jpg" ], "ocr_qwen2_vl_72b": "Clockwise\n\ntan\u03b8 = sin\u03b8 / cos\u03b8\nsin\u03b8 = cos\u03b8 / cot\u03b8\ncos\u03b8 = cot\u03b8 / cosec\u03b8\ncot\u03b8 = cosec\u03b8 / sec\u03b8\n\nAnti-Clockwise\n\ntan\u03b8 = sec\u03b8 / cosec\u03b8\nsin\u03b8 = tan\u03b8" } ], "image_num": 9, "text_num": 457, "token_num": 5641 }, { "images": [ "sample_100_images/W5h5HZ0yjyo@355.46_369.32#1.jpg", null, "sample_100_images/W5h5HZ0yjyo@369.68_385.9#1.jpg", null, "sample_100_images/W5h5HZ0yjyo@386.66_399.38#1.jpg", null, "sample_100_images/W5h5HZ0yjyo@399.68_410.46#1.jpg", null, "sample_100_images/W5h5HZ0yjyo@411.12_429.04#1.jpg", null, "sample_100_images/W5h5HZ0yjyo@429.46_447.48#1.jpg", null ], "texts": [ null, " So those represent the coefficients. So we're going to put the coefficients of my dividend right here. Okay. Now this is my divisor and that's going to tell me what number, what K value, I'm going to put on the outside of this.", null, " So this is x minus k. Our k in this case is 3. So we're going to put a 3. The other way to think of the number out here is what makes the divisor equal to zero. So it's kind of like a factor.", null, " So you're saying x minus 3, what makes that equal 0? Just plain positive 3. Okay. So now how you do synthetic division is you're going to add straight down. So on the 2, we're going to just drop the 2 straight down.", null, " And then you're going to multiply by the K value. So 2 times 3 is 6. And I'm going to put that number right here. Now I'm going to add straight down: So negative 7 plus 6 is negative 1.", null, " Now I'm going to multiply by the k value. So negative 1 times 3 is negative 3. I'm going to write it right there. Then I'm going to add down. So this would be negative 3. Then negative 3 times 3 is negative 9. Then negative 4 plus negative 9 is negative 13.", null, " And when you get to that last number, I always like to put a box around it because that represents our remainder. So it always helps me separate that final number from the rest of the equation. So if you have a cubic like this and you're dividing by a power of x, that means your answer is going to be negative three." ], "text_ocr_list": [ null, "We can see these text from the image: **EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)**\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\[\n\\begin{array}{c|cccc}\n & 2 & -7 & 0 & -4 \\\\\n\\hline\n & & & & \\\\\n\\end{array}\n\\].\n So those represent the coefficients. So we're going to put the coefficients of my dividend right here. Okay. Now this is my divisor and that's going to tell me what number, what K value, I'm going to put on the outside of this.", null, "We can see these text from the image: EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\( x - k \\)\n\n\\( 2 \\quad -7 \\quad 0 \\quad -4 \\).\n So this is x minus k. Our k in this case is 3. So we're going to put a 3. The other way to think of the number out here is what makes the divisor equal to zero. So it's kind of like a factor.", null, "We can see these text from the image: EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\( \\rightarrow 3 \\)\n\\( 2 \\)\n\\( -7 \\)\n\\( 0 \\)\n\\( -4 \\).\n So you're saying x minus 3, what makes that equal 0? Just plain positive 3. Okay. So now how you do synthetic division is you're going to add straight down. So on the 2, we're going to just drop the 2 straight down.", null, "We can see these text from the image: EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\[\n\\begin{array}{c|cccc}\n3 & 2 & -7 & 0 & -4 \\\\\n\\hline\n & & 6 & -3 & -9 \\\\\n\\hline\n & 2 & -1 & -3 & -13 \\\\\n\\end{array}\n\\].\n And then you're going to multiply by the K value. So 2 times 3 is 6. And I'm going to put that number right here. Now I'm going to add straight down: So negative 7 plus 6 is negative 1.", null, "We can see these text from the image: EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\[\n\\begin{array}{c|cccc}\n3 & 2 & -7 & 0 & -4 \\\\\n & & 6 & -9 & -9 \\\\\n\\hline\n & 2 & -1 & -9 & -13 \\\\\n\\end{array}\n\\].\n Now I'm going to multiply by the k value. So negative 1 times 3 is negative 3. I'm going to write it right there. Then I'm going to add down. So this would be negative 3. Then negative 3 times 3 is negative 9. Then negative 4 plus negative 9 is negative 13.", null, "We can see these text from the image: EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\[\n\\begin{array}{c|cccc}\n3 & 2 & -7 & 0 & -4 \\\\\n\\hline\n& & 6 & -3 & -9 \\\\\n\\hline\n& 2 & -1 & -3 & -13 \\\\\n\\end{array}\n\\].\n And when you get to that last number, I always like to put a box around it because that represents our remainder. So it always helps me separate that final number from the rest of the equation. So if you have a cubic like this and you're dividing by a power of x, that means your answer is going to be negative three." ], "metadata": [ { "vid": "W5h5HZ0yjyo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dividing Polynomials_30.json#####audio#####doingASR#####FinishASR/W5h5HZ0yjyo/355.46_369.32.mp4", "refined_asr": " So those represent the coefficients. So we're going to put the coefficients of my dividend right here. Okay. Now this is my divisor and that's going to tell me what number, what K value, I'm going to put on the outside of this.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@355.46_369.32#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@355.46_369.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@355.46_369.32#2.jpg" ], "ocr_qwen2_vl_72b": "**EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)**\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\[\n\\begin{array}{c|cccc}\n & 2 & -7 & 0 & -4 \\\\\n\\hline\n & & & & \\\\\n\\end{array}\n\\]" }, { "vid": "W5h5HZ0yjyo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dividing Polynomials_30.json#####audio#####doingASR#####FinishASR/W5h5HZ0yjyo/369.68_385.9.mp4", "refined_asr": " So this is x minus k. Our k in this case is 3. So we're going to put a 3. The other way to think of the number out here is what makes the divisor equal to zero. So it's kind of like a factor.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@369.68_385.9#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@369.68_385.9#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@369.68_385.9#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@369.68_385.9#3.jpg" ], "ocr_qwen2_vl_72b": "EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\( x - k \\)\n\n\\( 2 \\quad -7 \\quad 0 \\quad -4 \\)" }, { "vid": "W5h5HZ0yjyo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dividing Polynomials_30.json#####audio#####doingASR#####FinishASR/W5h5HZ0yjyo/386.66_399.38.mp4", "refined_asr": " So you're saying x minus 3, what makes that equal 0? Just plain positive 3. Okay. So now how you do synthetic division is you're going to add straight down. So on the 2, we're going to just drop the 2 straight down.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@386.66_399.38#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@386.66_399.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@386.66_399.38#2.jpg" ], "ocr_qwen2_vl_72b": "EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\( \\rightarrow 3 \\)\n\\( 2 \\)\n\\( -7 \\)\n\\( 0 \\)\n\\( -4 \\)" }, { "vid": "W5h5HZ0yjyo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dividing Polynomials_30.json#####audio#####doingASR#####FinishASR/W5h5HZ0yjyo/399.68_410.46.mp4", "refined_asr": " And then you're going to multiply by the K value. So 2 times 3 is 6. And I'm going to put that number right here. Now I'm going to add straight down: So negative 7 plus 6 is negative 1.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@399.68_410.46#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@399.68_410.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@399.68_410.46#2.jpg" ], "ocr_qwen2_vl_72b": "EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\[\n\\begin{array}{c|cccc}\n3 & 2 & -7 & 0 & -4 \\\\\n\\hline\n & & 6 & -3 & -9 \\\\\n\\hline\n & 2 & -1 & -3 & -13 \\\\\n\\end{array}\n\\]" }, { "vid": "W5h5HZ0yjyo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dividing Polynomials_30.json#####audio#####doingASR#####FinishASR/W5h5HZ0yjyo/411.12_429.04.mp4", "refined_asr": " Now I'm going to multiply by the k value. So negative 1 times 3 is negative 3. I'm going to write it right there. Then I'm going to add down. So this would be negative 3. Then negative 3 times 3 is negative 9. Then negative 4 plus negative 9 is negative 13.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@411.12_429.04#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@411.12_429.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@411.12_429.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@411.12_429.04#3.jpg" ], "ocr_qwen2_vl_72b": "EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\[\n\\begin{array}{c|cccc}\n3 & 2 & -7 & 0 & -4 \\\\\n & & 6 & -9 & -9 \\\\\n\\hline\n & 2 & -1 & -9 & -13 \\\\\n\\end{array}\n\\]" }, { "vid": "W5h5HZ0yjyo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Algebra tutorial on Dividing Polynomials_30.json#####audio#####doingASR#####FinishASR/W5h5HZ0yjyo/429.46_447.48.mp4", "refined_asr": " And when you get to that last number, I always like to put a box around it because that represents our remainder. So it always helps me separate that final number from the rest of the equation. So if you have a cubic like this and you're dividing by a power of x, that means your answer is going to be negative three.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@429.46_447.48#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@429.46_447.48#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@429.46_447.48#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/W5h5HZ0yjyo/W5h5HZ0yjyo@429.46_447.48#3.jpg" ], "ocr_qwen2_vl_72b": "EXAMPLE 2 Use Synthetic Division to Divide by \\( x - a \\)\n\nWhat is \\( 2x^3 - 7x^2 - 4 \\) divided by \\( x - 3 \\)? Use synthetic division.\n\n\\[\n\\begin{array}{c|cccc}\n3 & 2 & -7 & 0 & -4 \\\\\n\\hline\n& & 6 & -3 & -9 \\\\\n\\hline\n& 2 & -1 & -3 & -13 \\\\\n\\end{array}\n\\]" } ], "image_num": 6, "text_num": 388, "token_num": 3844 }, { "images": [ "sample_100_images/V_xro1bcAuA@51155.3_51171.3#1.jpg", null, "sample_100_images/V_xro1bcAuA@51171.3_51185.3#1.jpg", "sample_100_images/V_xro1bcAuA@51171.3_51185.3#2.jpg", null, "sample_100_images/V_xro1bcAuA@51185.3_51201.3#1.jpg", "sample_100_images/V_xro1bcAuA@51185.3_51201.3#2.jpg", null, "sample_100_images/V_xro1bcAuA@51201.3_51217.3#1.jpg", null, "sample_100_images/V_xro1bcAuA@51217.3_51237.3#1.jpg", "sample_100_images/V_xro1bcAuA@51217.3_51237.3#2.jpg", "sample_100_images/V_xro1bcAuA@51217.3_51237.3#3.jpg", null, "sample_100_images/V_xro1bcAuA@51237.3_51251.3#1.jpg", "sample_100_images/V_xro1bcAuA@51237.3_51251.3#2.jpg", null, "sample_100_images/V_xro1bcAuA@51251.3_51263.3#1.jpg", null, "sample_100_images/V_xro1bcAuA@51263.3_51277.3#1.jpg", "sample_100_images/V_xro1bcAuA@51263.3_51277.3#2.jpg", null ], "texts": [ null, " So in the last video, we covered what we're going to broadly address. We saw some examples of what computer vision problems are\u2014essentially, anything that you're able to see can potentially be turned into a computer vision problem. And we're going to be exploring many machine learning applications.", null, null, " Or specifically PyTorch computer vision code. You see, I fixed that typo. Now let's talk about what the inputs and outputs are of a typical computer vision problem. So let's start with a multi-classification example. And say we wanted to take", null, null, " Photos of different images of food and recognizing what they were. So we're replicating the functionality of Nutrify. Take a photo of food and learn about it. We might start with a bunch of food images that have a certain height and width.", null, " So we have width equals 224, height equals 224, and then they have three color channels. Why three? Well, that's because we have a value for red, green, and blue. So if we look at this, if we go red, green, blue,", null, null, null, " Image format. So 24-bit RGB images. A lot of images, or digital images, have some value for a red pixel, a green pixel, and a blue pixel. And if you were to convert images into numbers, they get represented by some value of red, some value of green, and some value of blue.", null, null, " And some value of blue. That is exactly the same as how we'd represent these images. So for example, this pixel here might be a little bit more red, a little less blue, and a little less green because it's close to orange. And then we convert that into numbers. So", null, " What we're trying to do here is essentially what we're trying to do with all of the data that we have with machine learning, which is to represent it as numbers. So the typical image format to represent an image, since we're using computer vision, is:", null, null, " Trying to figure out what's in an image. The typical way to represent that is in a tensor that has a value for the height, width, and color channels. And so we might numerically encode these, in other words, represent our images as a tensor. And this" ], "text_ocr_list": [ null, "We can see these text from the image: What we're going to cover (broadly)\n\n- Getting a vision dataset to work with using torchvision.datasets\n- Architecture of a convolutional neural network (CNN) with PyTorch\n- An end-to-end multi-class image classification problem\n- Steps in modelling with CNNs in PyTorch\n - Creating a CNN model with PyTorch\n - Picking a loss and optimizer\n - Training a PyTorch computer vision model\n - Evaluating a model\n\n(we'll be cooking up lots of code!)\n\n93. Computer vision input and output shapes.\n So in the last video, we covered what we're going to broadly address. We saw some examples of what computer vision problems are\u2014essentially, anything that you're able to see can potentially be turned into a computer vision problem. And we're going to be exploring many machine learning applications.", null, null, "We can see these text from the image: - Getting a vision dataset to work with using torchvision.datasets\n- Architecture of a convolutional neural network (CNN) with PyTorch\n- An end-to-end multi-class image classification problem\n- Steps in modelling with CNNs in PyTorch\n - Creating a CNN model with PyTorch\n - Picking a loss and optimizer\n - Training a PyTorch computer vision model\n - Evaluating a model\n\n(we'll be cooking up lots of code!).\n Or specifically PyTorch computer vision code. You see, I fixed that typo. Now let's talk about what the inputs and outputs are of a typical computer vision problem. So let's start with a multi-classification example. And say we wanted to take", null, null, "We can see these text from the image: Computer vision inputs and outputs\n\nInputs\n\nMachine Learning Algorithm\n\nOutputs.\n Photos of different images of food and recognizing what they were. So we're replicating the functionality of Nutrify. Take a photo of food and learn about it. We might start with a bunch of food images that have a certain height and width.", null, "We can see these text from the image: Computer vision input\n\nW = 224\nH = 224\nC = 3\n\n(c = colour channels, R, G, B).\n So we have width equals 224, height equals 224, and then they have three color channels. Why three? Well, that's because we have a value for red, green, and blue. So if we look at this, if we go red, green, blue,", null, null, null, "We can see these text from the image: red green blue image format\n\n24-bit RGB images\n\nThe colors in RGB images (24-bit with 8-bits for each of the red, green and blue channels) are used to show multi-channel images. The colors are designed to reflect genuine colors (i.e. the green in an RGB image reflects green color in the specimen). There are several RGB functions in Fiji.\n\nhttps://imagej.net > imaging > color-image-processing\n\nColor Image Processing - ImageJ Wiki.\n Image format. So 24-bit RGB images. A lot of images, or digital images, have some value for a red pixel, a green pixel, and a blue pixel. And if you were to convert images into numbers, they get represented by some value of red, some value of green, and some value of blue.", null, null, "We can see these text from the image: Computer vision inputs and outputs\n\nW = 224\nH = 224\nC = 3\n\n(c = colour channels, R, G, B)\n\nNumerical encoding\n\n(normalized pixel values)\n\nInputs\n\nMachine Learning Algorithm\n\nOutputs\n\nActual output\n\nSushi \ud83c\udf63\nSteak \ud83e\udd69\nPizza \ud83c\udf55\n\nPredicted output (comes from looking at lots of these).\n And some value of blue. That is exactly the same as how we'd represent these images. So for example, this pixel here might be a little bit more red, a little less blue, and a little less green because it's close to orange. And then we convert that into numbers. So", null, "We can see these text from the image: Computer vision inputs and outputs\n\nW = 224\nH = 224\nC = 3\n\nInputs\n\nMachine Learning Algorithm\n\nOutputs\n\nNumerical encoding\n\nPredicted output.\n What we're trying to do here is essentially what we're trying to do with all of the data that we have with machine learning, which is to represent it as numbers. So the typical image format to represent an image, since we're using computer vision, is:", null, null, "We can see these text from the image: Computer vision inputs and outputs\n\nW = 224\nH = 224\nC = 3\n\nInputs\n\nMachine Learning Algorithm\n\nOutputs\n\nPredicted output.\n Trying to figure out what's in an image. The typical way to represent that is in a tensor that has a value for the height, width, and color channels. And so we might numerically encode these, in other words, represent our images as a tensor. And this" ], "metadata": [ { "vid": "V_xro1bcAuA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Tutorials on Neural Networks in Machine Learning_30.json#####audio#####doingASR#####FinishASR/V_xro1bcAuA/51155.3_51171.3.mp4", "refined_asr": " So in the last video, we covered what we're going to broadly address. We saw some examples of what computer vision problems are\u2014essentially, anything that you're able to see can potentially be turned into a computer vision problem. And we're going to be exploring many machine learning applications.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51155.3_51171.3#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51155.3_51171.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51155.3_51171.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51155.3_51171.3#3.jpg" ], "ocr_qwen2_vl_72b": "What we're going to cover (broadly)\n\n- Getting a vision dataset to work with using torchvision.datasets\n- Architecture of a convolutional neural network (CNN) with PyTorch\n- An end-to-end multi-class image classification problem\n- Steps in modelling with CNNs in PyTorch\n - Creating a CNN model with PyTorch\n - Picking a loss and optimizer\n - Training a PyTorch computer vision model\n - Evaluating a model\n\n(we'll be cooking up lots of code!)\n\n93. Computer vision input and output shapes" }, { "vid": "V_xro1bcAuA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Tutorials on Neural Networks in Machine Learning_30.json#####audio#####doingASR#####FinishASR/V_xro1bcAuA/51171.3_51185.3.mp4", "refined_asr": " Or specifically PyTorch computer vision code. You see, I fixed that typo. Now let's talk about what the inputs and outputs are of a typical computer vision problem. So let's start with a multi-classification example. And say we wanted to take", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51171.3_51185.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51171.3_51185.3#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51171.3_51185.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51171.3_51185.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51171.3_51185.3#3.jpg" ], "ocr_qwen2_vl_72b": "- Getting a vision dataset to work with using torchvision.datasets\n- Architecture of a convolutional neural network (CNN) with PyTorch\n- An end-to-end multi-class image classification problem\n- Steps in modelling with CNNs in PyTorch\n - Creating a CNN model with PyTorch\n - Picking a loss and optimizer\n - Training a PyTorch computer vision model\n - Evaluating a model\n\n(we'll be cooking up lots of code!)" }, { "vid": "V_xro1bcAuA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Tutorials on Neural Networks in Machine Learning_30.json#####audio#####doingASR#####FinishASR/V_xro1bcAuA/51185.3_51201.3.mp4", "refined_asr": " Photos of different images of food and recognizing what they were. So we're replicating the functionality of Nutrify. Take a photo of food and learn about it. We might start with a bunch of food images that have a certain height and width.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51185.3_51201.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51185.3_51201.3#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51185.3_51201.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51185.3_51201.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51185.3_51201.3#3.jpg" ], "ocr_qwen2_vl_72b": "Computer vision inputs and outputs\n\nInputs\n\nMachine Learning Algorithm\n\nOutputs" }, { "vid": "V_xro1bcAuA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Tutorials on Neural Networks in Machine Learning_30.json#####audio#####doingASR#####FinishASR/V_xro1bcAuA/51201.3_51217.3.mp4", "refined_asr": " So we have width equals 224, height equals 224, and then they have three color channels. Why three? Well, that's because we have a value for red, green, and blue. So if we look at this, if we go red, green, blue,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51201.3_51217.3#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51201.3_51217.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51201.3_51217.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51201.3_51217.3#3.jpg" ], "ocr_qwen2_vl_72b": "Computer vision input\n\nW = 224\nH = 224\nC = 3\n\n(c = colour channels, R, G, B)" }, { "vid": "V_xro1bcAuA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Tutorials on Neural Networks in Machine Learning_30.json#####audio#####doingASR#####FinishASR/V_xro1bcAuA/51217.3_51237.3.mp4", "refined_asr": " Image format. So 24-bit RGB images. A lot of images, or digital images, have some value for a red pixel, a green pixel, and a blue pixel. And if you were to convert images into numbers, they get represented by some value of red, some value of green, and some value of blue.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51217.3_51237.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51217.3_51237.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51217.3_51237.3#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51217.3_51237.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51217.3_51237.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51217.3_51237.3#3.jpg" ], "ocr_qwen2_vl_72b": "red green blue image format\n\n24-bit RGB images\n\nThe colors in RGB images (24-bit with 8-bits for each of the red, green and blue channels) are used to show multi-channel images. The colors are designed to reflect genuine colors (i.e. the green in an RGB image reflects green color in the specimen). There are several RGB functions in Fiji.\n\nhttps://imagej.net > imaging > color-image-processing\n\nColor Image Processing - ImageJ Wiki" }, { "vid": "V_xro1bcAuA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Tutorials on Neural Networks in Machine Learning_30.json#####audio#####doingASR#####FinishASR/V_xro1bcAuA/51237.3_51251.3.mp4", "refined_asr": " And some value of blue. That is exactly the same as how we'd represent these images. So for example, this pixel here might be a little bit more red, a little less blue, and a little less green because it's close to orange. And then we convert that into numbers. So", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51237.3_51251.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51237.3_51251.3#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51237.3_51251.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51237.3_51251.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51237.3_51251.3#3.jpg" ], "ocr_qwen2_vl_72b": "Computer vision inputs and outputs\n\nW = 224\nH = 224\nC = 3\n\n(c = colour channels, R, G, B)\n\nNumerical encoding\n\n(normalized pixel values)\n\nInputs\n\nMachine Learning Algorithm\n\nOutputs\n\nActual output\n\nSushi \ud83c\udf63\nSteak \ud83e\udd69\nPizza \ud83c\udf55\n\nPredicted output (comes from looking at lots of these)" }, { "vid": "V_xro1bcAuA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Tutorials on Neural Networks in Machine Learning_30.json#####audio#####doingASR#####FinishASR/V_xro1bcAuA/51251.3_51263.3.mp4", "refined_asr": " What we're trying to do here is essentially what we're trying to do with all of the data that we have with machine learning, which is to represent it as numbers. So the typical image format to represent an image, since we're using computer vision, is:", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51251.3_51263.3#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51251.3_51263.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51251.3_51263.3#2.jpg" ], "ocr_qwen2_vl_72b": "Computer vision inputs and outputs\n\nW = 224\nH = 224\nC = 3\n\nInputs\n\nMachine Learning Algorithm\n\nOutputs\n\nNumerical encoding\n\nPredicted output" }, { "vid": "V_xro1bcAuA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced Computer Engineering Tutorials on Neural Networks in Machine Learning_30.json#####audio#####doingASR#####FinishASR/V_xro1bcAuA/51263.3_51277.3.mp4", "refined_asr": " Trying to figure out what's in an image. The typical way to represent that is in a tensor that has a value for the height, width, and color channels. And so we might numerically encode these, in other words, represent our images as a tensor. And this", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51263.3_51277.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51263.3_51277.3#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51263.3_51277.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51263.3_51277.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/V_xro1bcAuA/V_xro1bcAuA@51263.3_51277.3#3.jpg" ], "ocr_qwen2_vl_72b": "Computer vision inputs and outputs\n\nW = 224\nH = 224\nC = 3\n\nInputs\n\nMachine Learning Algorithm\n\nOutputs\n\nPredicted output" } ], "image_num": 14, "text_num": 490, "token_num": 8554 }, { "images": [ "sample_100_images/68PlPkeOwng@17988.04_18006.04#1.jpg", "sample_100_images/68PlPkeOwng@17988.04_18006.04#2.jpg", "sample_100_images/68PlPkeOwng@17988.04_18006.04#3.jpg", null, "sample_100_images/68PlPkeOwng@18006.04_18025.04#1.jpg", "sample_100_images/68PlPkeOwng@18006.04_18025.04#2.jpg", "sample_100_images/68PlPkeOwng@18006.04_18025.04#3.jpg", null, "sample_100_images/68PlPkeOwng@18025.04_18040.04#1.jpg", "sample_100_images/68PlPkeOwng@18025.04_18040.04#2.jpg", null, "sample_100_images/68PlPkeOwng@18042.04_18062.04#1.jpg", "sample_100_images/68PlPkeOwng@18042.04_18062.04#2.jpg", "sample_100_images/68PlPkeOwng@18042.04_18062.04#3.jpg", null, "sample_100_images/68PlPkeOwng@18064.04_18085.04#1.jpg", "sample_100_images/68PlPkeOwng@18064.04_18085.04#2.jpg", "sample_100_images/68PlPkeOwng@18064.04_18085.04#3.jpg", null, "sample_100_images/68PlPkeOwng@18087.04_18112.04#1.jpg", "sample_100_images/68PlPkeOwng@18087.04_18112.04#2.jpg", "sample_100_images/68PlPkeOwng@18087.04_18112.04#3.jpg", "sample_100_images/68PlPkeOwng@18087.04_18112.04#4.jpg", null, "sample_100_images/68PlPkeOwng@18115.04_18140.04#1.jpg", "sample_100_images/68PlPkeOwng@18115.04_18140.04#2.jpg", "sample_100_images/68PlPkeOwng@18115.04_18140.04#3.jpg", "sample_100_images/68PlPkeOwng@18115.04_18140.04#4.jpg", null ], "texts": [ null, null, null, " We've committed to making information about the changing climate freely available to policymakers, businesses, and research institutions. The European Space Agency has been working closely with NASA and ESA on the development of new technologies that will help to address climate change. What you see here on this graph.", null, null, null, " This is the CO2 concentration in the atmosphere over the last 800,000 years. And you see that these values are going up and down in different phases. The blue lines indicate ice ages, and the orange lines indicate periods between ice ages.", null, null, " There were periods where it's much warmer. But you also see that over the last 800,000 years, the value was always below 300 parts per million. And suddenly, since the last century, it goes up very steeply toward 400 parts per million or even beyond.", null, null, null, " The recent climate summit, COP25, held in Madrid, made little progress toward an international agreement to cut greenhouse gas emissions. While some businesses and economies will have to adjust, that task only gets more difficult as time passes. Far greater adjustments will be forced upon everyone.", null, null, null, " The Earth is the only place we know that harbours life. But the stability that has enabled this web of life is fragile. Plants and animals interact for mutual benefit. Our benign environment results from the complex and varied creatures with which we share the planet.", null, null, null, null, " It's important that we look after our home. The space industry is at a turning point. NASA's Space Launch System is about to fly. With it, the new Orion spacecraft. A collaboration between NASA and ESA will target the Moon. A landing is scheduled for 2024.", null, null, null, null, " As far as the United States is concerned, low Earth orbit is now the province of corporations, with Boeing and SpaceX competing to break Russia's stranglehold on manned trips to the International Space Station. New technology is cutting costs and increasing launch frequency, allowing complex satellites to be so small." ], "text_ocr_list": [ null, null, null, "We can see these text from the image: No text or formulas present in the image..\n We've committed to making information about the changing climate freely available to policymakers, businesses, and research institutions. The European Space Agency has been working closely with NASA and ESA on the development of new technologies that will help to address climate change. What you see here on this graph.", null, null, null, "We can see these text from the image: - 800,000 years of change in global CO2 concentration\n- JOSEF ASCHBACHER\n- ESA's Director of Earth Observation Programmes.\n This is the CO2 concentration in the atmosphere over the last 800,000 years. And you see that these values are going up and down in different phases. The blue lines indicate ice ages, and the orange lines indicate periods between ice ages.", null, null, "We can see these text from the image: - 800,000 years of change in global CO2 concentration.\n There were periods where it's much warmer. But you also see that over the last 800,000 years, the value was always below 300 parts per million. And suddenly, since the last century, it goes up very steeply toward 400 parts per million or even beyond.", null, null, null, "We can see these text from the image: #TiempoDeActuar\n\nCOP25 CHILE MADRID 2019\n\n#TimeForAct.\n The recent climate summit, COP25, held in Madrid, made little progress toward an international agreement to cut greenhouse gas emissions. While some businesses and economies will have to adjust, that task only gets more difficult as time passes. Far greater adjustments will be forced upon everyone.", null, null, null, "We can see these text from the image: There is no text or formula to extract from the image..\n The Earth is the only place we know that harbours life. But the stability that has enabled this web of life is fragile. Plants and animals interact for mutual benefit. Our benign environment results from the complex and varied creatures with which we share the planet.", null, null, null, null, "We can see these text from the image: UP NEXT:\nANOTHER EPISODE.\n It's important that we look after our home. The space industry is at a turning point. NASA's Space Launch System is about to fly. With it, the new Orion spacecraft. A collaboration between NASA and ESA will target the Moon. A landing is scheduled for 2024.", null, null, null, null, "We can see these text from the image: There is no text or formula to extract from this image..\n As far as the United States is concerned, low Earth orbit is now the province of corporations, with Boeing and SpaceX competing to break Russia's stranglehold on manned trips to the International Space Station. New technology is cutting costs and increasing launch frequency, allowing complex satellites to be so small." ], "metadata": [ { "vid": "68PlPkeOwng.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Aerospace Engineering Course on Spacecraft and Payload Design Tutorial_30.json#####audio#####doingASR#####FinishASR/68PlPkeOwng/17988.04_18006.04.mp4", "refined_asr": " We've committed to making information about the changing climate freely available to policymakers, businesses, and research institutions. The European Space Agency has been working closely with NASA and ESA on the development of new technologies that will help to address climate change. What you see here on this graph.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@17988.04_18006.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@17988.04_18006.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@17988.04_18006.04#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@17988.04_18006.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@17988.04_18006.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@17988.04_18006.04#3.jpg" ], "ocr_qwen2_vl_72b": "No text or formulas present in the image." }, { "vid": "68PlPkeOwng.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Aerospace Engineering Course on Spacecraft and Payload Design Tutorial_30.json#####audio#####doingASR#####FinishASR/68PlPkeOwng/18006.04_18025.04.mp4", "refined_asr": " This is the CO2 concentration in the atmosphere over the last 800,000 years. And you see that these values are going up and down in different phases. The blue lines indicate ice ages, and the orange lines indicate periods between ice ages.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18006.04_18025.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18006.04_18025.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18006.04_18025.04#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18006.04_18025.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18006.04_18025.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18006.04_18025.04#3.jpg" ], "ocr_qwen2_vl_72b": "- 800,000 years of change in global CO2 concentration\n- JOSEF ASCHBACHER\n- ESA's Director of Earth Observation Programmes" }, { "vid": "68PlPkeOwng.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Aerospace Engineering Course on Spacecraft and Payload Design Tutorial_30.json#####audio#####doingASR#####FinishASR/68PlPkeOwng/18025.04_18040.04.mp4", "refined_asr": " There were periods where it's much warmer. But you also see that over the last 800,000 years, the value was always below 300 parts per million. And suddenly, since the last century, it goes up very steeply toward 400 parts per million or even beyond.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18025.04_18040.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18025.04_18040.04#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18025.04_18040.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18025.04_18040.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18025.04_18040.04#3.jpg" ], "ocr_qwen2_vl_72b": "- 800,000 years of change in global CO2 concentration" }, { "vid": "68PlPkeOwng.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Aerospace Engineering Course on Spacecraft and Payload Design Tutorial_30.json#####audio#####doingASR#####FinishASR/68PlPkeOwng/18042.04_18062.04.mp4", "refined_asr": " The recent climate summit, COP25, held in Madrid, made little progress toward an international agreement to cut greenhouse gas emissions. While some businesses and economies will have to adjust, that task only gets more difficult as time passes. Far greater adjustments will be forced upon everyone.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18042.04_18062.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18042.04_18062.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18042.04_18062.04#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18042.04_18062.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18042.04_18062.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18042.04_18062.04#3.jpg" ], "ocr_qwen2_vl_72b": "#TiempoDeActuar\n\nCOP25 CHILE MADRID 2019\n\n#TimeForAct" }, { "vid": "68PlPkeOwng.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Aerospace Engineering Course on Spacecraft and Payload Design Tutorial_30.json#####audio#####doingASR#####FinishASR/68PlPkeOwng/18064.04_18085.04.mp4", "refined_asr": " The Earth is the only place we know that harbours life. But the stability that has enabled this web of life is fragile. Plants and animals interact for mutual benefit. Our benign environment results from the complex and varied creatures with which we share the planet.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18064.04_18085.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18064.04_18085.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18064.04_18085.04#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18064.04_18085.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18064.04_18085.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18064.04_18085.04#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18064.04_18085.04#4.jpg" ], "ocr_qwen2_vl_72b": "There is no text or formula to extract from the image." }, { "vid": "68PlPkeOwng.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Aerospace Engineering Course on Spacecraft and Payload Design Tutorial_30.json#####audio#####doingASR#####FinishASR/68PlPkeOwng/18087.04_18112.04.mp4", "refined_asr": " It's important that we look after our home. The space industry is at a turning point. NASA's Space Launch System is about to fly. With it, the new Orion spacecraft. A collaboration between NASA and ESA will target the Moon. A landing is scheduled for 2024.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18087.04_18112.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18087.04_18112.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18087.04_18112.04#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18087.04_18112.04#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18087.04_18112.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18087.04_18112.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18087.04_18112.04#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18087.04_18112.04#4.jpg" ], "ocr_qwen2_vl_72b": "UP NEXT:\nANOTHER EPISODE" }, { "vid": "68PlPkeOwng.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Aerospace Engineering Course on Spacecraft and Payload Design Tutorial_30.json#####audio#####doingASR#####FinishASR/68PlPkeOwng/18115.04_18140.04.mp4", "refined_asr": " As far as the United States is concerned, low Earth orbit is now the province of corporations, with Boeing and SpaceX competing to break Russia's stranglehold on manned trips to the International Space Station. New technology is cutting costs and increasing launch frequency, allowing complex satellites to be so small.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18115.04_18140.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18115.04_18140.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18115.04_18140.04#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18115.04_18140.04#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18115.04_18140.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18115.04_18140.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18115.04_18140.04#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/68PlPkeOwng/68PlPkeOwng@18115.04_18140.04#4.jpg" ], "ocr_qwen2_vl_72b": "There is no text or formula to extract from this image." } ], "image_num": 22, "text_num": 442, "token_num": 13114 }, { "images": [ "sample_100_images/QSzJSPtmsO8@265.85999999999996_287.34#1.jpg", null, "sample_100_images/QSzJSPtmsO8@287.34_308.90000000000003#1.jpg", "sample_100_images/QSzJSPtmsO8@287.34_308.90000000000003#3.jpg", null, "sample_100_images/QSzJSPtmsO8@308.90000000000003_324.92#1.jpg", null, "sample_100_images/QSzJSPtmsO8@324.92_343.16#1.jpg", "sample_100_images/QSzJSPtmsO8@324.92_343.16#3.jpg", null, "sample_100_images/QSzJSPtmsO8@343.16_363.66#1.jpg", "sample_100_images/QSzJSPtmsO8@343.16_363.66#3.jpg", null, "sample_100_images/QSzJSPtmsO8@363.66_388.06#1.jpg", "sample_100_images/QSzJSPtmsO8@363.66_388.06#2.jpg", null, "sample_100_images/QSzJSPtmsO8@388.06_406.08000000000004#1.jpg", null ], "texts": [ null, " Correct. You have many computers counting it for you in parallel. With these two improvements, your job will complete in a few minutes. Correct? What do you think? I believe there is a problem. Okay, what problem? We have multiple counts. Each computer knows its own number.", null, null, " Each computer knows the line count of its local blocks but no one knows the final total number right we want to get the final total count hmm correct but that's an easy problem let's collect all the individual line counts from every computer at one place", null, " We can sum it up to get the final total yeah but we'll be moving data again and we wanted to avoid that right perfect you're getting the sense of it you are correct our aim is to avoid data movement", null, null, " And we avoided moving 20 terabytes of data. Now we're moving just a few numbers. That shouldn't be a problem. Hmm, you are right. Great, I think you've got the core of it. We've broken our line counting problem into two parts.", null, null, " Let me give names to both of these parts. The first part is the data count. This part is called the map function. We execute this part on each computer. In fact, we run a map function for each data block. So if we have 1000 blocks, we run 1000 map functions.", null, null, " One for each block. The map function will simply count the lines in the block. That's it. The second part is called the reduce function. We execute it on a single computer. The reduce function will simply sum up the individual totals and show us the final total.", null, " That's great, but I have some doubts. How does a map function read the lines from a block? Who collects data from each map computer and sends it to the reduce computer? How does it all work? Can you elaborate on this?" ], "text_ocr_list": [ null, "We can see these text from the image: Hadoop Tutorial\n\n1. You are not moving the data.\n2. You have many computers counting it for you in parallel.\n\nLearning Journal - www.Learningjournal.in.\n Correct. You have many computers counting it for you in parallel. With these two improvements, your job will complete in a few minutes. Correct? What do you think? I believe there is a problem. Okay, what problem? We have multiple counts. Each computer knows its own number.", null, null, "We can see these text from the image: Hadoop Tutorial\n\nEach computer knows the line-count of its local blocks.\n\nLearning Journal - www.Learningjournal.in.\n Each computer knows the line count of its local blocks but no one knows the final total number right we want to get the final total count hmm correct but that's an easy problem let's collect all the individual line counts from every computer at one place", null, "We can see these text from the image: Hadoop Tutorial\n\nWe can sum it up to get the final total.\n\nLearning Journal - www.Learningjournal.in.\n We can sum it up to get the final total yeah but we'll be moving data again and we wanted to avoid that right perfect you're getting the sense of it you are correct our aim is to avoid data movement", null, null, "We can see these text from the image: Hadoop Tutorial\n\nNow, we are moving just a few numbers. That shouldn't be a problem.\n\nLearning Journal - www.Learningjournal.in.\n And we avoided moving 20 terabytes of data. Now we're moving just a few numbers. That shouldn't be a problem. Hmm, you are right. Great, I think you've got the core of it. We've broken our line counting problem into two parts.", null, null, "We can see these text from the image: Hadoop Tutorial\n\nWhat is Map Reduce?\n\nMap Function\nReduce Function.\n Let me give names to both of these parts. The first part is the data count. This part is called the map function. We execute this part on each computer. In fact, we run a map function for each data block. So if we have 1000 blocks, we run 1000 map functions.", null, null, "We can see these text from the image: Remember this!\nOne Map function for one block.\n One for each block. The map function will simply count the lines in the block. That's it. The second part is called the reduce function. We execute it on a single computer. The reduce function will simply sum up the individual totals and show us the final total.", null, "We can see these text from the image: Hadoop Tutorial\n\n- How will a Map function read the lines from a block?\n- Who will collect data from each Map computer and send it to the Reduce computer?\n\nLearning Journal - www.Learningjournal.in.\n That's great, but I have some doubts. How does a map function read the lines from a block? Who collects data from each map computer and sends it to the reduce computer? How does it all work? Can you elaborate on this?" ], "metadata": [ { "vid": "QSzJSPtmsO8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on MapReduce in Big Data_30.json#####audio#####doingASR#####FinishASR/QSzJSPtmsO8/265.85999999999996_287.34.mp4", "refined_asr": " Correct. You have many computers counting it for you in parallel. With these two improvements, your job will complete in a few minutes. Correct? What do you think? I believe there is a problem. Okay, what problem? We have multiple counts. Each computer knows its own number.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@265.85999999999996_287.34#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@265.85999999999996_287.34#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@265.85999999999996_287.34#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@265.85999999999996_287.34#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@265.85999999999996_287.34#4.jpg" ], "ocr_qwen2_vl_72b": "Hadoop Tutorial\n\n1. You are not moving the data.\n2. You have many computers counting it for you in parallel.\n\nLearning Journal - www.Learningjournal.in" }, { "vid": "QSzJSPtmsO8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on MapReduce in Big Data_30.json#####audio#####doingASR#####FinishASR/QSzJSPtmsO8/287.34_308.90000000000003.mp4", "refined_asr": " Each computer knows the line count of its local blocks but no one knows the final total number right we want to get the final total count hmm correct but that's an easy problem let's collect all the individual line counts from every computer at one place", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@287.34_308.90000000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@287.34_308.90000000000003#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@287.34_308.90000000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@287.34_308.90000000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@287.34_308.90000000000003#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@287.34_308.90000000000003#4.jpg" ], "ocr_qwen2_vl_72b": "Hadoop Tutorial\n\nEach computer knows the line-count of its local blocks.\n\nLearning Journal - www.Learningjournal.in" }, { "vid": "QSzJSPtmsO8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on MapReduce in Big Data_30.json#####audio#####doingASR#####FinishASR/QSzJSPtmsO8/308.90000000000003_324.92.mp4", "refined_asr": " We can sum it up to get the final total yeah but we'll be moving data again and we wanted to avoid that right perfect you're getting the sense of it you are correct our aim is to avoid data movement", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@308.90000000000003_324.92#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@308.90000000000003_324.92#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@308.90000000000003_324.92#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@308.90000000000003_324.92#3.jpg" ], "ocr_qwen2_vl_72b": "Hadoop Tutorial\n\nWe can sum it up to get the final total.\n\nLearning Journal - www.Learningjournal.in" }, { "vid": "QSzJSPtmsO8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on MapReduce in Big Data_30.json#####audio#####doingASR#####FinishASR/QSzJSPtmsO8/324.92_343.16.mp4", "refined_asr": " And we avoided moving 20 terabytes of data. Now we're moving just a few numbers. That shouldn't be a problem. Hmm, you are right. Great, I think you've got the core of it. We've broken our line counting problem into two parts.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@324.92_343.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@324.92_343.16#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@324.92_343.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@324.92_343.16#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@324.92_343.16#3.jpg" ], "ocr_qwen2_vl_72b": "Hadoop Tutorial\n\nNow, we are moving just a few numbers. That shouldn't be a problem.\n\nLearning Journal - www.Learningjournal.in" }, { "vid": "QSzJSPtmsO8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on MapReduce in Big Data_30.json#####audio#####doingASR#####FinishASR/QSzJSPtmsO8/343.16_363.66.mp4", "refined_asr": " Let me give names to both of these parts. The first part is the data count. This part is called the map function. We execute this part on each computer. In fact, we run a map function for each data block. So if we have 1000 blocks, we run 1000 map functions.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@343.16_363.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@343.16_363.66#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@343.16_363.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@343.16_363.66#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@343.16_363.66#3.jpg" ], "ocr_qwen2_vl_72b": "Hadoop Tutorial\n\nWhat is Map Reduce?\n\nMap Function\nReduce Function" }, { "vid": "QSzJSPtmsO8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on MapReduce in Big Data_30.json#####audio#####doingASR#####FinishASR/QSzJSPtmsO8/363.66_388.06.mp4", "refined_asr": " One for each block. The map function will simply count the lines in the block. That's it. The second part is called the reduce function. We execute it on a single computer. The reduce function will simply sum up the individual totals and show us the final total.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@363.66_388.06#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@363.66_388.06#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@363.66_388.06#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@363.66_388.06#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@363.66_388.06#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@363.66_388.06#4.jpg" ], "ocr_qwen2_vl_72b": "Remember this!\nOne Map function for one block" }, { "vid": "QSzJSPtmsO8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on MapReduce in Big Data_30.json#####audio#####doingASR#####FinishASR/QSzJSPtmsO8/388.06_406.08000000000004.mp4", "refined_asr": " That's great, but I have some doubts. How does a map function read the lines from a block? Who collects data from each map computer and sends it to the reduce computer? How does it all work? Can you elaborate on this?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@388.06_406.08000000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@388.06_406.08000000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@388.06_406.08000000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/QSzJSPtmsO8/QSzJSPtmsO8@388.06_406.08000000000004#3.jpg" ], "ocr_qwen2_vl_72b": "Hadoop Tutorial\n\n- How will a Map function read the lines from a block?\n- Who will collect data from each Map computer and send it to the Reduce computer?\n\nLearning Journal - www.Learningjournal.in" } ], "image_num": 11, "text_num": 401, "token_num": 6737 }, { "images": [ "sample_100_images/KcecJfxbd-4@27221.42_27243.74#1.jpg", "sample_100_images/KcecJfxbd-4@27221.42_27243.74#3.jpg", null, "sample_100_images/KcecJfxbd-4@27243.74_27267.5#1.jpg", null, "sample_100_images/KcecJfxbd-4@27267.5_27285.42#1.jpg", null, "sample_100_images/KcecJfxbd-4@27285.420000000002_27304.940000000002#1.jpg", "sample_100_images/KcecJfxbd-4@27285.420000000002_27304.940000000002#2.jpg", null, "sample_100_images/KcecJfxbd-4@27304.940000000002_27328.7#1.jpg", "sample_100_images/KcecJfxbd-4@27304.940000000002_27328.7#3.jpg", null, "sample_100_images/KcecJfxbd-4@27328.7_27350.54#1.jpg", "sample_100_images/KcecJfxbd-4@27328.7_27350.54#2.jpg", "sample_100_images/KcecJfxbd-4@27328.7_27350.54#3.jpg", null ], "texts": [ null, null, " And if all other processes are running, you will not be able to access the cluster name nodes. Metadata on disk is very important for the name node to come up and maintain the cluster. Name node metadata in RAM is basically for servicing all client requests. Now, when we look at data nodes, as I mentioned, data nodes hold the actual data blocks.", null, " And they're sending these block reports every 10 seconds. So the metadata in the NameNode's RAM is constantly getting updated, and metadata on disk is also constantly getting updated based on any kind of write activity happening on the cluster. Now, the DataNode which is storing the block will also help in any kind of read activity. Whenever a client requests data, the DataNode serves it.", null, " When an application or an API wants to read the data, it would first connect to the NameNode. The NameNode would look into its metadata stored in RAM and confirm to the client which machines could be reached to access that data. That's where your client would try to read the data from the DataNodes.", null, null, " So, which is actually getting the data from DataNodes. And that's how your read-write requests are satisfied. Now, what are the two types of metadata in NameNode? Server holds, as I mentioned earlier, metadata in disk - very important to remember - Edit Log, NFS Image. Metadata in RAM, which is", null, null, " Information about your data nodes files being split into blocks blocks residing on data nodes and file permissions So I will share a very good link on this And you can always look for more detailed information about your metadata So you can search for HDFS metadata directories explained Now this is from Hortonworks however", null, null, null, " It talks about the metadata in disk which the NameNode manages and details about this. So have a look at this link if you're more interested in learning about metadata on disk. Coming back, let's look at the next question: What is the difference between federation and high availability? Now, these are the features which were introduced in Hadoop version 2. Both of" ], "text_ocr_list": [ null, null, "We can see these text from the image: 6\n\nExplain the architecture of HDFS.\n\nNamenode\n\nMetaData (Name, replicas, ...): /home/foo/data, 3, ...\n\nNameNode is the master servers that host metadata in disc and RAM.\nIt holds information about the various datanodes, their location, the size of each block, etc.\n\nMetadata in Disk\nEdit log\nFsimage\n\nMetadata in RAM\nMetadata (Name, replicas, ...): /home/foo/data, 3, ....\n And if all other processes are running, you will not be able to access the cluster name nodes. Metadata on disk is very important for the name node to come up and maintain the cluster. Name node metadata in RAM is basically for servicing all client requests. Now, when we look at data nodes, as I mentioned, data nodes hold the actual data blocks.", null, "We can see these text from the image: 6\n\nExplain the architecture of HDFS.\n\nMetadata ops\n\n- Datanodes hold the actual data blocks and send block reports to Namenode every 10 seconds.\n- Datanode stores and retrieves the blocks when asked by the Namenode. It reads and writes client's request and performs block creation, deletion and replication on instruction from the Namenode.\n And they're sending these block reports every 10 seconds. So the metadata in the NameNode's RAM is constantly getting updated, and metadata on disk is also constantly getting updated based on any kind of write activity happening on the cluster. Now, the DataNode which is storing the block will also help in any kind of read activity. Whenever a client requests data, the DataNode serves it.", null, "We can see these text from the image: 6\n\nExplain the architecture of HDFS.\n\nMetadata ops\n\n- Datanodes hold the actual data blocks and send block reports to Namenode every 10 seconds.\n- Datanode stores and retrieves the blocks when asked by the Namenode. It reads and writes client's request and performs block creation, deletion and replication on instruction from the Namenode\n\nClient\n\nRead\n\nWrite\n\nClient\n\nWrite\n\nRack 1\n\nRack 2.\n When an application or an API wants to read the data, it would first connect to the NameNode. The NameNode would look into its metadata stored in RAM and confirm to the client which machines could be reached to access that data. That's where your client would try to read the data from the DataNodes.", null, null, "We can see these text from the image: 6\n\nExplain the architecture of HDFS.\n\nMetadata ops\n\nNamenode\n\nDatanodes hold the actual data blocks and send block reports to Namenode every 10 seconds.\nDatanode stores and retrieves the blocks when asked by the Namenode. It reads and writes client's request and performs block creation, deletion and replication on instruction from the Namenode\n\nClient\n\nRead\n\nWrite\n\nClient\n\nRack 1\n\nRack 2\n\n\u00a9Simplilearn. All rights reserved..\n So, which is actually getting the data from DataNodes. And that's how your read-write requests are satisfied. Now, what are the two types of metadata in NameNode? Server holds, as I mentioned earlier, metadata in disk - very important to remember - Edit Log, NFS Image. Metadata in RAM, which is", null, null, "We can see these text from the image: 7\n\nWhat are the 2 types of metadata a Namenode server holds?\n\nMetadata in Disk\nEdit log\nFsimage\n\nMetadata in RAM\nMetadata (Name, replicas,...): /home/foo/data, 3, ....\n Information about your data nodes files being split into blocks blocks residing on data nodes and file permissions So I will share a very good link on this And you can always look for more detailed information about your metadata So you can search for HDFS metadata directories explained Now this is from Hortonworks however", null, null, null, "We can see these text from the image: - modifications up to a specific transaction ID.\n- edits \u2013 An edits file is a log that lists each file system change (file creation, deletion or modification) that was made after the most recent fsimage.\n\nCheckpointing is the process of merging the content of the most recent fsimage with all edits applied after that fsimage is merged in order to create a new fsimage. Checkpointing is triggered automatically by configuration policies or manually by HDFS administration commands.\n\n**NAMENODE**\n\nHere is an example of an HDFS metadata directory taken from a NameNode. This shows the output of running the tree command on the metadata directory, which is configured by setting dfs.namenode.name.dir in hdfs-site.xml.\n\n```\ndata/dfs/name\n\u251c\u2500\u2500 current\n\u2502 \u251c\u2500\u2500 VERSION\n\u2502 \u251c\u2500\u2500 edits_0000000000000000001-0000000000000000007\n\u2502 \u251c\u2500\u2500 edits_0000000000000000008-0000000000000000015\n\u2502 \u251c\u2500\u2500 edits_0000000000000000016-0000000000000000022\n\u2502 \u251c\u2500\u2500 edits_0000000000000000023-0000000000000000029\n\u2502 \u251c\u2500\u2500 edits_0000000000000000030-0000000000000000031\n\u2502 \u251c\u2500\u2500 edits_inprogress_0000000000000000032\n\u2502 \u251c\u2500\u2500 fsimage_0000000000000000030\n\u2502 \u251c\u2500\u2500 fsimage_0000000000000000030.md5\n\u2502 \u251c\u2500\u2500 fsimage_0000000000000000031\n\u2502 \u2514\u2500\u2500 fsimage_0000000000000000031.md5\n```.\n It talks about the metadata in disk which the NameNode manages and details about this. So have a look at this link if you're more interested in learning about metadata on disk. Coming back, let's look at the next question: What is the difference between federation and high availability? Now, these are the features which were introduced in Hadoop version 2. Both of" ], "metadata": [ { "vid": "KcecJfxbd-4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on Big Data and Data Warehousing in Computer Science_30.json#####audio#####doingASR#####FinishASR/KcecJfxbd-4/27221.42_27243.74.mp4", "refined_asr": " And if all other processes are running, you will not be able to access the cluster name nodes. Metadata on disk is very important for the name node to come up and maintain the cluster. Name node metadata in RAM is basically for servicing all client requests. Now, when we look at data nodes, as I mentioned, data nodes hold the actual data blocks.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27221.42_27243.74#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27221.42_27243.74#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27221.42_27243.74#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27221.42_27243.74#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27221.42_27243.74#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27221.42_27243.74#4.jpg" ], "ocr_qwen2_vl_72b": "6\n\nExplain the architecture of HDFS.\n\nNamenode\n\nMetaData (Name, replicas, ...): /home/foo/data, 3, ...\n\nNameNode is the master servers that host metadata in disc and RAM.\nIt holds information about the various datanodes, their location, the size of each block, etc.\n\nMetadata in Disk\nEdit log\nFsimage\n\nMetadata in RAM\nMetadata (Name, replicas, ...): /home/foo/data, 3, ..." }, { "vid": "KcecJfxbd-4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on Big Data and Data Warehousing in Computer Science_30.json#####audio#####doingASR#####FinishASR/KcecJfxbd-4/27243.74_27267.5.mp4", "refined_asr": " And they're sending these block reports every 10 seconds. So the metadata in the NameNode's RAM is constantly getting updated, and metadata on disk is also constantly getting updated based on any kind of write activity happening on the cluster. Now, the DataNode which is storing the block will also help in any kind of read activity. Whenever a client requests data, the DataNode serves it.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27243.74_27267.5#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27243.74_27267.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27243.74_27267.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27243.74_27267.5#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27243.74_27267.5#4.jpg" ], "ocr_qwen2_vl_72b": "6\n\nExplain the architecture of HDFS.\n\nMetadata ops\n\n- Datanodes hold the actual data blocks and send block reports to Namenode every 10 seconds.\n- Datanode stores and retrieves the blocks when asked by the Namenode. It reads and writes client's request and performs block creation, deletion and replication on instruction from the Namenode" }, { "vid": "KcecJfxbd-4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on Big Data and Data Warehousing in Computer Science_30.json#####audio#####doingASR#####FinishASR/KcecJfxbd-4/27267.5_27285.42.mp4", "refined_asr": " When an application or an API wants to read the data, it would first connect to the NameNode. The NameNode would look into its metadata stored in RAM and confirm to the client which machines could be reached to access that data. That's where your client would try to read the data from the DataNodes.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27267.5_27285.42#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27267.5_27285.42#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27267.5_27285.42#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27267.5_27285.42#3.jpg" ], "ocr_qwen2_vl_72b": "6\n\nExplain the architecture of HDFS.\n\nMetadata ops\n\n- Datanodes hold the actual data blocks and send block reports to Namenode every 10 seconds.\n- Datanode stores and retrieves the blocks when asked by the Namenode. It reads and writes client's request and performs block creation, deletion and replication on instruction from the Namenode\n\nClient\n\nRead\n\nWrite\n\nClient\n\nWrite\n\nRack 1\n\nRack 2" }, { "vid": "KcecJfxbd-4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on Big Data and Data Warehousing in Computer Science_30.json#####audio#####doingASR#####FinishASR/KcecJfxbd-4/27285.420000000002_27304.940000000002.mp4", "refined_asr": " So, which is actually getting the data from DataNodes. And that's how your read-write requests are satisfied. Now, what are the two types of metadata in NameNode? Server holds, as I mentioned earlier, metadata in disk - very important to remember - Edit Log, NFS Image. Metadata in RAM, which is", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27285.420000000002_27304.940000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27285.420000000002_27304.940000000002#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27285.420000000002_27304.940000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27285.420000000002_27304.940000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27285.420000000002_27304.940000000002#3.jpg" ], "ocr_qwen2_vl_72b": "6\n\nExplain the architecture of HDFS.\n\nMetadata ops\n\nNamenode\n\nDatanodes hold the actual data blocks and send block reports to Namenode every 10 seconds.\nDatanode stores and retrieves the blocks when asked by the Namenode. It reads and writes client's request and performs block creation, deletion and replication on instruction from the Namenode\n\nClient\n\nRead\n\nWrite\n\nClient\n\nRack 1\n\nRack 2\n\n\u00a9Simplilearn. All rights reserved." }, { "vid": "KcecJfxbd-4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on Big Data and Data Warehousing in Computer Science_30.json#####audio#####doingASR#####FinishASR/KcecJfxbd-4/27304.940000000002_27328.7.mp4", "refined_asr": " Information about your data nodes files being split into blocks blocks residing on data nodes and file permissions So I will share a very good link on this And you can always look for more detailed information about your metadata So you can search for HDFS metadata directories explained Now this is from Hortonworks however", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27304.940000000002_27328.7#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27304.940000000002_27328.7#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27304.940000000002_27328.7#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27304.940000000002_27328.7#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27304.940000000002_27328.7#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27304.940000000002_27328.7#4.jpg" ], "ocr_qwen2_vl_72b": "7\n\nWhat are the 2 types of metadata a Namenode server holds?\n\nMetadata in Disk\nEdit log\nFsimage\n\nMetadata in RAM\nMetadata (Name, replicas,...): /home/foo/data, 3, ..." }, { "vid": "KcecJfxbd-4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Advanced database topics tutorial on Big Data and Data Warehousing in Computer Science_30.json#####audio#####doingASR#####FinishASR/KcecJfxbd-4/27328.7_27350.54.mp4", "refined_asr": " It talks about the metadata in disk which the NameNode manages and details about this. So have a look at this link if you're more interested in learning about metadata on disk. Coming back, let's look at the next question: What is the difference between federation and high availability? Now, these are the features which were introduced in Hadoop version 2. Both of", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27328.7_27350.54#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27328.7_27350.54#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27328.7_27350.54#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27328.7_27350.54#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27328.7_27350.54#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27328.7_27350.54#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KcecJfxbd-4/KcecJfxbd-4@27328.7_27350.54#4.jpg" ], "ocr_qwen2_vl_72b": "- modifications up to a specific transaction ID.\n- edits \u2013 An edits file is a log that lists each file system change (file creation, deletion or modification) that was made after the most recent fsimage.\n\nCheckpointing is the process of merging the content of the most recent fsimage with all edits applied after that fsimage is merged in order to create a new fsimage. Checkpointing is triggered automatically by configuration policies or manually by HDFS administration commands.\n\n**NAMENODE**\n\nHere is an example of an HDFS metadata directory taken from a NameNode. This shows the output of running the tree command on the metadata directory, which is configured by setting dfs.namenode.name.dir in hdfs-site.xml.\n\n```\ndata/dfs/name\n\u251c\u2500\u2500 current\n\u2502 \u251c\u2500\u2500 VERSION\n\u2502 \u251c\u2500\u2500 edits_0000000000000000001-0000000000000000007\n\u2502 \u251c\u2500\u2500 edits_0000000000000000008-0000000000000000015\n\u2502 \u251c\u2500\u2500 edits_0000000000000000016-0000000000000000022\n\u2502 \u251c\u2500\u2500 edits_0000000000000000023-0000000000000000029\n\u2502 \u251c\u2500\u2500 edits_0000000000000000030-0000000000000000031\n\u2502 \u251c\u2500\u2500 edits_inprogress_0000000000000000032\n\u2502 \u251c\u2500\u2500 fsimage_0000000000000000030\n\u2502 \u251c\u2500\u2500 fsimage_0000000000000000030.md5\n\u2502 \u251c\u2500\u2500 fsimage_0000000000000000031\n\u2502 \u2514\u2500\u2500 fsimage_0000000000000000031.md5\n```" } ], "image_num": 11, "text_num": 438, "token_num": 6774 }, { "images": [ "sample_100_images/xIJb9BRsLvg@166.66_178.64#1.jpg", "sample_100_images/xIJb9BRsLvg@166.66_178.64#2.jpg", null, "sample_100_images/xIJb9BRsLvg@178.64_205.04#1.jpg", "sample_100_images/xIJb9BRsLvg@178.64_205.04#2.jpg", "sample_100_images/xIJb9BRsLvg@178.64_205.04#3.jpg", null, "sample_100_images/lIE5dEwCZVo@0.96_45.96#1.jpg", "sample_100_images/lIE5dEwCZVo@0.96_45.96#2.jpg", "sample_100_images/lIE5dEwCZVo@0.96_45.96#3.jpg", "sample_100_images/lIE5dEwCZVo@0.96_45.96#4.jpg", null, "sample_100_images/lIE5dEwCZVo@45.96_82.02#1.jpg", null ], "texts": [ null, null, " Wetlands are biologically diverse wonderlands. Are there any other animals in wetlands? Do you know about the world's most diverse wetlands? Are there any other marine animals in wetlands? Are there any wetlands in your area? What plants and animals can be found there?", null, null, null, " How does the wetland change throughout the year? See you later alligator. Subtitles by the Amara.org community.", null, null, null, null, " Consideration of frictionless perfect fluid versus treating viscous or imperfect fluids. A frictionless fluid has no existence in nature, but it is hypothesized by mathematicians in order to facilitate the investigation of important laws and principles that may be approximately true of viscous or natural fluid. If you consider this statement about the fluid", null, "of frictionless perfect fluid. other treating of viscous or imperfect fluids. The frictionless fluid has no existence in nature, but it is hypothesized by mathematicians in order to facilitate the investigation of important laws and principles that may be approximately true of viscous or natural fluid. If you consider this statement that the fluid" ], "text_ocr_list": [ null, null, "We can see these text from the image: There is no text or formula in the image to extract..\n Wetlands are biologically diverse wonderlands. Are there any other animals in wetlands? Do you know about the world's most diverse wetlands? Are there any other marine animals in wetlands? Are there any wetlands in your area? What plants and animals can be found there?", null, null, null, "We can see these text from the image: No text or formulas detected in the image..\n How does the wetland change throughout the year? See you later alligator. Subtitles by the Amara.org community.", null, null, null, null, "We can see these text from the image: NP-TEL\n\nNational Programme\n\non\n\nTechnology Enhanced Learning.\n Consideration of frictionless perfect fluid versus treating viscous or imperfect fluids. A frictionless fluid has no existence in nature, but it is hypothesized by mathematicians in order to facilitate the investigation of important laws and principles that may be approximately true of viscous or natural fluid. If you consider this statement about the fluid", null, "We can see these text from the image: Lecture 6\n\nTheoretical fluid dynamics being a difficult subject is for convenience, commonly divided into two branches, one treating of frictionless perfect fluid, other treating of viscous or imperfect fluids. The frictionless fluid has no existence in nature, but in hypothesized by mathematician in order to facilitate the investigation of important laws and principles that may be approximately true of viscous or natural fluid.\n\nProf. Albert F Zahm, 1912\nFirst aeronautical lab in US universityof frictionless perfect fluid. other treating of viscous or imperfect fluids. The frictionless fluid has no existence in nature, but it is hypothesized by mathematicians in order to facilitate the investigation of important laws and principles that may be approximately true of viscous or natural fluid. If you consider this statement that the fluid" ], "metadata": [ { "vid": "xIJb9BRsLvg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Animal habitats and ecosystems_100.json#####audio#####doingASR#####FinishASR/xIJb9BRsLvg/166.66_178.64.mp4", "refined_asr": " Wetlands are biologically diverse wonderlands. Are there any other animals in wetlands? Do you know about the world's most diverse wetlands? Are there any other marine animals in wetlands? Are there any wetlands in your area? What plants and animals can be found there?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@166.66_178.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@166.66_178.64#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@166.66_178.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@166.66_178.64#2.jpg" ], "ocr_qwen2_vl_72b": "There is no text or formula in the image to extract." }, { "vid": "xIJb9BRsLvg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Animal habitats and ecosystems_100.json#####audio#####doingASR#####FinishASR/xIJb9BRsLvg/178.64_205.04.mp4", "refined_asr": " How does the wetland change throughout the year? See you later alligator. Subtitles by the Amara.org community.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@178.64_205.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@178.64_205.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@178.64_205.04#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@178.64_205.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@178.64_205.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@178.64_205.04#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/xIJb9BRsLvg/xIJb9BRsLvg@178.64_205.04#4.jpg" ], "ocr_qwen2_vl_72b": "No text or formulas detected in the image." }, { "vid": "lIE5dEwCZVo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Aerospace Engineering Advanced Propulsion tutorial on Propellantless Propulsion_30.json#####audio#####doingASR#####FinishASR/lIE5dEwCZVo/0.96_45.96.mp4", "refined_asr": " Consideration of frictionless perfect fluid versus treating viscous or imperfect fluids. A frictionless fluid has no existence in nature, but it is hypothesized by mathematicians in order to facilitate the investigation of important laws and principles that may be approximately true of viscous or natural fluid. If you consider this statement about the fluid", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@0.96_45.96#7.jpg" ], "ocr_qwen2_vl_72b": "NP-TEL\n\nNational Programme\n\non\n\nTechnology Enhanced Learning" }, { "vid": "lIE5dEwCZVo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Aerospace Engineering Advanced Propulsion tutorial on Propellantless Propulsion_30.json#####audio#####doingASR#####FinishASR/lIE5dEwCZVo/45.96_82.02.mp4", "refined_asr": null, "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@45.96_82.02#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@45.96_82.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@45.96_82.02#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@45.96_82.02#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@45.96_82.02#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@45.96_82.02#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lIE5dEwCZVo/lIE5dEwCZVo@45.96_82.02#6.jpg" ], "ocr_qwen2_vl_72b": "Lecture 6\n\nTheoretical fluid dynamics being a difficult subject is for convenience, commonly divided into two branches, one treating of frictionless perfect fluid, other treating of viscous or imperfect fluids. The frictionless fluid has no existence in nature, but in hypothesized by mathematician in order to facilitate the investigation of important laws and principles that may be approximately true of viscous or natural fluid.\n\nProf. Albert F Zahm, 1912\nFirst aeronautical lab in US university" } ], "image_num": 10, "text_num": 238, "token_num": 5998 }, { "images": [ "sample_100_images/AN_3iGiTisM@2129.38_2153.38#1.jpg", null, "sample_100_images/AN_3iGiTisM@2153.38_2177.38#1.jpg", null, "sample_100_images/AN_3iGiTisM@2177.38_2191.38#1.jpg", null, "sample_100_images/AN_3iGiTisM@2191.38_2213.38#1.jpg", "sample_100_images/AN_3iGiTisM@2191.38_2213.38#3.jpg", null, "sample_100_images/AN_3iGiTisM@2213.38_2230.38#1.jpg", "sample_100_images/AN_3iGiTisM@2213.38_2230.38#2.jpg", null, "sample_100_images/AN_3iGiTisM@2230.38_2253.38#1.jpg", null, "sample_100_images/AN_3iGiTisM@2253.38_2271.38#1.jpg", "sample_100_images/AN_3iGiTisM@2253.38_2271.38#2.jpg", null ], "texts": [ null, " In the meantime, I'll give you the takeaway that just measuring the axis ratios doesn't go far enough to differentiate between these models. It looks like you can see here, these blue lines on the right-hand panel are the different runs that we did with different cross-sections of dark matter.", null, " And the variation between them is comparable to these red lines which represent different galaxies simulated with CDM. So the variation from galaxy to galaxy, just by virtue of its formation history, gives a similar scatter in axis ratios and the shape of the dark matter halo.", null, " To the variation induced by fiddling with the dark matter cross-section. So just measuring the cross-section, or rather, the shape alone, is not enough to do it. Which is probably good because here's all the different attempts that I could find at measuring the shape of the Milky Way's halo.", null, null, " And the radii that you see here we're not very good at this yet so it'll be useful to know where to direct our efforts in future. One place that we're looking now is at what happens in terms of the response to a bar evolving in the center of the galaxy.", null, null, " Of a galaxy like this in different dark matter models. So even in CDM, a bar will stir up the dark matter just by virtue of dynamical friction. And we found that several of the simulated galaxies with different models of dark matter form a bar.", null, " And so we can look at how this responds. I'm a student who's working on an analytic model based on this paper by Weinberg, to try to predict what happens to the evolution of the bar's pattern speed if you change the dark matter. And therefore, the phase-based distribution of the dark matter.", null, null, " And therefore the dynamical friction that it exerts on the bar. I'll skip that for now because I want to get to lumpiness. So I'm probably running out of time. Okay. So the last thing I'll talk about here is the way in which" ], "text_ocr_list": [ null, "We can see these text from the image: Our results suggest the response is more complicated\n\nSameie et al 2018\n\nVargya, Sanderson et al in prep.\n In the meantime, I'll give you the takeaway that just measuring the axis ratios doesn't go far enough to differentiate between these models. It looks like you can see here, these blue lines on the right-hand panel are the different runs that we did with different cross-sections of dark matter.", null, "We can see these text from the image: Our results suggest the response is more complicated\n\nSameie et al 2018\n\nVargya, Sanderson et al in prep.\n And the variation between them is comparable to these red lines which represent different galaxies simulated with CDM. So the variation from galaxy to galaxy, just by virtue of its formation history, gives a similar scatter in axis ratios and the shape of the dark matter halo.", null, "We can see these text from the image: Our results suggest the response is more complicated\n\nSameie et al 2018\n\nVargya, Sanderson et al in prep.\n To the variation induced by fiddling with the dark matter cross-section. So just measuring the cross-section, or rather, the shape alone, is not enough to do it. Which is probably good because here's all the different attempts that I could find at measuring the shape of the Milky Way's halo.", null, null, "We can see these text from the image: Our results suggest the response is more complicated\n\nSameie et al 2018\nVargya, Sanderson et al in prep.\n And the radii that you see here we're not very good at this yet so it'll be useful to know where to direct our efforts in future. One place that we're looking now is at what happens in terms of the response to a bar evolving in the center of the galaxy.", null, null, "We can see these text from the image: - Response of halo depends on\n- DM particle's interaction cross section\n- via the DM phase-space distribution function\n\n- Underdense\n- Overdense\n\n- Relative dark matter density (analytically distributed N-body)\n\n- Weinberg & Katz 2007.\n Of a galaxy like this in different dark matter models. So even in CDM, a bar will stir up the dark matter just by virtue of dynamical friction. And we found that several of the simulated galaxies with different models of dark matter form a bar.", null, "We can see these text from the image: - Response of halo depends on\n- DM particle's interaction cross section\n- via the DM phase-space distribution function\n\n- Underdense\n- Overdense\n\n- Relative dark matter density (analytically distributed N-body)\n\n- Weinberg & Katz 2007.\n And so we can look at how this responds. I'm a student who's working on an analytic model based on this paper by Weinberg, to try to predict what happens to the evolution of the bar's pattern speed if you change the dark matter. And therefore, the phase-based distribution of the dark matter.", null, null, "We can see these text from the image: In our simulations, the central DM and stars evolve together\n\nFlattening of the central stellar distribution appears to:\n(1) influence the DM shape more, and to larger radius, at higher \u03c3\n(2) occur earlier for higher \u03c3\n\nVargya, Sanderson et al in prep.\n And therefore the dynamical friction that it exerts on the bar. I'll skip that for now because I want to get to lumpiness. So I'm probably running out of time. Okay. So the last thing I'll talk about here is the way in which" ], "metadata": [ { "vid": "AN_3iGiTisM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Tutorial on Galactic Halo and Dark Matter in the Milky Way Structure and Dynamics_30.json#####audio#####doingASR#####FinishASR/AN_3iGiTisM/2129.38_2153.38.mp4", "refined_asr": " In the meantime, I'll give you the takeaway that just measuring the axis ratios doesn't go far enough to differentiate between these models. It looks like you can see here, these blue lines on the right-hand panel are the different runs that we did with different cross-sections of dark matter.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2129.38_2153.38#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2129.38_2153.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2129.38_2153.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2129.38_2153.38#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2129.38_2153.38#4.jpg" ], "ocr_qwen2_vl_72b": "Our results suggest the response is more complicated\n\nSameie et al 2018\n\nVargya, Sanderson et al in prep" }, { "vid": "AN_3iGiTisM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Tutorial on Galactic Halo and Dark Matter in the Milky Way Structure and Dynamics_30.json#####audio#####doingASR#####FinishASR/AN_3iGiTisM/2153.38_2177.38.mp4", "refined_asr": " And the variation between them is comparable to these red lines which represent different galaxies simulated with CDM. So the variation from galaxy to galaxy, just by virtue of its formation history, gives a similar scatter in axis ratios and the shape of the dark matter halo.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2153.38_2177.38#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2153.38_2177.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2153.38_2177.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2153.38_2177.38#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2153.38_2177.38#4.jpg" ], "ocr_qwen2_vl_72b": "Our results suggest the response is more complicated\n\nSameie et al 2018\n\nVargya, Sanderson et al in prep" }, { "vid": "AN_3iGiTisM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Tutorial on Galactic Halo and Dark Matter in the Milky Way Structure and Dynamics_30.json#####audio#####doingASR#####FinishASR/AN_3iGiTisM/2177.38_2191.38.mp4", "refined_asr": " To the variation induced by fiddling with the dark matter cross-section. So just measuring the cross-section, or rather, the shape alone, is not enough to do it. Which is probably good because here's all the different attempts that I could find at measuring the shape of the Milky Way's halo.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2177.38_2191.38#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2177.38_2191.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2177.38_2191.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2177.38_2191.38#3.jpg" ], "ocr_qwen2_vl_72b": "Our results suggest the response is more complicated\n\nSameie et al 2018\n\nVargya, Sanderson et al in prep" }, { "vid": "AN_3iGiTisM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Tutorial on Galactic Halo and Dark Matter in the Milky Way Structure and Dynamics_30.json#####audio#####doingASR#####FinishASR/AN_3iGiTisM/2191.38_2213.38.mp4", "refined_asr": " And the radii that you see here we're not very good at this yet so it'll be useful to know where to direct our efforts in future. One place that we're looking now is at what happens in terms of the response to a bar evolving in the center of the galaxy.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2191.38_2213.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2191.38_2213.38#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2191.38_2213.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2191.38_2213.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2191.38_2213.38#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2191.38_2213.38#4.jpg" ], "ocr_qwen2_vl_72b": "Our results suggest the response is more complicated\n\nSameie et al 2018\nVargya, Sanderson et al in prep" }, { "vid": "AN_3iGiTisM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Tutorial on Galactic Halo and Dark Matter in the Milky Way Structure and Dynamics_30.json#####audio#####doingASR#####FinishASR/AN_3iGiTisM/2213.38_2230.38.mp4", "refined_asr": " Of a galaxy like this in different dark matter models. So even in CDM, a bar will stir up the dark matter just by virtue of dynamical friction. And we found that several of the simulated galaxies with different models of dark matter form a bar.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2213.38_2230.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2213.38_2230.38#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2213.38_2230.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2213.38_2230.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2213.38_2230.38#3.jpg" ], "ocr_qwen2_vl_72b": "- Response of halo depends on\n- DM particle's interaction cross section\n- via the DM phase-space distribution function\n\n- Underdense\n- Overdense\n\n- Relative dark matter density (analytically distributed N-body)\n\n- Weinberg & Katz 2007" }, { "vid": "AN_3iGiTisM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Tutorial on Galactic Halo and Dark Matter in the Milky Way Structure and Dynamics_30.json#####audio#####doingASR#####FinishASR/AN_3iGiTisM/2230.38_2253.38.mp4", "refined_asr": " And so we can look at how this responds. I'm a student who's working on an analytic model based on this paper by Weinberg, to try to predict what happens to the evolution of the bar's pattern speed if you change the dark matter. And therefore, the phase-based distribution of the dark matter.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2230.38_2253.38#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2230.38_2253.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2230.38_2253.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2230.38_2253.38#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2230.38_2253.38#4.jpg" ], "ocr_qwen2_vl_72b": "- Response of halo depends on\n- DM particle's interaction cross section\n- via the DM phase-space distribution function\n\n- Underdense\n- Overdense\n\n- Relative dark matter density (analytically distributed N-body)\n\n- Weinberg & Katz 2007" }, { "vid": "AN_3iGiTisM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Tutorial on Galactic Halo and Dark Matter in the Milky Way Structure and Dynamics_30.json#####audio#####doingASR#####FinishASR/AN_3iGiTisM/2253.38_2271.38.mp4", "refined_asr": " And therefore the dynamical friction that it exerts on the bar. I'll skip that for now because I want to get to lumpiness. So I'm probably running out of time. Okay. So the last thing I'll talk about here is the way in which", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2253.38_2271.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2253.38_2271.38#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2253.38_2271.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2253.38_2271.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AN_3iGiTisM/AN_3iGiTisM@2253.38_2271.38#3.jpg" ], "ocr_qwen2_vl_72b": "In our simulations, the central DM and stars evolve together\n\nFlattening of the central stellar distribution appears to:\n(1) influence the DM shape more, and to larger radius, at higher \u03c3\n(2) occur earlier for higher \u03c3\n\nVargya, Sanderson et al in prep" } ], "image_num": 10, "text_num": 451, "token_num": 6211 }, { "images": [ "sample_100_images/vY5iXj-dgQg@4314.12_4334.22#1.jpg", null, "sample_100_images/vY5iXj-dgQg@4334.22_4350.22#1.jpg", null, "sample_100_images/vY5iXj-dgQg@4350.22_4366.22#1.jpg", null, "sample_100_images/vY5iXj-dgQg@4366.22_4383.22#1.jpg", "sample_100_images/vY5iXj-dgQg@4366.22_4383.22#2.jpg", null, "sample_100_images/vY5iXj-dgQg@4383.22_4402.22#1.jpg", "sample_100_images/vY5iXj-dgQg@4383.22_4402.22#2.jpg", null, "sample_100_images/vY5iXj-dgQg@4402.22_4423.22#1.jpg", null, "sample_100_images/vY5iXj-dgQg@4423.22_4440.22#1.jpg", null ], "texts": [ null, " If you have a plasma, an ionized plasma of electrons and protons, okay, if the central object is radiating at above the Eddington luminosity, what will happen with this pair? What happens when the luminosity of the central object is above the Eddington luminosity?", null, " Given the concepts that we've just described, what will happen with this pair? It will be like 'Oh no,' okay? So this is what happens, okay.", null, " The importance of this concept is that it's giving you an interesting physical scale for the luminosity relative to the gravity of the central object. Okay, in the sense that you know if the object is shining at above the Eddington luminosity.", null, null, " It will maybe impede further accretion. Okay so this is what this is the main importance. It's a very useful luminosity unit in black hole astrophysics. And the adding to luminosity gives you roughly the maximal luminosity.", null, null, " That can be powered by accretion. Now we define another object, another concept, which is very useful in black hole astrophysics, which is the Eddington accretion rate. So to define the Eddington accretion rate.", null, " We have to assume there is again a central engine which is shining exactly at the Eddington luminosity. Okay, this is our working hypothesis to derive the concept of the Eddington accretion rate. And we ask ourselves the question: What if this central engine were converting the infalling mass?", null, " Stars radiate energy with an efficiency eta let's say it can be 10% - 10% of the infalling mass being converted to luminosity, a mass to luminosity conversion. So the way we frame this is that we write the Eddington luminosity." ], "text_ocr_list": [ null, "We can see these text from the image: Eddington luminosity: importance\n\nL > L_Edd\n\nM\n\nphoton field\n\nICTP | International Centre for Theoretical Physics\n\nSAIFR | South American Institute for Fundamental Research.\n If you have a plasma, an ionized plasma of electrons and protons, okay, if the central object is radiating at above the Eddington luminosity, what will happen with this pair? What happens when the luminosity of the central object is above the Eddington luminosity?", null, "We can see these text from the image: Eddington luminosity: importance\n\nL > L_Edd\n\nphoton field\n\nM\n\nICTP | International Centre for Theoretical Physics\nSAIFR | South American Institute for Fundamental Research.\n Given the concepts that we've just described, what will happen with this pair? It will be like 'Oh no,' okay? So this is what happens, okay.", null, "We can see these text from the image: Eddington luminosity: importance\n\nL > L_Edd\n\nM\n\nphoton field\n\nICTP | International Centre for Theoretical Physics\nSAIFR | South American Institute for Fundamental Research.\n The importance of this concept is that it's giving you an interesting physical scale for the luminosity relative to the gravity of the central object. Okay, in the sense that you know if the object is shining at above the Eddington luminosity.", null, null, "We can see these text from the image: Eddington luminosity: importance\n\nL > L_Edd\n\nM\n\nphoton field\n\nICTP | International Centre for Theoretical Physics\nSAIFR | South American Institute for Fundamental Research.\n It will maybe impede further accretion. Okay so this is what this is the main importance. It's a very useful luminosity unit in black hole astrophysics. And the adding to luminosity gives you roughly the maximal luminosity.", null, null, "We can see these text from the image: Eddington luminosity: importance\n\nUseful luminosity unit in BH astrophysics\n\nA system radiating at \\( L > L_{\\text{Edd}} \\) can halt mass accretion due to strong radiation pressure\n\nRoughly maximal luminosity that can be powered by accretion (if spherical symmetry)\n\nICTP | International Centre for Theoretical Physics\n\nSAIFR | South American Institute for Fundamental Research.\n That can be powered by accretion. Now we define another object, another concept, which is very useful in black hole astrophysics, which is the Eddington accretion rate. So to define the Eddington accretion rate.", null, "We can see these text from the image: Eddington accretion rate\n\n- Assume an engine radiating at \\( L = L_{\\text{Edd}} \\)\n- If it were converting mass to radiative energy with efficiency \\( \\eta \\)\n\nICTP | International Centre for Theoretical Physics\nSAIFR | South American Institute for Fundamental Research.\n We have to assume there is again a central engine which is shining exactly at the Eddington luminosity. Okay, this is our working hypothesis to derive the concept of the Eddington accretion rate. And we ask ourselves the question: What if this central engine were converting the infalling mass?", null, "We can see these text from the image: - Assume an engine radiating at \\( L = L_{\\text{Edd}} \\)\n- If it were converting mass to radiative energy with efficiency \\( \\eta \\).\n Stars radiate energy with an efficiency eta let's say it can be 10% - 10% of the infalling mass being converted to luminosity, a mass to luminosity conversion. So the way we frame this is that we write the Eddington luminosity." ], "metadata": [ { "vid": "vY5iXj-dgQg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics tutorial on Active Galactic Nuclei and Quasars in Galaxy Formation and Evolution_30.json#####audio#####doingASR#####FinishASR/vY5iXj-dgQg/4314.12_4334.22.mp4", "refined_asr": " If you have a plasma, an ionized plasma of electrons and protons, okay, if the central object is radiating at above the Eddington luminosity, what will happen with this pair? What happens when the luminosity of the central object is above the Eddington luminosity?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4314.12_4334.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4314.12_4334.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4314.12_4334.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4314.12_4334.22#3.jpg" ], "ocr_qwen2_vl_72b": "Eddington luminosity: importance\n\nL > L_Edd\n\nM\n\nphoton field\n\nICTP | International Centre for Theoretical Physics\n\nSAIFR | South American Institute for Fundamental Research" }, { "vid": "vY5iXj-dgQg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics tutorial on Active Galactic Nuclei and Quasars in Galaxy Formation and Evolution_30.json#####audio#####doingASR#####FinishASR/vY5iXj-dgQg/4334.22_4350.22.mp4", "refined_asr": " Given the concepts that we've just described, what will happen with this pair? It will be like 'Oh no,' okay? So this is what happens, okay.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4334.22_4350.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4334.22_4350.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4334.22_4350.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4334.22_4350.22#3.jpg" ], "ocr_qwen2_vl_72b": "Eddington luminosity: importance\n\nL > L_Edd\n\nphoton field\n\nM\n\nICTP | International Centre for Theoretical Physics\nSAIFR | South American Institute for Fundamental Research" }, { "vid": "vY5iXj-dgQg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics tutorial on Active Galactic Nuclei and Quasars in Galaxy Formation and Evolution_30.json#####audio#####doingASR#####FinishASR/vY5iXj-dgQg/4350.22_4366.22.mp4", "refined_asr": " The importance of this concept is that it's giving you an interesting physical scale for the luminosity relative to the gravity of the central object. Okay, in the sense that you know if the object is shining at above the Eddington luminosity.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4350.22_4366.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4350.22_4366.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4350.22_4366.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4350.22_4366.22#3.jpg" ], "ocr_qwen2_vl_72b": "Eddington luminosity: importance\n\nL > L_Edd\n\nM\n\nphoton field\n\nICTP | International Centre for Theoretical Physics\nSAIFR | South American Institute for Fundamental Research" }, { "vid": "vY5iXj-dgQg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics tutorial on Active Galactic Nuclei and Quasars in Galaxy Formation and Evolution_30.json#####audio#####doingASR#####FinishASR/vY5iXj-dgQg/4366.22_4383.22.mp4", "refined_asr": " It will maybe impede further accretion. Okay so this is what this is the main importance. It's a very useful luminosity unit in black hole astrophysics. And the adding to luminosity gives you roughly the maximal luminosity.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4366.22_4383.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4366.22_4383.22#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4366.22_4383.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4366.22_4383.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4366.22_4383.22#3.jpg" ], "ocr_qwen2_vl_72b": "Eddington luminosity: importance\n\nL > L_Edd\n\nM\n\nphoton field\n\nICTP | International Centre for Theoretical Physics\nSAIFR | South American Institute for Fundamental Research" }, { "vid": "vY5iXj-dgQg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics tutorial on Active Galactic Nuclei and Quasars in Galaxy Formation and Evolution_30.json#####audio#####doingASR#####FinishASR/vY5iXj-dgQg/4383.22_4402.22.mp4", "refined_asr": " That can be powered by accretion. Now we define another object, another concept, which is very useful in black hole astrophysics, which is the Eddington accretion rate. So to define the Eddington accretion rate.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4383.22_4402.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4383.22_4402.22#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4383.22_4402.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4383.22_4402.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4383.22_4402.22#3.jpg" ], "ocr_qwen2_vl_72b": "Eddington luminosity: importance\n\nUseful luminosity unit in BH astrophysics\n\nA system radiating at \\( L > L_{\\text{Edd}} \\) can halt mass accretion due to strong radiation pressure\n\nRoughly maximal luminosity that can be powered by accretion (if spherical symmetry)\n\nICTP | International Centre for Theoretical Physics\n\nSAIFR | South American Institute for Fundamental Research" }, { "vid": "vY5iXj-dgQg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics tutorial on Active Galactic Nuclei and Quasars in Galaxy Formation and Evolution_30.json#####audio#####doingASR#####FinishASR/vY5iXj-dgQg/4402.22_4423.22.mp4", "refined_asr": " We have to assume there is again a central engine which is shining exactly at the Eddington luminosity. Okay, this is our working hypothesis to derive the concept of the Eddington accretion rate. And we ask ourselves the question: What if this central engine were converting the infalling mass?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4402.22_4423.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4402.22_4423.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4402.22_4423.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4402.22_4423.22#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4402.22_4423.22#4.jpg" ], "ocr_qwen2_vl_72b": "Eddington accretion rate\n\n- Assume an engine radiating at \\( L = L_{\\text{Edd}} \\)\n- If it were converting mass to radiative energy with efficiency \\( \\eta \\)\n\nICTP | International Centre for Theoretical Physics\nSAIFR | South American Institute for Fundamental Research" }, { "vid": "vY5iXj-dgQg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics tutorial on Active Galactic Nuclei and Quasars in Galaxy Formation and Evolution_30.json#####audio#####doingASR#####FinishASR/vY5iXj-dgQg/4423.22_4440.22.mp4", "refined_asr": " Stars radiate energy with an efficiency eta let's say it can be 10% - 10% of the infalling mass being converted to luminosity, a mass to luminosity conversion. So the way we frame this is that we write the Eddington luminosity.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4423.22_4440.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4423.22_4440.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4423.22_4440.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vY5iXj-dgQg/vY5iXj-dgQg@4423.22_4440.22#3.jpg" ], "ocr_qwen2_vl_72b": "- Assume an engine radiating at \\( L = L_{\\text{Edd}} \\)\n- If it were converting mass to radiative energy with efficiency \\( \\eta \\)" } ], "image_num": 9, "text_num": 387, "token_num": 5571 }, { "images": [ "sample_100_images/ZbXZN3k1j9U@401.36_417.14000000000004#1.jpg", "sample_100_images/ZbXZN3k1j9U@401.36_417.14000000000004#2.jpg", null, "sample_100_images/vlSOESXQI7o@0.0_21.7#1.jpg", "sample_100_images/vlSOESXQI7o@0.0_21.7#2.jpg", "sample_100_images/vlSOESXQI7o@0.0_21.7#3.jpg", null, "sample_100_images/vlSOESXQI7o@21.96_37.84#1.jpg", "sample_100_images/vlSOESXQI7o@21.96_37.84#2.jpg", null, "sample_100_images/vlSOESXQI7o@38.3_52.18#1.jpg", null, "sample_100_images/vlSOESXQI7o@52.42_65.06#1.jpg", "sample_100_images/vlSOESXQI7o@52.42_65.06#2.jpg", null, "sample_100_images/vlSOESXQI7o@65.96_85.06#1.jpg", "sample_100_images/vlSOESXQI7o@65.96_85.06#2.jpg", "sample_100_images/vlSOESXQI7o@65.96_85.06#3.jpg", null, "sample_100_images/vlSOESXQI7o@85.4_100.14000000000001#1.jpg", "sample_100_images/vlSOESXQI7o@85.4_100.14000000000001#2.jpg", null, "sample_100_images/vlSOESXQI7o@100.56_119.76#1.jpg", "sample_100_images/vlSOESXQI7o@100.56_119.76#2.jpg", "sample_100_images/vlSOESXQI7o@100.56_119.76#3.jpg", null ], "texts": [ null, null, " And that means that in order for such a reaction to actually take place, energy must be inputted into our system. And such a reaction is known as an endothermic or an endoergic nuclear reaction.", null, null, null, " But I am just an imaginary picture of an atom. Real atoms are too tiny to be seen with the naked eye; you can only see them under a very special kind of microscope. Actually, this high-powered microscope can be considered a machine. Okay, now you know that we are so small that we can only be seen with the aid of such machines.", null, null, " Now I have a question for you. Do you know the meaning of your name? Very good if you do. If not, ask your parents. Interestingly, I know the meaning of my name. It means 'that which cannot be cut' or 'that which is uncuttable.'", null, " Actually, it's a Greek word which means it is indivisible. Indivisible, which means it can't be divided further or it can't be cut further. Let me show you something. Let me ask this man to cut the apple in his hand in half.", null, null, " Now do it again. Please do it again. Keep doing it until you can't cut it any further. Yes, this is the smallest part which you can't cut any further. But it can also be cut under a microscopic machine. So if you get a part so small that it can't be divided further, it is called indivisible.", null, null, null, " It can't be cut further. It is uncuttable. It is an atom. Let me tell you a secret: everything is made up of atoms. You, this water, this table, this blackboard, this fan, these fruits, everything.", null, null, " Everything is made up of atoms. You're right. Everything is made of molecules but all molecules are made of atoms. Confused? Go to the blackboard and write any sentence. Good. Now this is a whole sentence. Now circle all of the words in this sentence.", null, null, null, "Everything is made up of atoms. Oh, you're right too. Everything is made of molecules, But all molecules are made of atoms. Confused, Go to the blackboard and write any sentence. Good, Now, this is a whole sentence. Now circle all of the words in this sentence." ], "text_ocr_list": [ null, null, "We can see these text from the image: - Nuclear Reactions\n - A nuclear reaction is a reaction taking place between two nuclei or between some smaller particle (\u03b1, \u03b2, ...) and the nucleus of an atom. Such reactions tend to cause transmutations.\n - A transmutation is a reaction in which the parent nucleus is different from the daughter nucleus.\n - Example: Consider the nuclear reaction that takes place when an alpha particle smashes into the nucleus of nitrogen:\n \\[\n _2^4\\alpha + _7^{14}N \\rightarrow _8^{17}O + _1^1H\n \\]\n - Reaction Energy\n - Just like any reaction, nuclear reactions can be exothermic (exergonic) or endothermic (endergonic).\n - Exothermic (exergonic): nuclear reactions release energy.\n - Endothermic (endergonic): nuclear reactions require input to actually take place.\n\n- General Equation for Nuclear Reactions:\n \\[\n a + X \\rightarrow X' + b\n \\]\n - \\(a\\): some small nucleus or particle\n - \\(X\\): parent nucleus or atom\n - \\(X'\\): daughter nucleus or atom\n - \\(b\\): particle or nucleus emitted\n\n- Reaction Energy:\n \\[\n Q = (m_a + m_X - m_b - m_{X'})c^2\n \\]\n - \\(Q\\) is known as the reaction energy or Q-value.\n - If energy is conserved, we see that \\(Q\\) represents energy during the nuclear reaction.\n - \\(K_{initial} = (K_X + K_b) - (K_x + K_a)\\)\n\n- AK LECTURES.\n And that means that in order for such a reaction to actually take place, energy must be inputted into our system. And such a reaction is known as an endothermic or an endoergic nuclear reaction.", null, null, null, "We can see these text from the image: No text or formulas present in the image..\n But I am just an imaginary picture of an atom. Real atoms are too tiny to be seen with the naked eye; you can only see them under a very special kind of microscope. Actually, this high-powered microscope can be considered a machine. Okay, now you know that we are so small that we can only be seen with the aid of such machines.", null, null, "We can see these text from the image: visit www.makemegenius.com for more science videos.\n Now I have a question for you. Do you know the meaning of your name? Very good if you do. If not, ask your parents. Interestingly, I know the meaning of my name. It means 'that which cannot be cut' or 'that which is uncuttable.'", null, "We can see these text from the image: visit www.makemegenius.com for more science videos.\n Actually, it's a Greek word which means it is indivisible. Indivisible, which means it can't be divided further or it can't be cut further. Let me show you something. Let me ask this man to cut the apple in his hand in half.", null, null, "We can see these text from the image: Atom means indivisible.\n Now do it again. Please do it again. Keep doing it until you can't cut it any further. Yes, this is the smallest part which you can't cut any further. But it can also be cut under a microscopic machine. So if you get a part so small that it can't be divided further, it is called indivisible.", null, null, null, "We can see these text from the image: visit www.makemegenius.com for more science videos.\n It can't be cut further. It is uncuttable. It is an atom. Let me tell you a secret: everything is made up of atoms. You, this water, this table, this blackboard, this fan, these fruits, everything.", null, null, "We can see these text from the image: Atom means indivisible.\n Everything is made up of atoms. You're right. Everything is made of molecules but all molecules are made of atoms. Confused? Go to the blackboard and write any sentence. Good. Now this is a whole sentence. Now circle all of the words in this sentence.", null, null, null, "We can see these text from the image: Everything is made of Molecules\n\nNOEverything is made up of atoms. Oh, you're right too. Everything is made of molecules, But all molecules are made of atoms. Confused, Go to the blackboard and write any sentence. Good, Now, this is a whole sentence. Now circle all of the words in this sentence." ], "metadata": [ { "vid": "ZbXZN3k1j9U.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lecture on Nuclear Reactions in Nuclear Reactors._30.json#####audio#####doingASR#####FinishASR/ZbXZN3k1j9U/401.36_417.14000000000004.mp4", "refined_asr": " And that means that in order for such a reaction to actually take place, energy must be inputted into our system. And such a reaction is known as an endothermic or an endoergic nuclear reaction.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZbXZN3k1j9U/ZbXZN3k1j9U@401.36_417.14000000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZbXZN3k1j9U/ZbXZN3k1j9U@401.36_417.14000000000004#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZbXZN3k1j9U/ZbXZN3k1j9U@401.36_417.14000000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZbXZN3k1j9U/ZbXZN3k1j9U@401.36_417.14000000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZbXZN3k1j9U/ZbXZN3k1j9U@401.36_417.14000000000004#3.jpg" ], "ocr_qwen2_vl_72b": "- Nuclear Reactions\n - A nuclear reaction is a reaction taking place between two nuclei or between some smaller particle (\u03b1, \u03b2, ...) and the nucleus of an atom. Such reactions tend to cause transmutations.\n - A transmutation is a reaction in which the parent nucleus is different from the daughter nucleus.\n - Example: Consider the nuclear reaction that takes place when an alpha particle smashes into the nucleus of nitrogen:\n \\[\n _2^4\\alpha + _7^{14}N \\rightarrow _8^{17}O + _1^1H\n \\]\n - Reaction Energy\n - Just like any reaction, nuclear reactions can be exothermic (exergonic) or endothermic (endergonic).\n - Exothermic (exergonic): nuclear reactions release energy.\n - Endothermic (endergonic): nuclear reactions require input to actually take place.\n\n- General Equation for Nuclear Reactions:\n \\[\n a + X \\rightarrow X' + b\n \\]\n - \\(a\\): some small nucleus or particle\n - \\(X\\): parent nucleus or atom\n - \\(X'\\): daughter nucleus or atom\n - \\(b\\): particle or nucleus emitted\n\n- Reaction Energy:\n \\[\n Q = (m_a + m_X - m_b - m_{X'})c^2\n \\]\n - \\(Q\\) is known as the reaction energy or Q-value.\n - If energy is conserved, we see that \\(Q\\) represents energy during the nuclear reaction.\n - \\(K_{initial} = (K_X + K_b) - (K_x + K_a)\\)\n\n- AK LECTURES" }, { "vid": "vlSOESXQI7o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atoms and molecules for kids_100.json#####audio#####doingASR#####FinishASR/vlSOESXQI7o/0.0_21.7.mp4", "refined_asr": " But I am just an imaginary picture of an atom. Real atoms are too tiny to be seen with the naked eye; you can only see them under a very special kind of microscope. Actually, this high-powered microscope can be considered a machine. Okay, now you know that we are so small that we can only be seen with the aid of such machines.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@0.0_21.7#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@0.0_21.7#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@0.0_21.7#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@0.0_21.7#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@0.0_21.7#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@0.0_21.7#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@0.0_21.7#4.jpg" ], "ocr_qwen2_vl_72b": "No text or formulas present in the image." }, { "vid": "vlSOESXQI7o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atoms and molecules for kids_100.json#####audio#####doingASR#####FinishASR/vlSOESXQI7o/21.96_37.84.mp4", "refined_asr": " Now I have a question for you. Do you know the meaning of your name? Very good if you do. If not, ask your parents. Interestingly, I know the meaning of my name. It means 'that which cannot be cut' or 'that which is uncuttable.'", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@21.96_37.84#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@21.96_37.84#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@21.96_37.84#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@21.96_37.84#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@21.96_37.84#3.jpg" ], "ocr_qwen2_vl_72b": "visit www.makemegenius.com for more science videos" }, { "vid": "vlSOESXQI7o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atoms and molecules for kids_100.json#####audio#####doingASR#####FinishASR/vlSOESXQI7o/38.3_52.18.mp4", "refined_asr": " Actually, it's a Greek word which means it is indivisible. Indivisible, which means it can't be divided further or it can't be cut further. Let me show you something. Let me ask this man to cut the apple in his hand in half.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@38.3_52.18#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@38.3_52.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@38.3_52.18#2.jpg" ], "ocr_qwen2_vl_72b": "visit www.makemegenius.com for more science videos" }, { "vid": "vlSOESXQI7o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atoms and molecules for kids_100.json#####audio#####doingASR#####FinishASR/vlSOESXQI7o/52.42_65.06.mp4", "refined_asr": " Now do it again. Please do it again. Keep doing it until you can't cut it any further. Yes, this is the smallest part which you can't cut any further. But it can also be cut under a microscopic machine. So if you get a part so small that it can't be divided further, it is called indivisible.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@52.42_65.06#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@52.42_65.06#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@52.42_65.06#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@52.42_65.06#2.jpg" ], "ocr_qwen2_vl_72b": "Atom means indivisible" }, { "vid": "vlSOESXQI7o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atoms and molecules for kids_100.json#####audio#####doingASR#####FinishASR/vlSOESXQI7o/65.96_85.06.mp4", "refined_asr": " It can't be cut further. It is uncuttable. It is an atom. Let me tell you a secret: everything is made up of atoms. You, this water, this table, this blackboard, this fan, these fruits, everything.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@65.96_85.06#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@65.96_85.06#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@65.96_85.06#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@65.96_85.06#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@65.96_85.06#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@65.96_85.06#3.jpg" ], "ocr_qwen2_vl_72b": "visit www.makemegenius.com for more science videos" }, { "vid": "vlSOESXQI7o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atoms and molecules for kids_100.json#####audio#####doingASR#####FinishASR/vlSOESXQI7o/85.4_100.14000000000001.mp4", "refined_asr": " Everything is made up of atoms. You're right. Everything is made of molecules but all molecules are made of atoms. Confused? Go to the blackboard and write any sentence. Good. Now this is a whole sentence. Now circle all of the words in this sentence.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@85.4_100.14000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@85.4_100.14000000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@85.4_100.14000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@85.4_100.14000000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@85.4_100.14000000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Atom means indivisible" }, { "vid": "vlSOESXQI7o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atoms and molecules for kids_100.json#####audio#####doingASR#####FinishASR/vlSOESXQI7o/100.56_119.76.mp4", "refined_asr": null, "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@100.56_119.76#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@100.56_119.76#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@100.56_119.76#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@100.56_119.76#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@100.56_119.76#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/vlSOESXQI7o/vlSOESXQI7o@100.56_119.76#3.jpg" ], "ocr_qwen2_vl_72b": "Everything is made of Molecules\n\nNO" } ], "image_num": 18, "text_num": 506, "token_num": 10874 }, { "images": [ "sample_100_images/KUnsabZDGzU@168.26000000000002_184.5#1.jpg", null, "sample_100_images/KUnsabZDGzU@184.98000000000002_204.72#1.jpg", null, "sample_100_images/KUnsabZDGzU@204.84_220.16#1.jpg", null, "sample_100_images/KUnsabZDGzU@220.38_240.04000000000002#1.jpg", null, "sample_100_images/KUnsabZDGzU@240.82000000000002_258.82#1.jpg", null, "sample_100_images/KUnsabZDGzU@259.18_282.98#1.jpg", null ], "texts": [ null, " So M equals twenty-six. And we'll plug back in to see if that's true. So fifteen plus eleven equals twenty-six. So fifteen equals twenty-six minus eleven, which is true. And you can think of these equations like balancing a scale.", null, " Both sides need to be exactly equal. So on the left, we have fifteen. And twenty-six minus eleven equals fifteen as well. So both sides are exactly equal. On to number three, where we have G plus thirteen equals nineteen. So we need to isolate the G.", null, " The variable we are adding thirteen to. So the opposite of adding thirteen would be subtracting thirteen. That way, those thirteens cancel out. But remember, whatever we do to one side, we have to do to the other. So, subtract thirteen from the right.", null, " That's nineteen over there. G is now isolated. And we have nineteen minus thirteen, which gives us six. So G equals six. And we'll plug back in: six plus thirteen equals nineteen. So that's true, and we have the correct answer.", null, " On to number four where we have twenty-six equals Y plus eight. So we're adding eight to the variable Y. We need to use the inverse or opposite operation. So subtract eight that way we isolate Y. Whatever we do to one side we do to the other.", null, " So let's subtract eight from the left as well. Y is now isolated. And twenty-six minus eight equals eighteen. So Y equals eighteen. Let's plug back in and see if this is true. Eighteen plus eight is twenty-six. So we are good to go with number four." ], "text_ocr_list": [ null, "We can see these text from the image: Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n\n4) 26 = y + 8.\n So M equals twenty-six. And we'll plug back in to see if that's true. So fifteen plus eleven equals twenty-six. So fifteen equals twenty-six minus eleven, which is true. And you can think of these equations like balancing a scale.", null, "We can see these text from the image: Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n\n4) 26 = y + 8.\n Both sides need to be exactly equal. So on the left, we have fifteen. And twenty-six minus eleven equals fifteen as well. So both sides are exactly equal. On to number three, where we have G plus thirteen equals nineteen. So we need to isolate the G.", null, "We can see these text from the image: Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n\n4) 26 = y + 8.\n The variable we are adding thirteen to. So the opposite of adding thirteen would be subtracting thirteen. That way, those thirteens cancel out. But remember, whatever we do to one side, we have to do to the other. So, subtract thirteen from the right.", null, "We can see these text from the image: Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n -13 -13\n g = 6\n\n4) 26 = y + 8\n -8 -8\n 18 = y.\n That's nineteen over there. G is now isolated. And we have nineteen minus thirteen, which gives us six. So G equals six. And we'll plug back in: six plus thirteen equals nineteen. So that's true, and we have the correct answer.", null, "We can see these text from the image: Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n -13 -13\n g = 6\n\n4) 26 = y + 8.\n On to number four where we have twenty-six equals Y plus eight. So we're adding eight to the variable Y. We need to use the inverse or opposite operation. So subtract eight that way we isolate Y. Whatever we do to one side we do to the other.", null, "We can see these text from the image: Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n -13 -13\n g = 6\n\n4) 26 = y + 8\n -8 -8.\n So let's subtract eight from the left as well. Y is now isolated. And twenty-six minus eight equals eighteen. So Y equals eighteen. Let's plug back in and see if this is true. Eighteen plus eight is twenty-six. So we are good to go with number four." ], "metadata": [ { "vid": "KUnsabZDGzU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra Tutorial on Solving One-Step Equations_30.json#####audio#####doingASR#####FinishASR/KUnsabZDGzU/168.26000000000002_184.5.mp4", "refined_asr": " So M equals twenty-six. And we'll plug back in to see if that's true. So fifteen plus eleven equals twenty-six. So fifteen equals twenty-six minus eleven, which is true. And you can think of these equations like balancing a scale.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@168.26000000000002_184.5#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@168.26000000000002_184.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@168.26000000000002_184.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@168.26000000000002_184.5#3.jpg" ], "ocr_qwen2_vl_72b": "Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n\n4) 26 = y + 8" }, { "vid": "KUnsabZDGzU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra Tutorial on Solving One-Step Equations_30.json#####audio#####doingASR#####FinishASR/KUnsabZDGzU/184.98000000000002_204.72.mp4", "refined_asr": " Both sides need to be exactly equal. So on the left, we have fifteen. And twenty-six minus eleven equals fifteen as well. So both sides are exactly equal. On to number three, where we have G plus thirteen equals nineteen. So we need to isolate the G.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@184.98000000000002_204.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@184.98000000000002_204.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@184.98000000000002_204.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@184.98000000000002_204.72#3.jpg" ], "ocr_qwen2_vl_72b": "Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n\n4) 26 = y + 8" }, { "vid": "KUnsabZDGzU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra Tutorial on Solving One-Step Equations_30.json#####audio#####doingASR#####FinishASR/KUnsabZDGzU/204.84_220.16.mp4", "refined_asr": " The variable we are adding thirteen to. So the opposite of adding thirteen would be subtracting thirteen. That way, those thirteens cancel out. But remember, whatever we do to one side, we have to do to the other. So, subtract thirteen from the right.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@204.84_220.16#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@204.84_220.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@204.84_220.16#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@204.84_220.16#3.jpg" ], "ocr_qwen2_vl_72b": "Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n\n4) 26 = y + 8" }, { "vid": "KUnsabZDGzU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra Tutorial on Solving One-Step Equations_30.json#####audio#####doingASR#####FinishASR/KUnsabZDGzU/220.38_240.04000000000002.mp4", "refined_asr": " That's nineteen over there. G is now isolated. And we have nineteen minus thirteen, which gives us six. So G equals six. And we'll plug back in: six plus thirteen equals nineteen. So that's true, and we have the correct answer.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@220.38_240.04000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@220.38_240.04000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@220.38_240.04000000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@220.38_240.04000000000002#3.jpg" ], "ocr_qwen2_vl_72b": "Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n -13 -13\n g = 6\n\n4) 26 = y + 8\n -8 -8\n 18 = y" }, { "vid": "KUnsabZDGzU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra Tutorial on Solving One-Step Equations_30.json#####audio#####doingASR#####FinishASR/KUnsabZDGzU/240.82000000000002_258.82.mp4", "refined_asr": " On to number four where we have twenty-six equals Y plus eight. So we're adding eight to the variable Y. We need to use the inverse or opposite operation. So subtract eight that way we isolate Y. Whatever we do to one side we do to the other.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@240.82000000000002_258.82#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@240.82000000000002_258.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@240.82000000000002_258.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@240.82000000000002_258.82#3.jpg" ], "ocr_qwen2_vl_72b": "Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n -13 -13\n g = 6\n\n4) 26 = y + 8" }, { "vid": "KUnsabZDGzU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra Tutorial on Solving One-Step Equations_30.json#####audio#####doingASR#####FinishASR/KUnsabZDGzU/259.18_282.98.mp4", "refined_asr": " So let's subtract eight from the left as well. Y is now isolated. And twenty-six minus eight equals eighteen. So Y equals eighteen. Let's plug back in and see if this is true. Eighteen plus eight is twenty-six. So we are good to go with number four.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@259.18_282.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@259.18_282.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@259.18_282.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@259.18_282.98#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/KUnsabZDGzU/KUnsabZDGzU@259.18_282.98#4.jpg" ], "ocr_qwen2_vl_72b": "Addition & Subtraction One-Step Equations\n\nIsolate the variable (get it by itself) by using the inverse operation. Remember, BALANCE.\n\n1) x - 4 = 7\n +4 +4\n x = 11\n\n2) 15 = m - 11\n +11 +11\n 26 = m\n\n3) g + 13 = 19\n -13 -13\n g = 6\n\n4) 26 = y + 8\n -8 -8" } ], "image_num": 6, "text_num": 381, "token_num": 3837 }, { "images": [ "sample_100_images/LpyzdO2fXtA@1150.72_1166.72#1.jpg", null, "sample_100_images/LpyzdO2fXtA@1166.72_1182.72#1.jpg", null, "sample_100_images/LpyzdO2fXtA@1182.72_1200.72#1.jpg", null, "sample_100_images/LpyzdO2fXtA@1200.72_1218.72#1.jpg", null, "sample_100_images/LpyzdO2fXtA@1218.72_1238.72#1.jpg", null, "sample_100_images/LpyzdO2fXtA@1238.72_1260.72#1.jpg", null, "sample_100_images/LpyzdO2fXtA@1260.72_1278.72#1.jpg", null, "sample_100_images/LpyzdO2fXtA@1278.72_1298.72#1.jpg", null ], "texts": [ null, " Divide by 2 to get the radius. Because remember, the radius is half of the diameter. 12 divided by 2 gives us a radius of 6 feet. Let's plug that in. That's going to give us 6 squared.", null, " Now 6 squared, that means 6 times 6. Not 6 times 2. When we square a number, we multiply that number by itself. So let's do 6 squared, which gives us 6 times 6.", null, " 36. So we end up with pi times 36. Now remember, just like in our first example, one way to write our area is to write it in terms of pi. And again, that means we can write our number.", null, " In front of the pi symbol and then put our unit of measure so we could put thirty-six pi and this is square feet for our unit of measure so that's one way we can write out our area of that circle", null, " Another option is we can use an approximate rounded version of pi. As I mentioned earlier, we're going to use 3.14. So we can use this to get an approximate area of this circle in decimal form. Again, we're going to use pi.", null, " Pi is approximately 3.14. Let's write out our formula, which is: area equals pi r squared. And then we can plug in: so area is approximately... I'm going to use the approximate symbol there because, again, we're using an approximation.", null, " We're using an approximation of pi, so let's plug in 3.14. Multiply that by our radius, which is 6, squared. Now we can calculate. First, let's do 6 squared.", null, " So 3.14 times 6 squared which equals 36. And now let's multiply. So 3.14 times 36 gives us an approximate area of 113. And 4 hundredths. And this is" ], "text_ocr_list": [ null, "We can see these text from the image: Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\nA = \u03c0r\u00b2\nA = \u03c0 x.\n Divide by 2 to get the radius. Because remember, the radius is half of the diameter. 12 divided by 2 gives us a radius of 6 feet. Let's plug that in. That's going to give us 6 squared.", null, "We can see these text from the image: Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2.\n Now 6 squared, that means 6 times 6. Not 6 times 2. When we square a number, we multiply that number by itself. So let's do 6 squared, which gives us 6 times 6.", null, "We can see these text from the image: Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36.\n 36. So we end up with pi times 36. Now remember, just like in our first example, one way to write our area is to write it in terms of pi. And again, that means we can write our number.", null, "We can see these text from the image: Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36\nA =.\n In front of the pi symbol and then put our unit of measure so we could put thirty-six pi and this is square feet for our unit of measure so that's one way we can write out our area of that circle", null, "We can see these text from the image: Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36\nA = 36\u03c0 ft\u00b2.\n Another option is we can use an approximate rounded version of pi. As I mentioned earlier, we're going to use 3.14. So we can use this to get an approximate area of this circle in decimal form. Again, we're going to use pi.", null, "We can see these text from the image: Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\n\u03c0 \u2248 3.\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36\nA = 36\u03c0 ft\u00b2.\n Pi is approximately 3.14. Let's write out our formula, which is: area equals pi r squared. And then we can plug in: so area is approximately... I'm going to use the approximate symbol there because, again, we're using an approximation.", null, "We can see these text from the image: Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\n\u03c0 \u2248 3.14\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36\nA = 36\u03c0 ft\u00b2.\n We're using an approximation of pi, so let's plug in 3.14. Multiply that by our radius, which is 6, squared. Now we can calculate. First, let's do 6 squared.", null, "We can see these text from the image: Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\n\u03c0 \u2248 3.14\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36\nA = 36\u03c0 ft\u00b2.\n So 3.14 times 6 squared which equals 36. And now let's multiply. So 3.14 times 36 gives us an approximate area of 113. And 4 hundredths. And this is" ], "metadata": [ { "vid": "LpyzdO2fXtA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Area of Rectangles and Squares in Elementary Arithmetic Math Course tutorial_30.json#####audio#####doingASR#####FinishASR/LpyzdO2fXtA/1150.72_1166.72.mp4", "refined_asr": " Divide by 2 to get the radius. Because remember, the radius is half of the diameter. 12 divided by 2 gives us a radius of 6 feet. Let's plug that in. That's going to give us 6 squared.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1150.72_1166.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1150.72_1166.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1150.72_1166.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1150.72_1166.72#3.jpg" ], "ocr_qwen2_vl_72b": "Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\nA = \u03c0r\u00b2\nA = \u03c0 x" }, { "vid": "LpyzdO2fXtA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Area of Rectangles and Squares in Elementary Arithmetic Math Course tutorial_30.json#####audio#####doingASR#####FinishASR/LpyzdO2fXtA/1166.72_1182.72.mp4", "refined_asr": " Now 6 squared, that means 6 times 6. Not 6 times 2. When we square a number, we multiply that number by itself. So let's do 6 squared, which gives us 6 times 6.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1166.72_1182.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1166.72_1182.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1166.72_1182.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1166.72_1182.72#3.jpg" ], "ocr_qwen2_vl_72b": "Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2" }, { "vid": "LpyzdO2fXtA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Area of Rectangles and Squares in Elementary Arithmetic Math Course tutorial_30.json#####audio#####doingASR#####FinishASR/LpyzdO2fXtA/1182.72_1200.72.mp4", "refined_asr": " 36. So we end up with pi times 36. Now remember, just like in our first example, one way to write our area is to write it in terms of pi. And again, that means we can write our number.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1182.72_1200.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1182.72_1200.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1182.72_1200.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1182.72_1200.72#3.jpg" ], "ocr_qwen2_vl_72b": "Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36" }, { "vid": "LpyzdO2fXtA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Area of Rectangles and Squares in Elementary Arithmetic Math Course tutorial_30.json#####audio#####doingASR#####FinishASR/LpyzdO2fXtA/1200.72_1218.72.mp4", "refined_asr": " In front of the pi symbol and then put our unit of measure so we could put thirty-six pi and this is square feet for our unit of measure so that's one way we can write out our area of that circle", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1200.72_1218.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1200.72_1218.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1200.72_1218.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1200.72_1218.72#3.jpg" ], "ocr_qwen2_vl_72b": "Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36\nA =" }, { "vid": "LpyzdO2fXtA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Area of Rectangles and Squares in Elementary Arithmetic Math Course tutorial_30.json#####audio#####doingASR#####FinishASR/LpyzdO2fXtA/1218.72_1238.72.mp4", "refined_asr": " Another option is we can use an approximate rounded version of pi. As I mentioned earlier, we're going to use 3.14. So we can use this to get an approximate area of this circle in decimal form. Again, we're going to use pi.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1218.72_1238.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1218.72_1238.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1218.72_1238.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1218.72_1238.72#3.jpg" ], "ocr_qwen2_vl_72b": "Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36\nA = 36\u03c0 ft\u00b2" }, { "vid": "LpyzdO2fXtA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Area of Rectangles and Squares in Elementary Arithmetic Math Course tutorial_30.json#####audio#####doingASR#####FinishASR/LpyzdO2fXtA/1238.72_1260.72.mp4", "refined_asr": " Pi is approximately 3.14. Let's write out our formula, which is: area equals pi r squared. And then we can plug in: so area is approximately... I'm going to use the approximate symbol there because, again, we're using an approximation.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1238.72_1260.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1238.72_1260.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1238.72_1260.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1238.72_1260.72#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1238.72_1260.72#4.jpg" ], "ocr_qwen2_vl_72b": "Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\n\u03c0 \u2248 3.\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36\nA = 36\u03c0 ft\u00b2" }, { "vid": "LpyzdO2fXtA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Area of Rectangles and Squares in Elementary Arithmetic Math Course tutorial_30.json#####audio#####doingASR#####FinishASR/LpyzdO2fXtA/1260.72_1278.72.mp4", "refined_asr": " We're using an approximation of pi, so let's plug in 3.14. Multiply that by our radius, which is 6, squared. Now we can calculate. First, let's do 6 squared.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1260.72_1278.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1260.72_1278.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1260.72_1278.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1260.72_1278.72#3.jpg" ], "ocr_qwen2_vl_72b": "Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\n\u03c0 \u2248 3.14\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36\nA = 36\u03c0 ft\u00b2" }, { "vid": "LpyzdO2fXtA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Area of Rectangles and Squares in Elementary Arithmetic Math Course tutorial_30.json#####audio#####doingASR#####FinishASR/LpyzdO2fXtA/1278.72_1298.72.mp4", "refined_asr": " So 3.14 times 6 squared which equals 36. And now let's multiply. So 3.14 times 36 gives us an approximate area of 113. And 4 hundredths. And this is", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1278.72_1298.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1278.72_1298.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1278.72_1298.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LpyzdO2fXtA/LpyzdO2fXtA@1278.72_1298.72#3.jpg" ], "ocr_qwen2_vl_72b": "Calculate the Area of a Circle\n\nA = \u03c0r\u00b2\n\n\u03c0 \u2248 3.14\n\nA = \u03c0r\u00b2\nA = \u03c0 \u00d7 6\u00b2\nA = \u03c0 \u00d7 36\nA = 36\u03c0 ft\u00b2" } ], "image_num": 8, "text_num": 438, "token_num": 5046 }, { "images": [ "sample_100_images/lTzHlU3OrXs@16591.24_16607.24#1.jpg", null, "sample_100_images/lTzHlU3OrXs@16607.24_16639.24#1.jpg", null, "sample_100_images/lTzHlU3OrXs@16639.24_16661.940000000002#1.jpg", null, "sample_100_images/lTzHlU3OrXs@16661.94_16679.539999999997#1.jpg", "sample_100_images/lTzHlU3OrXs@16661.94_16679.539999999997#2.jpg", null, "sample_100_images/lTzHlU3OrXs@16680.3_16699.219999999998#1.jpg", "sample_100_images/lTzHlU3OrXs@16680.3_16699.219999999998#3.jpg", null, "sample_100_images/lTzHlU3OrXs@16699.219999999998_16717.68#1.jpg", null, "sample_100_images/lTzHlU3OrXs@16717.68_16731.94#1.jpg", null ], "texts": [ null, " And you can see here that if you have 100 people clicking on ads and you click on one of the ads, it might go in there and say okay, this person clicked on this ad. What is the best set of ads based on clicking on this ad or these two ads afterwards?", null, " Based on where they are browsing to see what they're doing or what their interests are. And again, when we talk about states, we're usually not talking about float variables. When you're using a Q-table, you're typically talking about binary conditions, true or false. We'll take a closer look at that in a second.", null, " And episodes end when an agent ends up in a terminating state and can't take a new action. This might be if you're playing a video game and your character stepped into a trap and is now dead. Q-values are used to determine how good an action 'a' taken at a particular state 's' is. It's denoted as Q(s, a).", null, null, " And temporal difference is a formula used to find the Q-value by using the value of the current state and action and the previous state and action. And there's Bellman's equation which basically is the equation that kind of covers what we just looked at in all those different terms.", null, null, " The Bellman equation is used to determine the values of a particular state and deduce how good it is to be in that state. The optimal state will give us the highest optimal value. Factors influencing Q values include the current state and action, that's your S-A, so your current state and your action.", null, " Then you have your previous state and action which is your S prime I'm not sure how they reference that S prime A prime so this is what happened before then you have a reward for action so you have your R reward and you have your maximum expected future reward", null, " And you can see there's also a learning curve. There's a learning rate put in there and a discount rate. So when we're looking at these, just like any other model, we don't want to have an absolute final value." ], "text_ocr_list": [ null, "We can see these text from the image: What is Q-Learning? : Ad suggestion\n\nUsing Q-learning, we can make an Ad recommendation system which will suggest related products to our previous purchase. The reward will be if user clicks on the suggested product.\n And you can see here that if you have 100 people clicking on ads and you click on one of the ads, it might go in there and say okay, this person clicked on this ad. What is the best set of ads based on clicking on this ad or these two ads afterwards?", null, "We can see these text from the image: States\n\nThe State, S, represents the current position of an agent in an environment.\n Based on where they are browsing to see what they're doing or what their interests are. And again, when we talk about states, we're usually not talking about float variables. When you're using a Q-table, you're typically talking about binary conditions, true or false. We'll take a closer look at that in a second.", null, "We can see these text from the image: Episodes\n\nWhen an agent ends up in a terminating state and can't take a new action.\n And episodes end when an agent ends up in a terminating state and can't take a new action. This might be if you're playing a video game and your character stepped into a trap and is now dead. Q-values are used to determine how good an action 'a' taken at a particular state 's' is. It's denoted as Q(s, a).", null, null, "We can see these text from the image: Episodes\n\nWhen an agent ends up in a terminating state and can't take a new action\n\nQ-Values\n\nUsed to determine how good an Action, A, taken at a particular state, S, is Q(A,S)\n\nTemporal Difference\n\nA formula used to find the Q-Value by using the value of current state and action and previous state and action.\n And temporal difference is a formula used to find the Q-value by using the value of the current state and action and the previous state and action. And there's Bellman's equation which basically is the equation that kind of covers what we just looked at in all those different terms.", null, null, "We can see these text from the image: Bellman\u2019s Equation\n\nThe Bellman Equation is used to determine the value of a particular state and deduce how good it is to be in/ take that state. The optimal state will give us the highest optimal value.\n\nNew Q(S, A) = Q(S, A) + \u03b1 [R(S, A)].\n The Bellman equation is used to determine the values of a particular state and deduce how good it is to be in that state. The optimal state will give us the highest optimal value. Factors influencing Q values include the current state and action, that's your S-A, so your current state and your action.", null, "We can see these text from the image: Bellman\u2019s Equation\n\nFactors Influencing Q-values:\n- Current State and Action (S, A)\n- Previous State and Action (S', A')\n- Reward for Action, R\n- Maximum expected future reward\n\nNew Q(S, A) = Q(S, A) + \u03b1 [R(S, A) + \u03b3 Max Q'(S', A') - Q(S, A)]\n\nCurrent Q Value\nLearning Rate\nReward\n\nDiscount Rate\nMaximum Expected Future Reward.\n Then you have your previous state and action which is your S prime I'm not sure how they reference that S prime A prime so this is what happened before then you have a reward for action so you have your R reward and you have your maximum expected future reward", null, "We can see these text from the image: Bellman's Equation\n\nFactors Influencing Q-values:\n- Current State and Action (S, A)\n- Previous State and Action (S', A')\n- Reward for Action, R\n- Maximum expected future reward\n\nNew Q(S, A) = Q(S, A) + \u03b1 [R(S, A) + \u03b3 Max Q'(S', A') - Q(S, A)]\n\nCurrent Q Value\nLearning Rate\nReward\n\nDiscount Rate\nMaximum Expected Future Reward.\n And you can see there's also a learning curve. There's a learning rate put in there and a discount rate. So when we're looking at these, just like any other model, we don't want to have an absolute final value." ], "metadata": [ { "vid": "lTzHlU3OrXs.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence course videos on fundamentals of AI focusing on Logical Reasoning and Resolution_30.json#####audio#####doingASR#####FinishASR/lTzHlU3OrXs/16591.24_16607.24.mp4", "refined_asr": " And you can see here that if you have 100 people clicking on ads and you click on one of the ads, it might go in there and say okay, this person clicked on this ad. What is the best set of ads based on clicking on this ad or these two ads afterwards?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16591.24_16607.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16591.24_16607.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16591.24_16607.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16591.24_16607.24#3.jpg" ], "ocr_qwen2_vl_72b": "What is Q-Learning? : Ad suggestion\n\nUsing Q-learning, we can make an Ad recommendation system which will suggest related products to our previous purchase. The reward will be if user clicks on the suggested product" }, { "vid": "lTzHlU3OrXs.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence course videos on fundamentals of AI focusing on Logical Reasoning and Resolution_30.json#####audio#####doingASR#####FinishASR/lTzHlU3OrXs/16607.24_16639.24.mp4", "refined_asr": " Based on where they are browsing to see what they're doing or what their interests are. And again, when we talk about states, we're usually not talking about float variables. When you're using a Q-table, you're typically talking about binary conditions, true or false. We'll take a closer look at that in a second.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16607.24_16639.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16607.24_16639.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16607.24_16639.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16607.24_16639.24#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16607.24_16639.24#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16607.24_16639.24#5.jpg" ], "ocr_qwen2_vl_72b": "States\n\nThe State, S, represents the current position of an agent in an environment" }, { "vid": "lTzHlU3OrXs.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence course videos on fundamentals of AI focusing on Logical Reasoning and Resolution_30.json#####audio#####doingASR#####FinishASR/lTzHlU3OrXs/16639.24_16661.940000000002.mp4", "refined_asr": " And episodes end when an agent ends up in a terminating state and can't take a new action. This might be if you're playing a video game and your character stepped into a trap and is now dead. Q-values are used to determine how good an action 'a' taken at a particular state 's' is. It's denoted as Q(s, a).", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16639.24_16661.940000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16639.24_16661.940000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16639.24_16661.940000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16639.24_16661.940000000002#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16639.24_16661.940000000002#4.jpg" ], "ocr_qwen2_vl_72b": "Episodes\n\nWhen an agent ends up in a terminating state and can't take a new action" }, { "vid": "lTzHlU3OrXs.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence course videos on fundamentals of AI focusing on Logical Reasoning and Resolution_30.json#####audio#####doingASR#####FinishASR/lTzHlU3OrXs/16661.94_16679.539999999997.mp4", "refined_asr": " And temporal difference is a formula used to find the Q-value by using the value of the current state and action and the previous state and action. And there's Bellman's equation which basically is the equation that kind of covers what we just looked at in all those different terms.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16661.94_16679.539999999997#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16661.94_16679.539999999997#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16661.94_16679.539999999997#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16661.94_16679.539999999997#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16661.94_16679.539999999997#3.jpg" ], "ocr_qwen2_vl_72b": "Episodes\n\nWhen an agent ends up in a terminating state and can't take a new action\n\nQ-Values\n\nUsed to determine how good an Action, A, taken at a particular state, S, is Q(A,S)\n\nTemporal Difference\n\nA formula used to find the Q-Value by using the value of current state and action and previous state and action" }, { "vid": "lTzHlU3OrXs.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence course videos on fundamentals of AI focusing on Logical Reasoning and Resolution_30.json#####audio#####doingASR#####FinishASR/lTzHlU3OrXs/16680.3_16699.219999999998.mp4", "refined_asr": " The Bellman equation is used to determine the values of a particular state and deduce how good it is to be in that state. The optimal state will give us the highest optimal value. Factors influencing Q values include the current state and action, that's your S-A, so your current state and your action.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16680.3_16699.219999999998#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16680.3_16699.219999999998#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16680.3_16699.219999999998#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16680.3_16699.219999999998#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16680.3_16699.219999999998#3.jpg" ], "ocr_qwen2_vl_72b": "Bellman\u2019s Equation\n\nThe Bellman Equation is used to determine the value of a particular state and deduce how good it is to be in/ take that state. The optimal state will give us the highest optimal value.\n\nNew Q(S, A) = Q(S, A) + \u03b1 [R(S, A)]" }, { "vid": "lTzHlU3OrXs.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence course videos on fundamentals of AI focusing on Logical Reasoning and Resolution_30.json#####audio#####doingASR#####FinishASR/lTzHlU3OrXs/16699.219999999998_16717.68.mp4", "refined_asr": " Then you have your previous state and action which is your S prime I'm not sure how they reference that S prime A prime so this is what happened before then you have a reward for action so you have your R reward and you have your maximum expected future reward", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16699.219999999998_16717.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16699.219999999998_16717.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16699.219999999998_16717.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16699.219999999998_16717.68#3.jpg" ], "ocr_qwen2_vl_72b": "Bellman\u2019s Equation\n\nFactors Influencing Q-values:\n- Current State and Action (S, A)\n- Previous State and Action (S', A')\n- Reward for Action, R\n- Maximum expected future reward\n\nNew Q(S, A) = Q(S, A) + \u03b1 [R(S, A) + \u03b3 Max Q'(S', A') - Q(S, A)]\n\nCurrent Q Value\nLearning Rate\nReward\n\nDiscount Rate\nMaximum Expected Future Reward" }, { "vid": "lTzHlU3OrXs.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence course videos on fundamentals of AI focusing on Logical Reasoning and Resolution_30.json#####audio#####doingASR#####FinishASR/lTzHlU3OrXs/16717.68_16731.94.mp4", "refined_asr": " And you can see there's also a learning curve. There's a learning rate put in there and a discount rate. So when we're looking at these, just like any other model, we don't want to have an absolute final value.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16717.68_16731.94#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16717.68_16731.94#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16717.68_16731.94#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lTzHlU3OrXs/lTzHlU3OrXs@16717.68_16731.94#3.jpg" ], "ocr_qwen2_vl_72b": "Bellman's Equation\n\nFactors Influencing Q-values:\n- Current State and Action (S, A)\n- Previous State and Action (S', A')\n- Reward for Action, R\n- Maximum expected future reward\n\nNew Q(S, A) = Q(S, A) + \u03b1 [R(S, A) + \u03b3 Max Q'(S', A') - Q(S, A)]\n\nCurrent Q Value\nLearning Rate\nReward\n\nDiscount Rate\nMaximum Expected Future Reward" } ], "image_num": 9, "text_num": 460, "token_num": 5644 }, { "images": [ "sample_100_images/JP5Z2ZQURaE@674.04_688.0999999999999#1.jpg", "sample_100_images/JP5Z2ZQURaE@674.04_688.0999999999999#2.jpg", null, "sample_100_images/JP5Z2ZQURaE@688.0999999999999_704.8399999999999#1.jpg", "sample_100_images/JP5Z2ZQURaE@688.0999999999999_704.8399999999999#2.jpg", null, "sample_100_images/JP5Z2ZQURaE@704.8399999999999_717.86#1.jpg", "sample_100_images/JP5Z2ZQURaE@704.8399999999999_717.86#2.jpg", null, "sample_100_images/JP5Z2ZQURaE@717.86_731.4799999999999#1.jpg", "sample_100_images/JP5Z2ZQURaE@717.86_731.4799999999999#2.jpg", null, "sample_100_images/JP5Z2ZQURaE@731.4799999999999_743.12#1.jpg", "sample_100_images/JP5Z2ZQURaE@731.4799999999999_743.12#2.jpg", null, "sample_100_images/JP5Z2ZQURaE@743.42_758.24#1.jpg", "sample_100_images/JP5Z2ZQURaE@743.42_758.24#2.jpg", null, "sample_100_images/JP5Z2ZQURaE@758.24_773.36#1.jpg", "sample_100_images/JP5Z2ZQURaE@758.24_773.36#2.jpg", null, "sample_100_images/JP5Z2ZQURaE@773.4200000000001_789.5000000000001#1.jpg", "sample_100_images/JP5Z2ZQURaE@773.4200000000001_789.5000000000001#2.jpg", null ], "texts": [ null, null, " Had to wait a long time for this. This is the last part we're waiting for, to be honest. So almost a month to try and get these in. And that's the sort of thing that should be just basic commodity stuff. Don't even get me started on SRAM brake codes, which is almost unobtainable.", null, null, " Anyway, we got there in the end. We've got some lever hoods, so let's pull these ones off. It's always easier to do this while they're on the actual bike. So, I need to disconnect these brake hoses. Gloves on, and we have a ring spanner which is a little five-sided spanner like this.", null, null, " SRAM levers actually have what they call a 'connectamajig.' So they've got a little valve on this lever side here. The lever side isn't going to leak, but this side will. So what I'm doing is I'm pulling this up and getting this.", null, null, " Make sure it's tight, so any leakage here doesn't come out. Okay, so here's a little thing to look out for: you can always tell when it's been done from the factory because when they factory-install these, the little olives don't have the little T7. I think it is adjusted so that you can unscrew these olives.", null, null, " These are just actually bonded in so we need to cut these off Normally Canyon does have a system where you don't actually have to route it through the handlebars Just finish that off with a little bit of tape and that's the way it's nice and safe You know", null, null, " They leak through all of that. And here you can see what Canyon does with their bars. That is allowing what looks like internal cable routing, which is actually external and makes servicing quite a bit easier. I have to say, it would be really nice to see grease on these threads, though.", null, null, " Because this is an area of the bike where it gets very sweaty. The reason I've got the bike at this angle is so that nothing just falls on the floor as I deconstruct things. As I'm taking all this apart, I'm just looking for any damage. These little top caps on Canyons.", null, null, " They're normally very easily damaged and we often see this little O-ring here has been snapped or this little spacer here has been cracked. Again with these cross compression rings, we normally can see lots of damage on this. This little rubber seal often gets damaged and these can." ], "text_ocr_list": [ null, null, "We can see these text from the image: MSW.\n Had to wait a long time for this. This is the last part we're waiting for, to be honest. So almost a month to try and get these in. And that's the sort of thing that should be just basic commodity stuff. Don't even get me started on SRAM brake codes, which is almost unobtainable.", null, null, "We can see these text from the image: JAPDEC CYCLE WORKS.\n Anyway, we got there in the end. We've got some lever hoods, so let's pull these ones off. It's always easier to do this while they're on the actual bike. So, I need to disconnect these brake hoses. Gloves on, and we have a ring spanner which is a little five-sided spanner like this.", null, null, "We can see these text from the image: No text or formulas are present in the image..\n SRAM levers actually have what they call a 'connectamajig.' So they've got a little valve on this lever side here. The lever side isn't going to leak, but this side will. So what I'm doing is I'm pulling this up and getting this.", null, null, "We can see these text from the image: M.\n Make sure it's tight, so any leakage here doesn't come out. Okay, so here's a little thing to look out for: you can always tell when it's been done from the factory because when they factory-install these, the little olives don't have the little T7. I think it is adjusted so that you can unscrew these olives.", null, null, "We can see these text from the image: M.\n These are just actually bonded in so we need to cut these off Normally Canyon does have a system where you don't actually have to route it through the handlebars Just finish that off with a little bit of tape and that's the way it's nice and safe You know", null, null, "We can see these text from the image: No text or formulas present in the image..\n They leak through all of that. And here you can see what Canyon does with their bars. That is allowing what looks like internal cable routing, which is actually external and makes servicing quite a bit easier. I have to say, it would be really nice to see grease on these threads, though.", null, null, "We can see these text from the image: No text or formulas are present in the image..\n Because this is an area of the bike where it gets very sweaty. The reason I've got the bike at this angle is so that nothing just falls on the floor as I deconstruct things. As I'm taking all this apart, I'm just looking for any damage. These little top caps on Canyons.", null, null, "We can see these text from the image: CPIQ.\n They're normally very easily damaged and we often see this little O-ring here has been snapped or this little spacer here has been cracked. Again with these cross compression rings, we normally can see lots of damage on this. This little rubber seal often gets damaged and these can." ], "metadata": [ { "vid": "JP5Z2ZQURaE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bicycle safety and maintenance_100.json#####audio#####doingASR#####FinishASR/JP5Z2ZQURaE/674.04_688.0999999999999.mp4", "refined_asr": " Had to wait a long time for this. This is the last part we're waiting for, to be honest. So almost a month to try and get these in. And that's the sort of thing that should be just basic commodity stuff. Don't even get me started on SRAM brake codes, which is almost unobtainable.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@674.04_688.0999999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@674.04_688.0999999999999#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@674.04_688.0999999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@674.04_688.0999999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@674.04_688.0999999999999#3.jpg" ], "ocr_qwen2_vl_72b": "MSW" }, { "vid": "JP5Z2ZQURaE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bicycle safety and maintenance_100.json#####audio#####doingASR#####FinishASR/JP5Z2ZQURaE/688.0999999999999_704.8399999999999.mp4", "refined_asr": " Anyway, we got there in the end. We've got some lever hoods, so let's pull these ones off. It's always easier to do this while they're on the actual bike. So, I need to disconnect these brake hoses. Gloves on, and we have a ring spanner which is a little five-sided spanner like this.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@688.0999999999999_704.8399999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@688.0999999999999_704.8399999999999#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@688.0999999999999_704.8399999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@688.0999999999999_704.8399999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@688.0999999999999_704.8399999999999#3.jpg" ], "ocr_qwen2_vl_72b": "JAPDEC CYCLE WORKS" }, { "vid": "JP5Z2ZQURaE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bicycle safety and maintenance_100.json#####audio#####doingASR#####FinishASR/JP5Z2ZQURaE/704.8399999999999_717.86.mp4", "refined_asr": " SRAM levers actually have what they call a 'connectamajig.' So they've got a little valve on this lever side here. The lever side isn't going to leak, but this side will. So what I'm doing is I'm pulling this up and getting this.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@704.8399999999999_717.86#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@704.8399999999999_717.86#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@704.8399999999999_717.86#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@704.8399999999999_717.86#2.jpg" ], "ocr_qwen2_vl_72b": "No text or formulas are present in the image." }, { "vid": "JP5Z2ZQURaE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bicycle safety and maintenance_100.json#####audio#####doingASR#####FinishASR/JP5Z2ZQURaE/717.86_731.4799999999999.mp4", "refined_asr": " Make sure it's tight, so any leakage here doesn't come out. Okay, so here's a little thing to look out for: you can always tell when it's been done from the factory because when they factory-install these, the little olives don't have the little T7. I think it is adjusted so that you can unscrew these olives.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@717.86_731.4799999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@717.86_731.4799999999999#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@717.86_731.4799999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@717.86_731.4799999999999#2.jpg" ], "ocr_qwen2_vl_72b": "M" }, { "vid": "JP5Z2ZQURaE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bicycle safety and maintenance_100.json#####audio#####doingASR#####FinishASR/JP5Z2ZQURaE/731.4799999999999_743.12.mp4", "refined_asr": " These are just actually bonded in so we need to cut these off Normally Canyon does have a system where you don't actually have to route it through the handlebars Just finish that off with a little bit of tape and that's the way it's nice and safe You know", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@731.4799999999999_743.12#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@731.4799999999999_743.12#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@731.4799999999999_743.12#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@731.4799999999999_743.12#2.jpg" ], "ocr_qwen2_vl_72b": "M" }, { "vid": "JP5Z2ZQURaE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bicycle safety and maintenance_100.json#####audio#####doingASR#####FinishASR/JP5Z2ZQURaE/743.42_758.24.mp4", "refined_asr": " They leak through all of that. And here you can see what Canyon does with their bars. That is allowing what looks like internal cable routing, which is actually external and makes servicing quite a bit easier. I have to say, it would be really nice to see grease on these threads, though.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@743.42_758.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@743.42_758.24#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@743.42_758.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@743.42_758.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@743.42_758.24#3.jpg" ], "ocr_qwen2_vl_72b": "No text or formulas present in the image." }, { "vid": "JP5Z2ZQURaE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bicycle safety and maintenance_100.json#####audio#####doingASR#####FinishASR/JP5Z2ZQURaE/758.24_773.36.mp4", "refined_asr": " Because this is an area of the bike where it gets very sweaty. The reason I've got the bike at this angle is so that nothing just falls on the floor as I deconstruct things. As I'm taking all this apart, I'm just looking for any damage. These little top caps on Canyons.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@758.24_773.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@758.24_773.36#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@758.24_773.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@758.24_773.36#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@758.24_773.36#3.jpg" ], "ocr_qwen2_vl_72b": "No text or formulas are present in the image." }, { "vid": "JP5Z2ZQURaE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bicycle safety and maintenance_100.json#####audio#####doingASR#####FinishASR/JP5Z2ZQURaE/773.4200000000001_789.5000000000001.mp4", "refined_asr": " They're normally very easily damaged and we often see this little O-ring here has been snapped or this little spacer here has been cracked. Again with these cross compression rings, we normally can see lots of damage on this. This little rubber seal often gets damaged and these can.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@773.4200000000001_789.5000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@773.4200000000001_789.5000000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@773.4200000000001_789.5000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@773.4200000000001_789.5000000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JP5Z2ZQURaE/JP5Z2ZQURaE@773.4200000000001_789.5000000000001#3.jpg" ], "ocr_qwen2_vl_72b": "CPIQ" } ], "image_num": 16, "text_num": 572, "token_num": 9788 }, { "images": [ "sample_100_images/_WtvU4Xn2UE@1000.26_1017.1600000000001#1.jpg", "sample_100_images/_WtvU4Xn2UE@1000.26_1017.1600000000001#2.jpg", null, "sample_100_images/_WtvU4Xn2UE@1017.2600000000001_1032.66#1.jpg", "sample_100_images/_WtvU4Xn2UE@1017.2600000000001_1032.66#2.jpg", null, "sample_100_images/_WtvU4Xn2UE@1032.76_1049.5600000000002#1.jpg", null, "sample_100_images/_WtvU4Xn2UE@1049.66_1067.46#1.jpg", null, "sample_100_images/_WtvU4Xn2UE@1067.46_1080.3600000000001#1.jpg", "sample_100_images/_WtvU4Xn2UE@1067.46_1080.3600000000001#2.jpg", null, "sample_100_images/_WtvU4Xn2UE@1080.46_1092.3600000000001#1.jpg", null, "sample_100_images/_WtvU4Xn2UE@1092.46_1106.2600000000002#1.jpg", null, "sample_100_images/_WtvU4Xn2UE@1106.3600000000001_1123.2600000000002#1.jpg", "sample_100_images/_WtvU4Xn2UE@1106.3600000000001_1123.2600000000002#2.jpg", null, "sample_100_images/_WtvU4Xn2UE@1123.3600000000001_1138.16#1.jpg", null ], "texts": [ null, null, " So it really settled the great debate and it said that these distant spiral nebulae were actually these very large very distant systems much like our own Milky Way. And so here's a modern day image. We no longer refer to these as the Andromeda Nebula or spiral nebulae.", null, null, " They're spiral galaxies because they're much like our own Milky Way galaxy. This is a beautiful image. And this is what we think of when we think of a spiral galaxy. Like I said here's our Milky Way and here's the solar system about halfway out in the disk of the Milky Way.", null, " And what this essentially did is it increased the size of the universe - what we thought - by about 100,000 times. So, from our perspective, I mean, nothing really took place in the universe that day, but basically we realized that it was actually about 100,000 times larger.", null, " than we had thought up until that point which blows my mind Not only did we realize that these distant objects were large and very far away but they were traveling away from us So Edwin Hubble's most notable discovery is probably that these objects", null, null, " These objects were traveling away and the farther the galaxy was from us the faster it's traveling away from us which is plotted very nicely here where you have the distance of a galaxy and you have the speed at which it's traveling away from us", null, " And you can see all these points. The farther something is away the faster it's traveling away from us. Which sounds kind of crazy initially right? Are we at the center of the universe and everything's just traveling away from us?", null, " What he was able to deduce was that in fact the universe was expanding outward. And so it wasn't so much that we were sitting still and everything was traveling away from us but that space itself is expanding.", null, null, " And so we're traveling away from each other and everyone appears to be at the center of their own space with everything traveling away. And that's why you get a space telescope named after you when you do major discoveries like this about the nature of the universe.", null, " So very quickly, so far, there's a lot of stuff in the sky in these smudges that we used to refer to as nebulae, but we now call galaxies. These galaxies, the universe is extremely large and there are many galaxies within it." ], "text_ocr_list": [ null, null, "We can see these text from the image: What is the nature of spiral nebulae?\n\nAre they small objects like stars or gas clouds?\nOr are they enormous \u201cisland universes\u201d like our Milky Way!\n\nLarge and distant!\n\n2.5 million light years!.\n So it really settled the great debate and it said that these distant spiral nebulae were actually these very large very distant systems much like our own Milky Way. And so here's a modern day image. We no longer refer to these as the Andromeda Nebula or spiral nebulae.", null, null, "We can see these text from the image: What is the nature of spiral nebulae?\nThey are distant galaxies like our own Milky Way.\nAndromeda Nebula\nAndromeda Galaxy.\n They're spiral galaxies because they're much like our own Milky Way galaxy. This is a beautiful image. And this is what we think of when we think of a spiral galaxy. Like I said here's our Milky Way and here's the solar system about halfway out in the disk of the Milky Way.", null, "We can see these text from the image: What is the nature of spiral nebulae?\n\nThey are distant galaxies like our own Milky Way.\nThe universe is much larger than previously thought! x100,000!.\n And what this essentially did is it increased the size of the universe - what we thought - by about 100,000 times. So, from our perspective, I mean, nothing really took place in the universe that day, but basically we realized that it was actually about 100,000 times larger.", null, "We can see these text from the image: What is the nature of spiral nebulae?\n\nThey are distant galaxies like our own Milky Way.\nThe universe is much larger than previously thought! x100,000!.\n than we had thought up until that point which blows my mind Not only did we realize that these distant objects were large and very far away but they were traveling away from us So Edwin Hubble's most notable discovery is probably that these objects", null, null, "We can see these text from the image: What is the nature of spiral nebulae?\n\nThey are distant galaxies like our own Milky Way.\nAnd they are traveling away from us!!.\n These objects were traveling away and the farther the galaxy was from us the faster it's traveling away from us which is plotted very nicely here where you have the distance of a galaxy and you have the speed at which it's traveling away from us", null, "We can see these text from the image: What is the nature of spiral nebulae?\n\nThey are distant galaxies like our own Milky Way.\nAnd they are traveling away from us!!\n\nSpeed at which its traveling away from us\n\nDistance of a Galaxy.\n And you can see all these points. The farther something is away the faster it's traveling away from us. Which sounds kind of crazy initially right? Are we at the center of the universe and everything's just traveling away from us?", null, "We can see these text from the image: What is the nature of spiral nebulae?\nThey are distant galaxies like our own Milky Way.\nThe universe is expanding!\nEdwin Hubble.\n What he was able to deduce was that in fact the universe was expanding outward. And so it wasn't so much that we were sitting still and everything was traveling away from us but that space itself is expanding.", null, null, "We can see these text from the image: What is the nature of spiral nebulae?\nThey are distant galaxies like our own Milky Way.\nThe universe is expanding!\n\nEdwin Hubble.\n And so we're traveling away from each other and everyone appears to be at the center of their own space with everything traveling away. And that's why you get a space telescope named after you when you do major discoveries like this about the nature of the universe.", null, "We can see these text from the image: Quick Summary So Far\n\n- There is a lot of stuff in the sky including smudges called spiral nebulae\n- Spiral nebulae are actually distant galaxies\n- The universe is enormous: billions of light years across!\n- The universe is expanding and getting even bigger with time\n- Edwin Hubble is awesome.\n So very quickly, so far, there's a lot of stuff in the sky in these smudges that we used to refer to as nebulae, but we now call galaxies. These galaxies, the universe is extremely large and there are many galaxies within it." ], "metadata": [ { "vid": "_WtvU4Xn2UE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Galactic Astronomy Tutorial on Galaxy Formation, Evolution, Interactions and Mergers_30.json#####audio#####doingASR#####FinishASR/_WtvU4Xn2UE/1000.26_1017.1600000000001.mp4", "refined_asr": " So it really settled the great debate and it said that these distant spiral nebulae were actually these very large very distant systems much like our own Milky Way. And so here's a modern day image. We no longer refer to these as the Andromeda Nebula or spiral nebulae.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1000.26_1017.1600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1000.26_1017.1600000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1000.26_1017.1600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1000.26_1017.1600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1000.26_1017.1600000000001#3.jpg" ], "ocr_qwen2_vl_72b": "What is the nature of spiral nebulae?\n\nAre they small objects like stars or gas clouds?\nOr are they enormous \u201cisland universes\u201d like our Milky Way!\n\nLarge and distant!\n\n2.5 million light years!" }, { "vid": "_WtvU4Xn2UE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Galactic Astronomy Tutorial on Galaxy Formation, Evolution, Interactions and Mergers_30.json#####audio#####doingASR#####FinishASR/_WtvU4Xn2UE/1017.2600000000001_1032.66.mp4", "refined_asr": " They're spiral galaxies because they're much like our own Milky Way galaxy. This is a beautiful image. And this is what we think of when we think of a spiral galaxy. Like I said here's our Milky Way and here's the solar system about halfway out in the disk of the Milky Way.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1017.2600000000001_1032.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1017.2600000000001_1032.66#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1017.2600000000001_1032.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1017.2600000000001_1032.66#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1017.2600000000001_1032.66#3.jpg" ], "ocr_qwen2_vl_72b": "What is the nature of spiral nebulae?\nThey are distant galaxies like our own Milky Way.\nAndromeda Nebula\nAndromeda Galaxy" }, { "vid": "_WtvU4Xn2UE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Galactic Astronomy Tutorial on Galaxy Formation, Evolution, Interactions and Mergers_30.json#####audio#####doingASR#####FinishASR/_WtvU4Xn2UE/1032.76_1049.5600000000002.mp4", "refined_asr": " And what this essentially did is it increased the size of the universe - what we thought - by about 100,000 times. So, from our perspective, I mean, nothing really took place in the universe that day, but basically we realized that it was actually about 100,000 times larger.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1032.76_1049.5600000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1032.76_1049.5600000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1032.76_1049.5600000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1032.76_1049.5600000000002#3.jpg" ], "ocr_qwen2_vl_72b": "What is the nature of spiral nebulae?\n\nThey are distant galaxies like our own Milky Way.\nThe universe is much larger than previously thought! x100,000!" }, { "vid": "_WtvU4Xn2UE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Galactic Astronomy Tutorial on Galaxy Formation, Evolution, Interactions and Mergers_30.json#####audio#####doingASR#####FinishASR/_WtvU4Xn2UE/1049.66_1067.46.mp4", "refined_asr": " than we had thought up until that point which blows my mind Not only did we realize that these distant objects were large and very far away but they were traveling away from us So Edwin Hubble's most notable discovery is probably that these objects", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1049.66_1067.46#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1049.66_1067.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1049.66_1067.46#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1049.66_1067.46#3.jpg" ], "ocr_qwen2_vl_72b": "What is the nature of spiral nebulae?\n\nThey are distant galaxies like our own Milky Way.\nThe universe is much larger than previously thought! x100,000!" }, { "vid": "_WtvU4Xn2UE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Galactic Astronomy Tutorial on Galaxy Formation, Evolution, Interactions and Mergers_30.json#####audio#####doingASR#####FinishASR/_WtvU4Xn2UE/1067.46_1080.3600000000001.mp4", "refined_asr": " These objects were traveling away and the farther the galaxy was from us the faster it's traveling away from us which is plotted very nicely here where you have the distance of a galaxy and you have the speed at which it's traveling away from us", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1067.46_1080.3600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1067.46_1080.3600000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1067.46_1080.3600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1067.46_1080.3600000000001#2.jpg" ], "ocr_qwen2_vl_72b": "What is the nature of spiral nebulae?\n\nThey are distant galaxies like our own Milky Way.\nAnd they are traveling away from us!!" }, { "vid": "_WtvU4Xn2UE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Galactic Astronomy Tutorial on Galaxy Formation, Evolution, Interactions and Mergers_30.json#####audio#####doingASR#####FinishASR/_WtvU4Xn2UE/1080.46_1092.3600000000001.mp4", "refined_asr": " And you can see all these points. The farther something is away the faster it's traveling away from us. Which sounds kind of crazy initially right? Are we at the center of the universe and everything's just traveling away from us?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1080.46_1092.3600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1080.46_1092.3600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1080.46_1092.3600000000001#2.jpg" ], "ocr_qwen2_vl_72b": "What is the nature of spiral nebulae?\n\nThey are distant galaxies like our own Milky Way.\nAnd they are traveling away from us!!\n\nSpeed at which its traveling away from us\n\nDistance of a Galaxy" }, { "vid": "_WtvU4Xn2UE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Galactic Astronomy Tutorial on Galaxy Formation, Evolution, Interactions and Mergers_30.json#####audio#####doingASR#####FinishASR/_WtvU4Xn2UE/1092.46_1106.2600000000002.mp4", "refined_asr": " What he was able to deduce was that in fact the universe was expanding outward. And so it wasn't so much that we were sitting still and everything was traveling away from us but that space itself is expanding.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1092.46_1106.2600000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1092.46_1106.2600000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1092.46_1106.2600000000002#2.jpg" ], "ocr_qwen2_vl_72b": "What is the nature of spiral nebulae?\nThey are distant galaxies like our own Milky Way.\nThe universe is expanding!\nEdwin Hubble" }, { "vid": "_WtvU4Xn2UE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Galactic Astronomy Tutorial on Galaxy Formation, Evolution, Interactions and Mergers_30.json#####audio#####doingASR#####FinishASR/_WtvU4Xn2UE/1106.3600000000001_1123.2600000000002.mp4", "refined_asr": " And so we're traveling away from each other and everyone appears to be at the center of their own space with everything traveling away. And that's why you get a space telescope named after you when you do major discoveries like this about the nature of the universe.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1106.3600000000001_1123.2600000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1106.3600000000001_1123.2600000000002#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1106.3600000000001_1123.2600000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1106.3600000000001_1123.2600000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1106.3600000000001_1123.2600000000002#3.jpg" ], "ocr_qwen2_vl_72b": "What is the nature of spiral nebulae?\nThey are distant galaxies like our own Milky Way.\nThe universe is expanding!\n\nEdwin Hubble" }, { "vid": "_WtvU4Xn2UE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics Galactic Astronomy Tutorial on Galaxy Formation, Evolution, Interactions and Mergers_30.json#####audio#####doingASR#####FinishASR/_WtvU4Xn2UE/1123.3600000000001_1138.16.mp4", "refined_asr": " So very quickly, so far, there's a lot of stuff in the sky in these smudges that we used to refer to as nebulae, but we now call galaxies. These galaxies, the universe is extremely large and there are many galaxies within it.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1123.3600000000001_1138.16#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1123.3600000000001_1138.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1123.3600000000001_1138.16#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_WtvU4Xn2UE/_WtvU4Xn2UE@1123.3600000000001_1138.16#3.jpg" ], "ocr_qwen2_vl_72b": "Quick Summary So Far\n\n- There is a lot of stuff in the sky including smudges called spiral nebulae\n- Spiral nebulae are actually distant galaxies\n- The universe is enormous: billions of light years across!\n- The universe is expanding and getting even bigger with time\n- Edwin Hubble is awesome" } ], "image_num": 13, "text_num": 538, "token_num": 8026 }, { "images": [ "sample_100_images/u0W-NQ-4RTU@858.96_874.64#1.jpg", null, "sample_100_images/u0W-NQ-4RTU@874.64_898.32#1.jpg", "sample_100_images/u0W-NQ-4RTU@874.64_898.32#3.jpg", null, "sample_100_images/u0W-NQ-4RTU@899.12_918.64#1.jpg", null, "sample_100_images/u0W-NQ-4RTU@918.64_939.6800000000001#1.jpg", null, "sample_100_images/u0W-NQ-4RTU@939.6800000000001_963.12#1.jpg", null, "sample_100_images/u0W-NQ-4RTU@965.84_988.48#1.jpg", null ], "texts": [ null, " The journey to the center of the Earth movie, right? In a sense, that movie is kind of true. If you were to go all the way to the center of the Earth, there's a part that would be liquid. You could almost get a boat and ride on that, of course, if you could withstand the conditions.", null, null, " The temperatures but it is there is a water there is a liquid there so in the sense it's it is kind of not that wrong then you get here oh okay so it gives you the whole thing here oh so if I click on this it will give me all the so if I go like that", null, " It gives you the whole thing about the temperature of each one, the S-wave velocity. You see, the S-wave velocity, the P-wave velocity. So the S-wave velocity, as you're going down, the velocities keep changing, you see. So you can kind of like click.", null, " Here, the density is 3.94 at that point, the pressure is 20.28 gigapascals, and gravity is 18.9. So, the gravity at the surface of the Earth is what? Nine point? It should be 9.8 at the surface of the Earth. That's the acceleration due to gravity. Now, as you're,", null, " Going down, the gravity is increasing, the temperature is increasing. And notice the S-wave velocity and the P-wave velocity is also slowly increasing as I go down. So, like that, then like that. Well, the S-wave velocity is negative. That's weird. The P-wave velocity then 0.48.", null, " Like that. So the P-waves are always faster and the temperature keeps increasing. Now if I go here S-wave velocity is zero. What do you notice there? You see the gravity at that point is 9.8. So right there you see" ], "text_ocr_list": [ null, "We can see these text from the image: ### Table\n\n| Layer | Percentage (%) | Temperature (K) | Density (g/cm\u00b3) |\n|-------|----------------|-----------------|-----------------|\n| Crust | 0.5% | 500 K | 2.5 g/cm\u00b3 |\n| Mantle | 67.0% | 3,000 K | 4.5 g/cm\u00b3 |\n| Outer Core | 30.8% | 5,200 K | 10.9 g/cm\u00b3 |\n| Inner Core | 1.7% | 5,700 K | 12.9 g/cm\u00b3 |\n\n### Instructions\n\nThe cutaway diagram of the interior regions of the Earth shows the physical properties that exist there. Click on the diagram to investigate physical properties such as density and pressure at various depths. Press right or up arrow keys to increase circle radius and left or down arrow keys to decrease circle radius. Use the \"show labels\" checkbox if you want to see the labels.\n\n### Diagram Labels\n\n- crust\n- mantle\n- outer core\n- inner core.\n The journey to the center of the Earth movie, right? In a sense, that movie is kind of true. If you were to go all the way to the center of the Earth, there's a part that would be liquid. You could almost get a boat and ride on that, of course, if you could withstand the conditions.", null, null, "We can see these text from the image: **Table:**\n| | Percentage (%) | Temperature (K) | Density (g/cm\u00b3) |\n|--------|----------------|-----------------|-----------------|\n| Crust | 0.5% | 500 K | 2.5 g/cm\u00b3 |\n| Mantle | 67.0% | 3,000 K | 4.5 g/cm\u00b3 |\n| Outer Core | 30.8% | 5,200 K | 10.9 g/cm\u00b3 |\n| Inner Core | 1.7% | 5,700 K | 12.9 g/cm\u00b3 |\n\n**Earth Explorer**\n\n**Instructions:**\nThe cutaway diagram of the interior regions of the Earth shows the different regions that exist there. Click on the diagram to investigate physical properties, such as density and pressure, at various depths. Press right or up arrow keys to increase circle radius and left or down arrow keys to decrease circle radius. Use the \"show labels\" checkbox if you want to see the labels..\n The temperatures but it is there is a water there is a liquid there so in the sense it's it is kind of not that wrong then you get here oh okay so it gives you the whole thing here oh so if I click on this it will give me all the so if I go like that", null, "We can see these text from the image: - Mantle: 67.0%, 3,000 K, 4.5 g/cm\u00b3\n- Outer Core: 30.8%, 5,200 K, 10.9 g/cm\u00b3\n- Inner Core: 1.7%, 5,700 K, 12.9 g/cm\u00b3\n\n**Earth Explorer**\n\n**Instructions**\n- Data Analysis\n - The Radius is 5861 km\n - The Depth is 510 km.\n - The Density is 3.86 g/cm\u00b3.\n - The Pressure is 17.8 GPa.\n - The Gravity is 19.1 m/s\u00b2.\n - The Temperature is 1928 K.\n - The S-wave velocity is 5.25 km/s.\n - The P-wave velocity is 9.89 km/s..\n It gives you the whole thing about the temperature of each one, the S-wave velocity. You see, the S-wave velocity, the P-wave velocity. So the S-wave velocity, as you're going down, the velocities keep changing, you see. So you can kind of like click.", null, "We can see these text from the image: - Mantle: 67.0%, 3,000 K, 4.5 g/cm\u00b3\n- Outer Core: 30.8%, 5,200 K, 10.9 g/cm\u00b3\n- Inner Core: 1.7%, 5,700 K, 12.9 g/cm\u00b3\n\n**Earth Explorer**\n\n**Instructions**\n- Data Analysis\n - The Radius is 5798 km\n - The Depth is 573 km.\n - The Density is 3.94 g/cm\u00b3.\n - The Pressure is 20.28 GPa.\n - The Gravity is 18.9 m/s\u00b2.\n - The Temperature is 1010 K.\n - The S-wave velocity is 5.44 km/s.\n - The P-wave velocity is 10 km/s..\n Here, the density is 3.94 at that point, the pressure is 20.28 gigapascals, and gravity is 18.9. So, the gravity at the surface of the Earth is what? Nine point? It should be 9.8 at the surface of the Earth. That's the acceleration due to gravity. Now, as you're,", null, "We can see these text from the image: - Mantle: 67.0%, 3,000 K, 4.5 g/cm\u00b3\n- Outer Core: 30.8%, 5,200 K, 10.9 g/cm\u00b3\n- Inner Core: 1.7%, 5,700 K, 12.9 g/cm\u00b3\n\n**Earth Explorer**\n\n**Instructions**\n- Data Analysis:\n - The Radius is 5989 km\n - The Depth is 382 km.\n - The Density is 3.53 g/cm\u00b3.\n - The Pressure is 12.33 GPa.\n - The Gravity is 14.5 m/s\u00b2\n - The Temperature is 1784 K.\n - The S-wave velocity is 4.76 km/s.\n - The P-wave velocity is 8.87 km/s..\n Going down, the gravity is increasing, the temperature is increasing. And notice the S-wave velocity and the P-wave velocity is also slowly increasing as I go down. So, like that, then like that. Well, the S-wave velocity is negative. That's weird. The P-wave velocity then 0.48.", null, "We can see these text from the image: ### Table\n\n| Layer | Percentage | Temperature (K) | Density (g/cm\u00b3) |\n|-------|------------|-----------------|----------------|\n| Mantle | 67.0% | 3,000 K | 4.5 |\n| Outer Core | 30.8% | 5,200 K | 10.9 |\n| Inner Core | 1.7% | 5,700 K | 12.9 |\n\n### Earth Explorer\n\n#### Instructions\n**Data Analysis**\n- The Radius is 4555 km\n- The Depth is 1816 km.\n- The Density is 3.33 g/cm\u00b3.\n- The Pressure is 57.77 GPa.\n- The Gravity is 12.9 m/s\u00b2\n- The Temperature is 3613 K\n- The S-wave velocity is 1.31 km/s.\n- The P-wave velocity is 16.8 km/s..\n Like that. So the P-waves are always faster and the temperature keeps increasing. Now if I go here S-wave velocity is zero. What do you notice there? You see the gravity at that point is 9.8. So right there you see" ], "metadata": [ { "vid": "u0W-NQ-4RTU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Planetary Interiors in Planetary Geology_30.json#####audio#####doingASR#####FinishASR/u0W-NQ-4RTU/858.96_874.64.mp4", "refined_asr": " The journey to the center of the Earth movie, right? In a sense, that movie is kind of true. If you were to go all the way to the center of the Earth, there's a part that would be liquid. You could almost get a boat and ride on that, of course, if you could withstand the conditions.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@858.96_874.64#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@858.96_874.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@858.96_874.64#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@858.96_874.64#3.jpg" ], "ocr_qwen2_vl_72b": "### Table\n\n| Layer | Percentage (%) | Temperature (K) | Density (g/cm\u00b3) |\n|-------|----------------|-----------------|-----------------|\n| Crust | 0.5% | 500 K | 2.5 g/cm\u00b3 |\n| Mantle | 67.0% | 3,000 K | 4.5 g/cm\u00b3 |\n| Outer Core | 30.8% | 5,200 K | 10.9 g/cm\u00b3 |\n| Inner Core | 1.7% | 5,700 K | 12.9 g/cm\u00b3 |\n\n### Instructions\n\nThe cutaway diagram of the interior regions of the Earth shows the physical properties that exist there. Click on the diagram to investigate physical properties such as density and pressure at various depths. Press right or up arrow keys to increase circle radius and left or down arrow keys to decrease circle radius. Use the \"show labels\" checkbox if you want to see the labels.\n\n### Diagram Labels\n\n- crust\n- mantle\n- outer core\n- inner core" }, { "vid": "u0W-NQ-4RTU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Planetary Interiors in Planetary Geology_30.json#####audio#####doingASR#####FinishASR/u0W-NQ-4RTU/874.64_898.32.mp4", "refined_asr": " The temperatures but it is there is a water there is a liquid there so in the sense it's it is kind of not that wrong then you get here oh okay so it gives you the whole thing here oh so if I click on this it will give me all the so if I go like that", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@874.64_898.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@874.64_898.32#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@874.64_898.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@874.64_898.32#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@874.64_898.32#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@874.64_898.32#4.jpg" ], "ocr_qwen2_vl_72b": "**Table:**\n| | Percentage (%) | Temperature (K) | Density (g/cm\u00b3) |\n|--------|----------------|-----------------|-----------------|\n| Crust | 0.5% | 500 K | 2.5 g/cm\u00b3 |\n| Mantle | 67.0% | 3,000 K | 4.5 g/cm\u00b3 |\n| Outer Core | 30.8% | 5,200 K | 10.9 g/cm\u00b3 |\n| Inner Core | 1.7% | 5,700 K | 12.9 g/cm\u00b3 |\n\n**Earth Explorer**\n\n**Instructions:**\nThe cutaway diagram of the interior regions of the Earth shows the different regions that exist there. Click on the diagram to investigate physical properties, such as density and pressure, at various depths. Press right or up arrow keys to increase circle radius and left or down arrow keys to decrease circle radius. Use the \"show labels\" checkbox if you want to see the labels." }, { "vid": "u0W-NQ-4RTU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Planetary Interiors in Planetary Geology_30.json#####audio#####doingASR#####FinishASR/u0W-NQ-4RTU/899.12_918.64.mp4", "refined_asr": " It gives you the whole thing about the temperature of each one, the S-wave velocity. You see, the S-wave velocity, the P-wave velocity. So the S-wave velocity, as you're going down, the velocities keep changing, you see. So you can kind of like click.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@899.12_918.64#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@899.12_918.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@899.12_918.64#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@899.12_918.64#3.jpg" ], "ocr_qwen2_vl_72b": "- Mantle: 67.0%, 3,000 K, 4.5 g/cm\u00b3\n- Outer Core: 30.8%, 5,200 K, 10.9 g/cm\u00b3\n- Inner Core: 1.7%, 5,700 K, 12.9 g/cm\u00b3\n\n**Earth Explorer**\n\n**Instructions**\n- Data Analysis\n - The Radius is 5861 km\n - The Depth is 510 km.\n - The Density is 3.86 g/cm\u00b3.\n - The Pressure is 17.8 GPa.\n - The Gravity is 19.1 m/s\u00b2.\n - The Temperature is 1928 K.\n - The S-wave velocity is 5.25 km/s.\n - The P-wave velocity is 9.89 km/s." }, { "vid": "u0W-NQ-4RTU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Planetary Interiors in Planetary Geology_30.json#####audio#####doingASR#####FinishASR/u0W-NQ-4RTU/918.64_939.6800000000001.mp4", "refined_asr": " Here, the density is 3.94 at that point, the pressure is 20.28 gigapascals, and gravity is 18.9. So, the gravity at the surface of the Earth is what? Nine point? It should be 9.8 at the surface of the Earth. That's the acceleration due to gravity. Now, as you're,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@918.64_939.6800000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@918.64_939.6800000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@918.64_939.6800000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@918.64_939.6800000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@918.64_939.6800000000001#4.jpg" ], "ocr_qwen2_vl_72b": "- Mantle: 67.0%, 3,000 K, 4.5 g/cm\u00b3\n- Outer Core: 30.8%, 5,200 K, 10.9 g/cm\u00b3\n- Inner Core: 1.7%, 5,700 K, 12.9 g/cm\u00b3\n\n**Earth Explorer**\n\n**Instructions**\n- Data Analysis\n - The Radius is 5798 km\n - The Depth is 573 km.\n - The Density is 3.94 g/cm\u00b3.\n - The Pressure is 20.28 GPa.\n - The Gravity is 18.9 m/s\u00b2.\n - The Temperature is 1010 K.\n - The S-wave velocity is 5.44 km/s.\n - The P-wave velocity is 10 km/s." }, { "vid": "u0W-NQ-4RTU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Planetary Interiors in Planetary Geology_30.json#####audio#####doingASR#####FinishASR/u0W-NQ-4RTU/939.6800000000001_963.12.mp4", "refined_asr": " Going down, the gravity is increasing, the temperature is increasing. And notice the S-wave velocity and the P-wave velocity is also slowly increasing as I go down. So, like that, then like that. Well, the S-wave velocity is negative. That's weird. The P-wave velocity then 0.48.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@939.6800000000001_963.12#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@939.6800000000001_963.12#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@939.6800000000001_963.12#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@939.6800000000001_963.12#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@939.6800000000001_963.12#4.jpg" ], "ocr_qwen2_vl_72b": "- Mantle: 67.0%, 3,000 K, 4.5 g/cm\u00b3\n- Outer Core: 30.8%, 5,200 K, 10.9 g/cm\u00b3\n- Inner Core: 1.7%, 5,700 K, 12.9 g/cm\u00b3\n\n**Earth Explorer**\n\n**Instructions**\n- Data Analysis:\n - The Radius is 5989 km\n - The Depth is 382 km.\n - The Density is 3.53 g/cm\u00b3.\n - The Pressure is 12.33 GPa.\n - The Gravity is 14.5 m/s\u00b2\n - The Temperature is 1784 K.\n - The S-wave velocity is 4.76 km/s.\n - The P-wave velocity is 8.87 km/s." }, { "vid": "u0W-NQ-4RTU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Planetary Interiors in Planetary Geology_30.json#####audio#####doingASR#####FinishASR/u0W-NQ-4RTU/965.84_988.48.mp4", "refined_asr": " Like that. So the P-waves are always faster and the temperature keeps increasing. Now if I go here S-wave velocity is zero. What do you notice there? You see the gravity at that point is 9.8. So right there you see", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@965.84_988.48#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@965.84_988.48#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@965.84_988.48#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@965.84_988.48#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u0W-NQ-4RTU/u0W-NQ-4RTU@965.84_988.48#4.jpg" ], "ocr_qwen2_vl_72b": "### Table\n\n| Layer | Percentage | Temperature (K) | Density (g/cm\u00b3) |\n|-------|------------|-----------------|----------------|\n| Mantle | 67.0% | 3,000 K | 4.5 |\n| Outer Core | 30.8% | 5,200 K | 10.9 |\n| Inner Core | 1.7% | 5,700 K | 12.9 |\n\n### Earth Explorer\n\n#### Instructions\n**Data Analysis**\n- The Radius is 4555 km\n- The Depth is 1816 km.\n- The Density is 3.33 g/cm\u00b3.\n- The Pressure is 57.77 GPa.\n- The Gravity is 12.9 m/s\u00b2\n- The Temperature is 3613 K\n- The S-wave velocity is 1.31 km/s.\n- The P-wave velocity is 16.8 km/s." } ], "image_num": 7, "text_num": 414, "token_num": 4446 }, { "images": [ "sample_100_images/e-cTygNbEUE@6619.76_6630.66#1.jpg", "sample_100_images/e-cTygNbEUE@6619.76_6630.66#2.jpg", null, "sample_100_images/e-cTygNbEUE@6630.76_6642.5599999999995#1.jpg", "sample_100_images/e-cTygNbEUE@6630.76_6642.5599999999995#2.jpg", null, "sample_100_images/e-cTygNbEUE@6642.66_6658.5599999999995#1.jpg", "sample_100_images/e-cTygNbEUE@6642.66_6658.5599999999995#2.jpg", null, "sample_100_images/e-cTygNbEUE@6658.66_6671.46#1.jpg", "sample_100_images/e-cTygNbEUE@6658.66_6671.46#2.jpg", null, "sample_100_images/e-cTygNbEUE@6671.56_6692.46#1.jpg", "sample_100_images/e-cTygNbEUE@6671.56_6692.46#2.jpg", "sample_100_images/e-cTygNbEUE@6671.56_6692.46#3.jpg", null, "sample_100_images/e-cTygNbEUE@6692.56_6713.36#1.jpg", "sample_100_images/e-cTygNbEUE@6692.56_6713.36#2.jpg", "sample_100_images/e-cTygNbEUE@6692.56_6713.36#3.jpg", null, "sample_100_images/e-cTygNbEUE@6713.46_6726.26#1.jpg", "sample_100_images/e-cTygNbEUE@6713.46_6726.26#2.jpg", null, "sample_100_images/e-cTygNbEUE@6726.360000000001_6749.26#1.jpg", "sample_100_images/e-cTygNbEUE@6726.360000000001_6749.26#2.jpg", "sample_100_images/e-cTygNbEUE@6726.360000000001_6749.26#3.jpg", null, "sample_100_images/e-cTygNbEUE@6749.360000000001_6763.26#1.jpg", "sample_100_images/e-cTygNbEUE@6749.360000000001_6763.26#2.jpg", null ], "texts": [ null, null, " Because of the power rule and all that stuff. So have a look. Anyway let me just write this down again. V is equal to one minus that. It's just to the power of a half. So it's not because I'm taking V to be this.", null, null, " I'm actually working this out. So that's how I end up with v being equal to y to the one-half power. It's a coincidence that they are equal in this case. But anyway, have a look. I will differentiate that.", null, null, " So I get d v d x equals bring down the power first that can represent bring down the power first and the minus one So I will get one half and then y to the power of minus one half which is negative one half", null, null, " You see, right here, you do the power rule. You have the minus one. Yeah, and here you have that extra one. So, they cancel out. Very nice stuff. And then you'll see you negate the power, so they will cancel.", null, null, null, " Very nice stuff. Alright, enough talking. I forgot the dy dx though. There. Okay, so here we go. Multiply this equation by this extra part. This is my recommendation whenever you're doing the Bernoulli substitution. It's easier this way.", null, null, null, " In my opinion you don't have to deal with rational exponents sometimes or negative exponents sometimes. This right here is better anyway. This times that we get the following which is positive one-half Phi negative one-half dy dx this and that.", null, null, " It's plus one half. This and that is what? This is the first power and that, so it's Y to the positive one half power. Just make a note of that. And this and that cancel out, but we still have one half.", null, null, null, " So this is equal to one half like this very nice stuff. Now have a look what's this? That's precisely our d v d x very good d v d x plus one half and the y to the one half power is our phi.", null, null, " And this right here is simply equal to one half. It's linear in the D N X world. Like that. Okay, so integrating factor. So here we go. We will have to do this right here. Let me see if I should have the space right." ], "text_ocr_list": [ null, null, "We can see these text from the image: (12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(\\frac{dy}{dx} + y = y^{\\frac{1}{2}}\\)\n\n\\(v = y^{1-\\frac{1}{2}}\\).\n Because of the power rule and all that stuff. So have a look. Anyway let me just write this down again. V is equal to one minus that. It's just to the power of a half. So it's not because I'm taking V to be this.", null, null, "We can see these text from the image: 12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(\\frac{dy}{dx} + y = y^{\\frac{1}{2}}\\)\n\n\\(v = y^{-\\frac{1}{2}}, v = y^{\\frac{1}{2}}\\).\n I'm actually working this out. So that's how I end up with v being equal to y to the one-half power. It's a coincidence that they are equal in this case. But anyway, have a look. I will differentiate that.", null, null, "We can see these text from the image: 12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(\\frac{dy}{dx} + y = y^{\\frac{1}{2}}\\)\n\n\\(v = y^{1-\\frac{1}{2}}, v = y^{\\frac{1}{2}}, \\frac{dv}{dx} = \\).\n So I get d v d x equals bring down the power first that can represent bring down the power first and the minus one So I will get one half and then y to the power of minus one half which is negative one half", null, null, "We can see these text from the image: (12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(v = y^{-\\frac{1}{2}}\\), \\(v = \\sqrt{\\frac{dy}{dx}} = \\frac{1}{2}y^{-\\frac{1}{2}}\\).\n You see, right here, you do the power rule. You have the minus one. Yeah, and here you have that extra one. So, they cancel out. Very nice stuff. And then you'll see you negate the power, so they will cancel.", null, null, null, "We can see these text from the image: 12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(1st: \\frac{1}{\\sqrt{y} - y} dy = dx\\)\n\n\\(2nd: \\frac{dy}{dx} + y = y^{\\frac{1}{2}}\\)\n\n\\(V = y^{1-\\frac{1}{2}}, V = y^{\\frac{1}{2}}, \\frac{dV}{dx} = \\frac{1}{2} y^{-\\frac{1}{2}}\\).\n Very nice stuff. Alright, enough talking. I forgot the dy dx though. There. Okay, so here we go. Multiply this equation by this extra part. This is my recommendation whenever you're doing the Bernoulli substitution. It's easier this way.", null, null, null, "We can see these text from the image: (12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(1st: \\frac{1}{\\sqrt{y} - y} dy = dx\\)\n\n\\(2nd: dy = dx\\)\n\n\\((\\frac{dy}{dx} + y = y^{\\frac{1}{2}}) \\frac{1}{2} y^{-\\frac{1}{2}}\\)\n\n\\(v = y^{-\\frac{1}{2}}, v = y^{\\frac{1}{2}}, \\frac{dv}{dx} = \\frac{1}{2} y^{-\\frac{1}{2}} \\frac{dy}{dx}\\).\n In my opinion you don't have to deal with rational exponents sometimes or negative exponents sometimes. This right here is better anyway. This times that we get the following which is positive one-half Phi negative one-half dy dx this and that.", null, null, "We can see these text from the image: (12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(v = y^{-\\frac{1}{2}}\\)\n\n\\(\\frac{1}{2} y^{-\\frac{1}{2}} \\frac{dy}{dx} + \\frac{1}{2}\\).\n It's plus one half. This and that is what? This is the first power and that, so it's Y to the positive one half power. Just make a note of that. And this and that cancel out, but we still have one half.", null, null, null, "We can see these text from the image: 12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n1st: \\(\\frac{1}{\\sqrt{y} - y} dy = dx\\)\n\n2nd:\n\n\\((\\frac{dy}{dx} + y = y^{\\frac{1}{2}})\\frac{1}{2}y^{-\\frac{1}{2}}\\)\n\n\\(v = y^{-\\frac{1}{2}}, v = y^{\\frac{1}{2}}, \\frac{dv}{dx} = \\frac{1}{2}y^{-\\frac{1}{2}}\\frac{dy}{dx}\\)\n\n\\(\\frac{1}{2}y^{-\\frac{1}{2}}\\frac{dy}{dx} + \\frac{1}{2}y^{\\frac{1}{2}} = \\frac{1}{2}\\).\n So this is equal to one half like this very nice stuff. Now have a look what's this? That's precisely our d v d x very good d v d x plus one half and the y to the one half power is our phi.", null, null, "We can see these text from the image: 12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\nLet: \\(\\frac{1}{\\sqrt{y} - y} dy = dx\\)\n\n\\((\\frac{dy}{dx} + y = y^{\\frac{1}{2}})\\frac{1}{2}y^{-\\frac{1}{2}}\\)\n\n\\(v = y^{-\\frac{1}{2}}, v = y^{\\frac{1}{2}}\\), \\(\\frac{dv}{dx} = \\frac{1}{2}y^{-\\frac{1}{2}}\\frac{dy}{dx}\\)\n\n\\(\\frac{1}{2}y^{-\\frac{1}{2}}\\frac{dy}{dx} + \\frac{1}{2}y^{\\frac{1}{2}} = \\frac{1}{2}\\)\n\n\\(\\frac{dv}{dx} + \\frac{1}{2}v = \\frac{1}{2}\\).\n And this right here is simply equal to one half. It's linear in the D N X world. Like that. Okay, so integrating factor. So here we go. We will have to do this right here. Let me see if I should have the space right." ], "metadata": [ { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/6619.76_6630.66.mp4", "refined_asr": " Because of the power rule and all that stuff. So have a look. Anyway let me just write this down again. V is equal to one minus that. It's just to the power of a half. So it's not because I'm taking V to be this.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6619.76_6630.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6619.76_6630.66#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6619.76_6630.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6619.76_6630.66#2.jpg" ], "ocr_qwen2_vl_72b": "(12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(\\frac{dy}{dx} + y = y^{\\frac{1}{2}}\\)\n\n\\(v = y^{1-\\frac{1}{2}}\\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/6630.76_6642.5599999999995.mp4", "refined_asr": " I'm actually working this out. So that's how I end up with v being equal to y to the one-half power. It's a coincidence that they are equal in this case. But anyway, have a look. I will differentiate that.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6630.76_6642.5599999999995#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6630.76_6642.5599999999995#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6630.76_6642.5599999999995#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6630.76_6642.5599999999995#2.jpg" ], "ocr_qwen2_vl_72b": "12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(\\frac{dy}{dx} + y = y^{\\frac{1}{2}}\\)\n\n\\(v = y^{-\\frac{1}{2}}, v = y^{\\frac{1}{2}}\\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/6642.66_6658.5599999999995.mp4", "refined_asr": " So I get d v d x equals bring down the power first that can represent bring down the power first and the minus one So I will get one half and then y to the power of minus one half which is negative one half", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6642.66_6658.5599999999995#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6642.66_6658.5599999999995#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6642.66_6658.5599999999995#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6642.66_6658.5599999999995#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6642.66_6658.5599999999995#3.jpg" ], "ocr_qwen2_vl_72b": "12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(\\frac{dy}{dx} + y = y^{\\frac{1}{2}}\\)\n\n\\(v = y^{1-\\frac{1}{2}}, v = y^{\\frac{1}{2}}, \\frac{dv}{dx} = \\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/6658.66_6671.46.mp4", "refined_asr": " You see, right here, you do the power rule. You have the minus one. Yeah, and here you have that extra one. So, they cancel out. Very nice stuff. And then you'll see you negate the power, so they will cancel.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6658.66_6671.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6658.66_6671.46#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6658.66_6671.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6658.66_6671.46#2.jpg" ], "ocr_qwen2_vl_72b": "(12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(v = y^{-\\frac{1}{2}}\\), \\(v = \\sqrt{\\frac{dy}{dx}} = \\frac{1}{2}y^{-\\frac{1}{2}}\\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/6671.56_6692.46.mp4", "refined_asr": " Very nice stuff. Alright, enough talking. I forgot the dy dx though. There. Okay, so here we go. Multiply this equation by this extra part. This is my recommendation whenever you're doing the Bernoulli substitution. It's easier this way.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6671.56_6692.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6671.56_6692.46#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6671.56_6692.46#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6671.56_6692.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6671.56_6692.46#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6671.56_6692.46#3.jpg" ], "ocr_qwen2_vl_72b": "12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(1st: \\frac{1}{\\sqrt{y} - y} dy = dx\\)\n\n\\(2nd: \\frac{dy}{dx} + y = y^{\\frac{1}{2}}\\)\n\n\\(V = y^{1-\\frac{1}{2}}, V = y^{\\frac{1}{2}}, \\frac{dV}{dx} = \\frac{1}{2} y^{-\\frac{1}{2}}\\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/6692.56_6713.36.mp4", "refined_asr": " In my opinion you don't have to deal with rational exponents sometimes or negative exponents sometimes. This right here is better anyway. This times that we get the following which is positive one-half Phi negative one-half dy dx this and that.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6692.56_6713.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6692.56_6713.36#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6692.56_6713.36#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6692.56_6713.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6692.56_6713.36#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6692.56_6713.36#3.jpg" ], "ocr_qwen2_vl_72b": "(12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(1st: \\frac{1}{\\sqrt{y} - y} dy = dx\\)\n\n\\(2nd: dy = dx\\)\n\n\\((\\frac{dy}{dx} + y = y^{\\frac{1}{2}}) \\frac{1}{2} y^{-\\frac{1}{2}}\\)\n\n\\(v = y^{-\\frac{1}{2}}, v = y^{\\frac{1}{2}}, \\frac{dv}{dx} = \\frac{1}{2} y^{-\\frac{1}{2}} \\frac{dy}{dx}\\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/6713.46_6726.26.mp4", "refined_asr": " It's plus one half. This and that is what? This is the first power and that, so it's Y to the positive one half power. Just make a note of that. And this and that cancel out, but we still have one half.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6713.46_6726.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6713.46_6726.26#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6713.46_6726.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6713.46_6726.26#2.jpg" ], "ocr_qwen2_vl_72b": "(12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n\\(v = y^{-\\frac{1}{2}}\\)\n\n\\(\\frac{1}{2} y^{-\\frac{1}{2}} \\frac{dy}{dx} + \\frac{1}{2}\\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/6726.360000000001_6749.26.mp4", "refined_asr": " So this is equal to one half like this very nice stuff. Now have a look what's this? That's precisely our d v d x very good d v d x plus one half and the y to the one half power is our phi.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6726.360000000001_6749.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6726.360000000001_6749.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6726.360000000001_6749.26#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6726.360000000001_6749.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6726.360000000001_6749.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6726.360000000001_6749.26#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6726.360000000001_6749.26#4.jpg" ], "ocr_qwen2_vl_72b": "12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\n1st: \\(\\frac{1}{\\sqrt{y} - y} dy = dx\\)\n\n2nd:\n\n\\((\\frac{dy}{dx} + y = y^{\\frac{1}{2}})\\frac{1}{2}y^{-\\frac{1}{2}}\\)\n\n\\(v = y^{-\\frac{1}{2}}, v = y^{\\frac{1}{2}}, \\frac{dv}{dx} = \\frac{1}{2}y^{-\\frac{1}{2}}\\frac{dy}{dx}\\)\n\n\\(\\frac{1}{2}y^{-\\frac{1}{2}}\\frac{dy}{dx} + \\frac{1}{2}y^{\\frac{1}{2}} = \\frac{1}{2}\\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/6749.360000000001_6763.26.mp4", "refined_asr": " And this right here is simply equal to one half. It's linear in the D N X world. Like that. Okay, so integrating factor. So here we go. We will have to do this right here. Let me see if I should have the space right.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6749.360000000001_6763.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6749.360000000001_6763.26#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6749.360000000001_6763.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@6749.360000000001_6763.26#2.jpg" ], "ocr_qwen2_vl_72b": "12) \\(\\frac{dy}{dx} = \\sqrt{y} - y\\)\n\nLet: \\(\\frac{1}{\\sqrt{y} - y} dy = dx\\)\n\n\\((\\frac{dy}{dx} + y = y^{\\frac{1}{2}})\\frac{1}{2}y^{-\\frac{1}{2}}\\)\n\n\\(v = y^{-\\frac{1}{2}}, v = y^{\\frac{1}{2}}\\), \\(\\frac{dv}{dx} = \\frac{1}{2}y^{-\\frac{1}{2}}\\frac{dy}{dx}\\)\n\n\\(\\frac{1}{2}y^{-\\frac{1}{2}}\\frac{dy}{dx} + \\frac{1}{2}y^{\\frac{1}{2}} = \\frac{1}{2}\\)\n\n\\(\\frac{dv}{dx} + \\frac{1}{2}v = \\frac{1}{2}\\)" } ], "image_num": 21, "text_num": 499, "token_num": 12595 }, { "images": [ "sample_100_images/e-cTygNbEUE@15397.84_15414.64#1.jpg", "sample_100_images/e-cTygNbEUE@15397.84_15414.64#2.jpg", null, "sample_100_images/e-cTygNbEUE@15415.48_15431.38#1.jpg", "sample_100_images/e-cTygNbEUE@15415.48_15431.38#2.jpg", null, "sample_100_images/e-cTygNbEUE@15431.88_15451.24#1.jpg", "sample_100_images/e-cTygNbEUE@15431.88_15451.24#2.jpg", "sample_100_images/e-cTygNbEUE@15431.88_15451.24#3.jpg", null, "sample_100_images/e-cTygNbEUE@15451.800000000001_15468.6#1.jpg", "sample_100_images/e-cTygNbEUE@15451.800000000001_15468.6#2.jpg", null, "sample_100_images/e-cTygNbEUE@15468.800000000001_15486.18#1.jpg", "sample_100_images/e-cTygNbEUE@15468.800000000001_15486.18#2.jpg", null, "sample_100_images/e-cTygNbEUE@15486.18_15502.68#1.jpg", "sample_100_images/e-cTygNbEUE@15486.18_15502.68#2.jpg", null, "sample_100_images/e-cTygNbEUE@15502.68_15517.18#1.jpg", "sample_100_images/e-cTygNbEUE@15502.68_15517.18#2.jpg", null ], "texts": [ null, null, " All right. So this right here is an integrating factor. Crazy stuff. Now multiply every term by this guy, which is right here: square root of x, e, and then we have 3 over 4, x to the 3 over 2.", null, null, " Like this. Okay so now as I said I'll try to fit in everything because this is going to be the most satisfying man. Some space actually oh what a bummer. This times this. Now do this in your head yeah.", null, null, null, " This time, this is going to give you the derivative of a product of the linear integrating factor, which is the square root of x. Let me just write it down better: Square root of x, e to the 3 over 4.", null, null, " x to the three-halves. Yeah, so we've done this so many times already. And then we're going to multiply by w, so the product of these two functions. And I know this is also a product, but then you do this times that.", null, null, " So be really careful. Here we have negative and then you have the square, all that stuff, square root of x, e to the 3 over 4, x to the 3 over 2. Right, so not so bad, not so bad.", null, null, " Now as usual you integrate both sides so you're going to just integrate this integrate that dx this right so this and that cancel very nice and you're going to just do a few things in your head", null, null, " W, it's going to be the following. Integrating this, guess what you can do. U sub. Let u equal to 3 over 4 x to the 3 over 2." ], "text_ocr_list": [ null, null, "We can see these text from the image: 22. \\( y = -x + \\frac{1}{2x}y + y^2, y(1) = 0 \\)\n\n(hint: \\( Y_1 = x^n \\))\n\n\\( Y_1 = x^n, Y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow Y_1 = x^{\\frac{1}{2}} \\)\n\nLet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2x}(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( -v^2 \\frac{dv}{dx} - (2\\sqrt{x} + 2\\sqrt{x})v = v^2 \\)\n\n\\( w = v^{-2}, w' = -2v^{-3} \\frac{dv}{dx} = -v^{-2} \\frac{dv}{dx} \\)\n\nRicatti:\n\n\\( y' = a(x) + b(x)y + c(x)y^2 \\)\n\nstart with a known solution\n\nthen\n\n\\( -v^2 \\frac{dv}{dx} + (2\\sqrt{x} + 2\\sqrt{x})v = -v^2 \\)\n\n\\( \\frac{dw}{dx} + (2\\sqrt{x} + 2\\sqrt{x})w = -1 \\).\n All right. So this right here is an integrating factor. Crazy stuff. Now multiply every term by this guy, which is right here: square root of x, e, and then we have 3 over 4, x to the 3 over 2.", null, null, "We can see these text from the image: 22) \\( y = -x + \\frac{1}{2x}y + y^2, y(1) = 0 \\)\n\n(hint: \\( y_1 = x^n \\))\n\n\\( y_1 = x^n, y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^{n-1} + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2x}(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( -v^2 \\left( \\frac{dv}{dx} - \\left( \\frac{1}{2x} + 2\\sqrt{x} \\right)v = v^2 \\right) \\)\n\n\\( w = v^{-2}, w' = -2v^{-3} \\frac{dw}{dx} = -V^{-2} \\frac{dv}{dx} \\)\n\nRicatti: \\( y' = a(x) + b(x)y + c(x)y^2 \\)\n\nstart with a known solution \\( y_1 \\)\n\nthen \\( y = y_1 + v \\ldots \\)\n\n\\( -v^2 \\frac{dv}{dx} + \\left( \\frac{1}{2x} + 2\\sqrt{x} \\right)v^2 = -1 \\)\n\n\\( \\sqrt{x}e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\left( \\frac{dw}{dx} + \\left( \\frac{1}{2x} + 2\\sqrt{x} \\right)w = -1 \\right) \\)\n\n\\( M(x) = e^{\\frac{1}{2}x^{\\frac{1}{2}} + \\frac{4}{3}x^{\\frac{3}{2}}} \\)\n\n\\( = e^{\\frac{1}{2}x^{\\frac{1}{2}}} e^{\\frac{4}{3}x^{\\frac{3}{2}}} \\)\n\n\\( = \\sqrt{x} e^{\\frac{4}{3}x^{\\frac{3}{2}}} \\).\n Like this. Okay so now as I said I'll try to fit in everything because this is going to be the most satisfying man. Some space actually oh what a bummer. This times this. Now do this in your head yeah.", null, null, null, "We can see these text from the image: 22. \\( y = -x + \\frac{1}{2}xy + y^2, y(1) = 0 \\)\n\n(hint: \\( y_1 = x^n \\))\n\n\\( y_1 = x^n, y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^n + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2}x(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2}x + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( -v(\\frac{dv}{dx} - (2x + 2\\sqrt{x}))v = v^2 \\)\n\n\\( w = v^{1-2}, w = v^{-1}, \\frac{dw}{dx} = -v^{-2}\\frac{dv}{dx} \\)\n\nRiccati:\n\nstart with a\n\nthen\n\n\\( -V^2\\frac{dV}{dx} + (2x + 2\\sqrt{x})V = \\sqrt{x}e^{3x^2} \\)\n\n\\( \\left( \\frac{dw}{dx} + (2x + 2\\sqrt{x})w = \\sqrt{x}e^{3x^2} \\right) \\).\n This time, this is going to give you the derivative of a product of the linear integrating factor, which is the square root of x. Let me just write it down better: Square root of x, e to the 3 over 4.", null, null, "We can see these text from the image: 22. \\( y' = -x + \\frac{1}{2x}y + y^2, y(1) = 0 \\)\n\n(hint: \\( Y_1 = x^n \\))\n\n\\( Y_1 = x^n, Y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^{n-1} + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2}, Y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2x}(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + \\frac{1}{2}v + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( -v\\left(\\frac{dv}{dx}\\right) - (\\frac{1}{2x} + 2\\sqrt{x})v = v^2 \\)\n\n\\( w = v^{-2}, w = v^{-1} - 1, \\frac{dw}{dx} = -v^{-2}\\frac{dv}{dx} \\)\n\nRicatti: \\( y' = a(x) + b(x)y + c(x)y^2 \\)\n\nstart with a known solution\n\nthen \\( y = y_1 + \\frac{1}{w} \\)\n\n\\( -v^2\\frac{dv}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})v^2 = -1 \\)\n\n\\( \\sqrt{x}e^{\\frac{3}{2}x^2}\\left(\\frac{dw}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})w = -1\\right) \\)\n\n\\( \\frac{d}{dx}(\\sqrt{x}e^{\\frac{3}{2}x^2}) = e^{\\frac{3}{2}x^2} \\).\n x to the three-halves. Yeah, so we've done this so many times already. And then we're going to multiply by w, so the product of these two functions. And I know this is also a product, but then you do this times that.", null, null, "We can see these text from the image: 22. \\( y' = -x + \\frac{1}{2x}y + y^2 \\), \\( y(1) = 0 \\)\n\n(hint: \\( Y_1 = x^n \\))\n\n\\( Y_1 = x^n \\), \\( Y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^{n-1} + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow Y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v \\), \\( dy = \\frac{1}{2\\sqrt{x}} + dv \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2x}(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{dv}{dx} - (\\frac{1}{2\\sqrt{x}} + 2\\sqrt{x})v = v^2 \\)\n\n\\( W = V^{-2} \\), \\( W = V^{-1} \\cdot \\frac{dW}{dx} = -V^{-2}\\frac{dv}{dx} \\)\n\nRicatti: \\( y' = a(x) + b(x)y + c(x)y^2 \\)\n\nstart with a known sol. \\( Y_1 \\)\n\nthen \\( y = Y_1 + v \\)\n\n\\( -V^2\\frac{dv}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})V^2 = -1 \\)\n\n\\( \\sqrt{x}e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\left( \\frac{dw}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})w = -1 \\right) \\)\n\n\\( M(x) = e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\)\n\n\\( = e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\).\n So be really careful. Here we have negative and then you have the square, all that stuff, square root of x, e to the 3 over 4, x to the 3 over 2. Right, so not so bad, not so bad.", null, null, "We can see these text from the image: 22) \\( y = -x + \\frac{1}{2}xy + y^2, y(1) = 0 \\)\n\n(hint: \\( Y_1 = x^n \\))\n\n\\( Y_1 = x^n, Y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^{n-1} + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow Y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2}x(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2}x + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2}x + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{dv}{dx} - (\\frac{1}{2\\sqrt{x}} + 2\\sqrt{x})v = v^2 \\)\n\n\\( w = v^{-2}, w' = -2v^{-3}\\frac{dv}{dx} = -V^{-2}\\frac{dv}{dx} \\)\n\nRicatti:\n\nstart with a known\n\nthen \\( y = \\)\n\n\\( -V^2\\frac{dv}{dx} + (\\frac{1}{2\\sqrt{x}} + 2\\sqrt{x})V = -1 \\)\n\n\\( \\frac{d}{dx}(\\sqrt{x}e^{\\frac{3}{2}x^{\\frac{3}{2}}}) = e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\)\n\n\\( \\frac{d}{dx}(\\sqrt{x}e^{\\frac{3}{2}x^{\\frac{3}{2}}}) = -\\sqrt{x}e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\).\n Now as usual you integrate both sides so you're going to just integrate this integrate that dx this right so this and that cancel very nice and you're going to just do a few things in your head", null, null, "We can see these text from the image: 22) \\( y = -x + \\frac{1}{2x}y + y^2, y(1) = 0 \\)\n\n(hint: \\( Y_1 = x^n \\))\n\n\\( Y_1 = x^n, Y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow Y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2x}(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + \\frac{1}{2x}v + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + \\frac{1}{2x}v + x + 2\\sqrt{x}v + v^2 - \\frac{1}{2\\sqrt{x}} \\)\n\n\\( \\frac{dv}{dx} = -x + \\frac{1}{2x}v + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{dv}{dx} = -x + \\frac{1}{2x}v + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{dv}{dx} = -x + \\frac{1}{2x}v + x + 2\\sqrt{x}v + v^2 \\)\n\nRiccati:\n\nstart with a known solution\n\nthen \\( y = \\frac{1}{u} \\)\n\n\\( -V^2 \\frac{dv}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})v = -1 \\)\n\n\\( \\sqrt{x}e^{\\frac{3}{4}x^2} \\left( \\frac{dw}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})w \\right) = -1 \\)\n\n\\( \\frac{d}{dx} \\left( \\sqrt{x}e^{\\frac{3}{4}x^2} w \\right) = -\\sqrt{x}e^{\\frac{3}{4}x^2} \\)\n\n\\( w = \\)\n\n\\( W = V^{-2}, W = V^{-1} \\)\n\n\\( \\frac{dw}{dx} = -V^{-2} \\frac{dv}{dx} \\).\n W, it's going to be the following. Integrating this, guess what you can do. U sub. Let u equal to 3 over 4 x to the 3 over 2." ], "metadata": [ { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/15397.84_15414.64.mp4", "refined_asr": " All right. So this right here is an integrating factor. Crazy stuff. Now multiply every term by this guy, which is right here: square root of x, e, and then we have 3 over 4, x to the 3 over 2.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15397.84_15414.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15397.84_15414.64#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15397.84_15414.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15397.84_15414.64#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15397.84_15414.64#3.jpg" ], "ocr_qwen2_vl_72b": "22. \\( y = -x + \\frac{1}{2x}y + y^2, y(1) = 0 \\)\n\n(hint: \\( Y_1 = x^n \\))\n\n\\( Y_1 = x^n, Y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow Y_1 = x^{\\frac{1}{2}} \\)\n\nLet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2x}(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( -v^2 \\frac{dv}{dx} - (2\\sqrt{x} + 2\\sqrt{x})v = v^2 \\)\n\n\\( w = v^{-2}, w' = -2v^{-3} \\frac{dv}{dx} = -v^{-2} \\frac{dv}{dx} \\)\n\nRicatti:\n\n\\( y' = a(x) + b(x)y + c(x)y^2 \\)\n\nstart with a known solution\n\nthen\n\n\\( -v^2 \\frac{dv}{dx} + (2\\sqrt{x} + 2\\sqrt{x})v = -v^2 \\)\n\n\\( \\frac{dw}{dx} + (2\\sqrt{x} + 2\\sqrt{x})w = -1 \\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/15415.48_15431.38.mp4", "refined_asr": " Like this. Okay so now as I said I'll try to fit in everything because this is going to be the most satisfying man. Some space actually oh what a bummer. This times this. Now do this in your head yeah.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15415.48_15431.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15415.48_15431.38#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15415.48_15431.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15415.48_15431.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15415.48_15431.38#3.jpg" ], "ocr_qwen2_vl_72b": "22) \\( y = -x + \\frac{1}{2x}y + y^2, y(1) = 0 \\)\n\n(hint: \\( y_1 = x^n \\))\n\n\\( y_1 = x^n, y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^{n-1} + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2x}(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( -v^2 \\left( \\frac{dv}{dx} - \\left( \\frac{1}{2x} + 2\\sqrt{x} \\right)v = v^2 \\right) \\)\n\n\\( w = v^{-2}, w' = -2v^{-3} \\frac{dw}{dx} = -V^{-2} \\frac{dv}{dx} \\)\n\nRicatti: \\( y' = a(x) + b(x)y + c(x)y^2 \\)\n\nstart with a known solution \\( y_1 \\)\n\nthen \\( y = y_1 + v \\ldots \\)\n\n\\( -v^2 \\frac{dv}{dx} + \\left( \\frac{1}{2x} + 2\\sqrt{x} \\right)v^2 = -1 \\)\n\n\\( \\sqrt{x}e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\left( \\frac{dw}{dx} + \\left( \\frac{1}{2x} + 2\\sqrt{x} \\right)w = -1 \\right) \\)\n\n\\( M(x) = e^{\\frac{1}{2}x^{\\frac{1}{2}} + \\frac{4}{3}x^{\\frac{3}{2}}} \\)\n\n\\( = e^{\\frac{1}{2}x^{\\frac{1}{2}}} e^{\\frac{4}{3}x^{\\frac{3}{2}}} \\)\n\n\\( = \\sqrt{x} e^{\\frac{4}{3}x^{\\frac{3}{2}}} \\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/15431.88_15451.24.mp4", "refined_asr": " This time, this is going to give you the derivative of a product of the linear integrating factor, which is the square root of x. Let me just write it down better: Square root of x, e to the 3 over 4.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15431.88_15451.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15431.88_15451.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15431.88_15451.24#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15431.88_15451.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15431.88_15451.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15431.88_15451.24#3.jpg" ], "ocr_qwen2_vl_72b": "22. \\( y = -x + \\frac{1}{2}xy + y^2, y(1) = 0 \\)\n\n(hint: \\( y_1 = x^n \\))\n\n\\( y_1 = x^n, y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^n + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2}x(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2}x + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( -v(\\frac{dv}{dx} - (2x + 2\\sqrt{x}))v = v^2 \\)\n\n\\( w = v^{1-2}, w = v^{-1}, \\frac{dw}{dx} = -v^{-2}\\frac{dv}{dx} \\)\n\nRiccati:\n\nstart with a\n\nthen\n\n\\( -V^2\\frac{dV}{dx} + (2x + 2\\sqrt{x})V = \\sqrt{x}e^{3x^2} \\)\n\n\\( \\left( \\frac{dw}{dx} + (2x + 2\\sqrt{x})w = \\sqrt{x}e^{3x^2} \\right) \\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/15451.800000000001_15468.6.mp4", "refined_asr": " x to the three-halves. Yeah, so we've done this so many times already. And then we're going to multiply by w, so the product of these two functions. And I know this is also a product, but then you do this times that.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15451.800000000001_15468.6#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15451.800000000001_15468.6#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15451.800000000001_15468.6#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15451.800000000001_15468.6#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15451.800000000001_15468.6#3.jpg" ], "ocr_qwen2_vl_72b": "22. \\( y' = -x + \\frac{1}{2x}y + y^2, y(1) = 0 \\)\n\n(hint: \\( Y_1 = x^n \\))\n\n\\( Y_1 = x^n, Y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^{n-1} + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2}, Y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2x}(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + \\frac{1}{2}v + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( -v\\left(\\frac{dv}{dx}\\right) - (\\frac{1}{2x} + 2\\sqrt{x})v = v^2 \\)\n\n\\( w = v^{-2}, w = v^{-1} - 1, \\frac{dw}{dx} = -v^{-2}\\frac{dv}{dx} \\)\n\nRicatti: \\( y' = a(x) + b(x)y + c(x)y^2 \\)\n\nstart with a known solution\n\nthen \\( y = y_1 + \\frac{1}{w} \\)\n\n\\( -v^2\\frac{dv}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})v^2 = -1 \\)\n\n\\( \\sqrt{x}e^{\\frac{3}{2}x^2}\\left(\\frac{dw}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})w = -1\\right) \\)\n\n\\( \\frac{d}{dx}(\\sqrt{x}e^{\\frac{3}{2}x^2}) = e^{\\frac{3}{2}x^2} \\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/15468.800000000001_15486.18.mp4", "refined_asr": " So be really careful. Here we have negative and then you have the square, all that stuff, square root of x, e to the 3 over 4, x to the 3 over 2. Right, so not so bad, not so bad.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15468.800000000001_15486.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15468.800000000001_15486.18#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15468.800000000001_15486.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15468.800000000001_15486.18#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15468.800000000001_15486.18#3.jpg" ], "ocr_qwen2_vl_72b": "22. \\( y' = -x + \\frac{1}{2x}y + y^2 \\), \\( y(1) = 0 \\)\n\n(hint: \\( Y_1 = x^n \\))\n\n\\( Y_1 = x^n \\), \\( Y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^{n-1} + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow Y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v \\), \\( dy = \\frac{1}{2\\sqrt{x}} + dv \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2x}(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{dv}{dx} - (\\frac{1}{2\\sqrt{x}} + 2\\sqrt{x})v = v^2 \\)\n\n\\( W = V^{-2} \\), \\( W = V^{-1} \\cdot \\frac{dW}{dx} = -V^{-2}\\frac{dv}{dx} \\)\n\nRicatti: \\( y' = a(x) + b(x)y + c(x)y^2 \\)\n\nstart with a known sol. \\( Y_1 \\)\n\nthen \\( y = Y_1 + v \\)\n\n\\( -V^2\\frac{dv}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})V^2 = -1 \\)\n\n\\( \\sqrt{x}e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\left( \\frac{dw}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})w = -1 \\right) \\)\n\n\\( M(x) = e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\)\n\n\\( = e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/15486.18_15502.68.mp4", "refined_asr": " Now as usual you integrate both sides so you're going to just integrate this integrate that dx this right so this and that cancel very nice and you're going to just do a few things in your head", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15486.18_15502.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15486.18_15502.68#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15486.18_15502.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15486.18_15502.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15486.18_15502.68#3.jpg" ], "ocr_qwen2_vl_72b": "22) \\( y = -x + \\frac{1}{2}xy + y^2, y(1) = 0 \\)\n\n(hint: \\( Y_1 = x^n \\))\n\n\\( Y_1 = x^n, Y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2}x^{n-1} + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow Y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2}x(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2}x + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2}x + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{dv}{dx} - (\\frac{1}{2\\sqrt{x}} + 2\\sqrt{x})v = v^2 \\)\n\n\\( w = v^{-2}, w' = -2v^{-3}\\frac{dv}{dx} = -V^{-2}\\frac{dv}{dx} \\)\n\nRicatti:\n\nstart with a known\n\nthen \\( y = \\)\n\n\\( -V^2\\frac{dv}{dx} + (\\frac{1}{2\\sqrt{x}} + 2\\sqrt{x})V = -1 \\)\n\n\\( \\frac{d}{dx}(\\sqrt{x}e^{\\frac{3}{2}x^{\\frac{3}{2}}}) = e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\)\n\n\\( \\frac{d}{dx}(\\sqrt{x}e^{\\frac{3}{2}x^{\\frac{3}{2}}}) = -\\sqrt{x}e^{\\frac{3}{2}x^{\\frac{3}{2}}} \\)" }, { "vid": "e-cTygNbEUE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on First Order Exact Differential Equations_30.json#####audio#####doingASR#####FinishASR/e-cTygNbEUE/15502.68_15517.18.mp4", "refined_asr": " W, it's going to be the following. Integrating this, guess what you can do. U sub. Let u equal to 3 over 4 x to the 3 over 2.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15502.68_15517.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15502.68_15517.18#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15502.68_15517.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15502.68_15517.18#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/e-cTygNbEUE/e-cTygNbEUE@15502.68_15517.18#3.jpg" ], "ocr_qwen2_vl_72b": "22) \\( y = -x + \\frac{1}{2x}y + y^2, y(1) = 0 \\)\n\n(hint: \\( Y_1 = x^n \\))\n\n\\( Y_1 = x^n, Y_1' = nx^{n-1} \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + (x^n)^2 \\)\n\n\\( nx^{n-1} = -x + \\frac{1}{2x}x^n + x^{2n} \\)\n\nSuggest that \\( n = \\frac{1}{2} \\Rightarrow Y_1 = x^{\\frac{1}{2}} \\)\n\nlet \\( y = \\sqrt{x} + v, \\frac{dy}{dx} = \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2x}(\\sqrt{x} + v) + (\\sqrt{x} + v)^2 \\)\n\n\\( \\frac{1}{2\\sqrt{x}} + \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + \\frac{1}{2x}v + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{dv}{dx} = -x + \\frac{1}{2\\sqrt{x}} + \\frac{1}{2x}v + x + 2\\sqrt{x}v + v^2 - \\frac{1}{2\\sqrt{x}} \\)\n\n\\( \\frac{dv}{dx} = -x + \\frac{1}{2x}v + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{dv}{dx} = -x + \\frac{1}{2x}v + x + 2\\sqrt{x}v + v^2 \\)\n\n\\( \\frac{dv}{dx} = -x + \\frac{1}{2x}v + x + 2\\sqrt{x}v + v^2 \\)\n\nRiccati:\n\nstart with a known solution\n\nthen \\( y = \\frac{1}{u} \\)\n\n\\( -V^2 \\frac{dv}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})v = -1 \\)\n\n\\( \\sqrt{x}e^{\\frac{3}{4}x^2} \\left( \\frac{dw}{dx} + (\\frac{1}{2x} + 2\\sqrt{x})w \\right) = -1 \\)\n\n\\( \\frac{d}{dx} \\left( \\sqrt{x}e^{\\frac{3}{4}x^2} w \\right) = -\\sqrt{x}e^{\\frac{3}{4}x^2} \\)\n\n\\( w = \\)\n\n\\( W = V^{-2}, W = V^{-1} \\)\n\n\\( \\frac{dw}{dx} = -V^{-2} \\frac{dv}{dx} \\)" } ], "image_num": 15, "text_num": 360, "token_num": 9000 }, { "images": [ "sample_100_images/5BXMCud20rg@23620.66_23637.2#1.jpg", null, "sample_100_images/5BXMCud20rg@23637.2_23658.239999999998#1.jpg", "sample_100_images/5BXMCud20rg@23637.2_23658.239999999998#3.jpg", null, "sample_100_images/5BXMCud20rg@23658.239999999998_23673.399999999998#1.jpg", null, "sample_100_images/5BXMCud20rg@23673.399999999998_23687.36#1.jpg", "sample_100_images/5BXMCud20rg@23673.399999999998_23687.36#2.jpg", null, "sample_100_images/5BXMCud20rg@23687.36_23704.28#1.jpg", "sample_100_images/5BXMCud20rg@23687.36_23704.28#2.jpg", null, "sample_100_images/5BXMCud20rg@23704.28_23718.219999999998#1.jpg", "sample_100_images/5BXMCud20rg@23704.28_23718.219999999998#2.jpg", null, "sample_100_images/5BXMCud20rg@23718.219999999998_23751.22#1.jpg", "sample_100_images/5BXMCud20rg@23718.219999999998_23751.22#3.jpg", "sample_100_images/5BXMCud20rg@23718.219999999998_23751.22#4.jpg", null ], "texts": [ null, " So, again, this type of building wouldn't necessarily lend itself to a component design approach. We have a lot of complexity with the exterior frames that really act more like moment frames. We have a lot of load sharing that would happen between, say, solid shaft walls and other structural elements.", null, null, " These exterior frames. Understanding how the force distribution would work in a project like that also requires a finite element analysis to get the right forces and stresses for the project. These continuous lintels that I talked about earlier in terms of how to make those work.", null, " And eliminate control joints. You know, maybe that's just one elevation of this fire station. You know, we have these apparatus bays that have, you know, pretty big openings. And the rest of the masonry project is going to be pretty standard.", null, null, " So maybe that's going to be one elevation that we look at with a finite element analysis And we can utilize that whether it's three doors five doors seven doors I can look at what the actual stresses and forces are and what the movements are going to be like.", null, null, " Like you know for that elevation, I also can do a comparison and contrast. If I had isolated columns, if I had something like this, you know, what is going to be my deflection on that elevation? If I have in-plane loads, I'm going to have all of the load that's going to be resolved.", null, null, " With very thin, tall piers, I'm going to have stresses that are a lot higher. I'm going to have a much higher load demand on those piers. And then, I can also do a compare and contrast between the two of them.", null, null, null, " I can see which one's going to deflect a lot more and which one, again, allows us to see what the relative stresses are between the two different solutions. Other elements that I think are fairly complicated and certainly can utilize tools like finite element analysis, anytime we have shaft walls. What I encourage designers to do is one." ], "text_ocr_list": [ null, "We can see these text from the image: Complex walls - how to approach this building\n\n- Again, choosing to model buildings like this without FE will lead to many inefficiencies and poor analysis and design\n- The only way to appropriately analyze and appropriately allocate the gravity and lateral loads to each lintel/pier/column/wall and design is with a finite element analysis program.\n So, again, this type of building wouldn't necessarily lend itself to a component design approach. We have a lot of complexity with the exterior frames that really act more like moment frames. We have a lot of load sharing that would happen between, say, solid shaft walls and other structural elements.", null, null, "We can see these text from the image: Complex walls\n- how to approach this building\n\n- Again, choosing to model buildings like this without FE will lead to many inefficiencies and poor analysis and design\n- The only way to appropriately analyze and appropriately allocate the gravity and lateral loads to each lintel/pier/column/wall and design is with a finite element analysis program.\n These exterior frames. Understanding how the force distribution would work in a project like that also requires a finite element analysis to get the right forces and stresses for the project. These continuous lintels that I talked about earlier in terms of how to make those work.", null, "We can see these text from the image: Buildings with a lot of openings or large openings are best modeled/analyzed/designed with FE\n\nsuggest no control joint be used on this area, utilize continuous masonry lintel\n\nFORSE.\n And eliminate control joints. You know, maybe that's just one elevation of this fire station. You know, we have these apparatus bays that have, you know, pretty big openings. And the rest of the masonry project is going to be pretty standard.", null, null, "We can see these text from the image: Buildings with a lot of openings or large openings are best modeled/analyzed/designed with FE\n\nsuggest no control joint be used on this area, utilize continuous masonry lintel\n\nFORSE.\n So maybe that's going to be one elevation that we look at with a finite element analysis And we can utilize that whether it's three doors five doors seven doors I can look at what the actual stresses and forces are and what the movements are going to be like.", null, null, "We can see these text from the image: Example - Single Story Wall with Openings\n\nEliminate CJ and Use Masonry Lintels\n\nreasonable deflection and stress.\n Like you know for that elevation, I also can do a comparison and contrast. If I had isolated columns, if I had something like this, you know, what is going to be my deflection on that elevation? If I have in-plane loads, I'm going to have all of the load that's going to be resolved.", null, null, "We can see these text from the image: Example - Single Story Wall with Openings if Using CJ and Steel Lintels\n\nvery little capacity with the isolated masonry piers to support steel beam lintels (NOT IDEAL).\n With very thin, tall piers, I'm going to have stresses that are a lot higher. I'm going to have a much higher load demand on those piers. And then, I can also do a compare and contrast between the two of them.", null, null, null, "We can see these text from the image: Example - Single Story Wall with Openings\n\nCompare Masonry Lintels vs. CJ and Steel Lintels\n\nFront\n\nFORSE Consulting, LLC.\n I can see which one's going to deflect a lot more and which one, again, allows us to see what the relative stresses are between the two different solutions. Other elements that I think are fairly complicated and certainly can utilize tools like finite element analysis, anytime we have shaft walls. What I encourage designers to do is one." ], "metadata": [ { "vid": "5BXMCud20rg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Civil Engineering Structural Engineering tutorials on Design of Masonry Structures_30.json#####audio#####doingASR#####FinishASR/5BXMCud20rg/23620.66_23637.2.mp4", "refined_asr": " So, again, this type of building wouldn't necessarily lend itself to a component design approach. We have a lot of complexity with the exterior frames that really act more like moment frames. We have a lot of load sharing that would happen between, say, solid shaft walls and other structural elements.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23620.66_23637.2#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23620.66_23637.2#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23620.66_23637.2#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23620.66_23637.2#3.jpg" ], "ocr_qwen2_vl_72b": "Complex walls - how to approach this building\n\n- Again, choosing to model buildings like this without FE will lead to many inefficiencies and poor analysis and design\n- The only way to appropriately analyze and appropriately allocate the gravity and lateral loads to each lintel/pier/column/wall and design is with a finite element analysis program" }, { "vid": "5BXMCud20rg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Civil Engineering Structural Engineering tutorials on Design of Masonry Structures_30.json#####audio#####doingASR#####FinishASR/5BXMCud20rg/23637.2_23658.239999999998.mp4", "refined_asr": " These exterior frames. Understanding how the force distribution would work in a project like that also requires a finite element analysis to get the right forces and stresses for the project. These continuous lintels that I talked about earlier in terms of how to make those work.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23637.2_23658.239999999998#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23637.2_23658.239999999998#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23637.2_23658.239999999998#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23637.2_23658.239999999998#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23637.2_23658.239999999998#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23637.2_23658.239999999998#4.jpg" ], "ocr_qwen2_vl_72b": "Complex walls\n- how to approach this building\n\n- Again, choosing to model buildings like this without FE will lead to many inefficiencies and poor analysis and design\n- The only way to appropriately analyze and appropriately allocate the gravity and lateral loads to each lintel/pier/column/wall and design is with a finite element analysis program" }, { "vid": "5BXMCud20rg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Civil Engineering Structural Engineering tutorials on Design of Masonry Structures_30.json#####audio#####doingASR#####FinishASR/5BXMCud20rg/23658.239999999998_23673.399999999998.mp4", "refined_asr": " And eliminate control joints. You know, maybe that's just one elevation of this fire station. You know, we have these apparatus bays that have, you know, pretty big openings. And the rest of the masonry project is going to be pretty standard.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23658.239999999998_23673.399999999998#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23658.239999999998_23673.399999999998#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23658.239999999998_23673.399999999998#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23658.239999999998_23673.399999999998#3.jpg" ], "ocr_qwen2_vl_72b": "Buildings with a lot of openings or large openings are best modeled/analyzed/designed with FE\n\nsuggest no control joint be used on this area, utilize continuous masonry lintel\n\nFORSE" }, { "vid": "5BXMCud20rg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Civil Engineering Structural Engineering tutorials on Design of Masonry Structures_30.json#####audio#####doingASR#####FinishASR/5BXMCud20rg/23673.399999999998_23687.36.mp4", "refined_asr": " So maybe that's going to be one elevation that we look at with a finite element analysis And we can utilize that whether it's three doors five doors seven doors I can look at what the actual stresses and forces are and what the movements are going to be like.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23673.399999999998_23687.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23673.399999999998_23687.36#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23673.399999999998_23687.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23673.399999999998_23687.36#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23673.399999999998_23687.36#3.jpg" ], "ocr_qwen2_vl_72b": "Buildings with a lot of openings or large openings are best modeled/analyzed/designed with FE\n\nsuggest no control joint be used on this area, utilize continuous masonry lintel\n\nFORSE" }, { "vid": "5BXMCud20rg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Civil Engineering Structural Engineering tutorials on Design of Masonry Structures_30.json#####audio#####doingASR#####FinishASR/5BXMCud20rg/23687.36_23704.28.mp4", "refined_asr": " Like you know for that elevation, I also can do a comparison and contrast. If I had isolated columns, if I had something like this, you know, what is going to be my deflection on that elevation? If I have in-plane loads, I'm going to have all of the load that's going to be resolved.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23687.36_23704.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23687.36_23704.28#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23687.36_23704.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23687.36_23704.28#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23687.36_23704.28#3.jpg" ], "ocr_qwen2_vl_72b": "Example - Single Story Wall with Openings\n\nEliminate CJ and Use Masonry Lintels\n\nreasonable deflection and stress" }, { "vid": "5BXMCud20rg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Civil Engineering Structural Engineering tutorials on Design of Masonry Structures_30.json#####audio#####doingASR#####FinishASR/5BXMCud20rg/23704.28_23718.219999999998.mp4", "refined_asr": " With very thin, tall piers, I'm going to have stresses that are a lot higher. I'm going to have a much higher load demand on those piers. And then, I can also do a compare and contrast between the two of them.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23704.28_23718.219999999998#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23704.28_23718.219999999998#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23704.28_23718.219999999998#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23704.28_23718.219999999998#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23704.28_23718.219999999998#3.jpg" ], "ocr_qwen2_vl_72b": "Example - Single Story Wall with Openings if Using CJ and Steel Lintels\n\nvery little capacity with the isolated masonry piers to support steel beam lintels (NOT IDEAL)" }, { "vid": "5BXMCud20rg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Civil Engineering Structural Engineering tutorials on Design of Masonry Structures_30.json#####audio#####doingASR#####FinishASR/5BXMCud20rg/23718.219999999998_23751.22.mp4", "refined_asr": " I can see which one's going to deflect a lot more and which one, again, allows us to see what the relative stresses are between the two different solutions. Other elements that I think are fairly complicated and certainly can utilize tools like finite element analysis, anytime we have shaft walls. What I encourage designers to do is one.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23718.219999999998_23751.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23718.219999999998_23751.22#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23718.219999999998_23751.22#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23718.219999999998_23751.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23718.219999999998_23751.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23718.219999999998_23751.22#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23718.219999999998_23751.22#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5BXMCud20rg/5BXMCud20rg@23718.219999999998_23751.22#5.jpg" ], "ocr_qwen2_vl_72b": "Example - Single Story Wall with Openings\n\nCompare Masonry Lintels vs. CJ and Steel Lintels\n\nFront\n\nFORSE Consulting, LLC" } ], "image_num": 13, "text_num": 442, "token_num": 7930 }, { "images": [ "sample_100_images/2WPA1L9uJqo@1586.1200000000001_1607.1200000000001#1.jpg", "sample_100_images/2WPA1L9uJqo@1586.1200000000001_1607.1200000000001#2.jpg", null, "sample_100_images/2WPA1L9uJqo@1607.1200000000001_1624.1200000000001#1.jpg", null, "sample_100_images/2WPA1L9uJqo@1624.1200000000001_1641.1200000000001#1.jpg", "sample_100_images/2WPA1L9uJqo@1624.1200000000001_1641.1200000000001#2.jpg", null, "sample_100_images/2WPA1L9uJqo@1641.1200000000001_1658.1200000000001#1.jpg", "sample_100_images/2WPA1L9uJqo@1641.1200000000001_1658.1200000000001#2.jpg", null, "sample_100_images/2WPA1L9uJqo@1658.1200000000001_1676.1200000000001#1.jpg", null, "sample_100_images/2WPA1L9uJqo@1676.1200000000001_1692.1200000000001#1.jpg", null, "sample_100_images/2WPA1L9uJqo@1692.1200000000001_1709.1200000000001#1.jpg", null ], "texts": [ null, null, " Corresponds to something real that can be determined experimentally. In 1926, the physicist Max Born suggested that the modulus squared of the wavefunction represents the probability of finding a particle at a particular location in space. In other words, if at some instant in time t", null, " If a measurement is made to locate the particle associated with the wave function Psi then the probability that the particle will be found at a coordinate between X and X plus DX is given by the following expression. As we've just seen the modulus squared of the wave function represents this probability density.", null, null, " It is calculated by multiplying the wavefunction by its complex conjugate. If we then want to find the probability that the particle is located within a given region, we simply need to integrate the modulus squared. So, for example, the probability of finding a particle.", null, null, " The probability density of a particle being located between two points a and b is given by the integral of the modulus squared evaluated with a and b as the limits of the integral. The justification of Born's postulate can be found in the following considerations since the motion of a particle is connected", null, " With the propagation of an associated wave function, these two entities must be associated in space. That is, the particle must be at some location where the waves have an appreciable amplitude. And therefore, the probability density must have an appreciable value where the wave function has an appreciable value.", null, " Furthermore, since the probability density is a measurable quantity, it should be a real, positive-valued number, whereas the wavefunction is in general complex. And therefore, it's obviously not possible to equate the probability simply with the wavefunction itself.", null, " However since the modulus squared of the wave function is always real and non-negative Born had reason to believe that it referred to a measurable quantity so what's this all telling us? Born's interpretation introduces a kind of indeterminacy into quantum mechanics." ], "text_ocr_list": [ null, null, "We can see these text from the image: Complex Numbers\n\nz = a + ib\n\n|z|^2 = a^2 + b^2\n\nA complex quantity cannot be measured by any actual physical instrument\n\n\u03a8(x, t) is not a measurable quantity\n\nRepresents a real number\n\nPerhaps |\u03a8(x, t)|^2 corresponds to something real that can be determined experimentally.\n Corresponds to something real that can be determined experimentally. In 1926, the physicist Max Born suggested that the modulus squared of the wavefunction represents the probability of finding a particle at a particular location in space. In other words, if at some instant in time t", null, "We can see these text from the image: Max Born\u2019s Interpretation\n\nx\nx + dx\n\nProbability of finding particle between x and x + dx\n\nMax Born\n(1882-1970).\n If a measurement is made to locate the particle associated with the wave function Psi then the probability that the particle will be found at a coordinate between X and X plus DX is given by the following expression. As we've just seen the modulus squared of the wave function represents this probability density.", null, null, "We can see these text from the image: Max Born\u2019s Interpretation\n\nProbability of finding particle between \\( x \\) and \\( x + dx \\)\n\n\\[ P(x,t) \\, dx = |\\Psi(x,t)|^2 \\, dx \\]\n\n\\[ |\\Psi(x,t)|^2 = \\Psi^*(x,t) \\Psi(x,t) \\]\n\nMax Born (1882-1970).\n It is calculated by multiplying the wavefunction by its complex conjugate. If we then want to find the probability that the particle is located within a given region, we simply need to integrate the modulus squared. So, for example, the probability of finding a particle.", null, null, "We can see these text from the image: - Max Born\u2019s Interpretation\n- Probability of finding particle between \\( x \\) and \\( x + dx \\)\n- \\( P(x,t) \\, dx = |\\Psi(x,t)|^2 \\, dx \\)\n- \\( |\\Psi(x,t)|^2 = \\Psi^*(x,t) \\Psi(x,t) \\)\n- \\( a \\leq x \\leq b \\).\n The probability density of a particle being located between two points a and b is given by the integral of the modulus squared evaluated with a and b as the limits of the integral. The justification of Born's postulate can be found in the following considerations since the motion of a particle is connected", null, "We can see these text from the image: Justification\n\nWave an entities associate.\n With the propagation of an associated wave function, these two entities must be associated in space. That is, the particle must be at some location where the waves have an appreciable amplitude. And therefore, the probability density must have an appreciable value where the wave function has an appreciable value.", null, "We can see these text from the image: Justification\n\nWave and Particle entities must be associated in space\n\nParticle must be at some location where the waves have an appreciable amplitude\n\nThe probability density \\( P(x, t) \\) is a measurable quantity.\n Furthermore, since the probability density is a measurable quantity, it should be a real, positive-valued number, whereas the wavefunction is in general complex. And therefore, it's obviously not possible to equate the probability simply with the wavefunction itself.", null, "We can see these text from the image: Justification\n\nWave and Particle entities must be associated in space\n\nParticle must be at some location where the waves have an appreciable amplitude\n\n\u03a8*(x, t)\u03a8(x, t) is always real and non-negative\n\nIt should be a real positive valued number\n\nThe probability density P(x, t) is a measurable quantity.\n However since the modulus squared of the wave function is always real and non-negative Born had reason to believe that it referred to a measurable quantity so what's this all telling us? Born's interpretation introduces a kind of indeterminacy into quantum mechanics." ], "metadata": [ { "vid": "2WPA1L9uJqo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lessons on Schrodinger Equation in Quantum Mechanics_30.json#####audio#####doingASR#####FinishASR/2WPA1L9uJqo/1586.1200000000001_1607.1200000000001.mp4", "refined_asr": " Corresponds to something real that can be determined experimentally. In 1926, the physicist Max Born suggested that the modulus squared of the wavefunction represents the probability of finding a particle at a particular location in space. In other words, if at some instant in time t", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1586.1200000000001_1607.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1586.1200000000001_1607.1200000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1586.1200000000001_1607.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1586.1200000000001_1607.1200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1586.1200000000001_1607.1200000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1586.1200000000001_1607.1200000000001#4.jpg" ], "ocr_qwen2_vl_72b": "Complex Numbers\n\nz = a + ib\n\n|z|^2 = a^2 + b^2\n\nA complex quantity cannot be measured by any actual physical instrument\n\n\u03a8(x, t) is not a measurable quantity\n\nRepresents a real number\n\nPerhaps |\u03a8(x, t)|^2 corresponds to something real that can be determined experimentally" }, { "vid": "2WPA1L9uJqo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lessons on Schrodinger Equation in Quantum Mechanics_30.json#####audio#####doingASR#####FinishASR/2WPA1L9uJqo/1607.1200000000001_1624.1200000000001.mp4", "refined_asr": " If a measurement is made to locate the particle associated with the wave function Psi then the probability that the particle will be found at a coordinate between X and X plus DX is given by the following expression. As we've just seen the modulus squared of the wave function represents this probability density.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1607.1200000000001_1624.1200000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1607.1200000000001_1624.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1607.1200000000001_1624.1200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1607.1200000000001_1624.1200000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Max Born\u2019s Interpretation\n\nx\nx + dx\n\nProbability of finding particle between x and x + dx\n\nMax Born\n(1882-1970)" }, { "vid": "2WPA1L9uJqo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lessons on Schrodinger Equation in Quantum Mechanics_30.json#####audio#####doingASR#####FinishASR/2WPA1L9uJqo/1624.1200000000001_1641.1200000000001.mp4", "refined_asr": " It is calculated by multiplying the wavefunction by its complex conjugate. If we then want to find the probability that the particle is located within a given region, we simply need to integrate the modulus squared. So, for example, the probability of finding a particle.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1624.1200000000001_1641.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1624.1200000000001_1641.1200000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1624.1200000000001_1641.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1624.1200000000001_1641.1200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1624.1200000000001_1641.1200000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Max Born\u2019s Interpretation\n\nProbability of finding particle between \\( x \\) and \\( x + dx \\)\n\n\\[ P(x,t) \\, dx = |\\Psi(x,t)|^2 \\, dx \\]\n\n\\[ |\\Psi(x,t)|^2 = \\Psi^*(x,t) \\Psi(x,t) \\]\n\nMax Born (1882-1970)" }, { "vid": "2WPA1L9uJqo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lessons on Schrodinger Equation in Quantum Mechanics_30.json#####audio#####doingASR#####FinishASR/2WPA1L9uJqo/1641.1200000000001_1658.1200000000001.mp4", "refined_asr": " The probability density of a particle being located between two points a and b is given by the integral of the modulus squared evaluated with a and b as the limits of the integral. The justification of Born's postulate can be found in the following considerations since the motion of a particle is connected", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1641.1200000000001_1658.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1641.1200000000001_1658.1200000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1641.1200000000001_1658.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1641.1200000000001_1658.1200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1641.1200000000001_1658.1200000000001#3.jpg" ], "ocr_qwen2_vl_72b": "- Max Born\u2019s Interpretation\n- Probability of finding particle between \\( x \\) and \\( x + dx \\)\n- \\( P(x,t) \\, dx = |\\Psi(x,t)|^2 \\, dx \\)\n- \\( |\\Psi(x,t)|^2 = \\Psi^*(x,t) \\Psi(x,t) \\)\n- \\( a \\leq x \\leq b \\)" }, { "vid": "2WPA1L9uJqo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lessons on Schrodinger Equation in Quantum Mechanics_30.json#####audio#####doingASR#####FinishASR/2WPA1L9uJqo/1658.1200000000001_1676.1200000000001.mp4", "refined_asr": " With the propagation of an associated wave function, these two entities must be associated in space. That is, the particle must be at some location where the waves have an appreciable amplitude. And therefore, the probability density must have an appreciable value where the wave function has an appreciable value.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1658.1200000000001_1676.1200000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1658.1200000000001_1676.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1658.1200000000001_1676.1200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1658.1200000000001_1676.1200000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Justification\n\nWave an entities associate" }, { "vid": "2WPA1L9uJqo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lessons on Schrodinger Equation in Quantum Mechanics_30.json#####audio#####doingASR#####FinishASR/2WPA1L9uJqo/1676.1200000000001_1692.1200000000001.mp4", "refined_asr": " Furthermore, since the probability density is a measurable quantity, it should be a real, positive-valued number, whereas the wavefunction is in general complex. And therefore, it's obviously not possible to equate the probability simply with the wavefunction itself.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1676.1200000000001_1692.1200000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1676.1200000000001_1692.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1676.1200000000001_1692.1200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1676.1200000000001_1692.1200000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Justification\n\nWave and Particle entities must be associated in space\n\nParticle must be at some location where the waves have an appreciable amplitude\n\nThe probability density \\( P(x, t) \\) is a measurable quantity" }, { "vid": "2WPA1L9uJqo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lessons on Schrodinger Equation in Quantum Mechanics_30.json#####audio#####doingASR#####FinishASR/2WPA1L9uJqo/1692.1200000000001_1709.1200000000001.mp4", "refined_asr": " However since the modulus squared of the wave function is always real and non-negative Born had reason to believe that it referred to a measurable quantity so what's this all telling us? Born's interpretation introduces a kind of indeterminacy into quantum mechanics.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1692.1200000000001_1709.1200000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1692.1200000000001_1709.1200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1692.1200000000001_1709.1200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/2WPA1L9uJqo/2WPA1L9uJqo@1692.1200000000001_1709.1200000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Justification\n\nWave and Particle entities must be associated in space\n\nParticle must be at some location where the waves have an appreciable amplitude\n\n\u03a8*(x, t)\u03a8(x, t) is always real and non-negative\n\nIt should be a real positive valued number\n\nThe probability density P(x, t) is a measurable quantity" } ], "image_num": 10, "text_num": 422, "token_num": 6182 }, { "images": [ "sample_100_images/pvrIagjEk4c@275.14_290.06#1.jpg", null, "sample_100_images/pvrIagjEk4c@293.18_312.32#1.jpg", "sample_100_images/pvrIagjEk4c@293.18_312.32#2.jpg", null, "sample_100_images/pvrIagjEk4c@312.32_335.18#1.jpg", "sample_100_images/pvrIagjEk4c@312.32_335.18#2.jpg", null, "sample_100_images/pvrIagjEk4c@335.18_352.78000000000003#1.jpg", null, "sample_100_images/pvrIagjEk4c@352.78000000000003_371.82#1.jpg", "sample_100_images/pvrIagjEk4c@352.78000000000003_371.82#2.jpg", null, "sample_100_images/pvrIagjEk4c@376.44_392.09999999999997#1.jpg", "sample_100_images/pvrIagjEk4c@376.44_392.09999999999997#2.jpg", null, "sample_100_images/pvrIagjEk4c@392.09999999999997_415.46000000000004#1.jpg", "sample_100_images/pvrIagjEk4c@392.09999999999997_415.46000000000004#3.jpg", null ], "texts": [ null, " In physics, engineering, and science, we're going to start thinking about what are some of those properties of those PDEs that are particularly useful. So maybe one of the things I'll start with is I'll just kind of describe some of the canonical PDEs.", null, null, " And all of these canonical PDEs that I'm going to describe have some common properties. They are going to be linear, and I'm going to define what I mean in a minute. So they're linear, they are second order, and they are what is known as homogeneous.", null, null, " Depending on whether it's homogeneous. Okay, so I'm going to define what each of these means. A linear PDE just means that the partial differential equation can be written in terms of linear operators. There are no non-linear terms like u squared or u times u sub x, and things like that.", null, " Let me make a quick sidebar. That oftentimes we use subscripts like u sub t to denote partial u over partial t. And things like u sub x or u sub xx would equal the second partial derivative of u with respect to x squared.", null, null, " Okay so oftentimes partial derivatives are denoted by subscripts and I'm going to do the same thing here. So linear basically means that there are no non-linear terms. There's no things like products or cross products no products between my various u sub x terms.", null, null, " So I don't have any u squareds or u times u x's or u x times u x x's. Those would be nonlinear. And I'm going to talk actually about linear more later. So just remember that second order means that there's no derivatives higher than two.", null, null, " Higher than second order. No derivatives higher than second order. Okay, so I could technically have a fourth derivative in X, a fourth order diffusion term. There are physical systems that have fourth order diffusion. But at least in these canonical PDEs." ], "text_ocr_list": [ null, "We can see these text from the image: - PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x,t) \\)\n- function space time\n- \\( \\frac{\\partial^2 u}{\\partial t^2} = c^2 \\frac{\\partial^2 u}{\\partial x^2} \\)\n- ODE: \\( \\frac{d}{dt}x(t) = x(t) \\).\n In physics, engineering, and science, we're going to start thinking about what are some of those properties of those PDEs that are particularly useful. So maybe one of the things I'll start with is I'll just kind of describe some of the canonical PDEs.", null, null, "We can see these text from the image: - PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- Canonical PDEs\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x,t) \\)\n- function space time\n- \\( \\frac{\\partial^2 u}{\\partial t^2} = c^2 \\frac{\\partial^2 u}{\\partial x^2} \\)\n- ODE: \\( \\frac{d}{dt} x(t) = x(t) \\).\n And all of these canonical PDEs that I'm going to describe have some common properties. They are going to be linear, and I'm going to define what I mean in a minute. So they're linear, they are second order, and they are what is known as homogeneous.", null, null, "We can see these text from the image: - PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- u(x, t) function space time\n- ODE: \\(\\frac{d}{dt} x(t) = x(t)\\)\n- Canonical PDEs\n- Linear 2nd Order Homogeneous.\n Depending on whether it's homogeneous. Okay, so I'm going to define what each of these means. A linear PDE just means that the partial differential equation can be written in terms of linear operators. There are no non-linear terms like u squared or u times u sub x, and things like that.", null, "We can see these text from the image: - PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x, t) \\)\n- \\( \\frac{\\partial^2 u}{\\partial t^2} = c^2 \\frac{\\partial^2 u}{\\partial x^2} \\)\n- ODE: \\( \\frac{d}{dt} x(t) = \\lambda x(t) \\)\n- Canonical PDEs\n- Linear:\n- 2nd Order:\n- Homogeneous:.\n Let me make a quick sidebar. That oftentimes we use subscripts like u sub t to denote partial u over partial t. And things like u sub x or u sub xx would equal the second partial derivative of u with respect to x squared.", null, null, "We can see these text from the image: - PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x,t) \\)\n- function space time\n- \\( \\frac{\\partial^2 u}{\\partial t^2} = c^2 \\frac{\\partial^2 u}{\\partial x^2} \\)\n- ODE: \\( \\frac{d}{dt}x(t) = f(x(t)) \\)\n- \\( u_t = \\frac{\\partial u}{\\partial t} \\)\n- \\( u_{xx} = \\frac{\\partial^2 u}{\\partial x^2} \\)\n- Canonical PDEs\n- Linear:\n- 2nd Order:\n- Homogeneous:.\n Okay so oftentimes partial derivatives are denoted by subscripts and I'm going to do the same thing here. So linear basically means that there are no non-linear terms. There's no things like products or cross products no products between my various u sub x terms.", null, null, "We can see these text from the image: - PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x,t) \\)\n- function space time\n- ODE: \\( \\frac{d}{dt} x(t) = x(t) \\)\n- \\( u_t = \\frac{\\partial u}{\\partial t} \\)\n- \\( u_{xx} = \\frac{\\partial^2 u}{\\partial x^2} \\)\n- Canonical PDEs\n- Linear: No products between \\( u, u_x, u_{xx} \\)\n- 2nd Order\n- Homogeneous.\n So I don't have any u squareds or u times u x's or u x times u x x's. Those would be nonlinear. And I'm going to talk actually about linear more later. So just remember that second order means that there's no derivatives higher than two.", null, null, "We can see these text from the image: - Partial Differential Equations (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x,t) \\)\n- \\( \\frac{\\partial^2 u}{\\partial t^2} = c^2 \\frac{\\partial^2 u}{\\partial x^2} \\)\n- ODE: \\( \\frac{d}{dt} x(t) = f(x(t)) \\)\n- \\( u_t = \\frac{\\partial u}{\\partial t} \\)\n- \\( u_{xx} = \\frac{\\partial^2 u}{\\partial x^2} \\)\n- Canonical PDEs\n- Linear: No products between \\( u, u_x, u_{xx} \\)\n- 2nd Order: No derivative\n- Homogeneous:.\n Higher than second order. No derivatives higher than second order. Okay, so I could technically have a fourth derivative in X, a fourth order diffusion term. There are physical systems that have fourth order diffusion. But at least in these canonical PDEs." ], "metadata": [ { "vid": "pvrIagjEk4c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus course on Partial Differential Equations in Fourier Series and PDEs_30.json#####audio#####doingASR#####FinishASR/pvrIagjEk4c/275.14_290.06.mp4", "refined_asr": " In physics, engineering, and science, we're going to start thinking about what are some of those properties of those PDEs that are particularly useful. So maybe one of the things I'll start with is I'll just kind of describe some of the canonical PDEs.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@275.14_290.06#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@275.14_290.06#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@275.14_290.06#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@275.14_290.06#3.jpg" ], "ocr_qwen2_vl_72b": "- PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x,t) \\)\n- function space time\n- \\( \\frac{\\partial^2 u}{\\partial t^2} = c^2 \\frac{\\partial^2 u}{\\partial x^2} \\)\n- ODE: \\( \\frac{d}{dt}x(t) = x(t) \\)" }, { "vid": "pvrIagjEk4c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus course on Partial Differential Equations in Fourier Series and PDEs_30.json#####audio#####doingASR#####FinishASR/pvrIagjEk4c/293.18_312.32.mp4", "refined_asr": " And all of these canonical PDEs that I'm going to describe have some common properties. They are going to be linear, and I'm going to define what I mean in a minute. So they're linear, they are second order, and they are what is known as homogeneous.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@293.18_312.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@293.18_312.32#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@293.18_312.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@293.18_312.32#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@293.18_312.32#3.jpg" ], "ocr_qwen2_vl_72b": "- PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- Canonical PDEs\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x,t) \\)\n- function space time\n- \\( \\frac{\\partial^2 u}{\\partial t^2} = c^2 \\frac{\\partial^2 u}{\\partial x^2} \\)\n- ODE: \\( \\frac{d}{dt} x(t) = x(t) \\)" }, { "vid": "pvrIagjEk4c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus course on Partial Differential Equations in Fourier Series and PDEs_30.json#####audio#####doingASR#####FinishASR/pvrIagjEk4c/312.32_335.18.mp4", "refined_asr": " Depending on whether it's homogeneous. Okay, so I'm going to define what each of these means. A linear PDE just means that the partial differential equation can be written in terms of linear operators. There are no non-linear terms like u squared or u times u sub x, and things like that.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@312.32_335.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@312.32_335.18#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@312.32_335.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@312.32_335.18#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@312.32_335.18#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@312.32_335.18#4.jpg" ], "ocr_qwen2_vl_72b": "- PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- u(x, t) function space time\n- ODE: \\(\\frac{d}{dt} x(t) = x(t)\\)\n- Canonical PDEs\n- Linear 2nd Order Homogeneous" }, { "vid": "pvrIagjEk4c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus course on Partial Differential Equations in Fourier Series and PDEs_30.json#####audio#####doingASR#####FinishASR/pvrIagjEk4c/335.18_352.78000000000003.mp4", "refined_asr": " Let me make a quick sidebar. That oftentimes we use subscripts like u sub t to denote partial u over partial t. And things like u sub x or u sub xx would equal the second partial derivative of u with respect to x squared.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@335.18_352.78000000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@335.18_352.78000000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@335.18_352.78000000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@335.18_352.78000000000003#3.jpg" ], "ocr_qwen2_vl_72b": "- PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x, t) \\)\n- \\( \\frac{\\partial^2 u}{\\partial t^2} = c^2 \\frac{\\partial^2 u}{\\partial x^2} \\)\n- ODE: \\( \\frac{d}{dt} x(t) = \\lambda x(t) \\)\n- Canonical PDEs\n- Linear:\n- 2nd Order:\n- Homogeneous:" }, { "vid": "pvrIagjEk4c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus course on Partial Differential Equations in Fourier Series and PDEs_30.json#####audio#####doingASR#####FinishASR/pvrIagjEk4c/352.78000000000003_371.82.mp4", "refined_asr": " Okay so oftentimes partial derivatives are denoted by subscripts and I'm going to do the same thing here. So linear basically means that there are no non-linear terms. There's no things like products or cross products no products between my various u sub x terms.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@352.78000000000003_371.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@352.78000000000003_371.82#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@352.78000000000003_371.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@352.78000000000003_371.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@352.78000000000003_371.82#3.jpg" ], "ocr_qwen2_vl_72b": "- PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x,t) \\)\n- function space time\n- \\( \\frac{\\partial^2 u}{\\partial t^2} = c^2 \\frac{\\partial^2 u}{\\partial x^2} \\)\n- ODE: \\( \\frac{d}{dt}x(t) = f(x(t)) \\)\n- \\( u_t = \\frac{\\partial u}{\\partial t} \\)\n- \\( u_{xx} = \\frac{\\partial^2 u}{\\partial x^2} \\)\n- Canonical PDEs\n- Linear:\n- 2nd Order:\n- Homogeneous:" }, { "vid": "pvrIagjEk4c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus course on Partial Differential Equations in Fourier Series and PDEs_30.json#####audio#####doingASR#####FinishASR/pvrIagjEk4c/376.44_392.09999999999997.mp4", "refined_asr": " So I don't have any u squareds or u times u x's or u x times u x x's. Those would be nonlinear. And I'm going to talk actually about linear more later. So just remember that second order means that there's no derivatives higher than two.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@376.44_392.09999999999997#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@376.44_392.09999999999997#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@376.44_392.09999999999997#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@376.44_392.09999999999997#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@376.44_392.09999999999997#3.jpg" ], "ocr_qwen2_vl_72b": "- PARTIAL DIFFERENTIAL EQUATIONS (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x,t) \\)\n- function space time\n- ODE: \\( \\frac{d}{dt} x(t) = x(t) \\)\n- \\( u_t = \\frac{\\partial u}{\\partial t} \\)\n- \\( u_{xx} = \\frac{\\partial^2 u}{\\partial x^2} \\)\n- Canonical PDEs\n- Linear: No products between \\( u, u_x, u_{xx} \\)\n- 2nd Order\n- Homogeneous" }, { "vid": "pvrIagjEk4c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus course on Partial Differential Equations in Fourier Series and PDEs_30.json#####audio#####doingASR#####FinishASR/pvrIagjEk4c/392.09999999999997_415.46000000000004.mp4", "refined_asr": " Higher than second order. No derivatives higher than second order. Okay, so I could technically have a fourth derivative in X, a fourth order diffusion term. There are physical systems that have fourth order diffusion. But at least in these canonical PDEs.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@392.09999999999997_415.46000000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@392.09999999999997_415.46000000000004#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@392.09999999999997_415.46000000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@392.09999999999997_415.46000000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@392.09999999999997_415.46000000000004#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pvrIagjEk4c/pvrIagjEk4c@392.09999999999997_415.46000000000004#4.jpg" ], "ocr_qwen2_vl_72b": "- Partial Differential Equations (PDEs)\n- A partial differential equation relates multivariate functions and their partial derivatives\n- \\( u(x,t) \\)\n- \\( \\frac{\\partial^2 u}{\\partial t^2} = c^2 \\frac{\\partial^2 u}{\\partial x^2} \\)\n- ODE: \\( \\frac{d}{dt} x(t) = f(x(t)) \\)\n- \\( u_t = \\frac{\\partial u}{\\partial t} \\)\n- \\( u_{xx} = \\frac{\\partial^2 u}{\\partial x^2} \\)\n- Canonical PDEs\n- Linear: No products between \\( u, u_x, u_{xx} \\)\n- 2nd Order: No derivative\n- Homogeneous:" } ], "image_num": 12, "text_num": 422, "token_num": 7334 }, { "images": [ "sample_100_images/-aCF0_wfVwY@69.0_88.0#1.jpg", null, "sample_100_images/-aCF0_wfVwY@88.0_112.0#1.jpg", null, "sample_100_images/-aCF0_wfVwY@113.0_132.0#1.jpg", null, "sample_100_images/-aCF0_wfVwY@132.0_151.0#1.jpg", null, "sample_100_images/-aCF0_wfVwY@151.0_173.0#1.jpg", null, "sample_100_images/-aCF0_wfVwY@173.0_194.0#1.jpg", null, "sample_100_images/-aCF0_wfVwY@194.0_215.0#1.jpg", null ], "texts": [ null, " Before building ARIMA models, we want to check whether the time series is stationary. This means verifying if the time series has its statistical properties remaining constant across time. The reason we favor ARIMA models for time series forecasting is that they assume stationarity. A stationary time series is easier to model and make predictions for.", null, " Here's an example of a stationary time series plot. You can see that it has no clear trend. The data just oscillates around a mean of zero with constant variance. So the ARIMA model can actually be broken down into three components: AR, I, and MA, with their corresponding parameters p, d, and q.", null, " These three parameters take non-negative integer values indicating which specific ARIMA model is used. For different time series, we might be able to fit them with any of the combinations of these three components of ARIMA. That's why we say ARIMA is a general class of models.", null, " Alright, now let's take a closer look at each of these three components. AR stands for Autoregressive, it means that the time series can be linearly regressed on its own past values. The parameter corresponding to the AR part is p.", null, " As you can see in this formula, P represents the number of past values included in the AR model. Here, C is a constant, phi1 to phiP are parameters, and epsilonT is white noise. Next, the I part, which stands for integrated.", null, " As mentioned earlier, we want the time series to be stationary for ARIMA models. So if it's not stationary, the time series can be differenced to become stationary by computing the differences between consecutive observations. For example, the first-order differencing is calculated as this:", null, " The corresponding parameter D represents the number of times the time series is differenced. The third part, MA or moving average, means a time series can be regressed on the past forecast errors, as shown in this equation. The parameter Q represents the number of past forecast errors." ], "text_ocr_list": [ null, "We can see these text from the image: What is ARIMA\n\n- ARIMA (Auto-Regressive Integrated Moving Average) is a general class of statistical models for time series analysis forecasting\n- ARIMA uses a time series' past values and/or forecast errors to predict its future values\n- ARIMA model assumption - stationary: the time series has its statistical properties remain constant across time.\n Before building ARIMA models, we want to check whether the time series is stationary. This means verifying if the time series has its statistical properties remaining constant across time. The reason we favor ARIMA models for time series forecasting is that they assume stationarity. A stationary time series is easier to model and make predictions for.", null, "We can see these text from the image: What is ARIMA\n\n- ARIMA (Auto-Regressive Integrated Moving Average) is a general class of statistical models for time series analysis forecasting\n- ARIMA uses a time series' past values and/or forecast errors to predict its future values\n- ARIMA model assumption - stationary: the time series has its statistical properties remain constant across time.\n Here's an example of a stationary time series plot. You can see that it has no clear trend. The data just oscillates around a mean of zero with constant variance. So the ARIMA model can actually be broken down into three components: AR, I, and MA, with their corresponding parameters p, d, and q.", null, "We can see these text from the image: What is ARIMA\n\n- ARIMA (Auto-Regressive Integrated Moving Average) is a general class of statistical models for time series analysis forecasting\n- ARIMA uses a time series' past values and/or forecast errors to predict its future values\n- ARIMA model assumption - stationary: the time series has its statistical properties remain constant across time\n- Three components/parameters: AR + I + MA (p, d, q).\n These three parameters take non-negative integer values indicating which specific ARIMA model is used. For different time series, we might be able to fit them with any of the combinations of these three components of ARIMA. That's why we say ARIMA is a general class of models.", null, "We can see these text from the image: ARIMA(p, d, q).\n Alright, now let's take a closer look at each of these three components. AR stands for Autoregressive, it means that the time series can be linearly regressed on its own past values. The parameter corresponding to the AR part is p.", null, "We can see these text from the image: ARIMA(p, d, q)\n\n- AR (Auto-Regressive): the time series is linearly regressed on its own past values\n - p: the number of past values included in the AR model\n\n\\( y_t = c + \\phi_1 y_{t-1} + \\phi_2 y_{t-2} + ... + \\phi_p y_{t-p} + \\epsilon_t \\).\n As you can see in this formula, P represents the number of past values included in the AR model. Here, C is a constant, phi1 to phiP are parameters, and epsilonT is white noise. Next, the I part, which stands for integrated.", null, "We can see these text from the image: ARIMA(p, d, q)\n\n- AR (Auto-Regressive): the time series is linearly regressed on its own past values\n - p: the number of past values included in the AR model\n \\( y_t = c + \\phi_1 y_{t-1} + \\phi_2 y_{t-2} + ... + \\phi_p y_{t-p} + \\epsilon_t \\)\n\n- I (Integrated): if not stationary, the time series can be differenced to become stationary, i.e., compute the differences between consecutive observations.\n As mentioned earlier, we want the time series to be stationary for ARIMA models. So if it's not stationary, the time series can be differenced to become stationary by computing the differences between consecutive observations. For example, the first-order differencing is calculated as this:", null, "We can see these text from the image: ARIMA(p, d, q)\n\n- AR (Auto-Regressive): the time series is linearly regressed on its own past values\n - p: the number of past values included in the AR model\n y_t = c + \u03c6_1y_{t-1} + \u03c6_2y_{t-2} + ... + \u03c6_py_{t-p} + \u03b5_t\n\n- I (Integrated): if not stationary, the time series can be differenced to become stationary, i.e., compute the differences between consecutive observations\n - d: the number of times the time series is differenced\n \u2207y_t = y_t - y_{t-1}.\n The corresponding parameter D represents the number of times the time series is differenced. The third part, MA or moving average, means a time series can be regressed on the past forecast errors, as shown in this equation. The parameter Q represents the number of past forecast errors." ], "metadata": [ { "vid": "-aCF0_wfVwY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy Tutorial on Time Series Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/-aCF0_wfVwY/69.0_88.0.mp4", "refined_asr": " Before building ARIMA models, we want to check whether the time series is stationary. This means verifying if the time series has its statistical properties remaining constant across time. The reason we favor ARIMA models for time series forecasting is that they assume stationarity. A stationary time series is easier to model and make predictions for.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@69.0_88.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@69.0_88.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@69.0_88.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@69.0_88.0#3.jpg" ], "ocr_qwen2_vl_72b": "What is ARIMA\n\n- ARIMA (Auto-Regressive Integrated Moving Average) is a general class of statistical models for time series analysis forecasting\n- ARIMA uses a time series' past values and/or forecast errors to predict its future values\n- ARIMA model assumption - stationary: the time series has its statistical properties remain constant across time" }, { "vid": "-aCF0_wfVwY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy Tutorial on Time Series Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/-aCF0_wfVwY/88.0_112.0.mp4", "refined_asr": " Here's an example of a stationary time series plot. You can see that it has no clear trend. The data just oscillates around a mean of zero with constant variance. So the ARIMA model can actually be broken down into three components: AR, I, and MA, with their corresponding parameters p, d, and q.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@88.0_112.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@88.0_112.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@88.0_112.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@88.0_112.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@88.0_112.0#4.jpg" ], "ocr_qwen2_vl_72b": "What is ARIMA\n\n- ARIMA (Auto-Regressive Integrated Moving Average) is a general class of statistical models for time series analysis forecasting\n- ARIMA uses a time series' past values and/or forecast errors to predict its future values\n- ARIMA model assumption - stationary: the time series has its statistical properties remain constant across time" }, { "vid": "-aCF0_wfVwY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy Tutorial on Time Series Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/-aCF0_wfVwY/113.0_132.0.mp4", "refined_asr": " These three parameters take non-negative integer values indicating which specific ARIMA model is used. For different time series, we might be able to fit them with any of the combinations of these three components of ARIMA. That's why we say ARIMA is a general class of models.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@113.0_132.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@113.0_132.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@113.0_132.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@113.0_132.0#3.jpg" ], "ocr_qwen2_vl_72b": "What is ARIMA\n\n- ARIMA (Auto-Regressive Integrated Moving Average) is a general class of statistical models for time series analysis forecasting\n- ARIMA uses a time series' past values and/or forecast errors to predict its future values\n- ARIMA model assumption - stationary: the time series has its statistical properties remain constant across time\n- Three components/parameters: AR + I + MA (p, d, q)" }, { "vid": "-aCF0_wfVwY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy Tutorial on Time Series Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/-aCF0_wfVwY/132.0_151.0.mp4", "refined_asr": " Alright, now let's take a closer look at each of these three components. AR stands for Autoregressive, it means that the time series can be linearly regressed on its own past values. The parameter corresponding to the AR part is p.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@132.0_151.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@132.0_151.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@132.0_151.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@132.0_151.0#3.jpg" ], "ocr_qwen2_vl_72b": "ARIMA(p, d, q)" }, { "vid": "-aCF0_wfVwY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy Tutorial on Time Series Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/-aCF0_wfVwY/151.0_173.0.mp4", "refined_asr": " As you can see in this formula, P represents the number of past values included in the AR model. Here, C is a constant, phi1 to phiP are parameters, and epsilonT is white noise. Next, the I part, which stands for integrated.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@151.0_173.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@151.0_173.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@151.0_173.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@151.0_173.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@151.0_173.0#4.jpg" ], "ocr_qwen2_vl_72b": "ARIMA(p, d, q)\n\n- AR (Auto-Regressive): the time series is linearly regressed on its own past values\n - p: the number of past values included in the AR model\n\n\\( y_t = c + \\phi_1 y_{t-1} + \\phi_2 y_{t-2} + ... + \\phi_p y_{t-p} + \\epsilon_t \\)" }, { "vid": "-aCF0_wfVwY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy Tutorial on Time Series Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/-aCF0_wfVwY/173.0_194.0.mp4", "refined_asr": " As mentioned earlier, we want the time series to be stationary for ARIMA models. So if it's not stationary, the time series can be differenced to become stationary by computing the differences between consecutive observations. For example, the first-order differencing is calculated as this:", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@173.0_194.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@173.0_194.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@173.0_194.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@173.0_194.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@173.0_194.0#4.jpg" ], "ocr_qwen2_vl_72b": "ARIMA(p, d, q)\n\n- AR (Auto-Regressive): the time series is linearly regressed on its own past values\n - p: the number of past values included in the AR model\n \\( y_t = c + \\phi_1 y_{t-1} + \\phi_2 y_{t-2} + ... + \\phi_p y_{t-p} + \\epsilon_t \\)\n\n- I (Integrated): if not stationary, the time series can be differenced to become stationary, i.e., compute the differences between consecutive observations" }, { "vid": "-aCF0_wfVwY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy Tutorial on Time Series Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/-aCF0_wfVwY/194.0_215.0.mp4", "refined_asr": " The corresponding parameter D represents the number of times the time series is differenced. The third part, MA or moving average, means a time series can be regressed on the past forecast errors, as shown in this equation. The parameter Q represents the number of past forecast errors.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@194.0_215.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@194.0_215.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@194.0_215.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@194.0_215.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-aCF0_wfVwY/-aCF0_wfVwY@194.0_215.0#4.jpg" ], "ocr_qwen2_vl_72b": "ARIMA(p, d, q)\n\n- AR (Auto-Regressive): the time series is linearly regressed on its own past values\n - p: the number of past values included in the AR model\n y_t = c + \u03c6_1y_{t-1} + \u03c6_2y_{t-2} + ... + \u03c6_py_{t-p} + \u03b5_t\n\n- I (Integrated): if not stationary, the time series can be differenced to become stationary, i.e., compute the differences between consecutive observations\n - d: the number of times the time series is differenced\n \u2207y_t = y_t - y_{t-1}" } ], "image_num": 7, "text_num": 436, "token_num": 4468 }, { "images": [ "sample_100_images/VUU29a-I84U@2169.16_2183.3#1.jpg", null, "sample_100_images/VUU29a-I84U@2183.3_2197.76#1.jpg", null, "sample_100_images/VUU29a-I84U@2197.76_2212.28#1.jpg", null, "sample_100_images/VUU29a-I84U@2212.28_2226.34#1.jpg", null, "sample_100_images/VUU29a-I84U@2226.34_2240.0#1.jpg", null, "sample_100_images/VUU29a-I84U@2240.0_2255.28#1.jpg", null, "sample_100_images/VUU29a-I84U@2255.28_2271.2400000000002#1.jpg", null ], "texts": [ null, " Infinity. And so we're never going to be plugging in any negative values to this sequence. So this sequence is always going to produce a positive term, which means that our terms are always going to be greater than or equal to zero. They're never going to be negative.", null, " They're never going to drop below zero. And so we have a lower bound for n, specifically at zero. So our lower bound is n is equal to zero. We would say that our sequence, a sub n, is bounded below by zero.", null, " It is always greater than or equal to that value. But then what about an upper bound? Does our sequence have an upper bound? And the answer is yes. Our first term, a sub 1, which is equal to 1, right, that is the highest value of a term.", null, " That we will ever have. We already proved that our sequence is decreasing. And so, it's never going to come back up above 1. It's going to continue to decrease for all values of n. And so, 1 is the highest value that a term will ever be.", null, " And so we can say that 1 is the upper bound of our sequence. That is, the terms of our sequence will never be greater than m, and so they are less than or equal to that value of m, which is 1.", null, " So our sequence is less than or equal to one. And so we have a lower bound. What we find here is that our sequence is bounded below by zero and bounded above by one. And since it is bounded below and above we can say that it is bounded.", null, " And so not only is a sub n monotonic, but a sub n is also bounded. Therefore, since we've determined that it's both monotonic and bounded, that means this sequence does converge. Now, we weren't asked to find the specific value to which it converges." ], "text_ocr_list": [ null, "We can see these text from the image: Ex. Does the sequence {an} = {1/n^3} converge?\n\na1 = 1/1^3 = 1/1 = 1\n\na2 = 1/2^3 = 1/8 = 0.125\n\na3 = 1/3^3 = 1/27 \u2248 0.037...\n\na4 = 1/4^3 = 1/64 = 0.015625\n\n{an} is monotonic (decreasing).\n Infinity. And so we're never going to be plugging in any negative values to this sequence. So this sequence is always going to produce a positive term, which means that our terms are always going to be greater than or equal to zero. They're never going to be negative.", null, "We can see these text from the image: Ex. Does the sequence {an} = {1/n^3} converge?\n\na1 = 1/1^3 = 1/1 = 1\n\na2 = 1/2^3 = 1/8 = 0.125\n\na3 = 1/3^3 = 1/27 \u2248 0.037...\n\na4 = 1/4^3 = 1/64 = 0.015625\n\n{an} is monotonic (decreasing).\n They're never going to drop below zero. And so we have a lower bound for n, specifically at zero. So our lower bound is n is equal to zero. We would say that our sequence, a sub n, is bounded below by zero.", null, "We can see these text from the image: Ex. Does the sequence {a_n} = \\left\\{ \\frac{1}{n^3} \\right\\} converge?\n\na_1 = \\frac{1}{1^3} = \\frac{1}{1} = 1\n\na_2 = \\frac{1}{2^3} = \\frac{1}{8} = 0.125\n\na_3 = \\frac{1}{3^3} = \\frac{1}{27} \u2248 0.037...\n\na_4 = \\frac{1}{4^3} = \\frac{1}{64} = 0.015625\n\n{a_n} is monotonic (decreasing)\n\n0 \u2264 {a_n}.\n It is always greater than or equal to that value. But then what about an upper bound? Does our sequence have an upper bound? And the answer is yes. Our first term, a sub 1, which is equal to 1, right, that is the highest value of a term.", null, "We can see these text from the image: Ex. Does the sequence {a\u2099} = {1/n\u00b3} converge?\n\na\u2081 = 1/1\u00b3 = 1/1 = 1\n\na\u2082 = 1/2\u00b3 = 1/8 = 0.125\n\na\u2083 = 1/3\u00b3 = 1/27 \u2248 0.037...\n\na\u2084 = 1/4\u00b3 = 1/64 = 0.015625\n\n{a\u2099} is monotonic (decreasing)\n\n0 \u2264 {a\u2099}.\n That we will ever have. We already proved that our sequence is decreasing. And so, it's never going to come back up above 1. It's going to continue to decrease for all values of n. And so, 1 is the highest value that a term will ever be.", null, "We can see these text from the image: Ex. Does the sequence {a\u2099} = {1/n\u00b3} converge?\n\na\u2081 = 1/1\u00b3 = 1/1 = 1\n\na\u2082 = 1/2\u00b3 = 1/8 = 0.125\n\na\u2083 = 1/3\u00b3 = 1/27 \u2248 0.037...\n\na\u2084 = 1/4\u00b3 = 1/64 = 0.015625\n\n{a\u2099} is monotonic (decreasing)\n\n0 \u2264 {a\u2099}.\n And so we can say that 1 is the upper bound of our sequence. That is, the terms of our sequence will never be greater than m, and so they are less than or equal to that value of m, which is 1.", null, "We can see these text from the image: Ex. Does the sequence {an} = {1/n^3} converge?\n\na1 = 1/1^3 = 1/1 = 1\na2 = 1/2^3 = 1/8 = 0.125\na3 = 1/3^3 = 1/27 \u2248 0.037...\na4 = 1/4^3 = 1/64 = 0.015625\n\n{an} is monotonic (decreasing)\n\n0 \u2264 {an} \u2264 1.\n So our sequence is less than or equal to one. And so we have a lower bound. What we find here is that our sequence is bounded below by zero and bounded above by one. And since it is bounded below and above we can say that it is bounded.", null, "We can see these text from the image: Ex. Does the sequence \\(\\left\\{a_n\\right\\} = \\left\\{\\frac{1}{n^3}\\right\\}\\) converge?\n\n\\(a_1 = \\frac{1}{1^3} = \\frac{1}{1} = 1\\)\n\n\\(a_2 = \\frac{1}{2^3} = \\frac{1}{8} = 0.125\\)\n\n\\(a_3 = \\frac{1}{3^3} = \\frac{1}{27} \\approx 0.037...\\)\n\n\\(a_4 = \\frac{1}{4^3} = \\frac{1}{64} = 0.015625\\)\n\n\\(\\left\\{a_n\\right\\}\\) is monotonic (decreasing)\n\n\\(0 \\leq \\left\\{a_n\\right\\} \\leq 1\\).\n And so not only is a sub n monotonic, but a sub n is also bounded. Therefore, since we've determined that it's both monotonic and bounded, that means this sequence does converge. Now, we weren't asked to find the specific value to which it converges." ], "metadata": [ { "vid": "VUU29a-I84U.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Monotonic and Bounded Sequences in series and sequences_30.json#####audio#####doingASR#####FinishASR/VUU29a-I84U/2169.16_2183.3.mp4", "refined_asr": " Infinity. And so we're never going to be plugging in any negative values to this sequence. So this sequence is always going to produce a positive term, which means that our terms are always going to be greater than or equal to zero. They're never going to be negative.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2169.16_2183.3#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2169.16_2183.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2169.16_2183.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2169.16_2183.3#3.jpg" ], "ocr_qwen2_vl_72b": "Ex. Does the sequence {an} = {1/n^3} converge?\n\na1 = 1/1^3 = 1/1 = 1\n\na2 = 1/2^3 = 1/8 = 0.125\n\na3 = 1/3^3 = 1/27 \u2248 0.037...\n\na4 = 1/4^3 = 1/64 = 0.015625\n\n{an} is monotonic (decreasing)" }, { "vid": "VUU29a-I84U.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Monotonic and Bounded Sequences in series and sequences_30.json#####audio#####doingASR#####FinishASR/VUU29a-I84U/2183.3_2197.76.mp4", "refined_asr": " They're never going to drop below zero. And so we have a lower bound for n, specifically at zero. So our lower bound is n is equal to zero. We would say that our sequence, a sub n, is bounded below by zero.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2183.3_2197.76#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2183.3_2197.76#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2183.3_2197.76#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2183.3_2197.76#3.jpg" ], "ocr_qwen2_vl_72b": "Ex. Does the sequence {an} = {1/n^3} converge?\n\na1 = 1/1^3 = 1/1 = 1\n\na2 = 1/2^3 = 1/8 = 0.125\n\na3 = 1/3^3 = 1/27 \u2248 0.037...\n\na4 = 1/4^3 = 1/64 = 0.015625\n\n{an} is monotonic (decreasing)" }, { "vid": "VUU29a-I84U.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Monotonic and Bounded Sequences in series and sequences_30.json#####audio#####doingASR#####FinishASR/VUU29a-I84U/2197.76_2212.28.mp4", "refined_asr": " It is always greater than or equal to that value. But then what about an upper bound? Does our sequence have an upper bound? And the answer is yes. Our first term, a sub 1, which is equal to 1, right, that is the highest value of a term.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2197.76_2212.28#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2197.76_2212.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2197.76_2212.28#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2197.76_2212.28#3.jpg" ], "ocr_qwen2_vl_72b": "Ex. Does the sequence {a_n} = \\left\\{ \\frac{1}{n^3} \\right\\} converge?\n\na_1 = \\frac{1}{1^3} = \\frac{1}{1} = 1\n\na_2 = \\frac{1}{2^3} = \\frac{1}{8} = 0.125\n\na_3 = \\frac{1}{3^3} = \\frac{1}{27} \u2248 0.037...\n\na_4 = \\frac{1}{4^3} = \\frac{1}{64} = 0.015625\n\n{a_n} is monotonic (decreasing)\n\n0 \u2264 {a_n}" }, { "vid": "VUU29a-I84U.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Monotonic and Bounded Sequences in series and sequences_30.json#####audio#####doingASR#####FinishASR/VUU29a-I84U/2212.28_2226.34.mp4", "refined_asr": " That we will ever have. We already proved that our sequence is decreasing. And so, it's never going to come back up above 1. It's going to continue to decrease for all values of n. And so, 1 is the highest value that a term will ever be.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2212.28_2226.34#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2212.28_2226.34#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2212.28_2226.34#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2212.28_2226.34#3.jpg" ], "ocr_qwen2_vl_72b": "Ex. Does the sequence {a\u2099} = {1/n\u00b3} converge?\n\na\u2081 = 1/1\u00b3 = 1/1 = 1\n\na\u2082 = 1/2\u00b3 = 1/8 = 0.125\n\na\u2083 = 1/3\u00b3 = 1/27 \u2248 0.037...\n\na\u2084 = 1/4\u00b3 = 1/64 = 0.015625\n\n{a\u2099} is monotonic (decreasing)\n\n0 \u2264 {a\u2099}" }, { "vid": "VUU29a-I84U.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Monotonic and Bounded Sequences in series and sequences_30.json#####audio#####doingASR#####FinishASR/VUU29a-I84U/2226.34_2240.0.mp4", "refined_asr": " And so we can say that 1 is the upper bound of our sequence. That is, the terms of our sequence will never be greater than m, and so they are less than or equal to that value of m, which is 1.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2226.34_2240.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2226.34_2240.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2226.34_2240.0#2.jpg" ], "ocr_qwen2_vl_72b": "Ex. Does the sequence {a\u2099} = {1/n\u00b3} converge?\n\na\u2081 = 1/1\u00b3 = 1/1 = 1\n\na\u2082 = 1/2\u00b3 = 1/8 = 0.125\n\na\u2083 = 1/3\u00b3 = 1/27 \u2248 0.037...\n\na\u2084 = 1/4\u00b3 = 1/64 = 0.015625\n\n{a\u2099} is monotonic (decreasing)\n\n0 \u2264 {a\u2099}" }, { "vid": "VUU29a-I84U.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Monotonic and Bounded Sequences in series and sequences_30.json#####audio#####doingASR#####FinishASR/VUU29a-I84U/2240.0_2255.28.mp4", "refined_asr": " So our sequence is less than or equal to one. And so we have a lower bound. What we find here is that our sequence is bounded below by zero and bounded above by one. And since it is bounded below and above we can say that it is bounded.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2240.0_2255.28#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2240.0_2255.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2240.0_2255.28#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2240.0_2255.28#3.jpg" ], "ocr_qwen2_vl_72b": "Ex. Does the sequence {an} = {1/n^3} converge?\n\na1 = 1/1^3 = 1/1 = 1\na2 = 1/2^3 = 1/8 = 0.125\na3 = 1/3^3 = 1/27 \u2248 0.037...\na4 = 1/4^3 = 1/64 = 0.015625\n\n{an} is monotonic (decreasing)\n\n0 \u2264 {an} \u2264 1" }, { "vid": "VUU29a-I84U.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Monotonic and Bounded Sequences in series and sequences_30.json#####audio#####doingASR#####FinishASR/VUU29a-I84U/2255.28_2271.2400000000002.mp4", "refined_asr": " And so not only is a sub n monotonic, but a sub n is also bounded. Therefore, since we've determined that it's both monotonic and bounded, that means this sequence does converge. Now, we weren't asked to find the specific value to which it converges.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2255.28_2271.2400000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2255.28_2271.2400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2255.28_2271.2400000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VUU29a-I84U/VUU29a-I84U@2255.28_2271.2400000000002#3.jpg" ], "ocr_qwen2_vl_72b": "Ex. Does the sequence \\(\\left\\{a_n\\right\\} = \\left\\{\\frac{1}{n^3}\\right\\}\\) converge?\n\n\\(a_1 = \\frac{1}{1^3} = \\frac{1}{1} = 1\\)\n\n\\(a_2 = \\frac{1}{2^3} = \\frac{1}{8} = 0.125\\)\n\n\\(a_3 = \\frac{1}{3^3} = \\frac{1}{27} \\approx 0.037...\\)\n\n\\(a_4 = \\frac{1}{4^3} = \\frac{1}{64} = 0.015625\\)\n\n\\(\\left\\{a_n\\right\\}\\) is monotonic (decreasing)\n\n\\(0 \\leq \\left\\{a_n\\right\\} \\leq 1\\)" } ], "image_num": 7, "text_num": 408, "token_num": 4440 }, { "images": [ "sample_100_images/7HfRNqRCkBc@16224.68_16240.68#1.jpg", "sample_100_images/7HfRNqRCkBc@16224.68_16240.68#2.jpg", null, "sample_100_images/7HfRNqRCkBc@16240.68_16257.68#1.jpg", null, "sample_100_images/7HfRNqRCkBc@16257.68_16272.68#1.jpg", "sample_100_images/7HfRNqRCkBc@16257.68_16272.68#2.jpg", null, "sample_100_images/7HfRNqRCkBc@16272.68_16286.68#1.jpg", null, "sample_100_images/7HfRNqRCkBc@16286.68_16300.68#1.jpg", null, "sample_100_images/7HfRNqRCkBc@16300.68_16316.68#1.jpg", null, "sample_100_images/7HfRNqRCkBc@16316.68_16332.68#1.jpg", null, "sample_100_images/7HfRNqRCkBc@16332.68_16347.68#1.jpg", null ], "texts": [ null, null, " From h of t minus one, we will calculate y of t. And this is how a recurrent neural network works. Now, you must be thinking how to train a recurrent neural network. So a recurrent neural network uses the backpropagation algorithm for training, but backpropagation happens for every time step.", null, " That is why it is commonly called backpropagation through time. And I have discussed backpropagation in detail in the artificial neural network tutorial, so you can go through that. I won't be discussing backpropagation in detail here; I will just give you a brief introduction.", null, null, " Let's explore what it is now with back propagation and exploding gradients. Let's see these one by one. So in vanishing gradient, what happens when you use back propagation? You tend to calculate the error, which is nothing but the actual output that you already know minus the model output.", null, " The output that you got through your model and the square of that. So you figure out the error. With that error, what do you do? You tend to find out the change in error with respect to the change in weight or any variable. So we'll call it weight here.", null, " So the change in error with respect to weight, multiplied by the learning rate, will give you the change in weight. Then you need to add that change in weight to the old weight to get the new weight. Alright, so obviously what we're trying to do is we're trying to reduce the error.", null, " The error. So for that, we need to figure out what will be the change in error if my variables are changed, right? So that way, we can get the change in the variable and add it to our old variable to get the new variable. Now, over here, what can happen?", null, " If the value delta E by delta W - that is a gradient - is definitely smaller than one, like it is 0.00 something, then if you multiply that with the learning rate, which is definitely smaller than one, you get the change of weight, which is negligible. Alright.", null, " So there might be certain examples where you know you are trying to predict, say, the next word in a sentence and that sentence is pretty long. For example, if I say \"I went to France,\" dash, dash, dash, \"I went to France.\" Then there are certain words. Then I say:" ], "text_ocr_list": [ null, null, "We can see these text from the image: What Is Recurrent Neural Network?\n\n\\( h(t) = g_h(w_i x(t) + w_R h(t-1) + b_h) \\)\n\n\\( y(t) = g_y(w_y h(t) + b_y) \\)\n\nRecurrent Neural Network\n\nLooking for AI Online Training? Call us at IN: 9606058406 / US: 18338555775 or visit www.edureka.co.\n From h of t minus one, we will calculate y of t. And this is how a recurrent neural network works. Now, you must be thinking how to train a recurrent neural network. So a recurrent neural network uses the backpropagation algorithm for training, but backpropagation happens for every time step.", null, "We can see these text from the image: Training A Recurrent Neural Network\n\nRecurrent Neural Nets uses backpropagation algorithm, but it is applied for every time stamp. It is commonly known as Backpropagation Through Time (BTT)..\n That is why it is commonly called backpropagation through time. And I have discussed backpropagation in detail in the artificial neural network tutorial, so you can go through that. I won't be discussing backpropagation in detail here; I will just give you a brief introduction.", null, null, "We can see these text from the image: Training A Recurrent Neural Network\n\nRecurrent Neural Nets uses backpropagation algorithm, but it is applied for every time stamp. It is commonly known as Backpropagation Through Time (BTT).\n\nLet's look at the issues with Backpropagation:\n\n- Vanishing Gradient\n- Exploding Gradient.\n Let's explore what it is now with back propagation and exploding gradients. Let's see these one by one. So in vanishing gradient, what happens when you use back propagation? You tend to calculate the error, which is nothing but the actual output that you already know minus the model output.", null, "We can see these text from the image: Vanishing Gradient\n\nBackpropagation\n\nw = w + \u0394w\n\u0394w = n * (de/dw)\ne = (Actual Output - Model Output)^2\n\nif de/dw <<<< 1\n\n\u0394w <<<< 1\nw <<<< 1.\n The output that you got through your model and the square of that. So you figure out the error. With that error, what do you do? You tend to find out the change in error with respect to the change in weight or any variable. So we'll call it weight here.", null, "We can see these text from the image: Vanishing Gradient\n\nBackpropagation\n\nw = w + \u0394w\n\u0394w = \u03b7 * (de/dw)\ne = (Actual Output - Model Output)^2\n\nif de/dw <<< 1\n\u0394w <<< 1\nw <<< 1.\n So the change in error with respect to weight, multiplied by the learning rate, will give you the change in weight. Then you need to add that change in weight to the old weight to get the new weight. Alright, so obviously what we're trying to do is we're trying to reduce the error.", null, "We can see these text from the image: Vanishing Gradient\n\nBackpropagation\n\nw = w + \u0394w\n\u0394w = n * (de/dw)\ne = (Actual Output - Model Output)^2\n\nif de/dw <<< 1\n\n\u0394w <<< 1\nw <<< 1.\n The error. So for that, we need to figure out what will be the change in error if my variables are changed, right? So that way, we can get the change in the variable and add it to our old variable to get the new variable. Now, over here, what can happen?", null, "We can see these text from the image: Vanishing Gradient\n\nBackpropagation\n\nw = w + \u0394w\n\u0394w = n (de/dw)\ne = (Actual Output - Model Output)^2\n\nif de/dw <<< 1\n\u0394w <<< 1\nw <<< 1.\n If the value delta E by delta W - that is a gradient - is definitely smaller than one, like it is 0.00 something, then if you multiply that with the learning rate, which is definitely smaller than one, you get the change of weight, which is negligible. Alright.", null, "We can see these text from the image: Vanishing Gradient\n\nBackpropagation\n\nw = w + \u0394w\n\u0394w = n (de/dw)\ne = (Actual Output - Model Output)^2\n\nif (de/dw) <<<< 1\n\u0394w <<<< 1\nw <<<< 1.\n So there might be certain examples where you know you are trying to predict, say, the next word in a sentence and that sentence is pretty long. For example, if I say \"I went to France,\" dash, dash, dash, \"I went to France.\" Then there are certain words. Then I say:" ], "metadata": [ { "vid": "7HfRNqRCkBc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial Intelligence course fundamentals - Knowledge Representation and Scripts tutorial_30.json#####audio#####doingASR#####FinishASR/7HfRNqRCkBc/16224.68_16240.68.mp4", "refined_asr": " From h of t minus one, we will calculate y of t. And this is how a recurrent neural network works. Now, you must be thinking how to train a recurrent neural network. So a recurrent neural network uses the backpropagation algorithm for training, but backpropagation happens for every time step.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16224.68_16240.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16224.68_16240.68#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16224.68_16240.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16224.68_16240.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16224.68_16240.68#3.jpg" ], "ocr_qwen2_vl_72b": "What Is Recurrent Neural Network?\n\n\\( h(t) = g_h(w_i x(t) + w_R h(t-1) + b_h) \\)\n\n\\( y(t) = g_y(w_y h(t) + b_y) \\)\n\nRecurrent Neural Network\n\nLooking for AI Online Training? Call us at IN: 9606058406 / US: 18338555775 or visit www.edureka.co" }, { "vid": "7HfRNqRCkBc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial Intelligence course fundamentals - Knowledge Representation and Scripts tutorial_30.json#####audio#####doingASR#####FinishASR/7HfRNqRCkBc/16240.68_16257.68.mp4", "refined_asr": " That is why it is commonly called backpropagation through time. And I have discussed backpropagation in detail in the artificial neural network tutorial, so you can go through that. I won't be discussing backpropagation in detail here; I will just give you a brief introduction.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16240.68_16257.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16240.68_16257.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16240.68_16257.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16240.68_16257.68#3.jpg" ], "ocr_qwen2_vl_72b": "Training A Recurrent Neural Network\n\nRecurrent Neural Nets uses backpropagation algorithm, but it is applied for every time stamp. It is commonly known as Backpropagation Through Time (BTT)." }, { "vid": "7HfRNqRCkBc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial Intelligence course fundamentals - Knowledge Representation and Scripts tutorial_30.json#####audio#####doingASR#####FinishASR/7HfRNqRCkBc/16257.68_16272.68.mp4", "refined_asr": " Let's explore what it is now with back propagation and exploding gradients. Let's see these one by one. So in vanishing gradient, what happens when you use back propagation? You tend to calculate the error, which is nothing but the actual output that you already know minus the model output.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16257.68_16272.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16257.68_16272.68#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16257.68_16272.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16257.68_16272.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16257.68_16272.68#3.jpg" ], "ocr_qwen2_vl_72b": "Training A Recurrent Neural Network\n\nRecurrent Neural Nets uses backpropagation algorithm, but it is applied for every time stamp. It is commonly known as Backpropagation Through Time (BTT).\n\nLet's look at the issues with Backpropagation:\n\n- Vanishing Gradient\n- Exploding Gradient" }, { "vid": "7HfRNqRCkBc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial Intelligence course fundamentals - Knowledge Representation and Scripts tutorial_30.json#####audio#####doingASR#####FinishASR/7HfRNqRCkBc/16272.68_16286.68.mp4", "refined_asr": " The output that you got through your model and the square of that. So you figure out the error. With that error, what do you do? You tend to find out the change in error with respect to the change in weight or any variable. So we'll call it weight here.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16272.68_16286.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16272.68_16286.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16272.68_16286.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16272.68_16286.68#3.jpg" ], "ocr_qwen2_vl_72b": "Vanishing Gradient\n\nBackpropagation\n\nw = w + \u0394w\n\u0394w = n * (de/dw)\ne = (Actual Output - Model Output)^2\n\nif de/dw <<<< 1\n\n\u0394w <<<< 1\nw <<<< 1" }, { "vid": "7HfRNqRCkBc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial Intelligence course fundamentals - Knowledge Representation and Scripts tutorial_30.json#####audio#####doingASR#####FinishASR/7HfRNqRCkBc/16286.68_16300.68.mp4", "refined_asr": " So the change in error with respect to weight, multiplied by the learning rate, will give you the change in weight. Then you need to add that change in weight to the old weight to get the new weight. Alright, so obviously what we're trying to do is we're trying to reduce the error.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16286.68_16300.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16286.68_16300.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16286.68_16300.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16286.68_16300.68#3.jpg" ], "ocr_qwen2_vl_72b": "Vanishing Gradient\n\nBackpropagation\n\nw = w + \u0394w\n\u0394w = \u03b7 * (de/dw)\ne = (Actual Output - Model Output)^2\n\nif de/dw <<< 1\n\u0394w <<< 1\nw <<< 1" }, { "vid": "7HfRNqRCkBc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial Intelligence course fundamentals - Knowledge Representation and Scripts tutorial_30.json#####audio#####doingASR#####FinishASR/7HfRNqRCkBc/16300.68_16316.68.mp4", "refined_asr": " The error. So for that, we need to figure out what will be the change in error if my variables are changed, right? So that way, we can get the change in the variable and add it to our old variable to get the new variable. Now, over here, what can happen?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16300.68_16316.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16300.68_16316.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16300.68_16316.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16300.68_16316.68#3.jpg" ], "ocr_qwen2_vl_72b": "Vanishing Gradient\n\nBackpropagation\n\nw = w + \u0394w\n\u0394w = n * (de/dw)\ne = (Actual Output - Model Output)^2\n\nif de/dw <<< 1\n\n\u0394w <<< 1\nw <<< 1" }, { "vid": "7HfRNqRCkBc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial Intelligence course fundamentals - Knowledge Representation and Scripts tutorial_30.json#####audio#####doingASR#####FinishASR/7HfRNqRCkBc/16316.68_16332.68.mp4", "refined_asr": " If the value delta E by delta W - that is a gradient - is definitely smaller than one, like it is 0.00 something, then if you multiply that with the learning rate, which is definitely smaller than one, you get the change of weight, which is negligible. Alright.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16316.68_16332.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16316.68_16332.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16316.68_16332.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16316.68_16332.68#3.jpg" ], "ocr_qwen2_vl_72b": "Vanishing Gradient\n\nBackpropagation\n\nw = w + \u0394w\n\u0394w = n (de/dw)\ne = (Actual Output - Model Output)^2\n\nif de/dw <<< 1\n\u0394w <<< 1\nw <<< 1" }, { "vid": "7HfRNqRCkBc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial Intelligence course fundamentals - Knowledge Representation and Scripts tutorial_30.json#####audio#####doingASR#####FinishASR/7HfRNqRCkBc/16332.68_16347.68.mp4", "refined_asr": " So there might be certain examples where you know you are trying to predict, say, the next word in a sentence and that sentence is pretty long. For example, if I say \"I went to France,\" dash, dash, dash, \"I went to France.\" Then there are certain words. Then I say:", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16332.68_16347.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16332.68_16347.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16332.68_16347.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7HfRNqRCkBc/7HfRNqRCkBc@16332.68_16347.68#3.jpg" ], "ocr_qwen2_vl_72b": "Vanishing Gradient\n\nBackpropagation\n\nw = w + \u0394w\n\u0394w = n (de/dw)\ne = (Actual Output - Model Output)^2\n\nif (de/dw) <<<< 1\n\u0394w <<<< 1\nw <<<< 1" } ], "image_num": 10, "text_num": 515, "token_num": 6275 }, { "images": [ "sample_100_images/MUTTdQWZ86c@2076.7200000000003_2107.6800000000003#1.jpg", null, "sample_100_images/MUTTdQWZ86c@2107.6800000000003_2124.04#1.jpg", null, "sample_100_images/MUTTdQWZ86c@2124.04_2145.4#1.jpg", null, "sample_100_images/MUTTdQWZ86c@2145.4_2166.26#1.jpg", null, "sample_100_images/MUTTdQWZ86c@2166.26_2190.6800000000003#1.jpg", null, "sample_100_images/MUTTdQWZ86c@2190.6800000000003_2210.52#1.jpg", null, "sample_100_images/MUTTdQWZ86c@2210.52_2229.58#1.jpg", null ], "texts": [ null, " The curves illustrate various behaviors. Allow me to briefly explain that this is evaluated nuclear data file. It was produced with worldwide collaboration. All cross-sections are documented here. Experimental cross-sections are plotted, and then a fitted curve is generated. The cross-sections are measured in barns.", null, " And this is a logarithmic scale on which it is shown on the x-axis. The energy is also shown in the logarithmic scale and it is MeV scale. Therefore, 10 to the power minus 6 multiplied by MeV will be 10 to the power minus.", null, " 6 multiplied by MeV will make it 1 electron volt. So, roughly speaking, here is the 1 electron volt, and the less than 1 electron volts are here, and the more than 1 electron volt are there. This is 10 million electron volts. The regions are called the cold region for", null, " The neutrons in these energy ranges are called cold neutrons. The neutrons in the energy range of 0.1 electron volt and less are called thermal neutrons, and the epithermal neutrons are those lying within the 0.1 to 1 electron volt energy range.", null, " The fast electrons, which have energy higher than 1 electron volt, are in the tens, hundreds, and thousands of electrons. So, 10 to the power of minus 6 multiplied by MeV is 1 electron volt. Less than a million, 10 to the power of 7 electron volts, is what's possibly measured in this graph.", null, " The highest possible neutron energy. As you can see, these cross sections are different for the different types of reactions and I will later discuss them. But the regions are according to the energy and their broader names are grouped as cold, thermal, epithermal, and fast.", null, " Neutrons. The resonances are these sharp changes in the scattering cross section. They occur at specific high energies and show a constant plateau. The fixed values are also shown, along with a 1 over V behavior, which is called the absorption-related behavior." ], "text_ocr_list": [ null, "We can see these text from the image: Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV).\n The curves illustrate various behaviors. Allow me to briefly explain that this is evaluated nuclear data file. It was produced with worldwide collaboration. All cross-sections are documented here. Experimental cross-sections are plotted, and then a fitted curve is generated. The cross-sections are measured in barns.", null, "We can see these text from the image: Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV).\n And this is a logarithmic scale on which it is shown on the x-axis. The energy is also shown in the logarithmic scale and it is MeV scale. Therefore, 10 to the power minus 6 multiplied by MeV will be 10 to the power minus.", null, "We can see these text from the image: Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58\nPrincipal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV).\n 6 multiplied by MeV will make it 1 electron volt. So, roughly speaking, here is the 1 electron volt, and the less than 1 electron volts are here, and the more than 1 electron volt are there. This is 10 million electron volts. The regions are called the cold region for", null, "We can see these text from the image: Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\nEnergy (MeV).\n The neutrons in these energy ranges are called cold neutrons. The neutrons in the energy range of 0.1 electron volt and less are called thermal neutrons, and the epithermal neutrons are those lying within the 0.1 to 1 electron volt energy range.", null, "We can see these text from the image: Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV).\n The fast electrons, which have energy higher than 1 electron volt, are in the tens, hundreds, and thousands of electrons. So, 10 to the power of minus 6 multiplied by MeV is 1 electron volt. Less than a million, 10 to the power of 7 electron volts, is what's possibly measured in this graph.", null, "We can see these text from the image: Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV).\n The highest possible neutron energy. As you can see, these cross sections are different for the different types of reactions and I will later discuss them. But the regions are according to the energy and their broader names are grouped as cold, thermal, epithermal, and fast.", null, "We can see these text from the image: Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV).\n Neutrons. The resonances are these sharp changes in the scattering cross section. They occur at specific high energies and show a constant plateau. The fixed values are also shown, along with a 1 over V behavior, which is called the absorption-related behavior." ], "metadata": [ { "vid": "MUTTdQWZ86c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lecture on Neutron Sources in Research and Technology_30.json#####audio#####doingASR#####FinishASR/MUTTdQWZ86c/2076.7200000000003_2107.6800000000003.mp4", "refined_asr": " The curves illustrate various behaviors. Allow me to briefly explain that this is evaluated nuclear data file. It was produced with worldwide collaboration. All cross-sections are documented here. Experimental cross-sections are plotted, and then a fitted curve is generated. The cross-sections are measured in barns.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2076.7200000000003_2107.6800000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2076.7200000000003_2107.6800000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2076.7200000000003_2107.6800000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2076.7200000000003_2107.6800000000003#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2076.7200000000003_2107.6800000000003#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2076.7200000000003_2107.6800000000003#5.jpg" ], "ocr_qwen2_vl_72b": "Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV)" }, { "vid": "MUTTdQWZ86c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lecture on Neutron Sources in Research and Technology_30.json#####audio#####doingASR#####FinishASR/MUTTdQWZ86c/2107.6800000000003_2124.04.mp4", "refined_asr": " And this is a logarithmic scale on which it is shown on the x-axis. The energy is also shown in the logarithmic scale and it is MeV scale. Therefore, 10 to the power minus 6 multiplied by MeV will be 10 to the power minus.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2107.6800000000003_2124.04#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2107.6800000000003_2124.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2107.6800000000003_2124.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2107.6800000000003_2124.04#3.jpg" ], "ocr_qwen2_vl_72b": "Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV)" }, { "vid": "MUTTdQWZ86c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lecture on Neutron Sources in Research and Technology_30.json#####audio#####doingASR#####FinishASR/MUTTdQWZ86c/2124.04_2145.4.mp4", "refined_asr": " 6 multiplied by MeV will make it 1 electron volt. So, roughly speaking, here is the 1 electron volt, and the less than 1 electron volts are here, and the more than 1 electron volt are there. This is 10 million electron volts. The regions are called the cold region for", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2124.04_2145.4#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2124.04_2145.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2124.04_2145.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2124.04_2145.4#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2124.04_2145.4#4.jpg" ], "ocr_qwen2_vl_72b": "Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58\nPrincipal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV)" }, { "vid": "MUTTdQWZ86c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lecture on Neutron Sources in Research and Technology_30.json#####audio#####doingASR#####FinishASR/MUTTdQWZ86c/2145.4_2166.26.mp4", "refined_asr": " The neutrons in these energy ranges are called cold neutrons. The neutrons in the energy range of 0.1 electron volt and less are called thermal neutrons, and the epithermal neutrons are those lying within the 0.1 to 1 electron volt energy range.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2145.4_2166.26#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2145.4_2166.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2145.4_2166.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2145.4_2166.26#3.jpg" ], "ocr_qwen2_vl_72b": "Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\nEnergy (MeV)" }, { "vid": "MUTTdQWZ86c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lecture on Neutron Sources in Research and Technology_30.json#####audio#####doingASR#####FinishASR/MUTTdQWZ86c/2166.26_2190.6800000000003.mp4", "refined_asr": " The fast electrons, which have energy higher than 1 electron volt, are in the tens, hundreds, and thousands of electrons. So, 10 to the power of minus 6 multiplied by MeV is 1 electron volt. Less than a million, 10 to the power of 7 electron volts, is what's possibly measured in this graph.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2166.26_2190.6800000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2166.26_2190.6800000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2166.26_2190.6800000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2166.26_2190.6800000000003#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2166.26_2190.6800000000003#4.jpg" ], "ocr_qwen2_vl_72b": "Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV)" }, { "vid": "MUTTdQWZ86c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lecture on Neutron Sources in Research and Technology_30.json#####audio#####doingASR#####FinishASR/MUTTdQWZ86c/2190.6800000000003_2210.52.mp4", "refined_asr": " The highest possible neutron energy. As you can see, these cross sections are different for the different types of reactions and I will later discuss them. But the regions are according to the energy and their broader names are grouped as cold, thermal, epithermal, and fast.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2190.6800000000003_2210.52#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2190.6800000000003_2210.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2190.6800000000003_2210.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2190.6800000000003_2210.52#3.jpg" ], "ocr_qwen2_vl_72b": "Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV)" }, { "vid": "MUTTdQWZ86c.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Atomic and Nuclear Physics Lecture on Neutron Sources in Research and Technology_30.json#####audio#####doingASR#####FinishASR/MUTTdQWZ86c/2210.52_2229.58.mp4", "refined_asr": " Neutrons. The resonances are these sharp changes in the scattering cross section. They occur at specific high energies and show a constant plateau. The fixed values are also shown, along with a 1 over V behavior, which is called the absorption-related behavior.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2210.52_2229.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2210.52_2229.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2210.52_2229.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/MUTTdQWZ86c/MUTTdQWZ86c@2210.52_2229.58#3.jpg" ], "ocr_qwen2_vl_72b": "Energy dependence of cross sections\n\nNote:\n- Resonances at high-energy\n- Constant plateau of scattering cross-section\n- Strong (1/v) dependence of absorption \u2013 related to the time spent near the nucleus (probability of capture).\n\nENDF/B-VII NI-58 Principal cross sections\n\nCross section (barns)\n\nCold\nThermal\nEpithermal\nFast\n\ntotal\nabsorption\nelastic\ngamma production\n\nEnergy (MeV)" } ], "image_num": 7, "text_num": 455, "token_num": 4487 }, { "images": [ "sample_100_images/LxO-6rlihSg@2941.28_2955.6800000000003#1.jpg", "sample_100_images/LxO-6rlihSg@2941.28_2955.6800000000003#2.jpg", null, "sample_100_images/LxO-6rlihSg@2955.6800000000003_2967.4#1.jpg", "sample_100_images/LxO-6rlihSg@2955.6800000000003_2967.4#2.jpg", null, "sample_100_images/LxO-6rlihSg@2967.4_2977.88#1.jpg", "sample_100_images/LxO-6rlihSg@2967.4_2977.88#2.jpg", null, "sample_100_images/LxO-6rlihSg@2977.88_2991.2200000000003#1.jpg", "sample_100_images/LxO-6rlihSg@2977.88_2991.2200000000003#2.jpg", null, "sample_100_images/LxO-6rlihSg@2991.2200000000003_3004.1800000000003#1.jpg", "sample_100_images/LxO-6rlihSg@2991.2200000000003_3004.1800000000003#2.jpg", null, "sample_100_images/LxO-6rlihSg@3004.18_3017.22#1.jpg", "sample_100_images/LxO-6rlihSg@3004.18_3017.22#2.jpg", null, "sample_100_images/LxO-6rlihSg@3017.22_3028.48#1.jpg", "sample_100_images/LxO-6rlihSg@3017.22_3028.48#2.jpg", null, "sample_100_images/LxO-6rlihSg@3028.48_3041.34#1.jpg", "sample_100_images/LxO-6rlihSg@3028.48_3041.34#2.jpg", null, "sample_100_images/LxO-6rlihSg@3041.34_3051.5#1.jpg", null, "sample_100_images/LxO-6rlihSg@3051.5_3065.84#1.jpg", "sample_100_images/LxO-6rlihSg@3051.5_3065.84#2.jpg", null ], "texts": [ null, null, " So what do you think the histogram graph would look like if you took a full frame photo of just a zebra skin\u2014so black and white? Obviously the graph's going to look a little bit like this one where you've got a whole lot of black and then kind of some shades near black, but then nothing until you get.", null, null, " Consider the highlights: some bright areas and a lot of white. You can't just look at a histogram on its own and conclude that a photo is underexposed or overexposed, because it depends on the image. Perhaps it's meant to feature a very bright, white subject.", null, null, " But generally speaking, you just want your histogram graph safely between those two ends. So to look at your histogram graph, you've normally got a bit of black and a bit of white. And that's probably in review mode or playback mode when you're looking at a picture you've taken.", null, null, " And then you either hit Info or Display a couple of times. For a Nikon, it's up or down. It just cycles through different pages of information and one of them will finally bring up your histogram. Don't worry about the one with all the different colors; you're just looking at that white.", null, null, " One, that's the overall brightness graph. Okay, so now's probably a good time to go outside and do our first practical session. So pick up your camera, make sure you're in P mode, and then just go and find something to take a photo of. It doesn't really matter what it is.", null, null, " Take a picture. Initially, it should come out being mid-brightness on average. And then work out on your camera where that little plus minus brightness button is. Press and hold it in, scroll, do something quite dramatic like plus one or plus two. Take the picture.", null, null, " Same picture again, and you should definitely see the picture then look a whole lot brighter. And do a darker one as well, you know, minus two, minus three. Check that out. Maybe bring up your histogram graph as well and watch all that information shift.", null, null, " The trick is to move the slider towards the darker end or up to the brighter end, trying to get it just about right\u2014safely between the two extremes. If you think that's all too easy and you've done all that before, a good extension can be to try and take a photo of something that's mostly white, like a sheet of white paper.", null, " Or something. If you take a photo of that, initially it's going to come out as a mid-gray sheet of paper, mid-brightness. And you'll be able to look at it on the back and just say: that's too dark, I'll go plus.", null, null, " One take it again and then it comes back looking brighter and you might think that's okay. But actually, if you checked your histogram graph, you'd probably find that the first photo, the mid-brightness spike, means the histogram will just be basically one big spike in the middle, indicating lots of grey." ], "text_ocr_list": [ null, null, "We can see these text from the image: Histograms.\n So what do you think the histogram graph would look like if you took a full frame photo of just a zebra skin\u2014so black and white? Obviously the graph's going to look a little bit like this one where you've got a whole lot of black and then kind of some shades near black, but then nothing until you get.", null, null, "We can see these text from the image: Histograms.\n Consider the highlights: some bright areas and a lot of white. You can't just look at a histogram on its own and conclude that a photo is underexposed or overexposed, because it depends on the image. Perhaps it's meant to feature a very bright, white subject.", null, null, "We can see these text from the image: Histograms.\n But generally speaking, you just want your histogram graph safely between those two ends. So to look at your histogram graph, you've normally got a bit of black and a bit of white. And that's probably in review mode or playback mode when you're looking at a picture you've taken.", null, null, "We can see these text from the image: Histograms.\n And then you either hit Info or Display a couple of times. For a Nikon, it's up or down. It just cycles through different pages of information and one of them will finally bring up your histogram. Don't worry about the one with all the different colors; you're just looking at that white.", null, null, "We can see these text from the image: CHRIS BRAY\n\nPHOTOGRAPHY\n\nTOURS - COURSES.\n One, that's the overall brightness graph. Okay, so now's probably a good time to go outside and do our first practical session. So pick up your camera, make sure you're in P mode, and then just go and find something to take a photo of. It doesn't really matter what it is.", null, null, "We can see these text from the image: Practical Session:.\n Take a picture. Initially, it should come out being mid-brightness on average. And then work out on your camera where that little plus minus brightness button is. Press and hold it in, scroll, do something quite dramatic like plus one or plus two. Take the picture.", null, null, "We can see these text from the image: Practical Session:\n\n-2 . . 1 . . 0 . . 1 . . +2.\n Same picture again, and you should definitely see the picture then look a whole lot brighter. And do a darker one as well, you know, minus two, minus three. Check that out. Maybe bring up your histogram graph as well and watch all that information shift.", null, null, "We can see these text from the image: Practical Session:.\n The trick is to move the slider towards the darker end or up to the brighter end, trying to get it just about right\u2014safely between the two extremes. If you think that's all too easy and you've done all that before, a good extension can be to try and take a photo of something that's mostly white, like a sheet of white paper.", null, "We can see these text from the image: Practical Session:.\n Or something. If you take a photo of that, initially it's going to come out as a mid-gray sheet of paper, mid-brightness. And you'll be able to look at it on the back and just say: that's too dark, I'll go plus.", null, null, "We can see these text from the image: Practical Session:.\n One take it again and then it comes back looking brighter and you might think that's okay. But actually, if you checked your histogram graph, you'd probably find that the first photo, the mid-brightness spike, means the histogram will just be basically one big spike in the middle, indicating lots of grey." ], "metadata": [ { "vid": "LxO-6rlihSg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic digital photography for kids_100.json#####audio#####doingASR#####FinishASR/LxO-6rlihSg/2941.28_2955.6800000000003.mp4", "refined_asr": " So what do you think the histogram graph would look like if you took a full frame photo of just a zebra skin\u2014so black and white? Obviously the graph's going to look a little bit like this one where you've got a whole lot of black and then kind of some shades near black, but then nothing until you get.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2941.28_2955.6800000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2941.28_2955.6800000000003#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2941.28_2955.6800000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2941.28_2955.6800000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2941.28_2955.6800000000003#3.jpg" ], "ocr_qwen2_vl_72b": "Histograms" }, { "vid": "LxO-6rlihSg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic digital photography for kids_100.json#####audio#####doingASR#####FinishASR/LxO-6rlihSg/2955.6800000000003_2967.4.mp4", "refined_asr": " Consider the highlights: some bright areas and a lot of white. You can't just look at a histogram on its own and conclude that a photo is underexposed or overexposed, because it depends on the image. Perhaps it's meant to feature a very bright, white subject.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2955.6800000000003_2967.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2955.6800000000003_2967.4#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2955.6800000000003_2967.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2955.6800000000003_2967.4#2.jpg" ], "ocr_qwen2_vl_72b": "Histograms" }, { "vid": "LxO-6rlihSg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic digital photography for kids_100.json#####audio#####doingASR#####FinishASR/LxO-6rlihSg/2967.4_2977.88.mp4", "refined_asr": " But generally speaking, you just want your histogram graph safely between those two ends. So to look at your histogram graph, you've normally got a bit of black and a bit of white. And that's probably in review mode or playback mode when you're looking at a picture you've taken.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2967.4_2977.88#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2967.4_2977.88#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2967.4_2977.88#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2967.4_2977.88#2.jpg" ], "ocr_qwen2_vl_72b": "Histograms" }, { "vid": "LxO-6rlihSg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic digital photography for kids_100.json#####audio#####doingASR#####FinishASR/LxO-6rlihSg/2977.88_2991.2200000000003.mp4", "refined_asr": " And then you either hit Info or Display a couple of times. For a Nikon, it's up or down. It just cycles through different pages of information and one of them will finally bring up your histogram. Don't worry about the one with all the different colors; you're just looking at that white.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2977.88_2991.2200000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2977.88_2991.2200000000003#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2977.88_2991.2200000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2977.88_2991.2200000000003#2.jpg" ], "ocr_qwen2_vl_72b": "Histograms" }, { "vid": "LxO-6rlihSg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic digital photography for kids_100.json#####audio#####doingASR#####FinishASR/LxO-6rlihSg/2991.2200000000003_3004.1800000000003.mp4", "refined_asr": " One, that's the overall brightness graph. Okay, so now's probably a good time to go outside and do our first practical session. So pick up your camera, make sure you're in P mode, and then just go and find something to take a photo of. It doesn't really matter what it is.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2991.2200000000003_3004.1800000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2991.2200000000003_3004.1800000000003#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2991.2200000000003_3004.1800000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@2991.2200000000003_3004.1800000000003#2.jpg" ], "ocr_qwen2_vl_72b": "CHRIS BRAY\n\nPHOTOGRAPHY\n\nTOURS - COURSES" }, { "vid": "LxO-6rlihSg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic digital photography for kids_100.json#####audio#####doingASR#####FinishASR/LxO-6rlihSg/3004.18_3017.22.mp4", "refined_asr": " Take a picture. Initially, it should come out being mid-brightness on average. And then work out on your camera where that little plus minus brightness button is. Press and hold it in, scroll, do something quite dramatic like plus one or plus two. Take the picture.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3004.18_3017.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3004.18_3017.22#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3004.18_3017.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3004.18_3017.22#2.jpg" ], "ocr_qwen2_vl_72b": "Practical Session:" }, { "vid": "LxO-6rlihSg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic digital photography for kids_100.json#####audio#####doingASR#####FinishASR/LxO-6rlihSg/3017.22_3028.48.mp4", "refined_asr": " Same picture again, and you should definitely see the picture then look a whole lot brighter. And do a darker one as well, you know, minus two, minus three. Check that out. Maybe bring up your histogram graph as well and watch all that information shift.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3017.22_3028.48#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3017.22_3028.48#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3017.22_3028.48#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3017.22_3028.48#2.jpg" ], "ocr_qwen2_vl_72b": "Practical Session:\n\n-2 . . 1 . . 0 . . 1 . . +2" }, { "vid": "LxO-6rlihSg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic digital photography for kids_100.json#####audio#####doingASR#####FinishASR/LxO-6rlihSg/3028.48_3041.34.mp4", "refined_asr": " The trick is to move the slider towards the darker end or up to the brighter end, trying to get it just about right\u2014safely between the two extremes. If you think that's all too easy and you've done all that before, a good extension can be to try and take a photo of something that's mostly white, like a sheet of white paper.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3028.48_3041.34#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3028.48_3041.34#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3028.48_3041.34#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3028.48_3041.34#2.jpg" ], "ocr_qwen2_vl_72b": "Practical Session:" }, { "vid": "LxO-6rlihSg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic digital photography for kids_100.json#####audio#####doingASR#####FinishASR/LxO-6rlihSg/3041.34_3051.5.mp4", "refined_asr": " Or something. If you take a photo of that, initially it's going to come out as a mid-gray sheet of paper, mid-brightness. And you'll be able to look at it on the back and just say: that's too dark, I'll go plus.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3041.34_3051.5#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3041.34_3051.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3041.34_3051.5#2.jpg" ], "ocr_qwen2_vl_72b": "Practical Session:" }, { "vid": "LxO-6rlihSg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic digital photography for kids_100.json#####audio#####doingASR#####FinishASR/LxO-6rlihSg/3051.5_3065.84.mp4", "refined_asr": " One take it again and then it comes back looking brighter and you might think that's okay. But actually, if you checked your histogram graph, you'd probably find that the first photo, the mid-brightness spike, means the histogram will just be basically one big spike in the middle, indicating lots of grey.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3051.5_3065.84#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3051.5_3065.84#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3051.5_3065.84#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3051.5_3065.84#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/LxO-6rlihSg/LxO-6rlihSg@3051.5_3065.84#3.jpg" ], "ocr_qwen2_vl_72b": "Practical Session:" } ], "image_num": 19, "text_num": 682, "token_num": 11626 }, { "images": [ "sample_100_images/leXuUp-uxGQ@1571.5800000000002_1585.8000000000002#1.jpg", null, "sample_100_images/leXuUp-uxGQ@1585.8000000000002_1595.6200000000001#1.jpg", null, "sample_100_images/leXuUp-uxGQ@1595.6200000000001_1608.9#1.jpg", "sample_100_images/leXuUp-uxGQ@1595.6200000000001_1608.9#2.jpg", null, "sample_100_images/leXuUp-uxGQ@1611.8400000000001_1624.16#1.jpg", null, "sample_100_images/leXuUp-uxGQ@1624.16_1638.28#1.jpg", null, "sample_100_images/leXuUp-uxGQ@1638.28_1654.0400000000002#1.jpg", "sample_100_images/leXuUp-uxGQ@1638.28_1654.0400000000002#2.jpg", null, "sample_100_images/leXuUp-uxGQ@1654.16_1672.46#1.jpg", null, "sample_100_images/leXuUp-uxGQ@1676.3400000000001_1690.4#1.jpg", null, "sample_100_images/leXuUp-uxGQ@1690.4_1699.26#1.jpg", null ], "texts": [ null, " And these wavelength lines are going to have some curvature in them. So figuring out where each one of those wavelength bins is, as a function of position on your detector, is one of the trickiest parts actually, of reducing a spectrum.", null, " So what we want now is we want the light from this trace object here as a function of wavelength. We want to know what the flux density is at each one of these wavelengths.", null, null, " Here we have 512 in this example. So each one of those, how bright is it? And this is called extracting the spectrum. So we take that two-dimensional data and we're going to get a one-dimensional answer, which is just the flux density as a function of wavelength.", null, " So I've kind of said this before. Just remember, all we're doing here is we're doing photometry, but we're just doing it many times at once. For each one of those wavelength bins, we have light.", null, " From our source, we have background light. And what we need to do is figure out how much light we're getting from our source and how much is contributed by the background. Remove that background component, and then we'll have how much is coming from our source.", null, null, " And we do that for each one of our wavelength bins. We get ourselves a 1D spectrum. So the first thing to do is extract the spectrum. Let's take the whole image and for each one of those columns, we can say well, this is one wavelength.", null, " So let's just integrate along that entire column and we'll get the total amount of flux. To do this, we just take the image that we've generated and we're just going to sum it along columns, which is axis zero.", null, " There we go. So this is the whole image now - the whole 2D image that we have just integrated. We've marginalized, if you will, over one axis. And then we're going to take the wavelength coordinates. We have those 512 measurements that we've got now.", null, " And at each one of these wavelengths we have some amount of total light on the image. So you can see there is this kind of curve here. That's our trace that we want. But on top of that there's all these emission lines." ], "text_ocr_list": [ null, "We can see these text from the image: What the data look like (cartoon version)\n\nIn [12]: image, sky = make_image()\nplt.imshow(image, vmin=0, vmax=0.5 * image.max())\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Spatial Coordinate');\n\nNote in real data:\n- the lines of constant wavelength are slightly tilted and curved with.\n And these wavelength lines are going to have some curvature in them. So figuring out where each one of those wavelength bins is, as a function of position on your detector, is one of the trickiest parts actually, of reducing a spectrum.", null, "We can see these text from the image: What the data look like (cartoon version)\n\nIn [12]: image, sky = make_image()\nplt.imshow(image, vmin=0, vmax=0.5 * image.max())\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Spatial Coordinate');\n\nNote in real data:\n- the lines of constant wavelength are slightly tilted and curved with.\n So what we want now is we want the light from this trace object here as a function of wavelength. We want to know what the flux density is at each one of these wavelengths.", null, null, "We can see these text from the image: What the data look like (cartoon version)\n\nIn [12]: image, sky = make_image()\nplt.imshow(image, vmin=0, vmax=0.5 * image.max())\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Spatial Coordinate');\n\nNote in real data:\n- the lines of constant wavelength are slightly tilted and curved with.\n Here we have 512 in this example. So each one of those, how bright is it? And this is called extracting the spectrum. So we take that two-dimensional data and we're going to get a one-dimensional answer, which is just the flux density as a function of wavelength.", null, "We can see these text from the image: Extracting a spectrum\n\n- spectroscopy is photometry but the band passes are smaller and you get many bands at once\n- same basic procedure for reducing the data, but wavelength dependencies need to be handled\n- extraction is also analogous to photometry, but again wavelength dependencies matter.\n So I've kind of said this before. Just remember, all we're doing here is we're doing photometry, but we're just doing it many times at once. For each one of those wavelength bins, we have light.", null, "We can see these text from the image: Extracting a spectrum\n\n- spectroscopy is photometry but the band passes are smaller and you get many bands at once\n- same basic procedure for reducing the data, but wavelength dependencies need to be handled\n- extraction is also analogous to photometry, but again wavelength dependencies matter.\n From our source, we have background light. And what we need to do is figure out how much light we're getting from our source and how much is contributed by the background. Remove that background component, and then we'll have how much is coming from our source.", null, null, "We can see these text from the image: Extracting a spectrum\n\n- spectroscopy is photometry but the band passes are smaller and you get many bands at once\n- same basic procedure for reducing the data, but wavelength dependencies need to be handled\n- extraction is also analogous to photometry, but again wavelength dependencies matter.\n And we do that for each one of our wavelength bins. We get ourselves a 1D spectrum. So the first thing to do is extract the spectrum. Let's take the whole image and for each one of those columns, we can say well, this is one wavelength.", null, "We can see these text from the image: ```python\nIn [ ]: spec =\nplt.plot(spec, lw=5)\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Relative Flux Density');\n```.\n So let's just integrate along that entire column and we'll get the total amount of flux. To do this, we just take the image that we've generated and we're just going to sum it along columns, which is axis zero.", null, "We can see these text from the image: ```python\nIn [35]:\nspec = image.sum(axis=0)\nplt.plot(spec, lw=5)\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Relative Flux Density');\n```\n\n**Relative Flux Density**\n\n**Wavelength Coordinate**.\n There we go. So this is the whole image now - the whole 2D image that we have just integrated. We've marginalized, if you will, over one axis. And then we're going to take the wavelength coordinates. We have those 512 measurements that we've got now.", null, "We can see these text from the image: ```python\nIn [35]:\nspec = image.sum(axis=0)\nplt.plot(spec, lw=5)\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Relative Flux Density');\n```\n\n**Figure Caption:**\n- **X-axis:** Wavelength Coordinate\n- **Y-axis:** Relative Flux Density.\n And at each one of these wavelengths we have some amount of total light on the image. So you can see there is this kind of curve here. That's our trace that we want. But on top of that there's all these emission lines." ], "metadata": [ { "vid": "leXuUp-uxGQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Spectral Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/leXuUp-uxGQ/1571.5800000000002_1585.8000000000002.mp4", "refined_asr": " And these wavelength lines are going to have some curvature in them. So figuring out where each one of those wavelength bins is, as a function of position on your detector, is one of the trickiest parts actually, of reducing a spectrum.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1571.5800000000002_1585.8000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1571.5800000000002_1585.8000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1571.5800000000002_1585.8000000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1571.5800000000002_1585.8000000000002#3.jpg" ], "ocr_qwen2_vl_72b": "What the data look like (cartoon version)\n\nIn [12]: image, sky = make_image()\nplt.imshow(image, vmin=0, vmax=0.5 * image.max())\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Spatial Coordinate');\n\nNote in real data:\n- the lines of constant wavelength are slightly tilted and curved with" }, { "vid": "leXuUp-uxGQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Spectral Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/leXuUp-uxGQ/1585.8000000000002_1595.6200000000001.mp4", "refined_asr": " So what we want now is we want the light from this trace object here as a function of wavelength. We want to know what the flux density is at each one of these wavelengths.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1585.8000000000002_1595.6200000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1585.8000000000002_1595.6200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1585.8000000000002_1595.6200000000001#2.jpg" ], "ocr_qwen2_vl_72b": "What the data look like (cartoon version)\n\nIn [12]: image, sky = make_image()\nplt.imshow(image, vmin=0, vmax=0.5 * image.max())\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Spatial Coordinate');\n\nNote in real data:\n- the lines of constant wavelength are slightly tilted and curved with" }, { "vid": "leXuUp-uxGQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Spectral Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/leXuUp-uxGQ/1595.6200000000001_1608.9.mp4", "refined_asr": " Here we have 512 in this example. So each one of those, how bright is it? And this is called extracting the spectrum. So we take that two-dimensional data and we're going to get a one-dimensional answer, which is just the flux density as a function of wavelength.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1595.6200000000001_1608.9#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1595.6200000000001_1608.9#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1595.6200000000001_1608.9#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1595.6200000000001_1608.9#2.jpg" ], "ocr_qwen2_vl_72b": "What the data look like (cartoon version)\n\nIn [12]: image, sky = make_image()\nplt.imshow(image, vmin=0, vmax=0.5 * image.max())\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Spatial Coordinate');\n\nNote in real data:\n- the lines of constant wavelength are slightly tilted and curved with" }, { "vid": "leXuUp-uxGQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Spectral Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/leXuUp-uxGQ/1611.8400000000001_1624.16.mp4", "refined_asr": " So I've kind of said this before. Just remember, all we're doing here is we're doing photometry, but we're just doing it many times at once. For each one of those wavelength bins, we have light.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1611.8400000000001_1624.16#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1611.8400000000001_1624.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1611.8400000000001_1624.16#2.jpg" ], "ocr_qwen2_vl_72b": "Extracting a spectrum\n\n- spectroscopy is photometry but the band passes are smaller and you get many bands at once\n- same basic procedure for reducing the data, but wavelength dependencies need to be handled\n- extraction is also analogous to photometry, but again wavelength dependencies matter" }, { "vid": "leXuUp-uxGQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Spectral Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/leXuUp-uxGQ/1624.16_1638.28.mp4", "refined_asr": " From our source, we have background light. And what we need to do is figure out how much light we're getting from our source and how much is contributed by the background. Remove that background component, and then we'll have how much is coming from our source.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1624.16_1638.28#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1624.16_1638.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1624.16_1638.28#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1624.16_1638.28#3.jpg" ], "ocr_qwen2_vl_72b": "Extracting a spectrum\n\n- spectroscopy is photometry but the band passes are smaller and you get many bands at once\n- same basic procedure for reducing the data, but wavelength dependencies need to be handled\n- extraction is also analogous to photometry, but again wavelength dependencies matter" }, { "vid": "leXuUp-uxGQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Spectral Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/leXuUp-uxGQ/1638.28_1654.0400000000002.mp4", "refined_asr": " And we do that for each one of our wavelength bins. We get ourselves a 1D spectrum. So the first thing to do is extract the spectrum. Let's take the whole image and for each one of those columns, we can say well, this is one wavelength.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1638.28_1654.0400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1638.28_1654.0400000000002#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1638.28_1654.0400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1638.28_1654.0400000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1638.28_1654.0400000000002#3.jpg" ], "ocr_qwen2_vl_72b": "Extracting a spectrum\n\n- spectroscopy is photometry but the band passes are smaller and you get many bands at once\n- same basic procedure for reducing the data, but wavelength dependencies need to be handled\n- extraction is also analogous to photometry, but again wavelength dependencies matter" }, { "vid": "leXuUp-uxGQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Spectral Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/leXuUp-uxGQ/1654.16_1672.46.mp4", "refined_asr": " So let's just integrate along that entire column and we'll get the total amount of flux. To do this, we just take the image that we've generated and we're just going to sum it along columns, which is axis zero.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1654.16_1672.46#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1654.16_1672.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1654.16_1672.46#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1654.16_1672.46#3.jpg" ], "ocr_qwen2_vl_72b": "```python\nIn [ ]: spec =\nplt.plot(spec, lw=5)\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Relative Flux Density');\n```" }, { "vid": "leXuUp-uxGQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Spectral Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/leXuUp-uxGQ/1676.3400000000001_1690.4.mp4", "refined_asr": " There we go. So this is the whole image now - the whole 2D image that we have just integrated. We've marginalized, if you will, over one axis. And then we're going to take the wavelength coordinates. We have those 512 measurements that we've got now.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1676.3400000000001_1690.4#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1676.3400000000001_1690.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1676.3400000000001_1690.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1676.3400000000001_1690.4#3.jpg" ], "ocr_qwen2_vl_72b": "```python\nIn [35]:\nspec = image.sum(axis=0)\nplt.plot(spec, lw=5)\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Relative Flux Density');\n```\n\n**Relative Flux Density**\n\n**Wavelength Coordinate**" }, { "vid": "leXuUp-uxGQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astronomy tutorial on Spectral Analysis in Observational Astronomy_30.json#####audio#####doingASR#####FinishASR/leXuUp-uxGQ/1690.4_1699.26.mp4", "refined_asr": " And at each one of these wavelengths we have some amount of total light on the image. So you can see there is this kind of curve here. That's our trace that we want. But on top of that there's all these emission lines.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1690.4_1699.26#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1690.4_1699.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/leXuUp-uxGQ/leXuUp-uxGQ@1690.4_1699.26#2.jpg" ], "ocr_qwen2_vl_72b": "```python\nIn [35]:\nspec = image.sum(axis=0)\nplt.plot(spec, lw=5)\nplt.xlabel('Wavelength Coordinate')\nplt.ylabel('Relative Flux Density');\n```\n\n**Figure Caption:**\n- **X-axis:** Wavelength Coordinate\n- **Y-axis:** Relative Flux Density" } ], "image_num": 11, "text_num": 520, "token_num": 6856 }, { "images": [ "sample_100_images/f9SJz4-UaQQ@885.0600000000001_902.02#1.jpg", null, "sample_100_images/f9SJz4-UaQQ@904.6_926.94#1.jpg", null, "sample_100_images/f9SJz4-UaQQ@926.94_959.28#1.jpg", "sample_100_images/f9SJz4-UaQQ@926.94_959.28#4.jpg", null, "sample_100_images/f9SJz4-UaQQ@959.28_976.24#1.jpg", null, "sample_100_images/f9SJz4-UaQQ@976.24_1000.1#1.jpg", null, "sample_100_images/f9SJz4-UaQQ@1000.1_1032.3600000000001#1.jpg", null ], "texts": [ null, " So 2x over 6x, that's going to be one third. So this is one third times the integral of 1 du, which is going to be one third ln u. And then that's going to be one third ln 3x squared plus 4.", null, " So now we're going to make this integral equal to the limit as b approaches infinity for the integral from 2 to b of 2x over 3x squared plus 4 dx. And so that's going to be the limit as b approaches infinity.", null, null, " And so we're going to get one-third times ln of 3x squared plus 4 evaluated from 2 to b. So we no longer need the absolute value expression because 3x squared plus 4 is always positive. So we now have the limit as b approaches infinity.", null, " And then this is one-third ln 3x squared plus 4 dx. We're going to take the integral of 3x squared plus 4. And then minus, now let's plug in 2. So one-third ln 2 squared which is 4 times 3.", null, " 12 plus 4, that's 16. Now, as b goes into infinity, what happens to ln(3b squared plus 4)? Well, infinity squared is infinity. If you multiply that by 3 then add 4, you're still going to get infinity. And the natural log of infinity is still infinity.", null, " Infinity. So this entire expression goes to infinity. Infinity minus some finite number is still going to be infinity. Therefore, since the integral from 2 to infinity of f of x dx, because it equals infinity - it doesn't equal a finite number - we could say that the integral diverges. And if the integral diverges, then the series also diverges." ], "text_ocr_list": [ null, "We can see these text from the image: \u222b 2x/u du = \u222b 2x/(3x^2 + 4) dx\n\nu = 3x^2 + 4\ndu = 6x dx\ndu/6x = dx.\n So 2x over 6x, that's going to be one third. So this is one third times the integral of 1 du, which is going to be one third ln u. And then that's going to be one third ln 3x squared plus 4.", null, "We can see these text from the image: \u222b 2x/u du = 1/3 \u222b 1/u du\n\nu = 3x^2 + 4\ndu = 6x dx\ndu/6x = dx\n\n\u222b from 2 to 8 (2x/(3x^2 + 4)) dx\n\n1/3 ln|u|\n1/3 ln(3x^2 + 4).\n So now we're going to make this integral equal to the limit as b approaches infinity for the integral from 2 to b of 2x over 3x squared plus 4 dx. And so that's going to be the limit as b approaches infinity.", null, null, "We can see these text from the image: \u222b 2x/u du/dx = 1/3 \u222b 1/u du\n\nu = 3x^2 + 4\ndu = 6x dx\ndu/6x = dx\n\n\u222b from 2 to \u221e 2x/(3x^2 + 4) dx\n\n1/3 ln|u|\n1/3 ln(3x^2 + 4)\n\n= lim b\u2192\u221e \u222b from 2 to b 2x/(3x^2 + 4) dx = lim b\u2192\u221e 1/3.\n And so we're going to get one-third times ln of 3x squared plus 4 evaluated from 2 to b. So we no longer need the absolute value expression because 3x squared plus 4 is always positive. So we now have the limit as b approaches infinity.", null, "We can see these text from the image: = lim 1\nb->\u221e 3\n\n= lim \u222b 2x dx = lim 1 ln(3x^2 + 4)\nb->\u221e 2 3x^2 + 4 b->\u221e 3 2.\n And then this is one-third ln 3x squared plus 4 dx. We're going to take the integral of 3x squared plus 4. And then minus, now let's plug in 2. So one-third ln 2 squared which is 4 times 3.", null, "We can see these text from the image: = lim 1/3 ln(3b^2 + 4) - 1/3 ln(16)\nb -> \u221e\n\n= lim \u222b[2 to b] (2x / (3x^2 + 4)) dx = lim 1/3 ln(3x^2 + 4)\nb -> \u221e b -> \u221e [2 to b].\n 12 plus 4, that's 16. Now, as b goes into infinity, what happens to ln(3b squared plus 4)? Well, infinity squared is infinity. If you multiply that by 3 then add 4, you're still going to get infinity. And the natural log of infinity is still infinity.", null, "We can see these text from the image: = lim b -> \u221e 1/3 ln (3b^2 + 4) - 1/3 ln (16).\n Infinity. So this entire expression goes to infinity. Infinity minus some finite number is still going to be infinity. Therefore, since the integral from 2 to infinity of f of x dx, because it equals infinity - it doesn't equal a finite number - we could say that the integral diverges. And if the integral diverges, then the series also diverges." ], "metadata": [ { "vid": "f9SJz4-UaQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Comparison Tests in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/f9SJz4-UaQQ/885.0600000000001_902.02.mp4", "refined_asr": " So 2x over 6x, that's going to be one third. So this is one third times the integral of 1 du, which is going to be one third ln u. And then that's going to be one third ln 3x squared plus 4.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@885.0600000000001_902.02#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@885.0600000000001_902.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@885.0600000000001_902.02#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@885.0600000000001_902.02#3.jpg" ], "ocr_qwen2_vl_72b": "\u222b 2x/u du = \u222b 2x/(3x^2 + 4) dx\n\nu = 3x^2 + 4\ndu = 6x dx\ndu/6x = dx" }, { "vid": "f9SJz4-UaQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Comparison Tests in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/f9SJz4-UaQQ/904.6_926.94.mp4", "refined_asr": " So now we're going to make this integral equal to the limit as b approaches infinity for the integral from 2 to b of 2x over 3x squared plus 4 dx. And so that's going to be the limit as b approaches infinity.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@904.6_926.94#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@904.6_926.94#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@904.6_926.94#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@904.6_926.94#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@904.6_926.94#4.jpg" ], "ocr_qwen2_vl_72b": "\u222b 2x/u du = 1/3 \u222b 1/u du\n\nu = 3x^2 + 4\ndu = 6x dx\ndu/6x = dx\n\n\u222b from 2 to 8 (2x/(3x^2 + 4)) dx\n\n1/3 ln|u|\n1/3 ln(3x^2 + 4)" }, { "vid": "f9SJz4-UaQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Comparison Tests in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/f9SJz4-UaQQ/926.94_959.28.mp4", "refined_asr": " And so we're going to get one-third times ln of 3x squared plus 4 evaluated from 2 to b. So we no longer need the absolute value expression because 3x squared plus 4 is always positive. So we now have the limit as b approaches infinity.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@926.94_959.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@926.94_959.28#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@926.94_959.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@926.94_959.28#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@926.94_959.28#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@926.94_959.28#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@926.94_959.28#5.jpg" ], "ocr_qwen2_vl_72b": "\u222b 2x/u du/dx = 1/3 \u222b 1/u du\n\nu = 3x^2 + 4\ndu = 6x dx\ndu/6x = dx\n\n\u222b from 2 to \u221e 2x/(3x^2 + 4) dx\n\n1/3 ln|u|\n1/3 ln(3x^2 + 4)\n\n= lim b\u2192\u221e \u222b from 2 to b 2x/(3x^2 + 4) dx = lim b\u2192\u221e 1/3" }, { "vid": "f9SJz4-UaQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Comparison Tests in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/f9SJz4-UaQQ/959.28_976.24.mp4", "refined_asr": " And then this is one-third ln 3x squared plus 4 dx. We're going to take the integral of 3x squared plus 4. And then minus, now let's plug in 2. So one-third ln 2 squared which is 4 times 3.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@959.28_976.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@959.28_976.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@959.28_976.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@959.28_976.24#3.jpg" ], "ocr_qwen2_vl_72b": "= lim 1\nb->\u221e 3\n\n= lim \u222b 2x dx = lim 1 ln(3x^2 + 4)\nb->\u221e 2 3x^2 + 4 b->\u221e 3 2" }, { "vid": "f9SJz4-UaQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Comparison Tests in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/f9SJz4-UaQQ/976.24_1000.1.mp4", "refined_asr": " 12 plus 4, that's 16. Now, as b goes into infinity, what happens to ln(3b squared plus 4)? Well, infinity squared is infinity. If you multiply that by 3 then add 4, you're still going to get infinity. And the natural log of infinity is still infinity.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@976.24_1000.1#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@976.24_1000.1#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@976.24_1000.1#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@976.24_1000.1#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@976.24_1000.1#4.jpg" ], "ocr_qwen2_vl_72b": "= lim 1/3 ln(3b^2 + 4) - 1/3 ln(16)\nb -> \u221e\n\n= lim \u222b[2 to b] (2x / (3x^2 + 4)) dx = lim 1/3 ln(3x^2 + 4)\nb -> \u221e b -> \u221e [2 to b]" }, { "vid": "f9SJz4-UaQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Comparison Tests in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/f9SJz4-UaQQ/1000.1_1032.3600000000001.mp4", "refined_asr": " Infinity. So this entire expression goes to infinity. Infinity minus some finite number is still going to be infinity. Therefore, since the integral from 2 to infinity of f of x dx, because it equals infinity - it doesn't equal a finite number - we could say that the integral diverges. And if the integral diverges, then the series also diverges.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@1000.1_1032.3600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@1000.1_1032.3600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@1000.1_1032.3600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@1000.1_1032.3600000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@1000.1_1032.3600000000001#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/f9SJz4-UaQQ/f9SJz4-UaQQ@1000.1_1032.3600000000001#5.jpg" ], "ocr_qwen2_vl_72b": "= lim b -> \u221e 1/3 ln (3b^2 + 4) - 1/3 ln (16)" } ], "image_num": 7, "text_num": 409, "token_num": 4441 }, { "images": [ "sample_100_images/U8m5ug9Q54M@2331.98_2345.98#1.jpg", "sample_100_images/U8m5ug9Q54M@2331.98_2345.98#2.jpg", null, "sample_100_images/U8m5ug9Q54M@2345.98_2359.98#1.jpg", null, "sample_100_images/U8m5ug9Q54M@2359.98_2383.98#1.jpg", null, "sample_100_images/U8m5ug9Q54M@2383.98_2400.98#1.jpg", null, "sample_100_images/U8m5ug9Q54M@2400.98_2414.98#1.jpg", "sample_100_images/U8m5ug9Q54M@2400.98_2414.98#2.jpg", null, "sample_100_images/U8m5ug9Q54M@2414.98_2430.98#1.jpg", "sample_100_images/U8m5ug9Q54M@2414.98_2430.98#2.jpg", null, "sample_100_images/U8m5ug9Q54M@2430.98_2443.98#1.jpg", null, "sample_100_images/U8m5ug9Q54M@2443.98_2457.98#1.jpg", "sample_100_images/U8m5ug9Q54M@2443.98_2457.98#2.jpg", null ], "texts": [ null, null, " To calculate our marginal likelihood, we are coming down to the important stuff. That is, the likelihood. Okay, so this is the part - this is the likelihood, which is the most important part and the toughest part to calculate. All this is pretty simple; you just have to understand the math behind it.", null, " So, in order to calculate our likelihood, what we're going to do is calculate the probability. Let's see how we can do that. We have this P of X when Y is equal to 1. Okay, what this means is when we have this email.", null, " So we already know that email belongs to spam. We already know that email belongs to non-spam. So what is the probability of that email belonging to this particular group? This is nothing but the probability that an email belongs to class one. So the probability that an email belongs to class zero, fine.", null, " So now let's see how 'X' would look like. Just to give you a brief overview before we move ahead, 'X' over here would be nothing but an email. So it will have multiple words, and somewhere over here in the middle, it might say 'get unlimited 50% off,' and so on, including many other words.", null, null, " These are called the features and based on these features we calculate whether this email belongs to a spam class or a non-spam class. So how does this work? How does this probability work? It takes each of these features. Let's take something like ultimatum.", null, null, " Okay, so it's going to be like the probability of the word ultimate belonging to spam. This would give me some value, say 0.9%, because it's a high probability that the word ultimate appears in a spam email. And at the same time, we'll also calculate the probability.", null, " That ultimate belongs to non-spam. So this is going to have a lesser probability. Obviously, you are not going to use 'ultimate' in your day-to-day activities or conversations. So that's how it's going to be.", null, null, " So let's now quickly see how we can calculate for this. So now we have 'X', right? So if 'X', this is a capital 'X', is nothing but a list of words, okay? And this is nothing but an email, okay? And then small 'x' represents the words which are there." ], "text_ocr_list": [ null, null, "We can see these text from the image: - Y_train\n- Y\n- X_test\n- Y\n- 100 emails\n- 40 spam\n- 60 not spam\n- P(y=1) = count all spam email / total no of email = 40 / 100\n- P(y=0) = count of all non spam / total no of email = 60 / 100\n- Mathematical form\n- 1/m \u03a3 (y^(i) = 1/0).\n To calculate our marginal likelihood, we are coming down to the important stuff. That is, the likelihood. Okay, so this is the part - this is the likelihood, which is the most important part and the toughest part to calculate. All this is pretty simple; you just have to understand the math behind it.", null, "We can see these text from the image: Looking for Python Online Training? Call us at IN: 9606058406 / US: 18338555775 or visit www.edureka.co.\n So, in order to calculate our likelihood, what we're going to do is calculate the probability. Let's see how we can do that. We have this P of X when Y is equal to 1. Okay, what this means is when we have this email.", null, "We can see these text from the image: - X-test\n- Y\n- Count all spam email = 40\n- Total No of email = 100\n- Count of all non-spam = 60\n- Total No of email = 100.\n So we already know that email belongs to spam. We already know that email belongs to non-spam. So what is the probability of that email belonging to this particular group? This is nothing but the probability that an email belongs to class one. So the probability that an email belongs to class zero, fine.", null, "We can see these text from the image: P(y=1) P(x|y=0) probability that email belong to class 0.\n So now let's see how 'X' would look like. Just to give you a brief overview before we move ahead, 'X' over here would be nothing but an email. So it will have multiple words, and somewhere over here in the middle, it might say 'get unlimited 50% off,' and so on, including many other words.", null, null, "We can see these text from the image: P(y=1)\nP(x|y=0) probability that email belong to class 0\n\nx = [get unlimited 50% off].\n These are called the features and based on these features we calculate whether this email belongs to a spam class or a non-spam class. So how does this work? How does this probability work? It takes each of these features. Let's take something like ultimatum.", null, null, "We can see these text from the image: P(x|y=1) probability that email belongs to spam\n\nP(x|y=0) probability that email belongs to class 0\n\nx = [get unlimited 50% off ...]\n\nP(w).\n Okay, so it's going to be like the probability of the word ultimate belonging to spam. This would give me some value, say 0.9%, because it's a high probability that the word ultimate appears in a spam email. And at the same time, we'll also calculate the probability.", null, "We can see these text from the image: - P(x|y=0) probability that email belongs to class 0\n- x = [get unlimited 50% off ...]\n- P(ultimate|y=1) = 0.90\n- P(ultimate|y=0).\n That ultimate belongs to non-spam. So this is going to have a lesser probability. Obviously, you are not going to use 'ultimate' in your day-to-day activities or conversations. So that's how it's going to be.", null, null, "We can see these text from the image: - \\( P(y=1) \\)\n- \\( P(y=0) \\)\n- \\( x = \\begin{bmatrix} \\text{get unlimited 50\\% off} \\\\ \\text{ultimate} \\end{bmatrix} \\)\n- \\( P(\\text{ultimate} | y=1) = 0.90 \\)\n- \\( P(\\text{ultimate} | y=0) = 0.20 \\).\n So let's now quickly see how we can calculate for this. So now we have 'X', right? So if 'X', this is a capital 'X', is nothing but a list of words, okay? And this is nothing but an email, okay? And then small 'x' represents the words which are there." ], "metadata": [ { "vid": "U8m5ug9Q54M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence tutorial on Sentiment Analysis using Lexicon-Based Methods in Natural Language Processing_30.json#####audio#####doingASR#####FinishASR/U8m5ug9Q54M/2331.98_2345.98.mp4", "refined_asr": " To calculate our marginal likelihood, we are coming down to the important stuff. That is, the likelihood. Okay, so this is the part - this is the likelihood, which is the most important part and the toughest part to calculate. All this is pretty simple; you just have to understand the math behind it.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2331.98_2345.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2331.98_2345.98#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2331.98_2345.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2331.98_2345.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2331.98_2345.98#3.jpg" ], "ocr_qwen2_vl_72b": "- Y_train\n- Y\n- X_test\n- Y\n- 100 emails\n- 40 spam\n- 60 not spam\n- P(y=1) = count all spam email / total no of email = 40 / 100\n- P(y=0) = count of all non spam / total no of email = 60 / 100\n- Mathematical form\n- 1/m \u03a3 (y^(i) = 1/0)" }, { "vid": "U8m5ug9Q54M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence tutorial on Sentiment Analysis using Lexicon-Based Methods in Natural Language Processing_30.json#####audio#####doingASR#####FinishASR/U8m5ug9Q54M/2345.98_2359.98.mp4", "refined_asr": " So, in order to calculate our likelihood, what we're going to do is calculate the probability. Let's see how we can do that. We have this P of X when Y is equal to 1. Okay, what this means is when we have this email.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2345.98_2359.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2345.98_2359.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2345.98_2359.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2345.98_2359.98#3.jpg" ], "ocr_qwen2_vl_72b": "Looking for Python Online Training? Call us at IN: 9606058406 / US: 18338555775 or visit www.edureka.co" }, { "vid": "U8m5ug9Q54M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence tutorial on Sentiment Analysis using Lexicon-Based Methods in Natural Language Processing_30.json#####audio#####doingASR#####FinishASR/U8m5ug9Q54M/2359.98_2383.98.mp4", "refined_asr": " So we already know that email belongs to spam. We already know that email belongs to non-spam. So what is the probability of that email belonging to this particular group? This is nothing but the probability that an email belongs to class one. So the probability that an email belongs to class zero, fine.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2359.98_2383.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2359.98_2383.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2359.98_2383.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2359.98_2383.98#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2359.98_2383.98#4.jpg" ], "ocr_qwen2_vl_72b": "- X-test\n- Y\n- Count all spam email = 40\n- Total No of email = 100\n- Count of all non-spam = 60\n- Total No of email = 100" }, { "vid": "U8m5ug9Q54M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence tutorial on Sentiment Analysis using Lexicon-Based Methods in Natural Language Processing_30.json#####audio#####doingASR#####FinishASR/U8m5ug9Q54M/2383.98_2400.98.mp4", "refined_asr": " So now let's see how 'X' would look like. Just to give you a brief overview before we move ahead, 'X' over here would be nothing but an email. So it will have multiple words, and somewhere over here in the middle, it might say 'get unlimited 50% off,' and so on, including many other words.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2383.98_2400.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2383.98_2400.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2383.98_2400.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2383.98_2400.98#3.jpg" ], "ocr_qwen2_vl_72b": "P(y=1) P(x|y=0) probability that email belong to class 0" }, { "vid": "U8m5ug9Q54M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence tutorial on Sentiment Analysis using Lexicon-Based Methods in Natural Language Processing_30.json#####audio#####doingASR#####FinishASR/U8m5ug9Q54M/2400.98_2414.98.mp4", "refined_asr": " These are called the features and based on these features we calculate whether this email belongs to a spam class or a non-spam class. So how does this work? How does this probability work? It takes each of these features. Let's take something like ultimatum.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2400.98_2414.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2400.98_2414.98#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2400.98_2414.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2400.98_2414.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2400.98_2414.98#3.jpg" ], "ocr_qwen2_vl_72b": "P(y=1)\nP(x|y=0) probability that email belong to class 0\n\nx = [get unlimited 50% off]" }, { "vid": "U8m5ug9Q54M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence tutorial on Sentiment Analysis using Lexicon-Based Methods in Natural Language Processing_30.json#####audio#####doingASR#####FinishASR/U8m5ug9Q54M/2414.98_2430.98.mp4", "refined_asr": " Okay, so it's going to be like the probability of the word ultimate belonging to spam. This would give me some value, say 0.9%, because it's a high probability that the word ultimate appears in a spam email. And at the same time, we'll also calculate the probability.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2414.98_2430.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2414.98_2430.98#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2414.98_2430.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2414.98_2430.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2414.98_2430.98#3.jpg" ], "ocr_qwen2_vl_72b": "P(x|y=1) probability that email belongs to spam\n\nP(x|y=0) probability that email belongs to class 0\n\nx = [get unlimited 50% off ...]\n\nP(w)" }, { "vid": "U8m5ug9Q54M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence tutorial on Sentiment Analysis using Lexicon-Based Methods in Natural Language Processing_30.json#####audio#####doingASR#####FinishASR/U8m5ug9Q54M/2430.98_2443.98.mp4", "refined_asr": " That ultimate belongs to non-spam. So this is going to have a lesser probability. Obviously, you are not going to use 'ultimate' in your day-to-day activities or conversations. So that's how it's going to be.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2430.98_2443.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2430.98_2443.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2430.98_2443.98#2.jpg" ], "ocr_qwen2_vl_72b": "- P(x|y=0) probability that email belongs to class 0\n- x = [get unlimited 50% off ...]\n- P(ultimate|y=1) = 0.90\n- P(ultimate|y=0)" }, { "vid": "U8m5ug9Q54M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Artificial intelligence tutorial on Sentiment Analysis using Lexicon-Based Methods in Natural Language Processing_30.json#####audio#####doingASR#####FinishASR/U8m5ug9Q54M/2443.98_2457.98.mp4", "refined_asr": " So let's now quickly see how we can calculate for this. So now we have 'X', right? So if 'X', this is a capital 'X', is nothing but a list of words, okay? And this is nothing but an email, okay? And then small 'x' represents the words which are there.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2443.98_2457.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2443.98_2457.98#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2443.98_2457.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2443.98_2457.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/U8m5ug9Q54M/U8m5ug9Q54M@2443.98_2457.98#3.jpg" ], "ocr_qwen2_vl_72b": "- \\( P(y=1) \\)\n- \\( P(y=0) \\)\n- \\( x = \\begin{bmatrix} \\text{get unlimited 50\\% off} \\\\ \\text{ultimate} \\end{bmatrix} \\)\n- \\( P(\\text{ultimate} | y=1) = 0.90 \\)\n- \\( P(\\text{ultimate} | y=0) = 0.20 \\)" } ], "image_num": 12, "text_num": 526, "token_num": 7438 }, { "images": [ "sample_100_images/72gMmt3082o@2377.94_2399.1#1.jpg", "sample_100_images/72gMmt3082o@2377.94_2399.1#2.jpg", null, "sample_100_images/72gMmt3082o@2399.1_2413.92#1.jpg", null, "sample_100_images/72gMmt3082o@2413.92_2431.6800000000003#1.jpg", null, "sample_100_images/72gMmt3082o@2431.6800000000003_2452.32#1.jpg", null, "sample_100_images/72gMmt3082o@2452.32_2467.56#1.jpg", null ], "texts": [ null, null, " Now there are a couple of ways. I'll show you a couple of simulations of how this can happen. This is more of a symmetric simulation where, again, you take the upper hemisphere of the star, begin to rotate it slowly with respect to the lower hemisphere.", null, " Twisting out the magnetic fields attached to the star. One thin line is the color J5, the little tangents. One thin line is J5, and the other one is highlighted there; it's like a thin flat line.", null, " And as you begin to twist strongly with a few fields, the magnification inflates. You can see that this particular thin line becomes more and more extended and eventually it opens up. There is an instability that when you impart magnetic twist, it radiates.", null, " Something like four or five. Then this configuration becomes unstable. So slowly the field inflates and then boom at some point it becomes unstable from a very thin turn sheets according to opposite magnetic classes this way. And then this turn sheet between them.", null, " And this fault is known as a magnetic collapse. It then becomes unstable to so-called tearing modes because this thin sheet begins to break up into threads, current threads. This is the reconnection process. You can see the simulation. So that's one way of doing it." ], "text_ocr_list": [ null, null, "We can see these text from the image: Magnetic reconnection.\n Now there are a couple of ways. I'll show you a couple of simulations of how this can happen. This is more of a symmetric simulation where, again, you take the upper hemisphere of the star, begin to rotate it slowly with respect to the lower hemisphere.", null, "We can see these text from the image: - Loss of magnetic equilibrium and reconnection\n\n(a) \\( t = t_{\\text{rec}} - 1000 \\)\n\n(b) \\( t = t_{\\text{rec}} - 500 \\)\n\n(c) \\( t = t_{\\text{rec}} - 30 \\)\n\n(d) \\( t = t_{\\text{rec}} \\)\n\n(e) \\( t = t_{\\text{rec}} + 30 \\)\n\n(f) \\( t = t_{\\text{rec}} + 250 \\)\n\n\\( \\psi = 3.1 \\)\n\n\\( \\psi = 4.4 \\)\n\n\\( \\psi = 5.5 \\)\n\nParfrey et al. (2013).\n Twisting out the magnetic fields attached to the star. One thin line is the color J5, the little tangents. One thin line is J5, and the other one is highlighted there; it's like a thin flat line.", null, "We can see these text from the image: - Loss of magnetic equilibrium and reconnection\n- (a) \\( t = t_{\\text{rec}} - 1000 \\)\n- (b) \\( t = t_{\\text{rec}} - 500 \\)\n- (c) \\( t = t_{\\text{rec}} - 30 \\)\n- (d) \\( t = t_{\\text{rec}} \\)\n- (e) \\( t = t_{\\text{rec}} + 30 \\)\n- (f) \\( t = t_{\\text{rec}} + 250 \\)\n- \\( \\psi = 3.1 \\)\n- \\( \\psi = 4.4 \\)\n- \\( \\psi = 5.5 \\)\n- Parfrey et al. (2013).\n And as you begin to twist strongly with a few fields, the magnification inflates. You can see that this particular thin line becomes more and more extended and eventually it opens up. There is an instability that when you impart magnetic twist, it radiates.", null, "We can see these text from the image: - Loss of magnetic equilibrium and reconnection\n- (a) \\( t = t_{rec} - 1000 \\)\n- (b) \\( t = t_{rec} - 500 \\)\n- (c) \\( t = t_{rec} - 30 \\)\n- (d) \\( t = t_{rec} \\)\n- (e) \\( t = t_{rec} + 30 \\)\n- (f) \\( t = t_{rec} + 250 \\)\n- \\( \\psi = 3.1 \\)\n- \\( \\psi = 4.4 \\)\n- \\( \\psi = 5.5 \\)\n- Parfrey et al. (2013).\n Something like four or five. Then this configuration becomes unstable. So slowly the field inflates and then boom at some point it becomes unstable from a very thin turn sheets according to opposite magnetic classes this way. And then this turn sheet between them.", null, "We can see these text from the image: - Loss of magnetic equilibrium and reconnection\n- (a) \\( t = t_{\\text{rec}} - 1000 \\)\n- (b) \\( t = t_{\\text{rec}} - 500 \\)\n- (c) \\( t = t_{\\text{rec}} - 30 \\)\n- (d) \\( t = t_{\\text{rec}} \\)\n- (e) \\( t = t_{\\text{rec}} + 30 \\)\n- (f) \\( t = t_{\\text{rec}} + 250 \\)\n- \\( \\psi = 3.1 \\)\n- \\( \\psi = 4.4 \\)\n- \\( \\psi = 5.5 \\)\n- \\( J_{\\phi} \\)\n- \\( c \\mu / r^4 \\)\n- Parfrey et al. (2013).\n And this fault is known as a magnetic collapse. It then becomes unstable to so-called tearing modes because this thin sheet begins to break up into threads, current threads. This is the reconnection process. You can see the simulation. So that's one way of doing it." ], "metadata": [ { "vid": "72gMmt3082o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics lecture on Magnetars in High Energy Astrophysics, focusing on Neutron Stars and Pulsars._30.json#####audio#####doingASR#####FinishASR/72gMmt3082o/2377.94_2399.1.mp4", "refined_asr": " Now there are a couple of ways. I'll show you a couple of simulations of how this can happen. This is more of a symmetric simulation where, again, you take the upper hemisphere of the star, begin to rotate it slowly with respect to the lower hemisphere.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2377.94_2399.1#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2377.94_2399.1#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2377.94_2399.1#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2377.94_2399.1#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2377.94_2399.1#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2377.94_2399.1#4.jpg" ], "ocr_qwen2_vl_72b": "Magnetic reconnection" }, { "vid": "72gMmt3082o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics lecture on Magnetars in High Energy Astrophysics, focusing on Neutron Stars and Pulsars._30.json#####audio#####doingASR#####FinishASR/72gMmt3082o/2399.1_2413.92.mp4", "refined_asr": " Twisting out the magnetic fields attached to the star. One thin line is the color J5, the little tangents. One thin line is J5, and the other one is highlighted there; it's like a thin flat line.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2399.1_2413.92#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2399.1_2413.92#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2399.1_2413.92#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2399.1_2413.92#3.jpg" ], "ocr_qwen2_vl_72b": "- Loss of magnetic equilibrium and reconnection\n\n(a) \\( t = t_{\\text{rec}} - 1000 \\)\n\n(b) \\( t = t_{\\text{rec}} - 500 \\)\n\n(c) \\( t = t_{\\text{rec}} - 30 \\)\n\n(d) \\( t = t_{\\text{rec}} \\)\n\n(e) \\( t = t_{\\text{rec}} + 30 \\)\n\n(f) \\( t = t_{\\text{rec}} + 250 \\)\n\n\\( \\psi = 3.1 \\)\n\n\\( \\psi = 4.4 \\)\n\n\\( \\psi = 5.5 \\)\n\nParfrey et al. (2013)" }, { "vid": "72gMmt3082o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics lecture on Magnetars in High Energy Astrophysics, focusing on Neutron Stars and Pulsars._30.json#####audio#####doingASR#####FinishASR/72gMmt3082o/2413.92_2431.6800000000003.mp4", "refined_asr": " And as you begin to twist strongly with a few fields, the magnification inflates. You can see that this particular thin line becomes more and more extended and eventually it opens up. There is an instability that when you impart magnetic twist, it radiates.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2413.92_2431.6800000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2413.92_2431.6800000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2413.92_2431.6800000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2413.92_2431.6800000000003#3.jpg" ], "ocr_qwen2_vl_72b": "- Loss of magnetic equilibrium and reconnection\n- (a) \\( t = t_{\\text{rec}} - 1000 \\)\n- (b) \\( t = t_{\\text{rec}} - 500 \\)\n- (c) \\( t = t_{\\text{rec}} - 30 \\)\n- (d) \\( t = t_{\\text{rec}} \\)\n- (e) \\( t = t_{\\text{rec}} + 30 \\)\n- (f) \\( t = t_{\\text{rec}} + 250 \\)\n- \\( \\psi = 3.1 \\)\n- \\( \\psi = 4.4 \\)\n- \\( \\psi = 5.5 \\)\n- Parfrey et al. (2013)" }, { "vid": "72gMmt3082o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics lecture on Magnetars in High Energy Astrophysics, focusing on Neutron Stars and Pulsars._30.json#####audio#####doingASR#####FinishASR/72gMmt3082o/2431.6800000000003_2452.32.mp4", "refined_asr": " Something like four or five. Then this configuration becomes unstable. So slowly the field inflates and then boom at some point it becomes unstable from a very thin turn sheets according to opposite magnetic classes this way. And then this turn sheet between them.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2431.6800000000003_2452.32#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2431.6800000000003_2452.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2431.6800000000003_2452.32#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2431.6800000000003_2452.32#3.jpg" ], "ocr_qwen2_vl_72b": "- Loss of magnetic equilibrium and reconnection\n- (a) \\( t = t_{rec} - 1000 \\)\n- (b) \\( t = t_{rec} - 500 \\)\n- (c) \\( t = t_{rec} - 30 \\)\n- (d) \\( t = t_{rec} \\)\n- (e) \\( t = t_{rec} + 30 \\)\n- (f) \\( t = t_{rec} + 250 \\)\n- \\( \\psi = 3.1 \\)\n- \\( \\psi = 4.4 \\)\n- \\( \\psi = 5.5 \\)\n- Parfrey et al. (2013)" }, { "vid": "72gMmt3082o.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Astrophysics lecture on Magnetars in High Energy Astrophysics, focusing on Neutron Stars and Pulsars._30.json#####audio#####doingASR#####FinishASR/72gMmt3082o/2452.32_2467.56.mp4", "refined_asr": " And this fault is known as a magnetic collapse. It then becomes unstable to so-called tearing modes because this thin sheet begins to break up into threads, current threads. This is the reconnection process. You can see the simulation. So that's one way of doing it.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2452.32_2467.56#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2452.32_2467.56#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2452.32_2467.56#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/72gMmt3082o/72gMmt3082o@2452.32_2467.56#3.jpg" ], "ocr_qwen2_vl_72b": "- Loss of magnetic equilibrium and reconnection\n- (a) \\( t = t_{\\text{rec}} - 1000 \\)\n- (b) \\( t = t_{\\text{rec}} - 500 \\)\n- (c) \\( t = t_{\\text{rec}} - 30 \\)\n- (d) \\( t = t_{\\text{rec}} \\)\n- (e) \\( t = t_{\\text{rec}} + 30 \\)\n- (f) \\( t = t_{\\text{rec}} + 250 \\)\n- \\( \\psi = 3.1 \\)\n- \\( \\psi = 4.4 \\)\n- \\( \\psi = 5.5 \\)\n- \\( J_{\\phi} \\)\n- \\( c \\mu / r^4 \\)\n- Parfrey et al. (2013)" } ], "image_num": 6, "text_num": 281, "token_num": 3737 }, { "images": [ "sample_100_images/ltdH0rW-1Jg@709.62_717.62#1.jpg", null, "sample_100_images/ltdH0rW-1Jg@717.62_728.62#1.jpg", "sample_100_images/ltdH0rW-1Jg@717.62_728.62#2.jpg", null, "sample_100_images/ltdH0rW-1Jg@728.62_741.16#1.jpg", "sample_100_images/ltdH0rW-1Jg@728.62_741.16#2.jpg", null, "sample_100_images/ltdH0rW-1Jg@741.16_752.6#1.jpg", "sample_100_images/ltdH0rW-1Jg@741.16_752.6#2.jpg", null, "sample_100_images/ltdH0rW-1Jg@752.6_764.6#1.jpg", "sample_100_images/ltdH0rW-1Jg@752.6_764.6#2.jpg", null, "sample_100_images/ltdH0rW-1Jg@764.6_809.88#1.jpg", "sample_100_images/ltdH0rW-1Jg@764.6_809.88#2.jpg", "sample_100_images/ltdH0rW-1Jg@764.6_809.88#3.jpg", "sample_100_images/ltdH0rW-1Jg@764.6_809.88#4.jpg", "sample_100_images/ltdH0rW-1Jg@764.6_809.88#5.jpg", "sample_100_images/ltdH0rW-1Jg@764.6_809.88#6.jpg", null, "sample_100_images/ltdH0rW-1Jg@810.22_823.12#1.jpg", "sample_100_images/ltdH0rW-1Jg@810.22_823.12#2.jpg", null, "sample_100_images/ltdH0rW-1Jg@823.52_835.4399999999999#1.jpg", "sample_100_images/ltdH0rW-1Jg@823.52_835.4399999999999#2.jpg", null, "sample_100_images/ltdH0rW-1Jg@835.9599999999999_852.5#1.jpg", "sample_100_images/ltdH0rW-1Jg@835.9599999999999_852.5#2.jpg", null ], "texts": [ null, " Okay, yep, here we go. Let's see, almost, almost. I think we should fill it up to about right there. Girls, what do you think's gonna happen? I think it's gonna explode. You think it's gonna explode?", null, null, " Yeah. Oh, I think so too. Alright, well, there's a lot right there. Alright ladies, you ready? I'm ready.", null, null, " I'm ready. One last ingredient: it's some really cool mints.", null, null, " Oh, you feel the earth shaking. Here we go. You might wanna watch out. Whoa! Wow! And it's erupting on the sides. That was awesome. It's the beach volcano. I'm outta here.", null, null, " I'm out.", null, null, null, null, null, null, " I'm out I don't know. I think I saw something in the sand. Oh can you see this? Oh wow look right there it's a ghost crab and he's buried himself in the ground.", null, null, " Kind of to protect himself. \"Hey girls, come look at this.\" Easy, easy, easy, easy, quiet. If you look right there, you see there's a ghost crab in the ground. Isn't that cool? And he's kind of trying to protect himself.", null, null, " And he's buried himself down in the sand. Isn't that really cool? You can see his eyes sticking straight up. And the rest of his body is in the sand. Let's see if we can try and move him.", null, null, " Alright y'all ready? Here we go. He's moving really quick. See? Oh, there he goes. Looks like he found his home. That's awesome. Oh, wow, my friends. Being at the beach is totally awesome." ], "text_ocr_list": [ null, "We can see these text from the image: No text or formulas detected in the image..\n Okay, yep, here we go. Let's see, almost, almost. I think we should fill it up to about right there. Girls, what do you think's gonna happen? I think it's gonna explode. You think it's gonna explode?", null, null, "We can see these text from the image: HAL.\n Yeah. Oh, I think so too. Alright, well, there's a lot right there. Alright ladies, you ready? I'm ready.", null, null, "We can see these text from the image: No text or formulas are present in the image..\n I'm ready. One last ingredient: it's some really cool mints.", null, null, "We can see these text from the image: HALI.\n Oh, you feel the earth shaking. Here we go. You might wanna watch out. Whoa! Wow! And it's erupting on the sides. That was awesome. It's the beach volcano. I'm outta here.", null, null, "We can see these text from the image: HAL.\n I'm out.", null, null, null, null, null, null, "We can see these text from the image: There is no text or formula present in the image..\n I'm out I don't know. I think I saw something in the sand. Oh can you see this? Oh wow look right there it's a ghost crab and he's buried himself in the ground.", null, null, "We can see these text from the image: HAL.\n Kind of to protect himself. \"Hey girls, come look at this.\" Easy, easy, easy, easy, quiet. If you look right there, you see there's a ghost crab in the ground. Isn't that cool? And he's kind of trying to protect himself.", null, null, "We can see these text from the image: There is no text or formula to extract from this image..\n And he's buried himself down in the sand. Isn't that really cool? You can see his eyes sticking straight up. And the rest of his body is in the sand. Let's see if we can try and move him.", null, null, "We can see these text from the image: There is no text or formula to extract from the image..\n Alright y'all ready? Here we go. He's moving really quick. See? Oh, there he goes. Looks like he found his home. That's awesome. Oh, wow, my friends. Being at the beach is totally awesome." ], "metadata": [ { "vid": "ltdH0rW-1Jg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Building sandcastles and sculptures_100.json#####audio#####doingASR#####FinishASR/ltdH0rW-1Jg/709.62_717.62.mp4", "refined_asr": " Okay, yep, here we go. Let's see, almost, almost. I think we should fill it up to about right there. Girls, what do you think's gonna happen? I think it's gonna explode. You think it's gonna explode?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@709.62_717.62#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@709.62_717.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@709.62_717.62#2.jpg" ], "ocr_qwen2_vl_72b": "No text or formulas detected in the image." }, { "vid": "ltdH0rW-1Jg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Building sandcastles and sculptures_100.json#####audio#####doingASR#####FinishASR/ltdH0rW-1Jg/717.62_728.62.mp4", "refined_asr": " Yeah. Oh, I think so too. Alright, well, there's a lot right there. Alright ladies, you ready? I'm ready.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@717.62_728.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@717.62_728.62#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@717.62_728.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@717.62_728.62#2.jpg" ], "ocr_qwen2_vl_72b": "HAL" }, { "vid": "ltdH0rW-1Jg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Building sandcastles and sculptures_100.json#####audio#####doingASR#####FinishASR/ltdH0rW-1Jg/728.62_741.16.mp4", "refined_asr": " I'm ready. One last ingredient: it's some really cool mints.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@728.62_741.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@728.62_741.16#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@728.62_741.16#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@728.62_741.16#2.jpg" ], "ocr_qwen2_vl_72b": "No text or formulas are present in the image." }, { "vid": "ltdH0rW-1Jg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Building sandcastles and sculptures_100.json#####audio#####doingASR#####FinishASR/ltdH0rW-1Jg/741.16_752.6.mp4", "refined_asr": " Oh, you feel the earth shaking. Here we go. You might wanna watch out. Whoa! Wow! And it's erupting on the sides. That was awesome. It's the beach volcano. I'm outta here.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@741.16_752.6#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@741.16_752.6#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@741.16_752.6#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@741.16_752.6#2.jpg" ], "ocr_qwen2_vl_72b": "HALI" }, { "vid": "ltdH0rW-1Jg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Building sandcastles and sculptures_100.json#####audio#####doingASR#####FinishASR/ltdH0rW-1Jg/752.6_764.6.mp4", "refined_asr": " I'm out.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@752.6_764.6#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@752.6_764.6#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@752.6_764.6#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@752.6_764.6#2.jpg" ], "ocr_qwen2_vl_72b": "HAL" }, { "vid": "ltdH0rW-1Jg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Building sandcastles and sculptures_100.json#####audio#####doingASR#####FinishASR/ltdH0rW-1Jg/764.6_809.88.mp4", "refined_asr": " I'm out I don't know. I think I saw something in the sand. Oh can you see this? Oh wow look right there it's a ghost crab and he's buried himself in the ground.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#6.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@764.6_809.88#7.jpg" ], "ocr_qwen2_vl_72b": "There is no text or formula present in the image." }, { "vid": "ltdH0rW-1Jg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Building sandcastles and sculptures_100.json#####audio#####doingASR#####FinishASR/ltdH0rW-1Jg/810.22_823.12.mp4", "refined_asr": " Kind of to protect himself. \"Hey girls, come look at this.\" Easy, easy, easy, easy, quiet. If you look right there, you see there's a ghost crab in the ground. Isn't that cool? And he's kind of trying to protect himself.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@810.22_823.12#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@810.22_823.12#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@810.22_823.12#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@810.22_823.12#2.jpg" ], "ocr_qwen2_vl_72b": "HAL" }, { "vid": "ltdH0rW-1Jg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Building sandcastles and sculptures_100.json#####audio#####doingASR#####FinishASR/ltdH0rW-1Jg/823.52_835.4399999999999.mp4", "refined_asr": " And he's buried himself down in the sand. Isn't that really cool? You can see his eyes sticking straight up. And the rest of his body is in the sand. Let's see if we can try and move him.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@823.52_835.4399999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@823.52_835.4399999999999#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@823.52_835.4399999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@823.52_835.4399999999999#2.jpg" ], "ocr_qwen2_vl_72b": "There is no text or formula to extract from this image." }, { "vid": "ltdH0rW-1Jg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Building sandcastles and sculptures_100.json#####audio#####doingASR#####FinishASR/ltdH0rW-1Jg/835.9599999999999_852.5.mp4", "refined_asr": " Alright y'all ready? Here we go. He's moving really quick. See? Oh, there he goes. Looks like he found his home. That's awesome. Oh, wow, my friends. Being at the beach is totally awesome.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@835.9599999999999_852.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@835.9599999999999_852.5#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@835.9599999999999_852.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@835.9599999999999_852.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ltdH0rW-1Jg/ltdH0rW-1Jg@835.9599999999999_852.5#3.jpg" ], "ocr_qwen2_vl_72b": "There is no text or formula to extract from the image." } ], "image_num": 21, "text_num": 404, "token_num": 12500 }, { "images": [ "sample_100_images/-lD0skTnqFo@88.7_101.22#1.jpg", "sample_100_images/-lD0skTnqFo@88.7_101.22#2.jpg", null, "sample_100_images/-lD0skTnqFo@101.32000000000001_118.10000000000001#1.jpg", "sample_100_images/-lD0skTnqFo@101.32000000000001_118.10000000000001#2.jpg", null, "sample_100_images/-lD0skTnqFo@118.7_124.18#1.jpg", null, "sample_100_images/-lD0skTnqFo@124.18_136.5#1.jpg", "sample_100_images/-lD0skTnqFo@124.18_136.5#2.jpg", null, "sample_100_images/-lD0skTnqFo@137.1_148.4#1.jpg", "sample_100_images/-lD0skTnqFo@137.1_148.4#2.jpg", null, "sample_100_images/-lD0skTnqFo@148.4_162.56#1.jpg", "sample_100_images/-lD0skTnqFo@148.4_162.56#2.jpg", null, "sample_100_images/-lD0skTnqFo@162.56_179.44#1.jpg", "sample_100_images/-lD0skTnqFo@162.56_179.44#2.jpg", null, "sample_100_images/-lD0skTnqFo@179.44_194.04000000000002#1.jpg", "sample_100_images/-lD0skTnqFo@179.44_194.04000000000002#2.jpg", null ], "texts": [ null, null, " And then I advance to my S6, and advance to my S7, and so on. That's because of the nature of this alternating series. There's this sort of back-and-forth process, and it kind of looks like it's going to converge.", null, null, " The actual value of the series is going to be stuck somewhere in the middle. Now if we look at all the even partial sums - the S2, the S4, the S6, the even sums is an increasing sequence. Well the odd partial sums - the S3, the S5, the S7 -", null, " The S6, the S8, the S9, the S10, the S11, the S12, the S13, the S14, the S15, the S16, and the S21.", null, null, " And then we start to get a decreasing sequence; it's getting smaller. So the final answer appears to be squished between these two different sides. Let's formalize this idea into a theorem. I'm going to say that the alternating series test says:", null, null, " If I begin with an alternating series, so minus one to the n, minus one times a sub n, some sequence that looks a lot like it did for the harmonic series, where it's going to be decreasing and where it's going to be positive.", null, null, " If these are true then we conclude that the series converges and that is our alternating series test. Let's see it for this specific example. So for this example this minus one to the n plus one divided by n well what do we have?", null, null, " It's certainly going to be the case that B sub n is positive. So B sub n in this example is just one over n. It's positive, it's decreasing as n gets larger, one over n gets smaller, and the limit of one over n is just zero. So this satisfies all three conditions.", null, null, " The conditions of the alternating series test lead us to conclude that it converges. Now, alternating series are particularly useful, and one reason for their utility is their ease of verification. Specifically, this final condition - that the limit of" ], "text_ocr_list": [ null, null, "We can see these text from the image: \\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} + \\frac{1}{7} - \\ldots\n\\]\n\n\\( S_7 \\)\n\n\\( s_2 \\)\n\n\\( s_4 s_6 \\)\n\n\\( s_3 \\)\n\n\\( s_1 \\).\n And then I advance to my S6, and advance to my S7, and so on. That's because of the nature of this alternating series. There's this sort of back-and-forth process, and it kind of looks like it's going to converge.", null, null, "We can see these text from the image: \\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} + \\frac{1}{7} - \\ldots\n\\]\n\n\\[\ns_7\n\\]\n\n\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n}\n\\].\n The actual value of the series is going to be stuck somewhere in the middle. Now if we look at all the even partial sums - the S2, the S4, the S6, the even sums is an increasing sequence. Well the odd partial sums - the S3, the S5, the S7 -", null, "We can see these text from the image: \\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} + \\frac{1}{7} - \\ldots\n\\]\n\n\\[\ns_7\n\\]\n\n\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n}\n\\].\n The S6, the S8, the S9, the S10, the S11, the S12, the S13, the S14, the S15, the S16, and the S21.", null, null, "We can see these text from the image: \\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} + \\frac{1}{7}\n\\]\n\n\\[\ns_7\n\\]\n\n\\[\n0 \\quad s_2 \\quad s_4 \\quad s_6 \\quad s_3 \\quad s_5 \\quad s_1\n\\]\n\n\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n}\n\\].\n And then we start to get a decreasing sequence; it's getting smaller. So the final answer appears to be squished between these two different sides. Let's formalize this idea into a theorem. I'm going to say that the alternating series test says:", null, null, "We can see these text from the image: Alternating Series Test\n\nFor \u2211 (-1)^(n-1) b_n with\nn=1\n\n1) b_n decreasing\n2) b_n positive\n3) lim b_n = 0\n n\u2192\u221e\n\nThen the series converges..\n If I begin with an alternating series, so minus one to the n, minus one times a sub n, some sequence that looks a lot like it did for the harmonic series, where it's going to be decreasing and where it's going to be positive.", null, null, "We can see these text from the image: Alternating Series Test\n\nFor \\(\\sum_{n=1}^{\\infty} (-1)^{n-1} b_n\\) with\n\n1) \\(b_n\\) decreasing\n2) \\(b_n\\) positive\n3) \\(\\lim_{n \\to \\infty} b_n = 0\\)\n\nThen the series converges.\n If these are true then we conclude that the series converges and that is our alternating series test. Let's see it for this specific example. So for this example this minus one to the n plus one divided by n well what do we have?", null, null, "We can see these text from the image: Alternating Series Test\n\nFor \u2211 (-1)^(n-1) b_n with\nn=1\n\n1) b_n decreasing\n2) b_n positive\n3) lim b_n = 0\n n\u2192\u221e\n\nThen the series converges\n\nEx: \u2211 (-1)^(n-1)\n n=1\n n.\n It's certainly going to be the case that B sub n is positive. So B sub n in this example is just one over n. It's positive, it's decreasing as n gets larger, one over n gets smaller, and the limit of one over n is just zero. So this satisfies all three conditions.", null, null, "We can see these text from the image: Alternating Series Test\n\nFor \u2211 (-1)^(n-1) b_n with\nn=1\n\n1) b_n decreasing\n2) b_n positive\n3) lim b_n = 0\nn\u2192\u221e\n\nThen the series converges\n\nEx: \u2211 (-1)^(n-1)\nn=1\nConverges!.\n The conditions of the alternating series test lead us to conclude that it converges. Now, alternating series are particularly useful, and one reason for their utility is their ease of verification. Specifically, this final condition - that the limit of" ], "metadata": [ { "vid": "-lD0skTnqFo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on The Alternating Series Test in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/-lD0skTnqFo/88.7_101.22.mp4", "refined_asr": " And then I advance to my S6, and advance to my S7, and so on. That's because of the nature of this alternating series. There's this sort of back-and-forth process, and it kind of looks like it's going to converge.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@88.7_101.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@88.7_101.22#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@88.7_101.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@88.7_101.22#2.jpg" ], "ocr_qwen2_vl_72b": "\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} + \\frac{1}{7} - \\ldots\n\\]\n\n\\( S_7 \\)\n\n\\( s_2 \\)\n\n\\( s_4 s_6 \\)\n\n\\( s_3 \\)\n\n\\( s_1 \\)" }, { "vid": "-lD0skTnqFo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on The Alternating Series Test in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/-lD0skTnqFo/101.32000000000001_118.10000000000001.mp4", "refined_asr": " The actual value of the series is going to be stuck somewhere in the middle. Now if we look at all the even partial sums - the S2, the S4, the S6, the even sums is an increasing sequence. Well the odd partial sums - the S3, the S5, the S7 -", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@101.32000000000001_118.10000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@101.32000000000001_118.10000000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@101.32000000000001_118.10000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@101.32000000000001_118.10000000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@101.32000000000001_118.10000000000001#3.jpg" ], "ocr_qwen2_vl_72b": "\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} + \\frac{1}{7} - \\ldots\n\\]\n\n\\[\ns_7\n\\]\n\n\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n}\n\\]" }, { "vid": "-lD0skTnqFo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on The Alternating Series Test in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/-lD0skTnqFo/118.7_124.18.mp4", "refined_asr": " The S6, the S8, the S9, the S10, the S11, the S12, the S13, the S14, the S15, the S16, and the S21.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@118.7_124.18#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@118.7_124.18#1.jpg" ], "ocr_qwen2_vl_72b": "\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} + \\frac{1}{7} - \\ldots\n\\]\n\n\\[\ns_7\n\\]\n\n\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n}\n\\]" }, { "vid": "-lD0skTnqFo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on The Alternating Series Test in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/-lD0skTnqFo/124.18_136.5.mp4", "refined_asr": " And then we start to get a decreasing sequence; it's getting smaller. So the final answer appears to be squished between these two different sides. Let's formalize this idea into a theorem. I'm going to say that the alternating series test says:", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@124.18_136.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@124.18_136.5#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@124.18_136.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@124.18_136.5#2.jpg" ], "ocr_qwen2_vl_72b": "\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} + \\frac{1}{7}\n\\]\n\n\\[\ns_7\n\\]\n\n\\[\n0 \\quad s_2 \\quad s_4 \\quad s_6 \\quad s_3 \\quad s_5 \\quad s_1\n\\]\n\n\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n}\n\\]" }, { "vid": "-lD0skTnqFo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on The Alternating Series Test in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/-lD0skTnqFo/137.1_148.4.mp4", "refined_asr": " If I begin with an alternating series, so minus one to the n, minus one times a sub n, some sequence that looks a lot like it did for the harmonic series, where it's going to be decreasing and where it's going to be positive.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@137.1_148.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@137.1_148.4#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@137.1_148.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@137.1_148.4#2.jpg" ], "ocr_qwen2_vl_72b": "Alternating Series Test\n\nFor \u2211 (-1)^(n-1) b_n with\nn=1\n\n1) b_n decreasing\n2) b_n positive\n3) lim b_n = 0\n n\u2192\u221e\n\nThen the series converges." }, { "vid": "-lD0skTnqFo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on The Alternating Series Test in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/-lD0skTnqFo/148.4_162.56.mp4", "refined_asr": " If these are true then we conclude that the series converges and that is our alternating series test. Let's see it for this specific example. So for this example this minus one to the n plus one divided by n well what do we have?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@148.4_162.56#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@148.4_162.56#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@148.4_162.56#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@148.4_162.56#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@148.4_162.56#3.jpg" ], "ocr_qwen2_vl_72b": "Alternating Series Test\n\nFor \\(\\sum_{n=1}^{\\infty} (-1)^{n-1} b_n\\) with\n\n1) \\(b_n\\) decreasing\n2) \\(b_n\\) positive\n3) \\(\\lim_{n \\to \\infty} b_n = 0\\)\n\nThen the series converges" }, { "vid": "-lD0skTnqFo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on The Alternating Series Test in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/-lD0skTnqFo/162.56_179.44.mp4", "refined_asr": " It's certainly going to be the case that B sub n is positive. So B sub n in this example is just one over n. It's positive, it's decreasing as n gets larger, one over n gets smaller, and the limit of one over n is just zero. So this satisfies all three conditions.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@162.56_179.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@162.56_179.44#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@162.56_179.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@162.56_179.44#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@162.56_179.44#3.jpg" ], "ocr_qwen2_vl_72b": "Alternating Series Test\n\nFor \u2211 (-1)^(n-1) b_n with\nn=1\n\n1) b_n decreasing\n2) b_n positive\n3) lim b_n = 0\n n\u2192\u221e\n\nThen the series converges\n\nEx: \u2211 (-1)^(n-1)\n n=1\n n" }, { "vid": "-lD0skTnqFo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on The Alternating Series Test in Series and Sequences_30.json#####audio#####doingASR#####FinishASR/-lD0skTnqFo/179.44_194.04000000000002.mp4", "refined_asr": " The conditions of the alternating series test lead us to conclude that it converges. Now, alternating series are particularly useful, and one reason for their utility is their ease of verification. Specifically, this final condition - that the limit of", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@179.44_194.04000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@179.44_194.04000000000002#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@179.44_194.04000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@179.44_194.04000000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-lD0skTnqFo/-lD0skTnqFo@179.44_194.04000000000002#3.jpg" ], "ocr_qwen2_vl_72b": "Alternating Series Test\n\nFor \u2211 (-1)^(n-1) b_n with\nn=1\n\n1) b_n decreasing\n2) b_n positive\n3) lim b_n = 0\nn\u2192\u221e\n\nThen the series converges\n\nEx: \u2211 (-1)^(n-1)\nn=1\nConverges!" } ], "image_num": 15, "text_num": 468, "token_num": 9108 }, { "images": [ "sample_100_images/El3UEMU9V5s@231.72_251.72#1.jpg", null, "sample_100_images/El3UEMU9V5s@251.72_261.72#1.jpg", null, "sample_100_images/El3UEMU9V5s@261.72_277.72#1.jpg", "sample_100_images/El3UEMU9V5s@261.72_277.72#2.jpg", null, "sample_100_images/El3UEMU9V5s@277.72_291.72#1.jpg", null, "sample_100_images/El3UEMU9V5s@291.72_311.72#1.jpg", null, "sample_100_images/El3UEMU9V5s@313.72_331.72#1.jpg", null, "sample_100_images/El3UEMU9V5s@331.72_353.72#1.jpg", null, "sample_100_images/El3UEMU9V5s@353.72_367.72#1.jpg", null ], "texts": [ null, " Till the tropopause, then they will diverge. Half of the part will go towards the North Pole and the other half will diverge towards the South Pole. That is why different cells will form in the Northern Hemisphere and in the Southern Hemisphere. So when the air rises at the zero-degree Equator, and basically at the ITCZ,", null, " Because in the ITCZ video we saw that the ITCZ is the area that produces the most low pressure and here the air will rise up. After diverging at the tropopause, it will start moving towards the north-easterly trade winds.", null, null, " Towards the poles. Now let's say this is 30-degree latitude and this is the 60-degree latitude. Now as the warm air rises up and diverges and reaches here, it will cool down. And when the air mass cools down, it tries to subside.", null, " It will subside towards the land. And when the air subsides towards the land mass, it splits into two parts: some will go this way, and some will go that way. So let's focus on this direction now, which is this way.", null, " The air is going here towards the equator. This will complete a cycle because after reaching the zero-degree equator, it will rise up again, converging with the south-easterly trade winds. So, these winds which are moving on the surface of the earth are known as north-easterly trade winds.", null, " Now these are also known as subtropical easterlies because they're moving in the subtropical region. This entire cell is known as the Hadley Cell. Why is it called the Hadley Cell? Because this theory was given by George Hadley.", null, " He gave this theory while describing how the trade winds emerge. Now let's come to the polar cell. Before we discuss the polar cell, you might have noticed that we have stepped the tropopause towards the North Pole. Why is that? The height of the tropopause is the lowest at the North Pole and the highest at the equator.", null, " The reason for this is that the air is denser at the North Pole because of this the tropopause is lower. And if you want to understand it in a very simple way let's say the temperature of the tropopause for example is minus 80 degrees Celsius." ], "text_ocr_list": [ null, "We can see these text from the image: - CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- TROPOPAUSE\n- N.P.\n- 0\u00b0\n- 30\u00b0.\n Till the tropopause, then they will diverge. Half of the part will go towards the North Pole and the other half will diverge towards the South Pole. That is why different cells will form in the Northern Hemisphere and in the Southern Hemisphere. So when the air rises at the zero-degree Equator, and basically at the ITCZ,", null, "We can see these text from the image: - CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- Tropopause\n- N.P..\n Because in the ITCZ video we saw that the ITCZ is the area that produces the most low pressure and here the air will rise up. After diverging at the tropopause, it will start moving towards the north-easterly trade winds.", null, null, "We can see these text from the image: - CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- Tropopause\n- N.P.\n- CivilCourify.\n Towards the poles. Now let's say this is 30-degree latitude and this is the 60-degree latitude. Now as the warm air rises up and diverges and reaches here, it will cool down. And when the air mass cools down, it tries to subside.", null, "We can see these text from the image: - CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- Tropopause\n- N.P.\n- 60\u00b0\n- 30\u00b0.\n It will subside towards the land. And when the air subsides towards the land mass, it splits into two parts: some will go this way, and some will go that way. So let's focus on this direction now, which is this way.", null, "We can see these text from the image: - CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- TROPOPAUSE\n- 90\u00b0\n- 60\u00b0\n- 30\u00b0\n- 0\u00b0.\n The air is going here towards the equator. This will complete a cycle because after reaching the zero-degree equator, it will rise up again, converging with the south-easterly trade winds. So, these winds which are moving on the surface of the earth are known as north-easterly trade winds.", null, "We can see these text from the image: - CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- Tropopause\n- 90\u00b0 N.P.\n- 60\u00b0\n- 30\u00b0 North-Easterly Trade Winds\n- Eq..\n Now these are also known as subtropical easterlies because they're moving in the subtropical region. This entire cell is known as the Hadley Cell. Why is it called the Hadley Cell? Because this theory was given by George Hadley.", null, "We can see these text from the image: - CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- TROPOPAUSE\n- HADLEY CELL\n- George Hadley\n- 30\u00b0 North-Easterly Trade Winds\n- Eq.\n He gave this theory while describing how the trade winds emerge. Now let's come to the polar cell. Before we discuss the polar cell, you might have noticed that we have stepped the tropopause towards the North Pole. Why is that? The height of the tropopause is the lowest at the North Pole and the highest at the equator.", null, "We can see these text from the image: - CELL FORMATION\n- POLAR CELL | FERREL CELL | HADLEY CELL\n- TROPOPAUSE\n- 90\u00b0 N.P.\n- 60\u00b0\n- 30\u00b0 North-Easterly Trade Winds\n- HADLEY CELL\n- George Hadley\n- Eq..\n The reason for this is that the air is denser at the North Pole because of this the tropopause is lower. And if you want to understand it in a very simple way let's say the temperature of the tropopause for example is minus 80 degrees Celsius." ], "metadata": [ { "vid": "El3UEMU9V5s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Atmospheric Circulation focusing on Hadley Cells_30.json#####audio#####doingASR#####FinishASR/El3UEMU9V5s/231.72_251.72.mp4", "refined_asr": " Till the tropopause, then they will diverge. Half of the part will go towards the North Pole and the other half will diverge towards the South Pole. That is why different cells will form in the Northern Hemisphere and in the Southern Hemisphere. So when the air rises at the zero-degree Equator, and basically at the ITCZ,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@231.72_251.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@231.72_251.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@231.72_251.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@231.72_251.72#3.jpg" ], "ocr_qwen2_vl_72b": "- CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- TROPOPAUSE\n- N.P.\n- 0\u00b0\n- 30\u00b0" }, { "vid": "El3UEMU9V5s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Atmospheric Circulation focusing on Hadley Cells_30.json#####audio#####doingASR#####FinishASR/El3UEMU9V5s/251.72_261.72.mp4", "refined_asr": " Because in the ITCZ video we saw that the ITCZ is the area that produces the most low pressure and here the air will rise up. After diverging at the tropopause, it will start moving towards the north-easterly trade winds.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@251.72_261.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@251.72_261.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@251.72_261.72#2.jpg" ], "ocr_qwen2_vl_72b": "- CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- Tropopause\n- N.P." }, { "vid": "El3UEMU9V5s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Atmospheric Circulation focusing on Hadley Cells_30.json#####audio#####doingASR#####FinishASR/El3UEMU9V5s/261.72_277.72.mp4", "refined_asr": " Towards the poles. Now let's say this is 30-degree latitude and this is the 60-degree latitude. Now as the warm air rises up and diverges and reaches here, it will cool down. And when the air mass cools down, it tries to subside.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@261.72_277.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@261.72_277.72#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@261.72_277.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@261.72_277.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@261.72_277.72#3.jpg" ], "ocr_qwen2_vl_72b": "- CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- Tropopause\n- N.P.\n- CivilCourify" }, { "vid": "El3UEMU9V5s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Atmospheric Circulation focusing on Hadley Cells_30.json#####audio#####doingASR#####FinishASR/El3UEMU9V5s/277.72_291.72.mp4", "refined_asr": " It will subside towards the land. And when the air subsides towards the land mass, it splits into two parts: some will go this way, and some will go that way. So let's focus on this direction now, which is this way.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@277.72_291.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@277.72_291.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@277.72_291.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@277.72_291.72#3.jpg" ], "ocr_qwen2_vl_72b": "- CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- Tropopause\n- N.P.\n- 60\u00b0\n- 30\u00b0" }, { "vid": "El3UEMU9V5s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Atmospheric Circulation focusing on Hadley Cells_30.json#####audio#####doingASR#####FinishASR/El3UEMU9V5s/291.72_311.72.mp4", "refined_asr": " The air is going here towards the equator. This will complete a cycle because after reaching the zero-degree equator, it will rise up again, converging with the south-easterly trade winds. So, these winds which are moving on the surface of the earth are known as north-easterly trade winds.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@291.72_311.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@291.72_311.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@291.72_311.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@291.72_311.72#3.jpg" ], "ocr_qwen2_vl_72b": "- CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- TROPOPAUSE\n- 90\u00b0\n- 60\u00b0\n- 30\u00b0\n- 0\u00b0" }, { "vid": "El3UEMU9V5s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Atmospheric Circulation focusing on Hadley Cells_30.json#####audio#####doingASR#####FinishASR/El3UEMU9V5s/313.72_331.72.mp4", "refined_asr": " Now these are also known as subtropical easterlies because they're moving in the subtropical region. This entire cell is known as the Hadley Cell. Why is it called the Hadley Cell? Because this theory was given by George Hadley.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@313.72_331.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@313.72_331.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@313.72_331.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@313.72_331.72#3.jpg" ], "ocr_qwen2_vl_72b": "- CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- Tropopause\n- 90\u00b0 N.P.\n- 60\u00b0\n- 30\u00b0 North-Easterly Trade Winds\n- Eq." }, { "vid": "El3UEMU9V5s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Atmospheric Circulation focusing on Hadley Cells_30.json#####audio#####doingASR#####FinishASR/El3UEMU9V5s/331.72_353.72.mp4", "refined_asr": " He gave this theory while describing how the trade winds emerge. Now let's come to the polar cell. Before we discuss the polar cell, you might have noticed that we have stepped the tropopause towards the North Pole. Why is that? The height of the tropopause is the lowest at the North Pole and the highest at the equator.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@331.72_353.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@331.72_353.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@331.72_353.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@331.72_353.72#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@331.72_353.72#4.jpg" ], "ocr_qwen2_vl_72b": "- CELL FORMATION\n- POLAR CELL\n- FERREL CELL\n- HADLEY CELL\n- TROPOPAUSE\n- HADLEY CELL\n- George Hadley\n- 30\u00b0 North-Easterly Trade Winds\n- Eq" }, { "vid": "El3UEMU9V5s.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Atmospheric Circulation focusing on Hadley Cells_30.json#####audio#####doingASR#####FinishASR/El3UEMU9V5s/353.72_367.72.mp4", "refined_asr": " The reason for this is that the air is denser at the North Pole because of this the tropopause is lower. And if you want to understand it in a very simple way let's say the temperature of the tropopause for example is minus 80 degrees Celsius.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@353.72_367.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@353.72_367.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@353.72_367.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/El3UEMU9V5s/El3UEMU9V5s@353.72_367.72#3.jpg" ], "ocr_qwen2_vl_72b": "- CELL FORMATION\n- POLAR CELL | FERREL CELL | HADLEY CELL\n- TROPOPAUSE\n- 90\u00b0 N.P.\n- 60\u00b0\n- 30\u00b0 North-Easterly Trade Winds\n- HADLEY CELL\n- George Hadley\n- Eq." } ], "image_num": 9, "text_num": 526, "token_num": 5710 }, { "images": [ "sample_100_images/p3TsVGo063Q@85.92_104.02000000000001#1.jpg", null, "sample_100_images/p3TsVGo063Q@104.08_114.68#1.jpg", null, "sample_100_images/p3TsVGo063Q@114.68_132.68#1.jpg", null, "sample_100_images/p3TsVGo063Q@132.68_144.04000000000002#1.jpg", null, "sample_100_images/p3TsVGo063Q@144.68_157.74#1.jpg", null, "sample_100_images/p3TsVGo063Q@157.74_171.62#1.jpg", null, "sample_100_images/p3TsVGo063Q@171.62_187.74#1.jpg", null, "sample_100_images/p3TsVGo063Q@187.74_200.98000000000002#1.jpg", null, "sample_100_images/p3TsVGo063Q@200.98000000000002_210.2#1.jpg", null ], "texts": [ null, " We can take and plug this value into the corresponding location in the other equation and go from there. So let's do that here. Let's take this and let's substitute it in. So over here, the second equation is going to be x plus y, but now we know that y is equal to 6x.", null, " So we don't want to plug the y in, we already have that. We want to stick this value in that we know y is equal to. So we put 6x here, because y is equal to 6x, and then we get 28.", null, " So on the left-hand side, what do we do? We can add those together and get 7x is equal to 28. And then we can solve this by dividing by 7, which gives us a cancellation. So x is equal to 28 over 7, which is 4. So actually, that's really important, because we", null, " Figured out. Notice that when we graph these guys, we get a point x comma y. That's the solution: x comma y. But here, we found half the solution. We know the x value of the intersection point, which is 4.", null, " Why do we find the corresponding Y value? Well we go and take our answer and we can substitute it back in. We can either substitute it into here or we can substitute it into here. Either way you're going to get the same answer. So let's take this and stick.", null, " If you plug it into here what you're going to get is y is equal to 6 times x but now we know x is equal to 4 so y is equal to 24 so basically this and this are really really important", null, " So, the point (4, 24) is the solution. If we were to graph these equations, they would intersect at one point: (4, 24). Now, let me add one more thing.", null, " So before we go on to the next problem, let's do something fun. We substituted in and solved for x, and then we took this value and substituted it in here. Just for grins, I want to take this and substitute it back into the other equation, just to show you what we'd get.", null, " It's x plus y so 4 plus y. So I get this x and this x and then this y and then this y and then this y. And then 4 plus y is equal to 28. We solve for y by subtracting 4 from both sides so it's going." ], "text_ocr_list": [ null, "We can see these text from the image: y = 6x\nx + y = 28\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\nMathTutorDVD.com.\n We can take and plug this value into the corresponding location in the other equation and go from there. So let's do that here. Let's take this and let's substitute it in. So over here, the second equation is going to be x plus y, but now we know that y is equal to 6x.", null, "We can see these text from the image: y = 6x\nx + y = 28\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\nMathTutorDVD.com.\n So we don't want to plug the y in, we already have that. We want to stick this value in that we know y is equal to. So we put 6x here, because y is equal to 6x, and then we get 28.", null, "We can see these text from the image: y = 6x\n\nx + y = 28\n\nx + 6x = 28\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com.\n So on the left-hand side, what do we do? We can add those together and get 7x is equal to 28. And then we can solve this by dividing by 7, which gives us a cancellation. So x is equal to 28 over 7, which is 4. So actually, that's really important, because we", null, "We can see these text from the image: y = 6x\n\nx + y = 28\n\nx + 6x = 28\n\n7x = 28\n\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com.\n Figured out. Notice that when we graph these guys, we get a point x comma y. That's the solution: x comma y. But here, we found half the solution. We know the x value of the intersection point, which is 4.", null, "We can see these text from the image: y = 6x\nx + y = 28\n\nx + 6x = 28\n7x = 28\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\nMathTutorDVD.com.\n Why do we find the corresponding Y value? Well we go and take our answer and we can substitute it back in. We can either substitute it into here or we can substitute it into here. Either way you're going to get the same answer. So let's take this and stick.", null, "We can see these text from the image: y = 6x\n\nx + y = 28\n\ny = \n\nx + 6x = 28\n\n7x = 28\n\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com.\n If you plug it into here what you're going to get is y is equal to 6 times x but now we know x is equal to 4 so y is equal to 24 so basically this and this are really really important", null, "We can see these text from the image: y = 6x\ny = 6 * 4\ny = 24\n\nx + y = 28\nx + 6x = 28\n7x = 28\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com.\n So, the point (4, 24) is the solution. If we were to graph these equations, they would intersect at one point: (4, 24). Now, let me add one more thing.", null, "We can see these text from the image: y = 6x\nx + y = 28\n\ny = 6 * 4\ny = 24\n\n(4, 24) Soln\n\nx + 6x = 28\n7x = 28\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com.\n So before we go on to the next problem, let's do something fun. We substituted in and solved for x, and then we took this value and substituted it in here. Just for grins, I want to take this and substitute it back into the other equation, just to show you what we'd get.", null, "We can see these text from the image: y = 6x\n\nx + y = 28\n\ny = 6 * 4\ny = 24\n\n(4, 24) Soln\n\nx + 6x = 28\n7x = 28\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com.\n It's x plus y so 4 plus y. So I get this x and this x and then this y and then this y and then this y. And then 4 plus y is equal to 28. We solve for y by subtracting 4 from both sides so it's going." ], "metadata": [ { "vid": "p3TsVGo063Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra tutorial on solving systems of equations by substitution_30.json#####audio#####doingASR#####FinishASR/p3TsVGo063Q/85.92_104.02000000000001.mp4", "refined_asr": " We can take and plug this value into the corresponding location in the other equation and go from there. So let's do that here. Let's take this and let's substitute it in. So over here, the second equation is going to be x plus y, but now we know that y is equal to 6x.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@85.92_104.02000000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@85.92_104.02000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@85.92_104.02000000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@85.92_104.02000000000001#3.jpg" ], "ocr_qwen2_vl_72b": "y = 6x\nx + y = 28\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\nMathTutorDVD.com" }, { "vid": "p3TsVGo063Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra tutorial on solving systems of equations by substitution_30.json#####audio#####doingASR#####FinishASR/p3TsVGo063Q/104.08_114.68.mp4", "refined_asr": " So we don't want to plug the y in, we already have that. We want to stick this value in that we know y is equal to. So we put 6x here, because y is equal to 6x, and then we get 28.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@104.08_114.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@104.08_114.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@104.08_114.68#2.jpg" ], "ocr_qwen2_vl_72b": "y = 6x\nx + y = 28\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\nMathTutorDVD.com" }, { "vid": "p3TsVGo063Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra tutorial on solving systems of equations by substitution_30.json#####audio#####doingASR#####FinishASR/p3TsVGo063Q/114.68_132.68.mp4", "refined_asr": " So on the left-hand side, what do we do? We can add those together and get 7x is equal to 28. And then we can solve this by dividing by 7, which gives us a cancellation. So x is equal to 28 over 7, which is 4. So actually, that's really important, because we", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@114.68_132.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@114.68_132.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@114.68_132.68#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@114.68_132.68#3.jpg" ], "ocr_qwen2_vl_72b": "y = 6x\n\nx + y = 28\n\nx + 6x = 28\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com" }, { "vid": "p3TsVGo063Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra tutorial on solving systems of equations by substitution_30.json#####audio#####doingASR#####FinishASR/p3TsVGo063Q/132.68_144.04000000000002.mp4", "refined_asr": " Figured out. Notice that when we graph these guys, we get a point x comma y. That's the solution: x comma y. But here, we found half the solution. We know the x value of the intersection point, which is 4.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@132.68_144.04000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@132.68_144.04000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@132.68_144.04000000000002#2.jpg" ], "ocr_qwen2_vl_72b": "y = 6x\n\nx + y = 28\n\nx + 6x = 28\n\n7x = 28\n\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com" }, { "vid": "p3TsVGo063Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra tutorial on solving systems of equations by substitution_30.json#####audio#####doingASR#####FinishASR/p3TsVGo063Q/144.68_157.74.mp4", "refined_asr": " Why do we find the corresponding Y value? Well we go and take our answer and we can substitute it back in. We can either substitute it into here or we can substitute it into here. Either way you're going to get the same answer. So let's take this and stick.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@144.68_157.74#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@144.68_157.74#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@144.68_157.74#2.jpg" ], "ocr_qwen2_vl_72b": "y = 6x\nx + y = 28\n\nx + 6x = 28\n7x = 28\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\nMathTutorDVD.com" }, { "vid": "p3TsVGo063Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra tutorial on solving systems of equations by substitution_30.json#####audio#####doingASR#####FinishASR/p3TsVGo063Q/157.74_171.62.mp4", "refined_asr": " If you plug it into here what you're going to get is y is equal to 6 times x but now we know x is equal to 4 so y is equal to 24 so basically this and this are really really important", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@157.74_171.62#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@157.74_171.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@157.74_171.62#2.jpg" ], "ocr_qwen2_vl_72b": "y = 6x\n\nx + y = 28\n\ny = \n\nx + 6x = 28\n\n7x = 28\n\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com" }, { "vid": "p3TsVGo063Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra tutorial on solving systems of equations by substitution_30.json#####audio#####doingASR#####FinishASR/p3TsVGo063Q/171.62_187.74.mp4", "refined_asr": " So, the point (4, 24) is the solution. If we were to graph these equations, they would intersect at one point: (4, 24). Now, let me add one more thing.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@171.62_187.74#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@171.62_187.74#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@171.62_187.74#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@171.62_187.74#3.jpg" ], "ocr_qwen2_vl_72b": "y = 6x\ny = 6 * 4\ny = 24\n\nx + y = 28\nx + 6x = 28\n7x = 28\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com" }, { "vid": "p3TsVGo063Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra tutorial on solving systems of equations by substitution_30.json#####audio#####doingASR#####FinishASR/p3TsVGo063Q/187.74_200.98000000000002.mp4", "refined_asr": " So before we go on to the next problem, let's do something fun. We substituted in and solved for x, and then we took this value and substituted it in here. Just for grins, I want to take this and substitute it back into the other equation, just to show you what we'd get.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@187.74_200.98000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@187.74_200.98000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@187.74_200.98000000000002#2.jpg" ], "ocr_qwen2_vl_72b": "y = 6x\nx + y = 28\n\ny = 6 * 4\ny = 24\n\n(4, 24) Soln\n\nx + 6x = 28\n7x = 28\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com" }, { "vid": "p3TsVGo063Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic Algebra tutorial on solving systems of equations by substitution_30.json#####audio#####doingASR#####FinishASR/p3TsVGo063Q/200.98000000000002_210.2.mp4", "refined_asr": " It's x plus y so 4 plus y. So I get this x and this x and then this y and then this y and then this y. And then 4 plus y is equal to 28. We solve for y by subtracting 4 from both sides so it's going.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@200.98000000000002_210.2#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@200.98000000000002_210.2#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/p3TsVGo063Q/p3TsVGo063Q@200.98000000000002_210.2#2.jpg" ], "ocr_qwen2_vl_72b": "y = 6x\n\nx + y = 28\n\ny = 6 * 4\ny = 24\n\n(4, 24) Soln\n\nx + 6x = 28\n7x = 28\nx = 4\n\nAlgebra 1 - Solve Systems of Equations by Substitution, Part 1\n\nMathTutorDVD.com" } ], "image_num": 9, "text_num": 556, "token_num": 5740 }, { "images": [ "sample_100_images/_MF8L7ZxwRE@16101.74_16118.539999999999#1.jpg", null, "sample_100_images/_MF8L7ZxwRE@16119.08_16137.26#1.jpg", "sample_100_images/_MF8L7ZxwRE@16119.08_16137.26#2.jpg", "sample_100_images/_MF8L7ZxwRE@16119.08_16137.26#3.jpg", null, "sample_100_images/_MF8L7ZxwRE@16137.26_16154.18#1.jpg", "sample_100_images/_MF8L7ZxwRE@16137.26_16154.18#2.jpg", null, "sample_100_images/_MF8L7ZxwRE@16154.18_16170.98#1.jpg", null, "sample_100_images/_MF8L7ZxwRE@16170.98_16187.9#1.jpg", null, "sample_100_images/_MF8L7ZxwRE@16187.9_16202.6#1.jpg", null, "sample_100_images/_MF8L7ZxwRE@16202.6_16219.52#1.jpg", null, "sample_100_images/_MF8L7ZxwRE@16219.52_16237.88#1.jpg", null ], "texts": [ null, " Print this 10 times, so I don't have to copy these five lines and paste them again 10 times, making them 50 lines. I can actually do this by simply copying this any number of times I want it to execute. So, if I wanted to execute it 10 times, I can copy this particular line 10 times.", null, null, null, " I hope that you are able to understand this and I hope that you can see what I mean by saying that you will avoid rewriting the same logic again and again if you use functions. Functions will also help you keep track of what you're doing in a program. So if we have our logic divided into,", null, null, " Different functions allow us to focus on one function and start debugging that function if it's not performing as expected. The third thing is to test and check logic independently. So, for example, I could have someone like Rohan Das write one function and someone like Jacob write another.", null, " Let's say you have three people working on three different functions. One person, say Sonali, writes one function. Another writes a second function, and the third writes yet another. These three can work independently on their respective functions. What I provide them is a function prototype. I specify the desired return value for each function and the parameters I intend to pass.", null, " Two or three or four values can be used as input to this function. Those values will be written inside the body of the function. If that makes sense, great! I hope you were able to understand this. Now, we'll talk about passing values to the functions.", null, " Functions is something that you will definitely want to learn. What does it mean? It simply means that there is a function that takes some values as an input and gives an output. You need to receive output by giving some inputs, and you will definitely want to use a function somewhere.", null, " You are supplying two input values and you are expecting an output value from your function. Let's see how these things work. We can pass values to a function and we can get a value in return from a function. For example, let's say 'sum' is a function and we have two values which we are supplying.", null, " Sum as an input. What are those two values? Those two values are 'a' and 'b'. If these values are three and four, I will expect my sum to return an integer, which is nothing but seven, because three plus four is seven. Similarly, if 'a' is five and 'b' is three, I would expect this sum to give me eight in return." ], "text_ocr_list": [ null, "We can see these text from the image: ```c\nint a;\ndisplay(); // Function call\nreturn 0;\n\n// Function definition\nvoid display()\n{\n printf(\"Hello world1\\n\");\n printf(\"Hello world2\\n\");\n printf(\"Hello world3\\n\");\n printf(\"Hello world4\\n\");\n printf(\"Hello world5\\n\");\n}\n```\n\n```sh\nHello world5\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5> cd \"d:\\MyData\\Business\\PWH C Tutorial\\Chapter 5\" ; if (!$?) { gcc 02_quick_quiz.c -o 02_quick_quiz } ; if (!$?) { iz }\nGood Morning\nGood Afternoon\nGood Night\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5>\n```.\n Print this 10 times, so I don't have to copy these five lines and paste them again 10 times, making them 50 lines. I can actually do this by simply copying this any number of times I want it to execute. So, if I wanted to execute it 10 times, I can copy this particular line 10 times.", null, null, null, "We can see these text from the image: ```c\n#include \n\nvoid display(); // Function prototype\n// void average(int, float, int); // Function prototype\n\nint main() {\n int a;\n display(); // Function call\n return 0;\n}\n\n// Function definition\nvoid display(){\n printf(\"Hello world1\\n\");\n printf(\"Hello world3\\n\");\n printf(\"Hello world4\\n\");\n printf(\"Hello world5\\n\");\n}\n```\n\n```bash\nHello world5\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5> cd \"d:\\MyData\\Business\\PWH C Tutorial\\Chapter 5\\\"; if (!$?) { gcc 02_quick_quiz.c -o 02_quick_quiz }; if (!$?)\niz\nGood Morning\nGood Afternoon\nGood Night\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5>\n```.\n I hope that you are able to understand this and I hope that you can see what I mean by saying that you will avoid rewriting the same logic again and again if you use functions. Functions will also help you keep track of what you're doing in a program. So if we have our logic divided into,", null, null, "We can see these text from the image: ```c\n// void average(int, float, int); // Function prototype\n\nint main(){\n int a;\n display(); // Function call\n return 0;\n}\n\n// Function definition\nvoid display(){\n printf(\"Hello world1\\n\");\n printf(\"Hello world2\\n\");\n printf(\"Hello world3\\n\");\n printf(\"Hello world4\\n\");\n printf(\"Hello world5\\n\");\n}\n```\n\n```bash\nHello world5\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5> cd \"d:\\MyData\\Business\\PWH C Tutorial\\Chapter 5\" ; if (!$?) { gcc 02_quick_quiz.c -o 02_quick_quiz } ; if (!$?) { .\\02_quick_quiz }\niz\nGood Morning\nGood Afternoon\nGood Night\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5>\n```.\n Different functions allow us to focus on one function and start debugging that function if it's not performing as expected. The third thing is to test and check logic independently. So, for example, I could have someone like Rohan Das write one function and someone like Jacob write another.", null, "We can see these text from the image: ```c\n#include \n\nvoid display(); // Function prototype\n\n// void average(int, float, int); // Function prototype\n\nint main() {\n display(); // Function call\n return 0;\n}\n\n// Function definition\nvoid display() {\n printf(\"Hello world1\\n\");\n printf(\"Hello world2\\n\");\n printf(\"Hello world3\\n\");\n printf(\"Hello world4\\n\");\n printf(\"Hello world5\\n\");\n}\n```\n\n```bash\nHello world5\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5> cd \"d:\\MyData\\Business\\PWH C Tutorial\\Chapter 5\"; if (!$?) { gcc 02_quick_quiz.c -o 02_quick_quiz }; if (!$?) { 02_quick_quiz }\niz\nGood Morning\nGood Afternoon\nGood Night\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5>\n```.\n Let's say you have three people working on three different functions. One person, say Sonali, writes one function. Another writes a second function, and the third writes yet another. These three can work independently on their respective functions. What I provide them is a function prototype. I specify the desired return value for each function and the parameters I intend to pass.", null, "We can see these text from the image: ```c\n#include \n\nvoid display(); // Function prototype\n\n// void average(int, float, int); // Function prototype\n\nint main(){\n int a;\n display(); // Function call\n return 0;\n}\n\n// Function definition\nvoid display(){\n printf(\"Hello world1\\n\");\n printf(\"Hello world2\\n\");\n printf(\"Hello world3\\n\");\n printf(\"Hello world4\\n\");\n printf(\"Hello world5\\n\");\n}\n```\n\n```sh\nHello world5\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5> cd \"d:\\MyData\\Business\\PWH C Tutorial\\Chapter 5\" ; if (!$?) { gcc 02_quick_quiz.c -o 02_quick_quiz } ; if (!$?) { .\\02_quick_quiz }\niz\nGood Morning\nGood Afternoon\nGood Night\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5>\n```.\n Two or three or four values can be used as input to this function. Those values will be written inside the body of the function. If that makes sense, great! I hope you were able to understand this. Now, we'll talk about passing values to the functions.", null, "We can see these text from the image: Passing values to functions\n\nWe can pass values to a function and can get a value in return from a function.\n\nint sum (int a, int b)\n\nThe above prototype means that sum is a function which takes values a (of type int) and b (of type int) and returns a value of type int.\n\nfunction definition of sum can be:\n\nint sum (int a, int b) {\n....\n Functions is something that you will definitely want to learn. What does it mean? It simply means that there is a function that takes some values as an input and gives an output. You need to receive output by giving some inputs, and you will definitely want to use a function somewhere.", null, "We can see these text from the image: Passing values to functions\n\nWe can pass values to a function and can get a value in return from a function.\n\nint sum(int a, int b)\n\nThe above prototype means that sum is a function which takes values a (of type int) and b (of type int) and returns a value of type int.\n\nfunction definition of sum can be:\n\nint sum(int a, int b) {\nint c;\nc = a + b;\nreturn c;\n}.\n You are supplying two input values and you are expecting an output value from your function. Let's see how these things work. We can pass values to a function and we can get a value in return from a function. For example, let's say 'sum' is a function and we have two values which we are supplying.", null, "We can see these text from the image: Passing values to functions\n\nWe can pass values to a function and can get a value in return from a function.\n\nint sum (int a, int b)\n\nThe above prototype means that sum is a function which takes values a (of type int) and b (of type int) and returns a value of type int.\n\nfunction definition of sum can be:\n\nint sum (int a, int b) {\nint c;\nc = a + b;\nreturn c;\n}.\n Sum as an input. What are those two values? Those two values are 'a' and 'b'. If these values are three and four, I will expect my sum to return an integer, which is nothing but seven, because three plus four is seven. Similarly, if 'a' is five and 'b' is three, I would expect this sum to give me eight in return." ], "metadata": [ { "vid": "_MF8L7ZxwRE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic programming tutorials for beginners_100.json#####audio#####doingASR#####FinishASR/_MF8L7ZxwRE/16101.74_16118.539999999999.mp4", "refined_asr": " Print this 10 times, so I don't have to copy these five lines and paste them again 10 times, making them 50 lines. I can actually do this by simply copying this any number of times I want it to execute. So, if I wanted to execute it 10 times, I can copy this particular line 10 times.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16101.74_16118.539999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16101.74_16118.539999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16101.74_16118.539999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16101.74_16118.539999999999#3.jpg" ], "ocr_qwen2_vl_72b": "```c\nint a;\ndisplay(); // Function call\nreturn 0;\n\n// Function definition\nvoid display()\n{\n printf(\"Hello world1\\n\");\n printf(\"Hello world2\\n\");\n printf(\"Hello world3\\n\");\n printf(\"Hello world4\\n\");\n printf(\"Hello world5\\n\");\n}\n```\n\n```sh\nHello world5\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5> cd \"d:\\MyData\\Business\\PWH C Tutorial\\Chapter 5\" ; if (!$?) { gcc 02_quick_quiz.c -o 02_quick_quiz } ; if (!$?) { iz }\nGood Morning\nGood Afternoon\nGood Night\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5>\n```" }, { "vid": "_MF8L7ZxwRE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic programming tutorials for beginners_100.json#####audio#####doingASR#####FinishASR/_MF8L7ZxwRE/16119.08_16137.26.mp4", "refined_asr": " I hope that you are able to understand this and I hope that you can see what I mean by saying that you will avoid rewriting the same logic again and again if you use functions. Functions will also help you keep track of what you're doing in a program. So if we have our logic divided into,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16119.08_16137.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16119.08_16137.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16119.08_16137.26#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16119.08_16137.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16119.08_16137.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16119.08_16137.26#3.jpg" ], "ocr_qwen2_vl_72b": "```c\n#include \n\nvoid display(); // Function prototype\n// void average(int, float, int); // Function prototype\n\nint main() {\n int a;\n display(); // Function call\n return 0;\n}\n\n// Function definition\nvoid display(){\n printf(\"Hello world1\\n\");\n printf(\"Hello world3\\n\");\n printf(\"Hello world4\\n\");\n printf(\"Hello world5\\n\");\n}\n```\n\n```bash\nHello world5\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5> cd \"d:\\MyData\\Business\\PWH C Tutorial\\Chapter 5\\\"; if (!$?) { gcc 02_quick_quiz.c -o 02_quick_quiz }; if (!$?)\niz\nGood Morning\nGood Afternoon\nGood Night\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5>\n```" }, { "vid": "_MF8L7ZxwRE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic programming tutorials for beginners_100.json#####audio#####doingASR#####FinishASR/_MF8L7ZxwRE/16137.26_16154.18.mp4", "refined_asr": " Different functions allow us to focus on one function and start debugging that function if it's not performing as expected. The third thing is to test and check logic independently. So, for example, I could have someone like Rohan Das write one function and someone like Jacob write another.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16137.26_16154.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16137.26_16154.18#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16137.26_16154.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16137.26_16154.18#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16137.26_16154.18#3.jpg" ], "ocr_qwen2_vl_72b": "```c\n// void average(int, float, int); // Function prototype\n\nint main(){\n int a;\n display(); // Function call\n return 0;\n}\n\n// Function definition\nvoid display(){\n printf(\"Hello world1\\n\");\n printf(\"Hello world2\\n\");\n printf(\"Hello world3\\n\");\n printf(\"Hello world4\\n\");\n printf(\"Hello world5\\n\");\n}\n```\n\n```bash\nHello world5\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5> cd \"d:\\MyData\\Business\\PWH C Tutorial\\Chapter 5\" ; if (!$?) { gcc 02_quick_quiz.c -o 02_quick_quiz } ; if (!$?) { .\\02_quick_quiz }\niz\nGood Morning\nGood Afternoon\nGood Night\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5>\n```" }, { "vid": "_MF8L7ZxwRE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic programming tutorials for beginners_100.json#####audio#####doingASR#####FinishASR/_MF8L7ZxwRE/16154.18_16170.98.mp4", "refined_asr": " Let's say you have three people working on three different functions. One person, say Sonali, writes one function. Another writes a second function, and the third writes yet another. These three can work independently on their respective functions. What I provide them is a function prototype. I specify the desired return value for each function and the parameters I intend to pass.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16154.18_16170.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16154.18_16170.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16154.18_16170.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16154.18_16170.98#3.jpg" ], "ocr_qwen2_vl_72b": "```c\n#include \n\nvoid display(); // Function prototype\n\n// void average(int, float, int); // Function prototype\n\nint main() {\n display(); // Function call\n return 0;\n}\n\n// Function definition\nvoid display() {\n printf(\"Hello world1\\n\");\n printf(\"Hello world2\\n\");\n printf(\"Hello world3\\n\");\n printf(\"Hello world4\\n\");\n printf(\"Hello world5\\n\");\n}\n```\n\n```bash\nHello world5\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5> cd \"d:\\MyData\\Business\\PWH C Tutorial\\Chapter 5\"; if (!$?) { gcc 02_quick_quiz.c -o 02_quick_quiz }; if (!$?) { 02_quick_quiz }\niz\nGood Morning\nGood Afternoon\nGood Night\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5>\n```" }, { "vid": "_MF8L7ZxwRE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic programming tutorials for beginners_100.json#####audio#####doingASR#####FinishASR/_MF8L7ZxwRE/16170.98_16187.9.mp4", "refined_asr": " Two or three or four values can be used as input to this function. Those values will be written inside the body of the function. If that makes sense, great! I hope you were able to understand this. Now, we'll talk about passing values to the functions.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16170.98_16187.9#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16170.98_16187.9#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16170.98_16187.9#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16170.98_16187.9#3.jpg" ], "ocr_qwen2_vl_72b": "```c\n#include \n\nvoid display(); // Function prototype\n\n// void average(int, float, int); // Function prototype\n\nint main(){\n int a;\n display(); // Function call\n return 0;\n}\n\n// Function definition\nvoid display(){\n printf(\"Hello world1\\n\");\n printf(\"Hello world2\\n\");\n printf(\"Hello world3\\n\");\n printf(\"Hello world4\\n\");\n printf(\"Hello world5\\n\");\n}\n```\n\n```sh\nHello world5\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5> cd \"d:\\MyData\\Business\\PWH C Tutorial\\Chapter 5\" ; if (!$?) { gcc 02_quick_quiz.c -o 02_quick_quiz } ; if (!$?) { .\\02_quick_quiz }\niz\nGood Morning\nGood Afternoon\nGood Night\nPS D:\\MyData\\Business\\code playground\\PWH C Tutorial\\Chapter 5>\n```" }, { "vid": "_MF8L7ZxwRE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic programming tutorials for beginners_100.json#####audio#####doingASR#####FinishASR/_MF8L7ZxwRE/16187.9_16202.6.mp4", "refined_asr": " Functions is something that you will definitely want to learn. What does it mean? It simply means that there is a function that takes some values as an input and gives an output. You need to receive output by giving some inputs, and you will definitely want to use a function somewhere.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16187.9_16202.6#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16187.9_16202.6#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16187.9_16202.6#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16187.9_16202.6#3.jpg" ], "ocr_qwen2_vl_72b": "Passing values to functions\n\nWe can pass values to a function and can get a value in return from a function.\n\nint sum (int a, int b)\n\nThe above prototype means that sum is a function which takes values a (of type int) and b (of type int) and returns a value of type int.\n\nfunction definition of sum can be:\n\nint sum (int a, int b) {\n..." }, { "vid": "_MF8L7ZxwRE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic programming tutorials for beginners_100.json#####audio#####doingASR#####FinishASR/_MF8L7ZxwRE/16202.6_16219.52.mp4", "refined_asr": " You are supplying two input values and you are expecting an output value from your function. Let's see how these things work. We can pass values to a function and we can get a value in return from a function. For example, let's say 'sum' is a function and we have two values which we are supplying.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16202.6_16219.52#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16202.6_16219.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16202.6_16219.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16202.6_16219.52#3.jpg" ], "ocr_qwen2_vl_72b": "Passing values to functions\n\nWe can pass values to a function and can get a value in return from a function.\n\nint sum(int a, int b)\n\nThe above prototype means that sum is a function which takes values a (of type int) and b (of type int) and returns a value of type int.\n\nfunction definition of sum can be:\n\nint sum(int a, int b) {\nint c;\nc = a + b;\nreturn c;\n}" }, { "vid": "_MF8L7ZxwRE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Basic programming tutorials for beginners_100.json#####audio#####doingASR#####FinishASR/_MF8L7ZxwRE/16219.52_16237.88.mp4", "refined_asr": " Sum as an input. What are those two values? Those two values are 'a' and 'b'. If these values are three and four, I will expect my sum to return an integer, which is nothing but seven, because three plus four is seven. Similarly, if 'a' is five and 'b' is three, I would expect this sum to give me eight in return.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16219.52_16237.88#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16219.52_16237.88#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16219.52_16237.88#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/_MF8L7ZxwRE/_MF8L7ZxwRE@16219.52_16237.88#3.jpg" ], "ocr_qwen2_vl_72b": "Passing values to functions\n\nWe can pass values to a function and can get a value in return from a function.\n\nint sum (int a, int b)\n\nThe above prototype means that sum is a function which takes values a (of type int) and b (of type int) and returns a value of type int.\n\nfunction definition of sum can be:\n\nint sum (int a, int b) {\nint c;\nc = a + b;\nreturn c;\n}" } ], "image_num": 11, "text_num": 547, "token_num": 6883 }, { "images": [ "sample_100_images/P3RXtoYCW4M@601.9_624.62#1.jpg", null, "sample_100_images/P3RXtoYCW4M@624.62_642.72#1.jpg", null, "sample_100_images/P3RXtoYCW4M@642.72_658.76#1.jpg", null, "sample_100_images/P3RXtoYCW4M@658.76_674.36#1.jpg", "sample_100_images/P3RXtoYCW4M@658.76_674.36#2.jpg", null, "sample_100_images/P3RXtoYCW4M@674.36_695.4200000000001#1.jpg", null, "sample_100_images/P3RXtoYCW4M@695.4200000000001_706.96#1.jpg", "sample_100_images/P3RXtoYCW4M@695.4200000000001_706.96#2.jpg", null, "sample_100_images/P3RXtoYCW4M@706.96_713.76#1.jpg", null, "sample_100_images/49LqdebQB3U@0.0_19.94#1.jpg", null ], "texts": [ null, " At the other end, this field also blends into molecular biology, which looks in the finest detail at how life arises out of the chemical processes inside cells. Within biochemistry, there are four main classes of molecules called biomolecules: Carbohydrates, which are used for structures and storing energy; lipids, which make up fats.", null, " Proteins, which are very large molecules made from amino acids, have a huge array of different functions in the body. Nucleic acids are used to convey genetic information. Research in biochemistry has had a huge impact on medicine, helping us understand infectious and genetic diseases.", null, " Improving organ and tissue transplantations, working out what's wrong with you with clinical diagnostics, and of course, understanding nutrition: looking at the functions of vitamins and minerals in our body. Biochemistry has also been very important for agriculture, studying soils and fertilizers.", null, null, " And pest controls. There's many other applications too. So that is my attempt to summarize all of chemistry in about 12 minutes. No simple task, as it's so incredibly complicated. It has always amazed me that something so complicated, so complex, could result in something as amazing as a human.", null, " It's built on a foundation of a huge number of simple chemical reactions. Your consciousness right now is a function of the chemistry going on in your brain cells oxygen being passed from your blood and sugars being metabolized inside them. Chemistry spans a huge mountain of complexity from a single atom to the cells that keep you alive.", null, null, " Life is alive and I find it endlessly fascinating. As with all my other videos, there's a post on my blog. And I'll be posting a video about it soon. So if you want to get hold of that, check out the links in the description.", null, " So, that's all for now. Otherwise, thanks again for watching, and for me, it's back to the drawing board.", null, "Okay, so you don't need to know a lot of math to be able to understand the solution to this calculus problem right here. Matter of fact, you'll be able to understand a very fundamental concept of calculus as long as you know some basic math and a very little bit of algebra." ], "text_ocr_list": [ null, "We can see these text from the image: - PESTICIDES\n- FERTILISERS\n- FLAVOURINGS\n- FRAGRANCES\n- ORGANIC CHEMISTRY\n- PLASTICS\n- LUBRICANTS\n- DRUGS\n- CATALYSTS\n- FERROCENE\n- AGRICULTURE\n- INORGANIC CHEMISTRY\n- ...EVERYTHING ELSE\n- ORGANOMETALLIC CHEMISTRY\n- BIOCHEMISTRY\n- CHEMISTRY OF LIFE\n- PROTEINS\n- WATER\n- IRON\n- FROM SIMPLE TO COMPLEX\n- DNA\n- MOLECULAR BIOLOGY THIS WAY\n- AREAS OF CHEMISTRY.\n At the other end, this field also blends into molecular biology, which looks in the finest detail at how life arises out of the chemical processes inside cells. Within biochemistry, there are four main classes of molecules called biomolecules: Carbohydrates, which are used for structures and storing energy; lipids, which make up fats.", null, "We can see these text from the image: - PESTICIDES\n- FERTILISERS\n- FLAVOURINGS\n- FRAGRANCES\n- ORGANIC CHEMISTRY\n- PLASTICS\n- LUBRICANTS\n- DRUGS\n- CATALYSTS\n- FERROCENE\n- AGRICULTURE\n- INORGANIC CHEMISTRY\n- ...EVERYTHING ELSE\n- MEDICINE\n- CARBOHYDRATES\n- ORGANOMETALLIC CHEMISTRY\n- BIOCHEMISTRY\n- CHEMISTRY OF LIFE\n- PROTEINS\n- BIOMOLECULES\n- IRON\n- WATER\n- FROM SIMPLE TO COMPLEX\n- DNA\n- LIPIDS\n- AREAS OF CHEMISTRY\n- MOLECULAR BIOLOGY THIS WAY.\n Proteins, which are very large molecules made from amino acids, have a huge array of different functions in the body. Nucleic acids are used to convey genetic information. Research in biochemistry has had a huge impact on medicine, helping us understand infectious and genetic diseases.", null, "We can see these text from the image: - PESTICIDES\n- FERTILISERS\n- FLAVOURINGS\n- FRAGRANCES\n- ORGANIC CHEMISTRY\n- PLASTICS\n- LUBRICANTS\n- DRUGS\n- CATALYSTS\n- V2O5\n- INORGANIC CHEMISTRY\n- ...EVERYTHING ELSE\n- CARBOHYDRATES\n- ORGANOMETALLIC CHEMISTRY\n- BIOCHEMISTRY\n- CHEMISTRY OF LIFE\n- TRANSFUSION MEDICINE\n- AREAS OF CHEMISTRY\n- PROTEINS\n- BIOMOLECULES\n- IRON\n- WATER\n- FROM SIMPLE TO COMPLEX\n- DNA\n- LIPIDS\n- FERROCENE\n- AGRICULTURE\n- PENICILLIN\n- OH\n- MOLECULAR BIOLOGY THIS WAY.\n Improving organ and tissue transplantations, working out what's wrong with you with clinical diagnostics, and of course, understanding nutrition: looking at the functions of vitamins and minerals in our body. Biochemistry has also been very important for agriculture, studying soils and fertilizers.", null, null, "We can see these text from the image: - PESTICIDES\n- FERTILISERS\n- FLAVOURINGS\n- FRAGRANCES\n- ORGANIC CHEMISTRY\n- PLASTICS\n- LUBRICANTS\n- DRUGS\n- CATALYSTS\n- V2O5\n- INORGANIC CHEMISTRY\n- ...EVERYTHING ELSE\n- ORGANOMETALLIC CHEMISTRY\n- CARBOHYDRATES\n- WATER\n- IRON\n- FROM SIMPLE TO COMPLEX\n- BIOCHEMISTRY\n- CHEMISTRY OF LIFE\n- NUTRITION\n- TRANSFUSION MEDICINE\n- AGRICULTURE\n- PROTEINS\n- BIOMOLECULES\n- DNA\n- LIPIDS\n- AREAS OF CHEMISTRY\n- PENICILLIN\n- OH\n- FERROCENE\n- MOLECULAR BIOLOGY THIS WAY.\n And pest controls. There's many other applications too. So that is my attempt to summarize all of chemistry in about 12 minutes. No simple task, as it's so incredibly complicated. It has always amazed me that something so complicated, so complex, could result in something as amazing as a human.", null, "We can see these text from the image: - TYPES OF REACTION\n - SYNTHESIS A + B \u2192 AB\n - DECOMPOSITION AB \u2192 A + B\n - SINGLE REPLACEMENT A + BC \u2192 AC + B\n - DOUBLE REPLACEMENT AB + CD \u2192 AD + CB\n\n- REDOX REACTION\n - OXIDATION A \u2192 AO\n - REDUCTION AO \u2192 A\n\n- ACIDS AND BASES\n - pH SCALE\n - ACIDIC\n - NEUTRAL\n - BASIC\n\n- CONSERVATION OF MASS AND ENERGY\n - NO ENERGY OR MASS IS CREATED OR DESTROYED, JUST CONVERTED\n\n- RULES OF CHEMISTRY\n - BONDING\n - COVALENT BOND\n - IONIC BONDING\n - HYDROGEN BONDING\n - VAN DER WAALS BONDING\n - MIXTURES\n - SOLUTIONS\n - SUSPENSIONS\n - COLLOIDS\n - CHEMICAL COMPOUNDS\n - H2O WATER\n - CO2 CARBON DIOXIDE\n - NH3 AMMONIA\n - EXPLODE\n - HYDROGEN OXYGEN\n - SIMILAR CHEMICAL PROPERTIES\n\n- THE PERIODIC TABLE\n - ELEMENTS LISTED BY ATOMIC NUMBER\n\n- ORGANIC MOLECULES MOSTLY MADE FROM A SMALL SET OF ATOMS\n - CARBON\n - HYDROGEN\n - OXYGEN\n - NITROGEN\n - PLUS SOME OTHERS\n\n- ORGANIC CHEMISTRY\n - PESTICIDES\n - FERTILISERS\n - FLAVOURINGS\n - FRAGRANCES\n - CARBOHYDRATES\n - PROTEINS\n - LIPIDS\n - DNA\n - BIOLOGICAL MOLECULES\n\n- AREAS OF CHEMISTRY\n - INORGANIC CHEMISTRY\n - ANALYTICAL CHEMISTRY\n - PHYSICAL CHEMISTRY\n - MATERIALS SCIENCE\n - ELECTROCHEMISTRY\n - COMPUTATIONAL CHEMISTRY\n - THEORETICAL CHEMISTRY\n - QUANTUM CHEMISTRY\n\n- KINETICS\n - REVERSIBLE REACTION\n - EQUILIBRIUM\n - PHASES\n - SOLID\n - LIQUID\n - GAS\n\n- ENERGY\n - HIGH PRESSURE\n - LOW PRESSURE\n - HIGH TEMPERATURE\n - LOW TEMPERATURE\n\n- MATTER\n - NOT BURSTY OR EXPLODY\n - BURSTY*\n\n- ORGANOMETALLIC CHEMISTRY\n - CATALYSTS\n - FERROCENE\n - AGROCHEMISTRY\n - INORGANIC CHEMISTRY\n - EVERYTHING ELSE\n\n- BIOCHEMISTRY\n - CHEMISTRY OF LIFE\n - NUTRITION\n - TRANSFUSION MEDICINE\n - MASS SPECTROMETRY\n - CHROMATOGRAPHY\n - PRECIPITATION\n - MYSTERY SUBSTANCE\n\n- DRUGS\n - ORIGINS\n - FERTILISER\n - METALWORKING\n - PLASTICS\n - FABRICS\n - MATERIALS\n - FIRE\n - GLASS\n\n- ATOM\n - PROTON\n - NEUTRON\n - ELECTRON\n\n- THE PERIODIC TABLE\n - ELEMENTS LISTED BY ATOMIC NUMBER\n\n- PHYSICS THIS WAY.\n It's built on a foundation of a huge number of simple chemical reactions. Your consciousness right now is a function of the chemistry going on in your brain cells oxygen being passed from your blood and sugars being metabolized inside them. Chemistry spans a huge mountain of complexity from a single atom to the cells that keep you alive.", null, null, "We can see these text from the image: - TYPES OF REACTION\n - SYNTHESIS A + B \u2192 AB\n - DECOMPOSITION AB \u2192 A + B\n - SINGLE REPLACEMENT A + BC \u2192 AC + B\n - DOUBLE REPLACEMENT AB + CD \u2192 AD + CB\n\n- REDOX REACTION\n - OXIDATION A \u2192 A+ + e-\n - REDUCTION B + e- \u2192 B-\n\n- ACIDS AND BASES\n - pH SCALE\n - ACIDIC < 7\n - NEUTRAL = 7\n - BASIC > 7\n\n- KINETICS\n - ENERGY LEVEL DIAGRAM\n - ACTIVATION ENERGY\n\n- RULES OF CHEMISTRY\n - CONSERVATION OF MASS AND ENERGY\n - NO ENERGY OR MASS IS CREATED OR DESTROYED, JUST CONVERTED\n\n- BONDING\n - COVALENT BOND\n - IONIC BONDING\n - HYDROGEN BONDING\n - VAN DER WAALS BONDING\n\n- MIXTURES\n - HOMOGENEOUS\n - HETEROGENEOUS\n\n- CRYSTAL\n - UNIT CELL\n\n- CHEMICAL COMPOUNDS\n - H2O WATER\n - CO2 CARBON DIOXIDE\n - NH3 AMMONIA\n\n- ORGANIC MOLECULES MOSTLY MADE FROM A SMALL SET OF ATOMS\n - CARBON\n - HYDROGEN\n - OXYGEN\n - NITROGEN\n - PLUS SOME OTHERS\n\n- ORGANIC CHEMISTRY\n - PESTICIDES\n - FERTILISERS\n - FLAVOURINGS\n - FRAGRANCES\n - DRUGS\n - LUBRICANTS\n - CATALYSTS\n - FERROCENE\n\n- INORGANIC CHEMISTRY\n - EVERYTHING ELSE\n\n- AREAS OF CHEMISTRY\n - BIOCHEMISTRY\n - CHEMISTRY OF LIFE\n - BIOMOLECULES\n - PROTEINS\n - CARBOHYDRATES\n - LIPIDS\n - DNA\n - TRANSFUSION MEDICINE\n - NUTRITION\n - AGRICULTURE\n - ANALYTICAL CHEMISTRY\n - MASS SPECTROMETRY\n - CHROMATOGRAPHY\n - PRECIPITATION\n - MYSTERY SUBSTANCE\n - COMPUTATIONAL CHEMISTRY\n - MATERIALS SCIENCE\n - PHYSICAL CHEMISTRY\n - ELECTROCHEMISTRY\n - THERMODYNAMICS\n - THEORETICAL CHEMISTRY\n - QUANTUM CHEMISTRY\n - THEORETICAL PHYSICS\n\n- MATTER\n - SOLID\n - LIQUID\n - GAS\n\n- EXPLODE\n - NOT BURST OR EXPLODY\n\n- THE PERIODIC TABLE\n - ELEMENTS LISTED BY ATOMIC NUMBER\n\n- ORGANOMETALLIC CHEMISTRY\n\n- ORIGINS\n - DRUGS\n - FERTILISER\n - METALWORKING\n - PLASTICS\n - FABRICS\n - MATERIALS\n - FIRE\n - GLASS\n - SELLING\n\n- AN ATOM\n - PROTON\n - NEUTRON\n - ELECTRON\n\n- PHASES\n - SOLID\n - LIQUID\n - GAS\n\n- EQUILIBRIUM\n - REACTANT\n - PRODUCT\n\n- FLUIDS\n - MADE BY HUMANS\n - DETERGENTS\n - FUELS\n - EMULSIFIERS\n - NEW MATERIALS.\n Life is alive and I find it endlessly fascinating. As with all my other videos, there's a post on my blog. And I'll be posting a video about it soon. So if you want to get hold of that, check out the links in the description.", null, "We can see these text from the image: CLICK HERE TO SUBSCRIBE\n\nwww.thegreatcoursesplus.com/domainofscience.\n So, that's all for now. Otherwise, thanks again for watching, and for me, it's back to the drawing board.", null, "We can see these text from the image: Which is Correct?\n\na) 1.5\nb) 6\nc) 50.2\nd) 19/3\n\n\u222b x\u00b2 dx\n2Okay, so you don't need to know a lot of math to be able to understand the solution to this calculus problem right here. Matter of fact, you'll be able to understand a very fundamental concept of calculus as long as you know some basic math and a very little bit of algebra." ], "metadata": [ { "vid": "P3RXtoYCW4M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemistry basics for beginners_100.json#####audio#####doingASR#####FinishASR/P3RXtoYCW4M/601.9_624.62.mp4", "refined_asr": " At the other end, this field also blends into molecular biology, which looks in the finest detail at how life arises out of the chemical processes inside cells. Within biochemistry, there are four main classes of molecules called biomolecules: Carbohydrates, which are used for structures and storing energy; lipids, which make up fats.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@601.9_624.62#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@601.9_624.62#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@601.9_624.62#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@601.9_624.62#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@601.9_624.62#4.jpg" ], "ocr_qwen2_vl_72b": "- PESTICIDES\n- FERTILISERS\n- FLAVOURINGS\n- FRAGRANCES\n- ORGANIC CHEMISTRY\n- PLASTICS\n- LUBRICANTS\n- DRUGS\n- CATALYSTS\n- FERROCENE\n- AGRICULTURE\n- INORGANIC CHEMISTRY\n- ...EVERYTHING ELSE\n- ORGANOMETALLIC CHEMISTRY\n- BIOCHEMISTRY\n- CHEMISTRY OF LIFE\n- PROTEINS\n- WATER\n- IRON\n- FROM SIMPLE TO COMPLEX\n- DNA\n- MOLECULAR BIOLOGY THIS WAY\n- AREAS OF CHEMISTRY" }, { "vid": "P3RXtoYCW4M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemistry basics for beginners_100.json#####audio#####doingASR#####FinishASR/P3RXtoYCW4M/624.62_642.72.mp4", "refined_asr": " Proteins, which are very large molecules made from amino acids, have a huge array of different functions in the body. Nucleic acids are used to convey genetic information. Research in biochemistry has had a huge impact on medicine, helping us understand infectious and genetic diseases.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@624.62_642.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@624.62_642.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@624.62_642.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@624.62_642.72#3.jpg" ], "ocr_qwen2_vl_72b": "- PESTICIDES\n- FERTILISERS\n- FLAVOURINGS\n- FRAGRANCES\n- ORGANIC CHEMISTRY\n- PLASTICS\n- LUBRICANTS\n- DRUGS\n- CATALYSTS\n- FERROCENE\n- AGRICULTURE\n- INORGANIC CHEMISTRY\n- ...EVERYTHING ELSE\n- MEDICINE\n- CARBOHYDRATES\n- ORGANOMETALLIC CHEMISTRY\n- BIOCHEMISTRY\n- CHEMISTRY OF LIFE\n- PROTEINS\n- BIOMOLECULES\n- IRON\n- WATER\n- FROM SIMPLE TO COMPLEX\n- DNA\n- LIPIDS\n- AREAS OF CHEMISTRY\n- MOLECULAR BIOLOGY THIS WAY" }, { "vid": "P3RXtoYCW4M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemistry basics for beginners_100.json#####audio#####doingASR#####FinishASR/P3RXtoYCW4M/642.72_658.76.mp4", "refined_asr": " Improving organ and tissue transplantations, working out what's wrong with you with clinical diagnostics, and of course, understanding nutrition: looking at the functions of vitamins and minerals in our body. Biochemistry has also been very important for agriculture, studying soils and fertilizers.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@642.72_658.76#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@642.72_658.76#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@642.72_658.76#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@642.72_658.76#3.jpg" ], "ocr_qwen2_vl_72b": "- PESTICIDES\n- FERTILISERS\n- FLAVOURINGS\n- FRAGRANCES\n- ORGANIC CHEMISTRY\n- PLASTICS\n- LUBRICANTS\n- DRUGS\n- CATALYSTS\n- V2O5\n- INORGANIC CHEMISTRY\n- ...EVERYTHING ELSE\n- CARBOHYDRATES\n- ORGANOMETALLIC CHEMISTRY\n- BIOCHEMISTRY\n- CHEMISTRY OF LIFE\n- TRANSFUSION MEDICINE\n- AREAS OF CHEMISTRY\n- PROTEINS\n- BIOMOLECULES\n- IRON\n- WATER\n- FROM SIMPLE TO COMPLEX\n- DNA\n- LIPIDS\n- FERROCENE\n- AGRICULTURE\n- PENICILLIN\n- OH\n- MOLECULAR BIOLOGY THIS WAY" }, { "vid": "P3RXtoYCW4M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemistry basics for beginners_100.json#####audio#####doingASR#####FinishASR/P3RXtoYCW4M/658.76_674.36.mp4", "refined_asr": " And pest controls. There's many other applications too. So that is my attempt to summarize all of chemistry in about 12 minutes. No simple task, as it's so incredibly complicated. It has always amazed me that something so complicated, so complex, could result in something as amazing as a human.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@658.76_674.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@658.76_674.36#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@658.76_674.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@658.76_674.36#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@658.76_674.36#3.jpg" ], "ocr_qwen2_vl_72b": "- PESTICIDES\n- FERTILISERS\n- FLAVOURINGS\n- FRAGRANCES\n- ORGANIC CHEMISTRY\n- PLASTICS\n- LUBRICANTS\n- DRUGS\n- CATALYSTS\n- V2O5\n- INORGANIC CHEMISTRY\n- ...EVERYTHING ELSE\n- ORGANOMETALLIC CHEMISTRY\n- CARBOHYDRATES\n- WATER\n- IRON\n- FROM SIMPLE TO COMPLEX\n- BIOCHEMISTRY\n- CHEMISTRY OF LIFE\n- NUTRITION\n- TRANSFUSION MEDICINE\n- AGRICULTURE\n- PROTEINS\n- BIOMOLECULES\n- DNA\n- LIPIDS\n- AREAS OF CHEMISTRY\n- PENICILLIN\n- OH\n- FERROCENE\n- MOLECULAR BIOLOGY THIS WAY" }, { "vid": "P3RXtoYCW4M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemistry basics for beginners_100.json#####audio#####doingASR#####FinishASR/P3RXtoYCW4M/674.36_695.4200000000001.mp4", "refined_asr": " It's built on a foundation of a huge number of simple chemical reactions. Your consciousness right now is a function of the chemistry going on in your brain cells oxygen being passed from your blood and sugars being metabolized inside them. Chemistry spans a huge mountain of complexity from a single atom to the cells that keep you alive.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@674.36_695.4200000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@674.36_695.4200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@674.36_695.4200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@674.36_695.4200000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@674.36_695.4200000000001#4.jpg" ], "ocr_qwen2_vl_72b": "- TYPES OF REACTION\n - SYNTHESIS A + B \u2192 AB\n - DECOMPOSITION AB \u2192 A + B\n - SINGLE REPLACEMENT A + BC \u2192 AC + B\n - DOUBLE REPLACEMENT AB + CD \u2192 AD + CB\n\n- REDOX REACTION\n - OXIDATION A \u2192 AO\n - REDUCTION AO \u2192 A\n\n- ACIDS AND BASES\n - pH SCALE\n - ACIDIC\n - NEUTRAL\n - BASIC\n\n- CONSERVATION OF MASS AND ENERGY\n - NO ENERGY OR MASS IS CREATED OR DESTROYED, JUST CONVERTED\n\n- RULES OF CHEMISTRY\n - BONDING\n - COVALENT BOND\n - IONIC BONDING\n - HYDROGEN BONDING\n - VAN DER WAALS BONDING\n - MIXTURES\n - SOLUTIONS\n - SUSPENSIONS\n - COLLOIDS\n - CHEMICAL COMPOUNDS\n - H2O WATER\n - CO2 CARBON DIOXIDE\n - NH3 AMMONIA\n - EXPLODE\n - HYDROGEN OXYGEN\n - SIMILAR CHEMICAL PROPERTIES\n\n- THE PERIODIC TABLE\n - ELEMENTS LISTED BY ATOMIC NUMBER\n\n- ORGANIC MOLECULES MOSTLY MADE FROM A SMALL SET OF ATOMS\n - CARBON\n - HYDROGEN\n - OXYGEN\n - NITROGEN\n - PLUS SOME OTHERS\n\n- ORGANIC CHEMISTRY\n - PESTICIDES\n - FERTILISERS\n - FLAVOURINGS\n - FRAGRANCES\n - CARBOHYDRATES\n - PROTEINS\n - LIPIDS\n - DNA\n - BIOLOGICAL MOLECULES\n\n- AREAS OF CHEMISTRY\n - INORGANIC CHEMISTRY\n - ANALYTICAL CHEMISTRY\n - PHYSICAL CHEMISTRY\n - MATERIALS SCIENCE\n - ELECTROCHEMISTRY\n - COMPUTATIONAL CHEMISTRY\n - THEORETICAL CHEMISTRY\n - QUANTUM CHEMISTRY\n\n- KINETICS\n - REVERSIBLE REACTION\n - EQUILIBRIUM\n - PHASES\n - SOLID\n - LIQUID\n - GAS\n\n- ENERGY\n - HIGH PRESSURE\n - LOW PRESSURE\n - HIGH TEMPERATURE\n - LOW TEMPERATURE\n\n- MATTER\n - NOT BURSTY OR EXPLODY\n - BURSTY*\n\n- ORGANOMETALLIC CHEMISTRY\n - CATALYSTS\n - FERROCENE\n - AGROCHEMISTRY\n - INORGANIC CHEMISTRY\n - EVERYTHING ELSE\n\n- BIOCHEMISTRY\n - CHEMISTRY OF LIFE\n - NUTRITION\n - TRANSFUSION MEDICINE\n - MASS SPECTROMETRY\n - CHROMATOGRAPHY\n - PRECIPITATION\n - MYSTERY SUBSTANCE\n\n- DRUGS\n - ORIGINS\n - FERTILISER\n - METALWORKING\n - PLASTICS\n - FABRICS\n - MATERIALS\n - FIRE\n - GLASS\n\n- ATOM\n - PROTON\n - NEUTRON\n - ELECTRON\n\n- THE PERIODIC TABLE\n - ELEMENTS LISTED BY ATOMIC NUMBER\n\n- PHYSICS THIS WAY" }, { "vid": "P3RXtoYCW4M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemistry basics for beginners_100.json#####audio#####doingASR#####FinishASR/P3RXtoYCW4M/695.4200000000001_706.96.mp4", "refined_asr": " Life is alive and I find it endlessly fascinating. As with all my other videos, there's a post on my blog. And I'll be posting a video about it soon. So if you want to get hold of that, check out the links in the description.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@695.4200000000001_706.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@695.4200000000001_706.96#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@695.4200000000001_706.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@695.4200000000001_706.96#2.jpg" ], "ocr_qwen2_vl_72b": "- TYPES OF REACTION\n - SYNTHESIS A + B \u2192 AB\n - DECOMPOSITION AB \u2192 A + B\n - SINGLE REPLACEMENT A + BC \u2192 AC + B\n - DOUBLE REPLACEMENT AB + CD \u2192 AD + CB\n\n- REDOX REACTION\n - OXIDATION A \u2192 A+ + e-\n - REDUCTION B + e- \u2192 B-\n\n- ACIDS AND BASES\n - pH SCALE\n - ACIDIC < 7\n - NEUTRAL = 7\n - BASIC > 7\n\n- KINETICS\n - ENERGY LEVEL DIAGRAM\n - ACTIVATION ENERGY\n\n- RULES OF CHEMISTRY\n - CONSERVATION OF MASS AND ENERGY\n - NO ENERGY OR MASS IS CREATED OR DESTROYED, JUST CONVERTED\n\n- BONDING\n - COVALENT BOND\n - IONIC BONDING\n - HYDROGEN BONDING\n - VAN DER WAALS BONDING\n\n- MIXTURES\n - HOMOGENEOUS\n - HETEROGENEOUS\n\n- CRYSTAL\n - UNIT CELL\n\n- CHEMICAL COMPOUNDS\n - H2O WATER\n - CO2 CARBON DIOXIDE\n - NH3 AMMONIA\n\n- ORGANIC MOLECULES MOSTLY MADE FROM A SMALL SET OF ATOMS\n - CARBON\n - HYDROGEN\n - OXYGEN\n - NITROGEN\n - PLUS SOME OTHERS\n\n- ORGANIC CHEMISTRY\n - PESTICIDES\n - FERTILISERS\n - FLAVOURINGS\n - FRAGRANCES\n - DRUGS\n - LUBRICANTS\n - CATALYSTS\n - FERROCENE\n\n- INORGANIC CHEMISTRY\n - EVERYTHING ELSE\n\n- AREAS OF CHEMISTRY\n - BIOCHEMISTRY\n - CHEMISTRY OF LIFE\n - BIOMOLECULES\n - PROTEINS\n - CARBOHYDRATES\n - LIPIDS\n - DNA\n - TRANSFUSION MEDICINE\n - NUTRITION\n - AGRICULTURE\n - ANALYTICAL CHEMISTRY\n - MASS SPECTROMETRY\n - CHROMATOGRAPHY\n - PRECIPITATION\n - MYSTERY SUBSTANCE\n - COMPUTATIONAL CHEMISTRY\n - MATERIALS SCIENCE\n - PHYSICAL CHEMISTRY\n - ELECTROCHEMISTRY\n - THERMODYNAMICS\n - THEORETICAL CHEMISTRY\n - QUANTUM CHEMISTRY\n - THEORETICAL PHYSICS\n\n- MATTER\n - SOLID\n - LIQUID\n - GAS\n\n- EXPLODE\n - NOT BURST OR EXPLODY\n\n- THE PERIODIC TABLE\n - ELEMENTS LISTED BY ATOMIC NUMBER\n\n- ORGANOMETALLIC CHEMISTRY\n\n- ORIGINS\n - DRUGS\n - FERTILISER\n - METALWORKING\n - PLASTICS\n - FABRICS\n - MATERIALS\n - FIRE\n - GLASS\n - SELLING\n\n- AN ATOM\n - PROTON\n - NEUTRON\n - ELECTRON\n\n- PHASES\n - SOLID\n - LIQUID\n - GAS\n\n- EQUILIBRIUM\n - REACTANT\n - PRODUCT\n\n- FLUIDS\n - MADE BY HUMANS\n - DETERGENTS\n - FUELS\n - EMULSIFIERS\n - NEW MATERIALS" }, { "vid": "P3RXtoYCW4M.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemistry basics for beginners_100.json#####audio#####doingASR#####FinishASR/P3RXtoYCW4M/706.96_713.76.mp4", "refined_asr": " So, that's all for now. Otherwise, thanks again for watching, and for me, it's back to the drawing board.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@706.96_713.76#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/P3RXtoYCW4M/P3RXtoYCW4M@706.96_713.76#1.jpg" ], "ocr_qwen2_vl_72b": "CLICK HERE TO SUBSCRIBE\n\nwww.thegreatcoursesplus.com/domainofscience" }, { "vid": "49LqdebQB3U.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus introduction for beginners_100.json#####audio#####doingASR#####FinishASR/49LqdebQB3U/0.0_19.94.mp4", "refined_asr": null, "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/49LqdebQB3U/49LqdebQB3U@0.0_19.94#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/49LqdebQB3U/49LqdebQB3U@0.0_19.94#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/49LqdebQB3U/49LqdebQB3U@0.0_19.94#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/49LqdebQB3U/49LqdebQB3U@0.0_19.94#3.jpg" ], "ocr_qwen2_vl_72b": "Which is Correct?\n\na) 1.5\nb) 6\nc) 50.2\nd) 19/3\n\n\u222b x\u00b2 dx\n2" } ], "image_num": 10, "text_num": 516, "token_num": 6276 }, { "images": [ "sample_100_images/SRjrUb4jkd4@1623.3000000000002_1671.6000000000001#1.jpg", null, "sample_100_images/SRjrUb4jkd4@1671.6000000000001_1689.5000000000002#1.jpg", null, "sample_100_images/SRjrUb4jkd4@1689.5000000000002_1708.8#1.jpg", "sample_100_images/SRjrUb4jkd4@1689.5000000000002_1708.8#3.jpg", null, "sample_100_images/SRjrUb4jkd4@1708.8_1738.24#1.jpg", null, "sample_100_images/SRjrUb4jkd4@1738.24_1768.8#1.jpg", "sample_100_images/SRjrUb4jkd4@1738.24_1768.8#3.jpg", null ], "texts": [ null, " How many nutrients or what are the loads that are going to maintain a healthy ecosystem and stop it from sort of switching to the unhealthy eutrophic conditions? So in this case, this table is showing the three main categories: the attachments, the serpentine, the Murray, and the Harvey.", null, " That's the column on the left-hand side. Have a healthy ecosystem and then they work out exactly how much is coming in. So in this case, 69 tons per annum from the Serpentine, 16 from the Murray, and 61 from the Harvey. So you can see that.", null, null, " It's the Serpentine River which is in the very top of the catchment and the Harvey River which is right at the bottom of the Peel Harvey They're the ones where the most activity needs to be done to reduce those nutrient loads So in this case it's the Serpentine up here coming in sort of", null, " Just to the east of Mandra and the Harvey River at the bottom that needs to have the most action to reduce the nutrient loads. So those load reductions can be done through a variety of ways: through educating the farmers, the graziers, and the residential population, to reduce the nutrient use by planting.", null, null, " The vegetation along the streams and rivers can take up the nutrients before they actually reach the wetlands. So there's a variety of things that can be done in that space. Another way to manage eutrophication is by actually pumping oxygen into the waterways. This is something that's done at a number of places in the Southwest." ], "text_ocr_list": [ null, "We can see these text from the image: **Load reduction targets**\n\n| Catchment | Total Phosphorus load target to estuary (t/yr) | Estimated current Total Phosphorus winter load (t/yr) | Estimated load reductions required to meet 0.1 mg/L (%) |\n|---|---|---|---|\n| Total Serpentine | 21 | 69 | 60 |\n| Dirk Brook Yangedi | - | - | - |\n| Nambeelup Brook | - | - | - |\n| Peel Main Drain | - | - | - |\n| Ungauged Lower Serpentine | - | - | - |\n| Total Murray | 16 | 16 | - |\n| Dandalup, Upper and Lower Murray | - | - | - |\n| Total Harvey | 38 | 61 | 40 |\n| Coastal Western Harvey | - | - | - |\n| Coastal Central | - | - | - |\n| East Harvey Peel Drains | - | - | - |\n| Grand Total | 75 | 145 | 48 |.\n How many nutrients or what are the loads that are going to maintain a healthy ecosystem and stop it from sort of switching to the unhealthy eutrophic conditions? So in this case, this table is showing the three main categories: the attachments, the serpentine, the Murray, and the Harvey.", null, "We can see these text from the image: **Load reduction targets**\n\n**Table 3: Load reduction targets by main catchments and critical reporting catchments (courtesy Christian Zammit)**\n\n| Catchment | Total Phosphorus load target to estuary (t/yr) | Estimated current Total Phosphorus winter load (t/yr) | Estimated load reductions required to meet 0.1 mg/L (%) |\n|----------------------------------|-----------------------------------------------|-------------------------------------------------------|---------------------------------------------------------|\n| **Total Serpentine** | 21 | 69 | 60 |\n| Dirk Brook Yangedi | - | - | - |\n| Nambeelup Brook | - | - | - |\n| Peel Main Drain | - | - | - |\n| Ungauged Lower Serpentine | - | - | - |\n| **Total Murray** | 16 | 16 | - |\n| Dandalup, Upper and Lower Murray | - | - | - |\n| **Total Harvey** | 38 | 61 | 40 |\n| Coastal Western Harvey | - | - | - |\n| Coastal Central | - | - | - |\n| East Harvey Peel Drains | - | - | - |\n| **Grand Total** | 75 | 145 | 48 |.\n That's the column on the left-hand side. Have a healthy ecosystem and then they work out exactly how much is coming in. So in this case, 69 tons per annum from the Serpentine, 16 from the Murray, and 61 from the Harvey. So you can see that.", null, null, "We can see these text from the image: **Load reduction targets**\n\n**Table 3: Load reduction targets by main catchments and critical reporting catchments (courtesy Christian Zammit)**\n\n| Catchment | Total Phosphorus load target to estuary (t/yr) | Estimated current Total Phosphorus winter load (t/yr) | Estimated load reductions required to meet 0.1 mg/L (%) |\n|-----------------------------------|-----------------------------------------------|------------------------------------------------------|--------------------------------------------------------|\n| Total Serpentine | 21 | 69 | 60 |\n| Dirk Brook Yangedi | - | - | - |\n| Nambeelup Brook | - | - | - |\n| Peel Main Drain | - | - | - |\n| Ungauged Lower Serpentine | - | - | - |\n| Total Murray | 16 | 16 | - |\n| Dandalup, Upper and Lower Murray | - | - | - |\n| Total Harvey | 38 | 61 | 40 |\n| Coastal Western Harvey | - | - | - |\n| Coastal Central | - | - | - |\n| East Harvey Peel Drains | - | - | - |\n| Grand Total | 75 | 145 | 48 |.\n It's the Serpentine River which is in the very top of the catchment and the Harvey River which is right at the bottom of the Peel Harvey They're the ones where the most activity needs to be done to reduce those nutrient loads So in this case it's the Serpentine up here coming in sort of", null, "We can see these text from the image: **Load reduction targets**\n\n**Table 3: Load reduction targets by catchment (courtesy Christian Zammit)**\n\n| Catchment | Estimated load reductions required to meet 0.1 mg/L (%) |\n|----------------------------|---------------------------------------------------------|\n| Total Serpentine | 60 |\n| Dirk Brook Yangedi | 82 |\n| Nambeelup Brook | 78 |\n| Peel Main Drain | 66 |\n| Ungauged Lower Serpentine | 60 |\n| Total Murray | - |\n| Dandalup, Upper and Lower Murray | - |\n| Total Harvey | 40 |\n| Coastal Western Harvey | 83 |\n| Coastal Central | 79 |\n| East Harvey Peel Drains | 47 |\n| Grand Total | 48 |\n\nSERPENTINE\nMURRAY\nHARVEY.\n Just to the east of Mandra and the Harvey River at the bottom that needs to have the most action to reduce the nutrient loads. So those load reductions can be done through a variety of ways: through educating the farmers, the graziers, and the residential population, to reduce the nutrient use by planting.", null, null, "We can see these text from the image: **Load reduction targets**\n\n- **Table 3: Load reduction targets by catchment (courtesy Christian Zammit)**\n - **Catchment**\n - Total Serpentine\n - Dirk Brook Yangedi\n - Nambeelup Brook\n - Peel Main Drain\n - Ungauged Lower Serpentine\n - Total Murray\n - Dandalup, Upper and Lower Murray\n - Total Harvey\n - Coastal Western Harvey\n - Coastal Central\n - East Harvey Peel Drains\n - Grand Total\n\n- **SERPENTINE**\n - **MURRAY**\n - **HARVEY**\n\n- **Estimated load reductions required to meet 0.1 mg/L (%)**\n - Total Serpentine: 60\n - Dirk Brook Yangedi: 82\n - Nambeelup Brook: 78\n - Peel Main Drain: 66\n - Ungauged Lower Serpentine: 60\n - Total Murray: -\n - Dandalup, Upper and Lower Murray: -\n - Total Harvey: 40\n - Coastal Western Harvey: 83\n - Coastal Central: 79\n - East Harvey Peel Drains: 47\n - Grand Total: 48.\n The vegetation along the streams and rivers can take up the nutrients before they actually reach the wetlands. So there's a variety of things that can be done in that space. Another way to manage eutrophication is by actually pumping oxygen into the waterways. This is something that's done at a number of places in the Southwest." ], "metadata": [ { "vid": "SRjrUb4jkd4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Oceanography tutorial on Marine Pollution focusing on Eutrophication_30.json#####audio#####doingASR#####FinishASR/SRjrUb4jkd4/1623.3000000000002_1671.6000000000001.mp4", "refined_asr": " How many nutrients or what are the loads that are going to maintain a healthy ecosystem and stop it from sort of switching to the unhealthy eutrophic conditions? So in this case, this table is showing the three main categories: the attachments, the serpentine, the Murray, and the Harvey.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1623.3000000000002_1671.6000000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1623.3000000000002_1671.6000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1623.3000000000002_1671.6000000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1623.3000000000002_1671.6000000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1623.3000000000002_1671.6000000000001#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1623.3000000000002_1671.6000000000001#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1623.3000000000002_1671.6000000000001#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1623.3000000000002_1671.6000000000001#7.jpg" ], "ocr_qwen2_vl_72b": "**Load reduction targets**\n\n| Catchment | Total Phosphorus load target to estuary (t/yr) | Estimated current Total Phosphorus winter load (t/yr) | Estimated load reductions required to meet 0.1 mg/L (%) |\n|---|---|---|---|\n| Total Serpentine | 21 | 69 | 60 |\n| Dirk Brook Yangedi | - | - | - |\n| Nambeelup Brook | - | - | - |\n| Peel Main Drain | - | - | - |\n| Ungauged Lower Serpentine | - | - | - |\n| Total Murray | 16 | 16 | - |\n| Dandalup, Upper and Lower Murray | - | - | - |\n| Total Harvey | 38 | 61 | 40 |\n| Coastal Western Harvey | - | - | - |\n| Coastal Central | - | - | - |\n| East Harvey Peel Drains | - | - | - |\n| Grand Total | 75 | 145 | 48 |" }, { "vid": "SRjrUb4jkd4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Oceanography tutorial on Marine Pollution focusing on Eutrophication_30.json#####audio#####doingASR#####FinishASR/SRjrUb4jkd4/1671.6000000000001_1689.5000000000002.mp4", "refined_asr": " That's the column on the left-hand side. Have a healthy ecosystem and then they work out exactly how much is coming in. So in this case, 69 tons per annum from the Serpentine, 16 from the Murray, and 61 from the Harvey. So you can see that.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1671.6000000000001_1689.5000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1671.6000000000001_1689.5000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1671.6000000000001_1689.5000000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1671.6000000000001_1689.5000000000002#3.jpg" ], "ocr_qwen2_vl_72b": "**Load reduction targets**\n\n**Table 3: Load reduction targets by main catchments and critical reporting catchments (courtesy Christian Zammit)**\n\n| Catchment | Total Phosphorus load target to estuary (t/yr) | Estimated current Total Phosphorus winter load (t/yr) | Estimated load reductions required to meet 0.1 mg/L (%) |\n|----------------------------------|-----------------------------------------------|-------------------------------------------------------|---------------------------------------------------------|\n| **Total Serpentine** | 21 | 69 | 60 |\n| Dirk Brook Yangedi | - | - | - |\n| Nambeelup Brook | - | - | - |\n| Peel Main Drain | - | - | - |\n| Ungauged Lower Serpentine | - | - | - |\n| **Total Murray** | 16 | 16 | - |\n| Dandalup, Upper and Lower Murray | - | - | - |\n| **Total Harvey** | 38 | 61 | 40 |\n| Coastal Western Harvey | - | - | - |\n| Coastal Central | - | - | - |\n| East Harvey Peel Drains | - | - | - |\n| **Grand Total** | 75 | 145 | 48 |" }, { "vid": "SRjrUb4jkd4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Oceanography tutorial on Marine Pollution focusing on Eutrophication_30.json#####audio#####doingASR#####FinishASR/SRjrUb4jkd4/1689.5000000000002_1708.8.mp4", "refined_asr": " It's the Serpentine River which is in the very top of the catchment and the Harvey River which is right at the bottom of the Peel Harvey They're the ones where the most activity needs to be done to reduce those nutrient loads So in this case it's the Serpentine up here coming in sort of", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1689.5000000000002_1708.8#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1689.5000000000002_1708.8#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1689.5000000000002_1708.8#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1689.5000000000002_1708.8#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1689.5000000000002_1708.8#3.jpg" ], "ocr_qwen2_vl_72b": "**Load reduction targets**\n\n**Table 3: Load reduction targets by main catchments and critical reporting catchments (courtesy Christian Zammit)**\n\n| Catchment | Total Phosphorus load target to estuary (t/yr) | Estimated current Total Phosphorus winter load (t/yr) | Estimated load reductions required to meet 0.1 mg/L (%) |\n|-----------------------------------|-----------------------------------------------|------------------------------------------------------|--------------------------------------------------------|\n| Total Serpentine | 21 | 69 | 60 |\n| Dirk Brook Yangedi | - | - | - |\n| Nambeelup Brook | - | - | - |\n| Peel Main Drain | - | - | - |\n| Ungauged Lower Serpentine | - | - | - |\n| Total Murray | 16 | 16 | - |\n| Dandalup, Upper and Lower Murray | - | - | - |\n| Total Harvey | 38 | 61 | 40 |\n| Coastal Western Harvey | - | - | - |\n| Coastal Central | - | - | - |\n| East Harvey Peel Drains | - | - | - |\n| Grand Total | 75 | 145 | 48 |" }, { "vid": "SRjrUb4jkd4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Oceanography tutorial on Marine Pollution focusing on Eutrophication_30.json#####audio#####doingASR#####FinishASR/SRjrUb4jkd4/1708.8_1738.24.mp4", "refined_asr": " Just to the east of Mandra and the Harvey River at the bottom that needs to have the most action to reduce the nutrient loads. So those load reductions can be done through a variety of ways: through educating the farmers, the graziers, and the residential population, to reduce the nutrient use by planting.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1708.8_1738.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1708.8_1738.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1708.8_1738.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1708.8_1738.24#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1708.8_1738.24#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1708.8_1738.24#5.jpg" ], "ocr_qwen2_vl_72b": "**Load reduction targets**\n\n**Table 3: Load reduction targets by catchment (courtesy Christian Zammit)**\n\n| Catchment | Estimated load reductions required to meet 0.1 mg/L (%) |\n|----------------------------|---------------------------------------------------------|\n| Total Serpentine | 60 |\n| Dirk Brook Yangedi | 82 |\n| Nambeelup Brook | 78 |\n| Peel Main Drain | 66 |\n| Ungauged Lower Serpentine | 60 |\n| Total Murray | - |\n| Dandalup, Upper and Lower Murray | - |\n| Total Harvey | 40 |\n| Coastal Western Harvey | 83 |\n| Coastal Central | 79 |\n| East Harvey Peel Drains | 47 |\n| Grand Total | 48 |\n\nSERPENTINE\nMURRAY\nHARVEY" }, { "vid": "SRjrUb4jkd4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Oceanography tutorial on Marine Pollution focusing on Eutrophication_30.json#####audio#####doingASR#####FinishASR/SRjrUb4jkd4/1738.24_1768.8.mp4", "refined_asr": " The vegetation along the streams and rivers can take up the nutrients before they actually reach the wetlands. So there's a variety of things that can be done in that space. Another way to manage eutrophication is by actually pumping oxygen into the waterways. This is something that's done at a number of places in the Southwest.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1738.24_1768.8#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1738.24_1768.8#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1738.24_1768.8#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1738.24_1768.8#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1738.24_1768.8#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1738.24_1768.8#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SRjrUb4jkd4/SRjrUb4jkd4@1738.24_1768.8#5.jpg" ], "ocr_qwen2_vl_72b": "**Load reduction targets**\n\n- **Table 3: Load reduction targets by catchment (courtesy Christian Zammit)**\n - **Catchment**\n - Total Serpentine\n - Dirk Brook Yangedi\n - Nambeelup Brook\n - Peel Main Drain\n - Ungauged Lower Serpentine\n - Total Murray\n - Dandalup, Upper and Lower Murray\n - Total Harvey\n - Coastal Western Harvey\n - Coastal Central\n - East Harvey Peel Drains\n - Grand Total\n\n- **SERPENTINE**\n - **MURRAY**\n - **HARVEY**\n\n- **Estimated load reductions required to meet 0.1 mg/L (%)**\n - Total Serpentine: 60\n - Dirk Brook Yangedi: 82\n - Nambeelup Brook: 78\n - Peel Main Drain: 66\n - Ungauged Lower Serpentine: 60\n - Total Murray: -\n - Dandalup, Upper and Lower Murray: -\n - Total Harvey: 40\n - Coastal Western Harvey: 83\n - Coastal Central: 79\n - East Harvey Peel Drains: 47\n - Grand Total: 48" } ], "image_num": 7, "text_num": 370, "token_num": 4402 }, { "images": [ "sample_100_images/TglD4Y6lmQk@813.1_840.46#1.jpg", "sample_100_images/TglD4Y6lmQk@813.1_840.46#2.jpg", "sample_100_images/TglD4Y6lmQk@813.1_840.46#3.jpg", "sample_100_images/TglD4Y6lmQk@813.1_840.46#4.jpg", null, "sample_100_images/TglD4Y6lmQk@840.46_861.26#1.jpg", "sample_100_images/TglD4Y6lmQk@840.46_861.26#2.jpg", "sample_100_images/TglD4Y6lmQk@840.46_861.26#3.jpg", null, "sample_100_images/TglD4Y6lmQk@861.26_889.26#1.jpg", "sample_100_images/TglD4Y6lmQk@861.26_889.26#2.jpg", "sample_100_images/TglD4Y6lmQk@861.26_889.26#3.jpg", "sample_100_images/TglD4Y6lmQk@861.26_889.26#4.jpg", null, "sample_100_images/TglD4Y6lmQk@889.26_925.5799999999999#1.jpg", "sample_100_images/TglD4Y6lmQk@889.26_925.5799999999999#2.jpg", "sample_100_images/TglD4Y6lmQk@889.26_925.5799999999999#3.jpg", "sample_100_images/TglD4Y6lmQk@889.26_925.5799999999999#4.jpg", "sample_100_images/TglD4Y6lmQk@889.26_925.5799999999999#5.jpg", null, "sample_100_images/TglD4Y6lmQk@928.4599999999999_945.66#1.jpg", "sample_100_images/TglD4Y6lmQk@928.4599999999999_945.66#2.jpg", null ], "texts": [ null, null, null, null, "\nof the function f of x. Yeah, well, the limit as x approaches two means we don't care about the value when x is exactly two, right? So the answer is not negative one, but rather, you can see how the curve is approaching the y-value of 1.5 when we come from the right. And if we approach from the left, the curve is also approaching the same y-value.", null, null, null, " 1.5. It's pretty much the Y value of this open circle especially when you have the left and right endpoints they're approaching to the same open circle. So it's that 1.5. And in this particular case you see how originally we have a curve but like we're just", null, null, null, null, " Missing that, like that little hole right here is actually called the removable discontinuity. So, removable discontinuity; and the reason it's called removable is that if you just fill in the open circle, guess what? The curve right here would be continuous. That's why it's called removable, meaning you can just fix it easily.", null, null, null, null, null, " You can fill in the circle and then no more discontinuous business. All right. Number three: Limit as x approaches infinity. We have a horizontal asymptote here, y is two, right? So it's two. And again, here it tells us horizontal, as though y equals two. And then, last question: Limit as x approaches negative infinity.", null, null, " All the way to the left, the Y, as you can see from the picture on the file, is approaching negative two. And guess what? If the question is asking us to find the horizontal asymptotes, we will indeed have two horizontal asymptotes. This is the first one." ], "text_ocr_list": [ null, null, null, null, "We can see these text from the image: - \\( f(2) = -1 \\)\n- \\(\\lim_{{x \\to 2}} f(x)\\).\n\nof the function f of x. Yeah, well, the limit as x approaches two means we don't care about the value when x is exactly two, right? So the answer is not negative one, but rather, you can see how the curve is approaching the y-value of 1.5 when we come from the right. And if we approach from the left, the curve is also approaching the same y-value.", null, null, null, "We can see these text from the image: - \\( f(2) = -1 \\)\n- \\(\\lim_{{x \\to 2}} f(x) =\\).\n 1.5. It's pretty much the Y value of this open circle especially when you have the left and right endpoints they're approaching to the same open circle. So it's that 1.5. And in this particular case you see how originally we have a curve but like we're just", null, null, null, null, "We can see these text from the image: - \\( f(2) = -1 \\)\n- \\(\\lim_{{x \\to 2}} f(x) = 1.5\\).\n Missing that, like that little hole right here is actually called the removable discontinuity. So, removable discontinuity; and the reason it's called removable is that if you just fill in the open circle, guess what? The curve right here would be continuous. That's why it's called removable, meaning you can just fix it easily.", null, null, null, null, null, "We can see these text from the image: - \\( f(2) = -1 \\)\n- \\(\\lim_{{x \\to 2}} f(x) = 1.5\\)\n- removable discontinuity.\n You can fill in the circle and then no more discontinuous business. All right. Number three: Limit as x approaches infinity. We have a horizontal asymptote here, y is two, right? So it's two. And again, here it tells us horizontal, as though y equals two. And then, last question: Limit as x approaches negative infinity.", null, null, "We can see these text from the image: - \\( f(2) \\)\n- \\(\\lim_{x \\to 2}\\)\n- removable\n- \\(\\lim_{x \\to -\\infty}\\)\n- \\(\\frac{0}{0}\\)\n- \\(\\frac{\\infty}{\\infty}\\).\n All the way to the left, the Y, as you can see from the picture on the file, is approaching negative two. And guess what? If the question is asking us to find the horizontal asymptotes, we will indeed have two horizontal asymptotes. This is the first one." ], "metadata": [ { "vid": "TglD4Y6lmQk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on Asymptotes and Graphical Limits in Differential Calculus_30.json#####audio#####doingASR#####FinishASR/TglD4Y6lmQk/813.1_840.46.mp4", "refined_asr": "\nof the function f of x. Yeah, well, the limit as x approaches two means we don't care about the value when x is exactly two, right? So the answer is not negative one, but rather, you can see how the curve is approaching the y-value of 1.5 when we come from the right. And if we approach from the left, the curve is also approaching the same y-value.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@813.1_840.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@813.1_840.46#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@813.1_840.46#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@813.1_840.46#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@813.1_840.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@813.1_840.46#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@813.1_840.46#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@813.1_840.46#4.jpg" ], "ocr_qwen2_vl_72b": "- \\( f(2) = -1 \\)\n- \\(\\lim_{{x \\to 2}} f(x)\\)" }, { "vid": "TglD4Y6lmQk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on Asymptotes and Graphical Limits in Differential Calculus_30.json#####audio#####doingASR#####FinishASR/TglD4Y6lmQk/840.46_861.26.mp4", "refined_asr": " 1.5. It's pretty much the Y value of this open circle especially when you have the left and right endpoints they're approaching to the same open circle. So it's that 1.5. And in this particular case you see how originally we have a curve but like we're just", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@840.46_861.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@840.46_861.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@840.46_861.26#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@840.46_861.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@840.46_861.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@840.46_861.26#3.jpg" ], "ocr_qwen2_vl_72b": "- \\( f(2) = -1 \\)\n- \\(\\lim_{{x \\to 2}} f(x) =\\)" }, { "vid": "TglD4Y6lmQk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on Asymptotes and Graphical Limits in Differential Calculus_30.json#####audio#####doingASR#####FinishASR/TglD4Y6lmQk/861.26_889.26.mp4", "refined_asr": " Missing that, like that little hole right here is actually called the removable discontinuity. So, removable discontinuity; and the reason it's called removable is that if you just fill in the open circle, guess what? The curve right here would be continuous. That's why it's called removable, meaning you can just fix it easily.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@861.26_889.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@861.26_889.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@861.26_889.26#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@861.26_889.26#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@861.26_889.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@861.26_889.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@861.26_889.26#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@861.26_889.26#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@861.26_889.26#5.jpg" ], "ocr_qwen2_vl_72b": "- \\( f(2) = -1 \\)\n- \\(\\lim_{{x \\to 2}} f(x) = 1.5\\)" }, { "vid": "TglD4Y6lmQk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on Asymptotes and Graphical Limits in Differential Calculus_30.json#####audio#####doingASR#####FinishASR/TglD4Y6lmQk/889.26_925.5799999999999.mp4", "refined_asr": " You can fill in the circle and then no more discontinuous business. All right. Number three: Limit as x approaches infinity. We have a horizontal asymptote here, y is two, right? So it's two. And again, here it tells us horizontal, as though y equals two. And then, last question: Limit as x approaches negative infinity.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#5.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@889.26_925.5799999999999#6.jpg" ], "ocr_qwen2_vl_72b": "- \\( f(2) = -1 \\)\n- \\(\\lim_{{x \\to 2}} f(x) = 1.5\\)\n- removable discontinuity" }, { "vid": "TglD4Y6lmQk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus Tutorial on Asymptotes and Graphical Limits in Differential Calculus_30.json#####audio#####doingASR#####FinishASR/TglD4Y6lmQk/928.4599999999999_945.66.mp4", "refined_asr": " All the way to the left, the Y, as you can see from the picture on the file, is approaching negative two. And guess what? If the question is asking us to find the horizontal asymptotes, we will indeed have two horizontal asymptotes. This is the first one.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@928.4599999999999_945.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@928.4599999999999_945.66#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@928.4599999999999_945.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@928.4599999999999_945.66#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/TglD4Y6lmQk/TglD4Y6lmQk@928.4599999999999_945.66#3.jpg" ], "ocr_qwen2_vl_72b": "- \\( f(2) \\)\n- \\(\\lim_{x \\to 2}\\)\n- removable\n- \\(\\lim_{x \\to -\\infty}\\)\n- \\(\\frac{0}{0}\\)\n- \\(\\frac{\\infty}{\\infty}\\)" } ], "image_num": 18, "text_num": 373, "token_num": 10741 }, { "images": [ "sample_100_images/-B5CjyJOkLg@2177.18_2211.82#1.jpg", "sample_100_images/-B5CjyJOkLg@2177.18_2211.82#2.jpg", "sample_100_images/-B5CjyJOkLg@2177.18_2211.82#3.jpg", "sample_100_images/-B5CjyJOkLg@2177.18_2211.82#5.jpg", null, "sample_100_images/-B5CjyJOkLg@2211.82_2240.26#1.jpg", "sample_100_images/-B5CjyJOkLg@2211.82_2240.26#3.jpg", "sample_100_images/-B5CjyJOkLg@2211.82_2240.26#4.jpg", null, "sample_100_images/-B5CjyJOkLg@2240.26_2250.2000000000003#1.jpg", null, "sample_100_images/-B5CjyJOkLg@2250.2200000000003_2271.26#1.jpg", "sample_100_images/-B5CjyJOkLg@2250.2200000000003_2271.26#2.jpg", null, "sample_100_images/-B5CjyJOkLg@2271.26_2295.1400000000003#1.jpg", null, "sample_100_images/-B5CjyJOkLg@2295.1400000000003_2315.24#1.jpg", null ], "texts": [ null, null, null, null, " Lower emission scenarios. And here is another video from NASA which hopefully will work. And you can see how that then evolves. So you can see in this, this is just a very short video. But you can see at this very high resolution, the ice loss at 8.5, moving very fast along this, particularly in this central west. And", null, null, null, " Southwest coast here. So if you see it again, you can see these are these outlet glaciers here. And you know, compared to where we were in AR5, you see, ice sheet modeling has advanced dramatically in that time. And so this is probably viewed as quite a sensitive model.", null, " It's 8.5. scenarios project a loss of up to about 33 centimeters from Greenland by 2100. So you know, it's significant.", null, null, " It's you know, it's perhaps could be viewed as a slightly more upper bound estimate. And the final thing I just wanted to touch on is the question about whether ice loss can be reversed from Greenland. And so there's an older paper here by Jeff Ridley, one of my colleagues at the", null, " The Hadley Centre In this paper they looked at the ice loss and then tried to determine if it would recover. So you see here that as time continues, this represents the ice loss. If they stop at a particular point, the question is whether the ice will recover or not.", null, " So this is the fraction of the initial ice sheet volume. You can see that up to a certain point, you can still recover the ice sheet. But once you've passed the threshold, it seems you can no longer return to the original ice sheet volume." ], "text_ocr_list": [ null, null, null, null, "We can see these text from the image: Met Office Hadley Centre\n\nHigh resolution projections of Greenland\n\nhttps://svs.gsfc.nasa.gov/vis/a000000/a004700/a04727/Greenland_RCP_85_2008_2300_comp_080p30.mp4\n\nProjects up to 33cm by 2100\n\nAschwanden et al., 2019.\n Lower emission scenarios. And here is another video from NASA which hopefully will work. And you can see how that then evolves. So you can see in this, this is just a very short video. But you can see at this very high resolution, the ice loss at 8.5, moving very fast along this, particularly in this central west. And", null, null, null, "We can see these text from the image: Year: 2046\n\nIce Sheet Velocity Magnitude\n\n0 200 400 600 800 1000 meters / year.\n Southwest coast here. So if you see it again, you can see these are these outlet glaciers here. And you know, compared to where we were in AR5, you see, ice sheet modeling has advanced dramatically in that time. And so this is probably viewed as quite a sensitive model.", null, "We can see these text from the image: - https://svs.gsfc.nasa.gov/vis/a000000/a004700/a04727/Greenland_RCP_85_2008_2300_comp_1080p30.mp4\n- Projects up to 33cm by 2100\nAschwanden et al., 2019.\n It's 8.5. scenarios project a loss of up to about 33 centimeters from Greenland by 2100. So you know, it's significant.", null, null, "We can see these text from the image: - https://svs.gsfc.nasa.gov/vis/a000000/a004700/a04727/Greenland_RCP85_2008_2300_comp_1080p30.mp4\n- Projects up to 33cm by 2100\n- Aschwanden et al., 2019.\n It's you know, it's perhaps could be viewed as a slightly more upper bound estimate. And the final thing I just wanted to touch on is the question about whether ice loss can be reversed from Greenland. And so there's an older paper here by Jeff Ridley, one of my colleagues at the", null, "We can see these text from the image: Can the ice loss be reversed?\n\n\u2022 Loss depends on global temperature change (colour gives global warming)\n\n\u2022 When reverted to a 20th century climate, ice sheet does not always recover\n\n\u2022 Beyond a size threshold the Greenland ice sheet is irreversible and there is committed sea level rise\n\nGregory et al., 2020\nRidley et al., 2010.\n The Hadley Centre In this paper they looked at the ice loss and then tried to determine if it would recover. So you see here that as time continues, this represents the ice loss. If they stop at a particular point, the question is whether the ice will recover or not.", null, "We can see these text from the image: - Can the ice loss be reversed?\n- Large and irreversible future decline of the Greenland ice-sheet\n- Loss depends on global temperature change (colour gives global warming)\n- When reverted to a 20th century climate, ice sheet does not always recover\n- Beyond a size threshold the Greenland ice sheet is irreversible and there is committed sea level rise\n- Gregory et al., 2020\n- Ridley et al., 2010.\n So this is the fraction of the initial ice sheet volume. You can see that up to a certain point, you can still recover the ice sheet. But once you've passed the threshold, it seems you can no longer return to the original ice sheet volume." ], "metadata": [ { "vid": "-B5CjyJOkLg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Climate System Components focusing on the Cryosphere_30.json#####audio#####doingASR#####FinishASR/-B5CjyJOkLg/2177.18_2211.82.mp4", "refined_asr": " Lower emission scenarios. And here is another video from NASA which hopefully will work. And you can see how that then evolves. So you can see in this, this is just a very short video. But you can see at this very high resolution, the ice loss at 8.5, moving very fast along this, particularly in this central west. And", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2177.18_2211.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2177.18_2211.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2177.18_2211.82#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2177.18_2211.82#5.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2177.18_2211.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2177.18_2211.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2177.18_2211.82#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2177.18_2211.82#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2177.18_2211.82#5.jpg" ], "ocr_qwen2_vl_72b": "Met Office Hadley Centre\n\nHigh resolution projections of Greenland\n\nhttps://svs.gsfc.nasa.gov/vis/a000000/a004700/a04727/Greenland_RCP_85_2008_2300_comp_080p30.mp4\n\nProjects up to 33cm by 2100\n\nAschwanden et al., 2019" }, { "vid": "-B5CjyJOkLg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Climate System Components focusing on the Cryosphere_30.json#####audio#####doingASR#####FinishASR/-B5CjyJOkLg/2211.82_2240.26.mp4", "refined_asr": " Southwest coast here. So if you see it again, you can see these are these outlet glaciers here. And you know, compared to where we were in AR5, you see, ice sheet modeling has advanced dramatically in that time. And so this is probably viewed as quite a sensitive model.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2211.82_2240.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2211.82_2240.26#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2211.82_2240.26#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2211.82_2240.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2211.82_2240.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2211.82_2240.26#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2211.82_2240.26#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2211.82_2240.26#5.jpg" ], "ocr_qwen2_vl_72b": "Year: 2046\n\nIce Sheet Velocity Magnitude\n\n0 200 400 600 800 1000 meters / year" }, { "vid": "-B5CjyJOkLg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Climate System Components focusing on the Cryosphere_30.json#####audio#####doingASR#####FinishASR/-B5CjyJOkLg/2240.26_2250.2000000000003.mp4", "refined_asr": " It's 8.5. scenarios project a loss of up to about 33 centimeters from Greenland by 2100. So you know, it's significant.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2240.26_2250.2000000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2240.26_2250.2000000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2240.26_2250.2000000000003#2.jpg" ], "ocr_qwen2_vl_72b": "- https://svs.gsfc.nasa.gov/vis/a000000/a004700/a04727/Greenland_RCP_85_2008_2300_comp_1080p30.mp4\n- Projects up to 33cm by 2100\nAschwanden et al., 2019" }, { "vid": "-B5CjyJOkLg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Climate System Components focusing on the Cryosphere_30.json#####audio#####doingASR#####FinishASR/-B5CjyJOkLg/2250.2200000000003_2271.26.mp4", "refined_asr": " It's you know, it's perhaps could be viewed as a slightly more upper bound estimate. And the final thing I just wanted to touch on is the question about whether ice loss can be reversed from Greenland. And so there's an older paper here by Jeff Ridley, one of my colleagues at the", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2250.2200000000003_2271.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2250.2200000000003_2271.26#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2250.2200000000003_2271.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2250.2200000000003_2271.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2250.2200000000003_2271.26#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2250.2200000000003_2271.26#4.jpg" ], "ocr_qwen2_vl_72b": "- https://svs.gsfc.nasa.gov/vis/a000000/a004700/a04727/Greenland_RCP85_2008_2300_comp_1080p30.mp4\n- Projects up to 33cm by 2100\n- Aschwanden et al., 2019" }, { "vid": "-B5CjyJOkLg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Climate System Components focusing on the Cryosphere_30.json#####audio#####doingASR#####FinishASR/-B5CjyJOkLg/2271.26_2295.1400000000003.mp4", "refined_asr": " The Hadley Centre In this paper they looked at the ice loss and then tried to determine if it would recover. So you see here that as time continues, this represents the ice loss. If they stop at a particular point, the question is whether the ice will recover or not.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2271.26_2295.1400000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2271.26_2295.1400000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2271.26_2295.1400000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2271.26_2295.1400000000003#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2271.26_2295.1400000000003#4.jpg" ], "ocr_qwen2_vl_72b": "Can the ice loss be reversed?\n\n\u2022 Loss depends on global temperature change (colour gives global warming)\n\n\u2022 When reverted to a 20th century climate, ice sheet does not always recover\n\n\u2022 Beyond a size threshold the Greenland ice sheet is irreversible and there is committed sea level rise\n\nGregory et al., 2020\nRidley et al., 2010" }, { "vid": "-B5CjyJOkLg.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Climate Science Tutorial on Climate System Components focusing on the Cryosphere_30.json#####audio#####doingASR#####FinishASR/-B5CjyJOkLg/2295.1400000000003_2315.24.mp4", "refined_asr": " So this is the fraction of the initial ice sheet volume. You can see that up to a certain point, you can still recover the ice sheet. But once you've passed the threshold, it seems you can no longer return to the original ice sheet volume.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2295.1400000000003_2315.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2295.1400000000003_2315.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2295.1400000000003_2315.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/-B5CjyJOkLg/-B5CjyJOkLg@2295.1400000000003_2315.24#3.jpg" ], "ocr_qwen2_vl_72b": "- Can the ice loss be reversed?\n- Large and irreversible future decline of the Greenland ice-sheet\n- Loss depends on global temperature change (colour gives global warming)\n- When reverted to a 20th century climate, ice sheet does not always recover\n- Beyond a size threshold the Greenland ice sheet is irreversible and there is committed sea level rise\n- Gregory et al., 2020\n- Ridley et al., 2010" } ], "image_num": 12, "text_num": 363, "token_num": 7275 }, { "images": [ "sample_100_images/pnw3E-1iuLY@1921.82_1942.82#1.jpg", "sample_100_images/pnw3E-1iuLY@1921.82_1942.82#2.jpg", null, "sample_100_images/pnw3E-1iuLY@1942.82_1964.82#1.jpg", null, "sample_100_images/pnw3E-1iuLY@1964.82_1983.82#1.jpg", null, "sample_100_images/pnw3E-1iuLY@1983.82_2004.82#1.jpg", null, "sample_100_images/pnw3E-1iuLY@2004.82_2023.82#1.jpg", null, "sample_100_images/pnw3E-1iuLY@2023.82_2042.82#1.jpg", null, "sample_100_images/pnw3E-1iuLY@2042.82_2060.82#1.jpg", null ], "texts": [ null, null, " This means that my y sub p will have to have this form here and my general solution will have the form here. And what's significant is that I still have the sinusoidal, the oscillatory behavior that the natural system demonstrated. But this t is acting like", null, " A magnifier a scaler it's acting like the amplitude on these trig functions. And so as time increases whereas the natural behavior of the system demonstrates steady oscillation with no increase or decrease in the amplitude these new terms out here on the end", null, " The amplitude of the oscillation introduced due to the forcing function will have an increasing amplitude. Specifically, because as time increases, the amplitude will increase linearly with time. So after two seconds, the amplitude will be twice as big. After three seconds, three times as big, and so on.", null, " So this causes an otherwise stable system to grow without bound. A system which naturally has bounded behavior now has unbounded behavior. And it's worth pointing out that we made no mention of the amplitude of the forcing function. At no point did I indicate.", null, " That g of t needed to have a really large magnitude. The magnitude, the amplitude of g of t was not significant. The only thing that was significant is that the frequency of g of t perfectly matched the natural behavior of the system. When that occurs.", null, " That leads to this phenomenon of resonance, which turns a bounded system into an unbounded system. I always say my favorite example of this is when you were jumping on the trampoline with your little brother or sister and your little brother or sister was jumping at some specific frequency.", null, " And you showed up and you started jumping. And so long as you and your little sibling were jumping at different frequencies, your amplitudes - how high you were jumping - was relatively unaffected. But if your timing was just right and you could make your frequency match," ], "text_ocr_list": [ null, null, "We can see these text from the image: EX 3. Consider the potential influence of a sinusoidal forcing function, say \\( g(t) = \\alpha \\sin(\\kappa t) \\) or \\( g(t) = \\beta \\cos(\\kappa t) \\) on solutions of \\( mx'' + \\gamma x' + kx = g(t) \\).\n\nAssume \\( \\gamma = 0 \\), no friction\n\n\\( mx'' + kx = g(t) \\)\n\n\\( m r^2 + k = 0 \\)\n\n\\( r^2 = -\\frac{k}{m} \\)\n\n\\( r = \\pm \\sqrt{\\frac{k}{m}} i \\)\n\n\\( y_h = C_1 \\cos(\\sqrt{\\frac{k}{m}} t) + C_2 \\sin(\\sqrt{\\frac{k}{m}} t) \\)\n\nAssume \\( k = \\sqrt{\\frac{k}{m}} \\)\n\nIf the external force perfectly matches the natural period/frequency/wavelength of the system\n\n\\( y_p = A \\cos(\\kappa t) + B \\sin(\\kappa t) \\)\n\n\\( y_p = t(A \\cos(\\kappa t) + B \\sin(\\kappa t)) \\).\n This means that my y sub p will have to have this form here and my general solution will have the form here. And what's significant is that I still have the sinusoidal, the oscillatory behavior that the natural system demonstrated. But this t is acting like", null, "We can see these text from the image: - Assume \\( g = 0 \\)\n- \\( mX'' + kX = g(t) \\)\n- \\( mr^2 + k = 0 \\)\n- \\( r^2 = -\\frac{k}{m} \\)\n- \\( r = \\pm \\sqrt{\\frac{k}{m}}i \\)\n\n- \\( y_h = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) \\)\n\n- Assume \\( k = \\sqrt{\\frac{k}{m}} \\)\n\n- The external force perfectly matches the natural period/frequency/wavelength of the system\n\n- \\( y_p = A \\cos(kt) + B \\sin(kt) \\)\n- \\( y_p = t(A \\cos(kt) + B \\sin(kt)) \\)\n\n- \\( y = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) + t(A \\cos(kt) + B \\sin(kt)) \\).\n A magnifier a scaler it's acting like the amplitude on these trig functions. And so as time increases whereas the natural behavior of the system demonstrates steady oscillation with no increase or decrease in the amplitude these new terms out here on the end", null, "We can see these text from the image: - Assume g = 0, no gravity\n\n- mx'' + kx = glt\n\n- mr^2 + k = 0\n\n- r^2 = -k/m\n\n- r = \u00b1 \u221a(k/m) i\n\n- y_h = C_1 cos(\u221a(k/m)t) + C_2 sin(\u221a(k/m)t)\n\n- Assume k = \u221a(k/m)\n\n- Assume the external force perfectly matches the natural period/frequency/wavelength of the system\n\n- y_p = A cos(kt) + B sin(kt)\n\n- y_p = t(A cos(kt) + B sin(kt))\n\n- y = C_1 cos(\u221a(k/m)t) + C_2 sin(\u221a(k/m)t) + t(A cos(kt) + B sin(kt)).\n The amplitude of the oscillation introduced due to the forcing function will have an increasing amplitude. Specifically, because as time increases, the amplitude will increase linearly with time. So after two seconds, the amplitude will be twice as big. After three seconds, three times as big, and so on.", null, "We can see these text from the image: Assume g = 0, no gravity\n\nmx'' + kx = gl(t)\n\nmr^2 + k = 0\n\nr^2 = -k/m\n\nr = \u00b1 \u221a(k/m) i\n\ny_h = C_1 cos(\u221a(k/m)t) + C_2 sin(\u221a(k/m)t)\n\nAssume k = \u221a(k/m)\n\nLTHE EXTERNAL FORCE PERFECTLY MATCHES THE NATURAL PERIOD/FREQUENCY/WAVELENGTH OF THE SYSTEM\n\ny_p = A cos(kt) + B sin(kt)\n\ny_p = t(A cos(kt) + B sin(kt))\n\ny = C_1 cos(\u221a(k/m)t) + C_2 sin(\u221a(k/m)t) + t(A cos(kt) + B sin(kt)).\n So this causes an otherwise stable system to grow without bound. A system which naturally has bounded behavior now has unbounded behavior. And it's worth pointing out that we made no mention of the amplitude of the forcing function. At no point did I indicate.", null, "We can see these text from the image: - Assume \\( g = 0 \\)\n- \\( mX'' + kX = glt \\)\n- \\( mr^2 + k = 0 \\)\n- \\( r^2 = -\\frac{k}{m} \\)\n- \\( r = \\pm \\sqrt{\\frac{k}{m}}i \\)\n- \\( y_h = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) \\)\n- Assume \\( k = \\sqrt{\\frac{k}{m}} \\)\n- The external force perfectly matches the natural period/frequency/wavelength of the system\n- \\( y_p = A \\cos(kt) + B \\sin(kt) \\)\n- \\( y_p = t(A \\cos(kt) + B \\sin(kt)) \\)\n- \\( y = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) + t(A \\cos(kt) + B \\sin(kt)) \\)\n- Solutions grow w/o bound.\n That g of t needed to have a really large magnitude. The magnitude, the amplitude of g of t was not significant. The only thing that was significant is that the frequency of g of t perfectly matched the natural behavior of the system. When that occurs.", null, "We can see these text from the image: - Assume g = 0, no friction\n- mx'' + kx = glt)\n- mr^2 + k = 0\n- r^2 = -k/m\n- r = \u00b1 \u221a(k/m) i\n\n- y_h = C_1 cos(\u221a(k/m) t) + C_2 sin(\u221a(k/m) t)\n\n- Assume K = \u221a(k/m)\n\n- LTHE EXTERNAL FORCE PERFECTLY MATCHES THE NATURAL PERIOD/FREQUENCY/WAVELENGTH OF THE SYSTEM\n\n- y_p = A cos(Kt) + B sin(Kt)\n- y_p = t (A cos(Kt) + B sin(Kt))\n\n- y = C_1 cos(\u221a(k/m) t) + C_2 sin(\u221a(k/m) t) + t (A cos(Kt) + B sin(Kt))\n- SOLUTIONS GROW W/O BOUND.\n That leads to this phenomenon of resonance, which turns a bounded system into an unbounded system. I always say my favorite example of this is when you were jumping on the trampoline with your little brother or sister and your little brother or sister was jumping at some specific frequency.", null, "We can see these text from the image: - Assume \\( g = 0 \\), no friction\n\n\\[ mX'' + kX = g(t) \\]\n\n\\[ mr^2 + k = 0 \\]\n\n\\[ r^2 = -\\frac{k}{m} \\]\n\n\\[ r = \\pm \\sqrt{\\frac{k}{m}}i \\]\n\n\\[ y_h = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) \\]\n\nAssume \\( k = \\sqrt{\\frac{k}{m}} \\)\n\nLet the external force perfectly matches the natural period/frequency/wavelength of the system\n\n\\[ y_p = A \\cos(kt) + B \\sin(kt) \\]\n\n\\[ y_p = t(A \\cos(kt) + B \\sin(kt)) \\]\n\n\\[ y = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) + t(A \\cos(kt) + B \\sin(kt)) \\]\n\nSolutions grow w/o bound.\n And you showed up and you started jumping. And so long as you and your little sibling were jumping at different frequencies, your amplitudes - how high you were jumping - was relatively unaffected. But if your timing was just right and you could make your frequency match," ], "metadata": [ { "vid": "pnw3E-1iuLY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Applications of Second Order Differential Equations to Mechanical Vibrations_30.json#####audio#####doingASR#####FinishASR/pnw3E-1iuLY/1921.82_1942.82.mp4", "refined_asr": " This means that my y sub p will have to have this form here and my general solution will have the form here. And what's significant is that I still have the sinusoidal, the oscillatory behavior that the natural system demonstrated. But this t is acting like", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1921.82_1942.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1921.82_1942.82#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1921.82_1942.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1921.82_1942.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1921.82_1942.82#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1921.82_1942.82#4.jpg" ], "ocr_qwen2_vl_72b": "EX 3. Consider the potential influence of a sinusoidal forcing function, say \\( g(t) = \\alpha \\sin(\\kappa t) \\) or \\( g(t) = \\beta \\cos(\\kappa t) \\) on solutions of \\( mx'' + \\gamma x' + kx = g(t) \\).\n\nAssume \\( \\gamma = 0 \\), no friction\n\n\\( mx'' + kx = g(t) \\)\n\n\\( m r^2 + k = 0 \\)\n\n\\( r^2 = -\\frac{k}{m} \\)\n\n\\( r = \\pm \\sqrt{\\frac{k}{m}} i \\)\n\n\\( y_h = C_1 \\cos(\\sqrt{\\frac{k}{m}} t) + C_2 \\sin(\\sqrt{\\frac{k}{m}} t) \\)\n\nAssume \\( k = \\sqrt{\\frac{k}{m}} \\)\n\nIf the external force perfectly matches the natural period/frequency/wavelength of the system\n\n\\( y_p = A \\cos(\\kappa t) + B \\sin(\\kappa t) \\)\n\n\\( y_p = t(A \\cos(\\kappa t) + B \\sin(\\kappa t)) \\)" }, { "vid": "pnw3E-1iuLY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Applications of Second Order Differential Equations to Mechanical Vibrations_30.json#####audio#####doingASR#####FinishASR/pnw3E-1iuLY/1942.82_1964.82.mp4", "refined_asr": " A magnifier a scaler it's acting like the amplitude on these trig functions. And so as time increases whereas the natural behavior of the system demonstrates steady oscillation with no increase or decrease in the amplitude these new terms out here on the end", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1942.82_1964.82#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1942.82_1964.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1942.82_1964.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1942.82_1964.82#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1942.82_1964.82#4.jpg" ], "ocr_qwen2_vl_72b": "- Assume \\( g = 0 \\)\n- \\( mX'' + kX = g(t) \\)\n- \\( mr^2 + k = 0 \\)\n- \\( r^2 = -\\frac{k}{m} \\)\n- \\( r = \\pm \\sqrt{\\frac{k}{m}}i \\)\n\n- \\( y_h = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) \\)\n\n- Assume \\( k = \\sqrt{\\frac{k}{m}} \\)\n\n- The external force perfectly matches the natural period/frequency/wavelength of the system\n\n- \\( y_p = A \\cos(kt) + B \\sin(kt) \\)\n- \\( y_p = t(A \\cos(kt) + B \\sin(kt)) \\)\n\n- \\( y = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) + t(A \\cos(kt) + B \\sin(kt)) \\)" }, { "vid": "pnw3E-1iuLY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Applications of Second Order Differential Equations to Mechanical Vibrations_30.json#####audio#####doingASR#####FinishASR/pnw3E-1iuLY/1964.82_1983.82.mp4", "refined_asr": " The amplitude of the oscillation introduced due to the forcing function will have an increasing amplitude. Specifically, because as time increases, the amplitude will increase linearly with time. So after two seconds, the amplitude will be twice as big. After three seconds, three times as big, and so on.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1964.82_1983.82#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1964.82_1983.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1964.82_1983.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1964.82_1983.82#3.jpg" ], "ocr_qwen2_vl_72b": "- Assume g = 0, no gravity\n\n- mx'' + kx = glt\n\n- mr^2 + k = 0\n\n- r^2 = -k/m\n\n- r = \u00b1 \u221a(k/m) i\n\n- y_h = C_1 cos(\u221a(k/m)t) + C_2 sin(\u221a(k/m)t)\n\n- Assume k = \u221a(k/m)\n\n- Assume the external force perfectly matches the natural period/frequency/wavelength of the system\n\n- y_p = A cos(kt) + B sin(kt)\n\n- y_p = t(A cos(kt) + B sin(kt))\n\n- y = C_1 cos(\u221a(k/m)t) + C_2 sin(\u221a(k/m)t) + t(A cos(kt) + B sin(kt))" }, { "vid": "pnw3E-1iuLY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Applications of Second Order Differential Equations to Mechanical Vibrations_30.json#####audio#####doingASR#####FinishASR/pnw3E-1iuLY/1983.82_2004.82.mp4", "refined_asr": " So this causes an otherwise stable system to grow without bound. A system which naturally has bounded behavior now has unbounded behavior. And it's worth pointing out that we made no mention of the amplitude of the forcing function. At no point did I indicate.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1983.82_2004.82#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1983.82_2004.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1983.82_2004.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1983.82_2004.82#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@1983.82_2004.82#4.jpg" ], "ocr_qwen2_vl_72b": "Assume g = 0, no gravity\n\nmx'' + kx = gl(t)\n\nmr^2 + k = 0\n\nr^2 = -k/m\n\nr = \u00b1 \u221a(k/m) i\n\ny_h = C_1 cos(\u221a(k/m)t) + C_2 sin(\u221a(k/m)t)\n\nAssume k = \u221a(k/m)\n\nLTHE EXTERNAL FORCE PERFECTLY MATCHES THE NATURAL PERIOD/FREQUENCY/WAVELENGTH OF THE SYSTEM\n\ny_p = A cos(kt) + B sin(kt)\n\ny_p = t(A cos(kt) + B sin(kt))\n\ny = C_1 cos(\u221a(k/m)t) + C_2 sin(\u221a(k/m)t) + t(A cos(kt) + B sin(kt))" }, { "vid": "pnw3E-1iuLY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Applications of Second Order Differential Equations to Mechanical Vibrations_30.json#####audio#####doingASR#####FinishASR/pnw3E-1iuLY/2004.82_2023.82.mp4", "refined_asr": " That g of t needed to have a really large magnitude. The magnitude, the amplitude of g of t was not significant. The only thing that was significant is that the frequency of g of t perfectly matched the natural behavior of the system. When that occurs.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2004.82_2023.82#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2004.82_2023.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2004.82_2023.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2004.82_2023.82#3.jpg" ], "ocr_qwen2_vl_72b": "- Assume \\( g = 0 \\)\n- \\( mX'' + kX = glt \\)\n- \\( mr^2 + k = 0 \\)\n- \\( r^2 = -\\frac{k}{m} \\)\n- \\( r = \\pm \\sqrt{\\frac{k}{m}}i \\)\n- \\( y_h = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) \\)\n- Assume \\( k = \\sqrt{\\frac{k}{m}} \\)\n- The external force perfectly matches the natural period/frequency/wavelength of the system\n- \\( y_p = A \\cos(kt) + B \\sin(kt) \\)\n- \\( y_p = t(A \\cos(kt) + B \\sin(kt)) \\)\n- \\( y = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) + t(A \\cos(kt) + B \\sin(kt)) \\)\n- Solutions grow w/o bound" }, { "vid": "pnw3E-1iuLY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Applications of Second Order Differential Equations to Mechanical Vibrations_30.json#####audio#####doingASR#####FinishASR/pnw3E-1iuLY/2023.82_2042.82.mp4", "refined_asr": " That leads to this phenomenon of resonance, which turns a bounded system into an unbounded system. I always say my favorite example of this is when you were jumping on the trampoline with your little brother or sister and your little brother or sister was jumping at some specific frequency.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2023.82_2042.82#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2023.82_2042.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2023.82_2042.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2023.82_2042.82#3.jpg" ], "ocr_qwen2_vl_72b": "- Assume g = 0, no friction\n- mx'' + kx = glt)\n- mr^2 + k = 0\n- r^2 = -k/m\n- r = \u00b1 \u221a(k/m) i\n\n- y_h = C_1 cos(\u221a(k/m) t) + C_2 sin(\u221a(k/m) t)\n\n- Assume K = \u221a(k/m)\n\n- LTHE EXTERNAL FORCE PERFECTLY MATCHES THE NATURAL PERIOD/FREQUENCY/WAVELENGTH OF THE SYSTEM\n\n- y_p = A cos(Kt) + B sin(Kt)\n- y_p = t (A cos(Kt) + B sin(Kt))\n\n- y = C_1 cos(\u221a(k/m) t) + C_2 sin(\u221a(k/m) t) + t (A cos(Kt) + B sin(Kt))\n- SOLUTIONS GROW W/O BOUND" }, { "vid": "pnw3E-1iuLY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Applications of Second Order Differential Equations to Mechanical Vibrations_30.json#####audio#####doingASR#####FinishASR/pnw3E-1iuLY/2042.82_2060.82.mp4", "refined_asr": " And you showed up and you started jumping. And so long as you and your little sibling were jumping at different frequencies, your amplitudes - how high you were jumping - was relatively unaffected. But if your timing was just right and you could make your frequency match,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2042.82_2060.82#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2042.82_2060.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2042.82_2060.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pnw3E-1iuLY/pnw3E-1iuLY@2042.82_2060.82#3.jpg" ], "ocr_qwen2_vl_72b": "- Assume \\( g = 0 \\), no friction\n\n\\[ mX'' + kX = g(t) \\]\n\n\\[ mr^2 + k = 0 \\]\n\n\\[ r^2 = -\\frac{k}{m} \\]\n\n\\[ r = \\pm \\sqrt{\\frac{k}{m}}i \\]\n\n\\[ y_h = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) \\]\n\nAssume \\( k = \\sqrt{\\frac{k}{m}} \\)\n\nLet the external force perfectly matches the natural period/frequency/wavelength of the system\n\n\\[ y_p = A \\cos(kt) + B \\sin(kt) \\]\n\n\\[ y_p = t(A \\cos(kt) + B \\sin(kt)) \\]\n\n\\[ y = C_1 \\cos(\\sqrt{\\frac{k}{m}}t) + C_2 \\sin(\\sqrt{\\frac{k}{m}}t) + t(A \\cos(kt) + B \\sin(kt)) \\]\n\nSolutions grow w/o bound" } ], "image_num": 8, "text_num": 397, "token_num": 5005 }, { "images": [ "sample_100_images/ftnpM_RO0Jc@2227.4_2247.4#1.jpg", "sample_100_images/ftnpM_RO0Jc@2227.4_2247.4#2.jpg", "sample_100_images/ftnpM_RO0Jc@2227.4_2247.4#3.jpg", null, "sample_100_images/ftnpM_RO0Jc@2248.4_2263.4#1.jpg", "sample_100_images/ftnpM_RO0Jc@2248.4_2263.4#2.jpg", null, "sample_100_images/ftnpM_RO0Jc@2263.4_2277.4#1.jpg", "sample_100_images/ftnpM_RO0Jc@2263.4_2277.4#2.jpg", null, "sample_100_images/ftnpM_RO0Jc@2277.4_2291.4#1.jpg", "sample_100_images/ftnpM_RO0Jc@2277.4_2291.4#2.jpg", null, "sample_100_images/ftnpM_RO0Jc@2291.4_2307.4#1.jpg", "sample_100_images/ftnpM_RO0Jc@2291.4_2307.4#2.jpg", null, "sample_100_images/ftnpM_RO0Jc@2307.4_2326.4#1.jpg", "sample_100_images/ftnpM_RO0Jc@2307.4_2326.4#2.jpg", "sample_100_images/ftnpM_RO0Jc@2307.4_2326.4#3.jpg", null, "sample_100_images/ftnpM_RO0Jc@2326.4_2339.4#1.jpg", "sample_100_images/ftnpM_RO0Jc@2326.4_2339.4#2.jpg", null, "sample_100_images/ftnpM_RO0Jc@2339.4_2355.4#1.jpg", "sample_100_images/ftnpM_RO0Jc@2339.4_2355.4#2.jpg", null ], "texts": [ null, null, null, " This is negative already. This is positive infinity. S has to be positive so that we can actually get negative infinity out of this. Right. So with that being said, S has to be greater than zero. So with that condition, we can actually say the first part.", null, null, " It has to be zero. And again, we need to have that condition. Anything that I write down right here is that we need to have that in order to make this work and work out the rest. Negative negative becomes positive. And then we just have that.", null, null, " That's pretty much it. So in the end, we have plus one over s times e to the negative s a. So finally, we will just write this down as e to the negative s a. And some people like to write this as e to the minus s a.", null, null, " And some people like to write this as e to the negative A S. It depends on how you want to write it but this is okay. And then over S. And again we have the condition so let's write that down.", null, null, " S has to be greater than zero in order for that to work. Just like that. So this right here is a unit step function. And now let me introduce you to something that's very similar to the unit step function, namely the window function.", null, null, null, " The pi sub a b t function. I should be able to do that right here. So I will do that right here. Number eight. So here we have the Laplace transform and inside is the capital pi a comma b.", null, null, " This is not the original pi function. It's just what I think is called the window function over there. Oh, by the way, this is also called the Heaviside function. But that's not because this side is Heaviside.", null, null, " That's why it's zero. Heaviside is just the name of the person. Anyway, I thought it wasn't funny. So we have pi a, b. And the input is just usually with a t, like this:" ], "text_ocr_list": [ null, null, null, "We can see these text from the image: 1. \\( \\mathcal{L}\\{U(t-a)\\} \\)\n2. \\( = \\int_{0}^{\\infty} e^{-st} U(t-a) dt \\)\n3. \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n4. \\( = \\left[ -\\frac{1}{s} e^{-st} \\right]_{a}^{\\infty} \\)\n5. \\( = \\left( \\frac{1}{s} e^{-s \\cdot \\infty} \\right) - \\left( \\frac{1}{s} e^{-sa} \\right) \\).\n This is negative already. This is positive infinity. S has to be positive so that we can actually get negative infinity out of this. Right. So with that being said, S has to be greater than zero. So with that condition, we can actually say the first part.", null, null, "We can see these text from the image: 1. \\( U(t-a) \\)\n2. \\( 0 \\) if \\( t-a < 0 \\)\n3. \\( 1 \\) if \\( t-a > 0 \\)\n4. \\( \\int_{0}^{\\infty} e^{-st} U(t-a) dt \\)\n5. \\( \\int_{a}^{\\infty} e^{-st} dt \\)\n6. \\( -\\frac{1}{s} e^{-st} \\bigg|_{a}^{\\infty} \\)\n7. \\( = \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 \\).\n It has to be zero. And again, we need to have that condition. Anything that I write down right here is that we need to have that in order to make this work and work out the rest. Negative negative becomes positive. And then we just have that.", null, null, "We can see these text from the image: - \\( L\\{U(t-a)\\} \\)\n- \\( U(t-a) = \\begin{cases} 0 & \\text{if } t-a < 0 \\\\ 1 & \\text{if } t-a > 0 \\end{cases} \\)\n- \\( = \\int_{0}^{\\infty} e^{-st} U(t-a) dt \\)\n- \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n- \\( = -\\frac{1}{s} e^{-st} \\bigg|_{a}^{\\infty} \\)\n- \\( = \\left( \\frac{1}{s} e^{-s \\cdot \\infty} \\right) - \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n- \\( = 0 + \\).\n That's pretty much it. So in the end, we have plus one over s times e to the negative s a. So finally, we will just write this down as e to the negative s a. And some people like to write this as e to the minus s a.", null, null, "We can see these text from the image: 1. \\( \\mathcal{L}\\{u(t-a)\\} \\)\n2. \\( u(t-a) = \\begin{cases} 0 & t < a \\\\ 1 & t \\geq a \\end{cases} \\)\n3. \\( = \\int_{0}^{\\infty} e^{-st} u(t-a) dt \\)\n4. \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n5. \\( = -\\frac{1}{s} e^{-st} \\Big|_{a}^{\\infty} \\)\n6. \\( = \\left( -\\frac{1}{s} e^{-s \\cdot \\infty} \\right) - \\left( -\\frac{1}{s} e^{-sa} \\right) \\)\n7. \\( = \\left( -\\frac{1}{s} \\cdot 0 \\right) + \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 + \\frac{1}{s} e^{-sa} \\)\n9. \\( = \\frac{1}{s} e^{-sa} \\).\n And some people like to write this as e to the negative A S. It depends on how you want to write it but this is okay. And then over S. And again we have the condition so let's write that down.", null, null, "We can see these text from the image: 1. \\( \\mathcal{L}\\{U(t-a)\\} \\)\n2. \\( U(t-a) = \\begin{cases} 0 & \\text{if } t < a \\\\ 1 & \\text{if } t \\geq a \\end{cases} \\)\n3. \\( = \\int_{0}^{\\infty} e^{-st} U(t-a) dt \\)\n4. \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n5. \\( = -\\frac{1}{s} e^{-st} \\bigg|_{a}^{\\infty} \\)\n6. \\( = \\left( -\\frac{1}{s} e^{-s \\cdot \\infty} \\right) - \\left( -\\frac{1}{s} e^{-sa} \\right) \\)\n7. \\( = \\left( \\frac{1}{s} e^{-s \\cdot \\infty} \\right) + \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 + \\frac{1}{s} e^{-sa} \\)\n9. \\( = \\frac{e^{-sa}}{s}, s > 0 \\).\n S has to be greater than zero in order for that to work. Just like that. So this right here is a unit step function. And now let me introduce you to something that's very similar to the unit step function, namely the window function.", null, null, null, "We can see these text from the image: 1. \\( \\mathcal{L}\\{u(t-a)\\} \\)\n2. \\( u(t-a) = \\begin{cases} 0 & t < a \\\\ 1 & t \\geq a \\end{cases} \\)\n3. \\( = \\int_{0}^{\\infty} e^{-st} u(t-a) dt \\)\n4. \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n5. \\( = \\left[ -\\frac{1}{s} e^{-st} \\right]_{a}^{\\infty} \\)\n6. \\( = \\left( -\\frac{1}{s} e^{-s(\\infty)} \\right) - \\left( -\\frac{1}{s} e^{-sa} \\right) \\)\n7. \\( = \\left( -\\frac{1}{s} \\cdot 0 \\right) + \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 + \\frac{1}{s} e^{-sa} \\)\n9. \\( = \\frac{e^{-sa}}{s}, s > 0 \\).\n The pi sub a b t function. I should be able to do that right here. So I will do that right here. Number eight. So here we have the Laplace transform and inside is the capital pi a comma b.", null, null, "We can see these text from the image: 1. \\( \\mathcal{L}\\{U(t-a)\\} \\)\n2. \\( U(t-a) = \\begin{cases} 0 & \\text{if } t - a < 0 \\\\ 1 & \\text{if } t - a > 0 \\end{cases} \\)\n3. \\( = \\int_{0}^{\\infty} e^{-st} U(t-a) dt \\)\n4. \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n5. \\( = \\left[ -\\frac{1}{s} e^{-st} \\right]_{a}^{\\infty} \\)\n6. \\( = \\left( -\\frac{1}{s} e^{-s \\cdot \\infty} \\right) - \\left( -\\frac{1}{s} e^{-sa} \\right) \\)\n7. \\( = \\left( \\frac{1}{s} e^{-s \\cdot \\infty} \\right) + \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 + \\frac{1}{s} e^{-sa} \\)\n9. \\( = \\frac{e^{-sa}}{s}, s > 0 \\).\n This is not the original pi function. It's just what I think is called the window function over there. Oh, by the way, this is also called the Heaviside function. But that's not because this side is Heaviside.", null, null, "We can see these text from the image: 1. \\( \\mathcal{L}\\{u(t-a)\\} \\)\n2. \\( u(t-a) = \\begin{cases} 0 & \\text{if } t < a \\\\ 1 & \\text{if } t \\geq a \\end{cases} \\)\n3. \\( = \\int_0^\\infty e^{-st} u(t-a) dt \\)\n4. \\( = \\int_a^\\infty e^{-st} dt \\)\n5. \\( = \\left[ -\\frac{1}{s} e^{-st} \\right]_a^\\infty \\)\n6. \\( = \\left( -\\frac{1}{s} e^{-s\\infty} \\right) - \\left( -\\frac{1}{s} e^{-sa} \\right) \\)\n7. \\( = \\left( \\frac{1}{s} e^{-s\\infty} \\right) + \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 + \\frac{1}{s} e^{-sa} \\)\n9. \\( = \\frac{e^{-sa}}{s}, s > 0 \\).\n That's why it's zero. Heaviside is just the name of the person. Anyway, I thought it wasn't funny. So we have pi a, b. And the input is just usually with a t, like this:" ], "metadata": [ { "vid": "ftnpM_RO0Jc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Laplace Transform - Definition and Properties_30.json#####audio#####doingASR#####FinishASR/ftnpM_RO0Jc/2227.4_2247.4.mp4", "refined_asr": " This is negative already. This is positive infinity. S has to be positive so that we can actually get negative infinity out of this. Right. So with that being said, S has to be greater than zero. So with that condition, we can actually say the first part.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2227.4_2247.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2227.4_2247.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2227.4_2247.4#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2227.4_2247.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2227.4_2247.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2227.4_2247.4#3.jpg" ], "ocr_qwen2_vl_72b": "1. \\( \\mathcal{L}\\{U(t-a)\\} \\)\n2. \\( = \\int_{0}^{\\infty} e^{-st} U(t-a) dt \\)\n3. \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n4. \\( = \\left[ -\\frac{1}{s} e^{-st} \\right]_{a}^{\\infty} \\)\n5. \\( = \\left( \\frac{1}{s} e^{-s \\cdot \\infty} \\right) - \\left( \\frac{1}{s} e^{-sa} \\right) \\)" }, { "vid": "ftnpM_RO0Jc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Laplace Transform - Definition and Properties_30.json#####audio#####doingASR#####FinishASR/ftnpM_RO0Jc/2248.4_2263.4.mp4", "refined_asr": " It has to be zero. And again, we need to have that condition. Anything that I write down right here is that we need to have that in order to make this work and work out the rest. Negative negative becomes positive. And then we just have that.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2248.4_2263.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2248.4_2263.4#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2248.4_2263.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2248.4_2263.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2248.4_2263.4#3.jpg" ], "ocr_qwen2_vl_72b": "1. \\( U(t-a) \\)\n2. \\( 0 \\) if \\( t-a < 0 \\)\n3. \\( 1 \\) if \\( t-a > 0 \\)\n4. \\( \\int_{0}^{\\infty} e^{-st} U(t-a) dt \\)\n5. \\( \\int_{a}^{\\infty} e^{-st} dt \\)\n6. \\( -\\frac{1}{s} e^{-st} \\bigg|_{a}^{\\infty} \\)\n7. \\( = \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 \\)" }, { "vid": "ftnpM_RO0Jc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Laplace Transform - Definition and Properties_30.json#####audio#####doingASR#####FinishASR/ftnpM_RO0Jc/2263.4_2277.4.mp4", "refined_asr": " That's pretty much it. So in the end, we have plus one over s times e to the negative s a. So finally, we will just write this down as e to the negative s a. And some people like to write this as e to the minus s a.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2263.4_2277.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2263.4_2277.4#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2263.4_2277.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2263.4_2277.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2263.4_2277.4#3.jpg" ], "ocr_qwen2_vl_72b": "- \\( L\\{U(t-a)\\} \\)\n- \\( U(t-a) = \\begin{cases} 0 & \\text{if } t-a < 0 \\\\ 1 & \\text{if } t-a > 0 \\end{cases} \\)\n- \\( = \\int_{0}^{\\infty} e^{-st} U(t-a) dt \\)\n- \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n- \\( = -\\frac{1}{s} e^{-st} \\bigg|_{a}^{\\infty} \\)\n- \\( = \\left( \\frac{1}{s} e^{-s \\cdot \\infty} \\right) - \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n- \\( = 0 + \\)" }, { "vid": "ftnpM_RO0Jc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Laplace Transform - Definition and Properties_30.json#####audio#####doingASR#####FinishASR/ftnpM_RO0Jc/2277.4_2291.4.mp4", "refined_asr": " And some people like to write this as e to the negative A S. It depends on how you want to write it but this is okay. And then over S. And again we have the condition so let's write that down.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2277.4_2291.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2277.4_2291.4#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2277.4_2291.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2277.4_2291.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2277.4_2291.4#3.jpg" ], "ocr_qwen2_vl_72b": "1. \\( \\mathcal{L}\\{u(t-a)\\} \\)\n2. \\( u(t-a) = \\begin{cases} 0 & t < a \\\\ 1 & t \\geq a \\end{cases} \\)\n3. \\( = \\int_{0}^{\\infty} e^{-st} u(t-a) dt \\)\n4. \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n5. \\( = -\\frac{1}{s} e^{-st} \\Big|_{a}^{\\infty} \\)\n6. \\( = \\left( -\\frac{1}{s} e^{-s \\cdot \\infty} \\right) - \\left( -\\frac{1}{s} e^{-sa} \\right) \\)\n7. \\( = \\left( -\\frac{1}{s} \\cdot 0 \\right) + \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 + \\frac{1}{s} e^{-sa} \\)\n9. \\( = \\frac{1}{s} e^{-sa} \\)" }, { "vid": "ftnpM_RO0Jc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Laplace Transform - Definition and Properties_30.json#####audio#####doingASR#####FinishASR/ftnpM_RO0Jc/2291.4_2307.4.mp4", "refined_asr": " S has to be greater than zero in order for that to work. Just like that. So this right here is a unit step function. And now let me introduce you to something that's very similar to the unit step function, namely the window function.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2291.4_2307.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2291.4_2307.4#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2291.4_2307.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2291.4_2307.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2291.4_2307.4#3.jpg" ], "ocr_qwen2_vl_72b": "1. \\( \\mathcal{L}\\{U(t-a)\\} \\)\n2. \\( U(t-a) = \\begin{cases} 0 & \\text{if } t < a \\\\ 1 & \\text{if } t \\geq a \\end{cases} \\)\n3. \\( = \\int_{0}^{\\infty} e^{-st} U(t-a) dt \\)\n4. \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n5. \\( = -\\frac{1}{s} e^{-st} \\bigg|_{a}^{\\infty} \\)\n6. \\( = \\left( -\\frac{1}{s} e^{-s \\cdot \\infty} \\right) - \\left( -\\frac{1}{s} e^{-sa} \\right) \\)\n7. \\( = \\left( \\frac{1}{s} e^{-s \\cdot \\infty} \\right) + \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 + \\frac{1}{s} e^{-sa} \\)\n9. \\( = \\frac{e^{-sa}}{s}, s > 0 \\)" }, { "vid": "ftnpM_RO0Jc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Laplace Transform - Definition and Properties_30.json#####audio#####doingASR#####FinishASR/ftnpM_RO0Jc/2307.4_2326.4.mp4", "refined_asr": " The pi sub a b t function. I should be able to do that right here. So I will do that right here. Number eight. So here we have the Laplace transform and inside is the capital pi a comma b.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2307.4_2326.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2307.4_2326.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2307.4_2326.4#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2307.4_2326.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2307.4_2326.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2307.4_2326.4#3.jpg" ], "ocr_qwen2_vl_72b": "1. \\( \\mathcal{L}\\{u(t-a)\\} \\)\n2. \\( u(t-a) = \\begin{cases} 0 & t < a \\\\ 1 & t \\geq a \\end{cases} \\)\n3. \\( = \\int_{0}^{\\infty} e^{-st} u(t-a) dt \\)\n4. \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n5. \\( = \\left[ -\\frac{1}{s} e^{-st} \\right]_{a}^{\\infty} \\)\n6. \\( = \\left( -\\frac{1}{s} e^{-s(\\infty)} \\right) - \\left( -\\frac{1}{s} e^{-sa} \\right) \\)\n7. \\( = \\left( -\\frac{1}{s} \\cdot 0 \\right) + \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 + \\frac{1}{s} e^{-sa} \\)\n9. \\( = \\frac{e^{-sa}}{s}, s > 0 \\)" }, { "vid": "ftnpM_RO0Jc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Laplace Transform - Definition and Properties_30.json#####audio#####doingASR#####FinishASR/ftnpM_RO0Jc/2326.4_2339.4.mp4", "refined_asr": " This is not the original pi function. It's just what I think is called the window function over there. Oh, by the way, this is also called the Heaviside function. But that's not because this side is Heaviside.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2326.4_2339.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2326.4_2339.4#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2326.4_2339.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2326.4_2339.4#2.jpg" ], "ocr_qwen2_vl_72b": "1. \\( \\mathcal{L}\\{U(t-a)\\} \\)\n2. \\( U(t-a) = \\begin{cases} 0 & \\text{if } t - a < 0 \\\\ 1 & \\text{if } t - a > 0 \\end{cases} \\)\n3. \\( = \\int_{0}^{\\infty} e^{-st} U(t-a) dt \\)\n4. \\( = \\int_{a}^{\\infty} e^{-st} dt \\)\n5. \\( = \\left[ -\\frac{1}{s} e^{-st} \\right]_{a}^{\\infty} \\)\n6. \\( = \\left( -\\frac{1}{s} e^{-s \\cdot \\infty} \\right) - \\left( -\\frac{1}{s} e^{-sa} \\right) \\)\n7. \\( = \\left( \\frac{1}{s} e^{-s \\cdot \\infty} \\right) + \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 + \\frac{1}{s} e^{-sa} \\)\n9. \\( = \\frac{e^{-sa}}{s}, s > 0 \\)" }, { "vid": "ftnpM_RO0Jc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Laplace Transform - Definition and Properties_30.json#####audio#####doingASR#####FinishASR/ftnpM_RO0Jc/2339.4_2355.4.mp4", "refined_asr": " That's why it's zero. Heaviside is just the name of the person. Anyway, I thought it wasn't funny. So we have pi a, b. And the input is just usually with a t, like this:", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2339.4_2355.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2339.4_2355.4#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2339.4_2355.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2339.4_2355.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ftnpM_RO0Jc/ftnpM_RO0Jc@2339.4_2355.4#3.jpg" ], "ocr_qwen2_vl_72b": "1. \\( \\mathcal{L}\\{u(t-a)\\} \\)\n2. \\( u(t-a) = \\begin{cases} 0 & \\text{if } t < a \\\\ 1 & \\text{if } t \\geq a \\end{cases} \\)\n3. \\( = \\int_0^\\infty e^{-st} u(t-a) dt \\)\n4. \\( = \\int_a^\\infty e^{-st} dt \\)\n5. \\( = \\left[ -\\frac{1}{s} e^{-st} \\right]_a^\\infty \\)\n6. \\( = \\left( -\\frac{1}{s} e^{-s\\infty} \\right) - \\left( -\\frac{1}{s} e^{-sa} \\right) \\)\n7. \\( = \\left( \\frac{1}{s} e^{-s\\infty} \\right) + \\left( \\frac{1}{s} e^{-sa} \\right) \\)\n8. \\( = 0 + \\frac{1}{s} e^{-sa} \\)\n9. \\( = \\frac{e^{-sa}}{s}, s > 0 \\)" } ], "image_num": 18, "text_num": 433, "token_num": 10801 }, { "images": [ "sample_100_images/msk7OtyR0cQ@856.5_871.5#1.jpg", "sample_100_images/msk7OtyR0cQ@856.5_871.5#2.jpg", null, "sample_100_images/msk7OtyR0cQ@871.5_892.5#1.jpg", "sample_100_images/msk7OtyR0cQ@871.5_892.5#2.jpg", "sample_100_images/msk7OtyR0cQ@871.5_892.5#3.jpg", null, "sample_100_images/msk7OtyR0cQ@892.5_912.5#1.jpg", "sample_100_images/msk7OtyR0cQ@892.5_912.5#2.jpg", "sample_100_images/msk7OtyR0cQ@892.5_912.5#3.jpg", null, "sample_100_images/msk7OtyR0cQ@912.5_930.5#1.jpg", "sample_100_images/msk7OtyR0cQ@912.5_930.5#2.jpg", "sample_100_images/msk7OtyR0cQ@912.5_930.5#3.jpg", null, "sample_100_images/msk7OtyR0cQ@930.5_945.5#1.jpg", "sample_100_images/msk7OtyR0cQ@930.5_945.5#2.jpg", null, "sample_100_images/msk7OtyR0cQ@945.5_963.5#1.jpg", "sample_100_images/msk7OtyR0cQ@945.5_963.5#2.jpg", "sample_100_images/msk7OtyR0cQ@945.5_963.5#3.jpg", null, "sample_100_images/msk7OtyR0cQ@963.5_977.5#1.jpg", "sample_100_images/msk7OtyR0cQ@963.5_977.5#2.jpg", null, "sample_100_images/msk7OtyR0cQ@977.5_992.5#1.jpg", null ], "texts": [ null, null, " We want to make sure that we don't contact the vital anatomy and we have adequate safety margins. The surgeon has to review this, confirm that it is indeed safe to proceed. This is a pre-positioning frame that attaches to the patient.", null, null, null, " The spherical metal markers here serve both as fiducial markers for localization and also the attachment points for the robot. We place the positioning frame on the head, leaving enough room for access to the lateral skull base for both the robot and the human operator.", null, null, null, " After this is attached, we take a CT scan again using that intraoperative CT scanner which will show us where the fiducial markers are in both CT space and we'll pick those up in the operating room and link them together so now we've found those markers.", null, null, null, " We're going to do a little bit of kinematics work with the robot to ensure that it doesn't fold over itself. And then we're going to attach the robot to the pre-positioning frame and we're going to start to drill. And it's interesting on this that the suction irrigator", null, null, " That's being held by the human being is there only so you can see what the robot's doing. Because the robot knows where the tip of that drill bit is in relation to the fiducial markers, which is then in relation to the patient's anatomy.", null, null, null, " And as we go down below the cortex, we will start to drill behind the facial nerve and do the translabyrinthine approach. The concept for this is that this would free up a surgeon to be able to do the vital work, the delicate work of the IAC: teasing off the tumor.", null, null, " To do this right now in the cadaver lab, once the robot is attached, takes approximately one and a half hours. But with better path planning, we think we can get that down to about 30 to 40 minutes. So imagine if you will.", null, " Imagine you're in the operating room, you attach this, and 30 minutes later, you're doing a nice dissection in the IAC with your neurosurgical colleagues. I think it would make for a quicker day. So hopefully, I've covered these three things." ], "text_ocr_list": [ null, null, "We can see these text from the image: Outlined slices are converted to three-dimensional target volume.\n We want to make sure that we don't contact the vital anatomy and we have adequate safety margins. The surgeon has to review this, confirm that it is indeed safe to proceed. This is a pre-positioning frame that attaches to the patient.", null, null, null, "We can see these text from the image: - Robot is attached to the patient via a rigid positioning frame.\n- Three legs on bottom side screw into the skull surface.\n- Attachment Points for Mounting Robot.\n The spherical metal markers here serve both as fiducial markers for localization and also the attachment points for the robot. We place the positioning frame on the head, leaving enough room for access to the lateral skull base for both the robot and the human operator.", null, null, null, "We can see these text from the image: Rigid fixation prevents relative motion between robot and patient, which could lead to positioning error..\n After this is attached, we take a CT scan again using that intraoperative CT scanner which will show us where the fiducial markers are in both CT space and we'll pick those up in the operating room and link them together so now we've found those markers.", null, null, null, "We can see these text from the image: - Intra-Operative Image Processing and Planning\n- Robot Joint Trajectories\n- 3. Plan safe robot trajectory.\n We're going to do a little bit of kinematics work with the robot to ensure that it doesn't fold over itself. And then we're going to attach the robot to the pre-positioning frame and we're going to start to drill. And it's interesting on this that the suction irrigator", null, null, "We can see these text from the image: 2x.\n That's being held by the human being is there only so you can see what the robot's doing. Because the robot knows where the tip of that drill bit is in relation to the fiducial markers, which is then in relation to the patient's anatomy.", null, null, null, "We can see these text from the image: 2x.\n And as we go down below the cortex, we will start to drill behind the facial nerve and do the translabyrinthine approach. The concept for this is that this would free up a surgeon to be able to do the vital work, the delicate work of the IAC: teasing off the tumor.", null, null, "We can see these text from the image: VANDERBILT UNIVERSITY.\n To do this right now in the cadaver lab, once the robot is attached, takes approximately one and a half hours. But with better path planning, we think we can get that down to about 30 to 40 minutes. So imagine if you will.", null, "We can see these text from the image: VANDERBILT UNIVERSITY.\n Imagine you're in the operating room, you attach this, and 30 minutes later, you're doing a nice dissection in the IAC with your neurosurgical colleagues. I think it would make for a quicker day. So hopefully, I've covered these three things." ], "metadata": [ { "vid": "msk7OtyR0cQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering Course on Image-Guided Surgery in Biomedical Imaging and Image Processing Applications_30.json#####audio#####doingASR#####FinishASR/msk7OtyR0cQ/856.5_871.5.mp4", "refined_asr": " We want to make sure that we don't contact the vital anatomy and we have adequate safety margins. The surgeon has to review this, confirm that it is indeed safe to proceed. This is a pre-positioning frame that attaches to the patient.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@856.5_871.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@856.5_871.5#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@856.5_871.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@856.5_871.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@856.5_871.5#3.jpg" ], "ocr_qwen2_vl_72b": "Outlined slices are converted to three-dimensional target volume" }, { "vid": "msk7OtyR0cQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering Course on Image-Guided Surgery in Biomedical Imaging and Image Processing Applications_30.json#####audio#####doingASR#####FinishASR/msk7OtyR0cQ/871.5_892.5.mp4", "refined_asr": " The spherical metal markers here serve both as fiducial markers for localization and also the attachment points for the robot. We place the positioning frame on the head, leaving enough room for access to the lateral skull base for both the robot and the human operator.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@871.5_892.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@871.5_892.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@871.5_892.5#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@871.5_892.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@871.5_892.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@871.5_892.5#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@871.5_892.5#4.jpg" ], "ocr_qwen2_vl_72b": "- Robot is attached to the patient via a rigid positioning frame.\n- Three legs on bottom side screw into the skull surface.\n- Attachment Points for Mounting Robot" }, { "vid": "msk7OtyR0cQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering Course on Image-Guided Surgery in Biomedical Imaging and Image Processing Applications_30.json#####audio#####doingASR#####FinishASR/msk7OtyR0cQ/892.5_912.5.mp4", "refined_asr": " After this is attached, we take a CT scan again using that intraoperative CT scanner which will show us where the fiducial markers are in both CT space and we'll pick those up in the operating room and link them together so now we've found those markers.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@892.5_912.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@892.5_912.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@892.5_912.5#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@892.5_912.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@892.5_912.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@892.5_912.5#3.jpg" ], "ocr_qwen2_vl_72b": "Rigid fixation prevents relative motion between robot and patient, which could lead to positioning error." }, { "vid": "msk7OtyR0cQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering Course on Image-Guided Surgery in Biomedical Imaging and Image Processing Applications_30.json#####audio#####doingASR#####FinishASR/msk7OtyR0cQ/912.5_930.5.mp4", "refined_asr": " We're going to do a little bit of kinematics work with the robot to ensure that it doesn't fold over itself. And then we're going to attach the robot to the pre-positioning frame and we're going to start to drill. And it's interesting on this that the suction irrigator", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@912.5_930.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@912.5_930.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@912.5_930.5#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@912.5_930.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@912.5_930.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@912.5_930.5#3.jpg" ], "ocr_qwen2_vl_72b": "- Intra-Operative Image Processing and Planning\n- Robot Joint Trajectories\n- 3. Plan safe robot trajectory" }, { "vid": "msk7OtyR0cQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering Course on Image-Guided Surgery in Biomedical Imaging and Image Processing Applications_30.json#####audio#####doingASR#####FinishASR/msk7OtyR0cQ/930.5_945.5.mp4", "refined_asr": " That's being held by the human being is there only so you can see what the robot's doing. Because the robot knows where the tip of that drill bit is in relation to the fiducial markers, which is then in relation to the patient's anatomy.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@930.5_945.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@930.5_945.5#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@930.5_945.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@930.5_945.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@930.5_945.5#3.jpg" ], "ocr_qwen2_vl_72b": "2x" }, { "vid": "msk7OtyR0cQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering Course on Image-Guided Surgery in Biomedical Imaging and Image Processing Applications_30.json#####audio#####doingASR#####FinishASR/msk7OtyR0cQ/945.5_963.5.mp4", "refined_asr": " And as we go down below the cortex, we will start to drill behind the facial nerve and do the translabyrinthine approach. The concept for this is that this would free up a surgeon to be able to do the vital work, the delicate work of the IAC: teasing off the tumor.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@945.5_963.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@945.5_963.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@945.5_963.5#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@945.5_963.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@945.5_963.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@945.5_963.5#3.jpg" ], "ocr_qwen2_vl_72b": "2x" }, { "vid": "msk7OtyR0cQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering Course on Image-Guided Surgery in Biomedical Imaging and Image Processing Applications_30.json#####audio#####doingASR#####FinishASR/msk7OtyR0cQ/963.5_977.5.mp4", "refined_asr": " To do this right now in the cadaver lab, once the robot is attached, takes approximately one and a half hours. But with better path planning, we think we can get that down to about 30 to 40 minutes. So imagine if you will.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@963.5_977.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@963.5_977.5#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@963.5_977.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@963.5_977.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@963.5_977.5#3.jpg" ], "ocr_qwen2_vl_72b": "VANDERBILT UNIVERSITY" }, { "vid": "msk7OtyR0cQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering Course on Image-Guided Surgery in Biomedical Imaging and Image Processing Applications_30.json#####audio#####doingASR#####FinishASR/msk7OtyR0cQ/977.5_992.5.mp4", "refined_asr": " Imagine you're in the operating room, you attach this, and 30 minutes later, you're doing a nice dissection in the IAC with your neurosurgical colleagues. I think it would make for a quicker day. So hopefully, I've covered these three things.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@977.5_992.5#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@977.5_992.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@977.5_992.5#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/msk7OtyR0cQ/msk7OtyR0cQ@977.5_992.5#3.jpg" ], "ocr_qwen2_vl_72b": "VANDERBILT UNIVERSITY" } ], "image_num": 19, "text_num": 500, "token_num": 11444 }, { "images": [ "sample_100_images/G9QoX51IDzI@537.54_557.4#1.jpg", "sample_100_images/G9QoX51IDzI@537.54_557.4#2.jpg", "sample_100_images/G9QoX51IDzI@537.54_557.4#3.jpg", null, "sample_100_images/G9QoX51IDzI@557.4_577.0799999999999#1.jpg", null, "sample_100_images/G9QoX51IDzI@577.9_590.52#1.jpg", "sample_100_images/G9QoX51IDzI@577.9_590.52#2.jpg", null, "sample_100_images/G9QoX51IDzI@590.52_606.52#1.jpg", "sample_100_images/G9QoX51IDzI@590.52_606.52#2.jpg", null, "sample_100_images/G9QoX51IDzI@606.52_621.52#1.jpg", null, "sample_100_images/G9QoX51IDzI@621.52_647.52#1.jpg", "sample_100_images/G9QoX51IDzI@621.52_647.52#2.jpg", "sample_100_images/G9QoX51IDzI@621.52_647.52#3.jpg", "sample_100_images/G9QoX51IDzI@621.52_647.52#4.jpg", null, "sample_100_images/G9QoX51IDzI@648.52_668.28#1.jpg", null, "sample_100_images/G9QoX51IDzI@668.84_686.1999999999999#1.jpg", null ], "texts": [ null, null, null, " This function returns the value seven as only that part is reserved. Listen up, this is important. Moving on to studying how a function is called. A function is called by simply using the function name followed by a list of actual parameters.", null, " If any information is missing, when a compiler encounters a function call, it transfers control from the current statement to the function definition. In the given example, the control is transferred to the 'add' function. The value of 'x' is assigned to 'a', and the value of 'y' is assigned to 'b'.", null, null, " The function is then executed till its last statement or until the return statement is encountered and the value is returned to the main function and assigned to Z. There are various ways in which we can call a function.", null, null, " We can place the actual arguments or the constants or a combination of them or another function call or expressions within the function call. When an expression is used in a call, it must be evaluated to a single value in order to pass it in the call as a parameter.", null, " A function which returns a value can be used in any expression like other variables. However, \"add a b equals 12\" is an invalid statement.", null, null, null, null, " Let's go to the function definition which is also known as function implementation. It includes the function name, type, and parameters combined into the function header. Meanwhile, local variables, function statements, and return statements are grouped into the function body. The general format is: the function type defines the type of value that a function is going to return to the calling function.", null, " If the return type is not mentioned, then C will assume it to be of type integer. If the function does not return anything, then the return type is void. The function name should be a valid identifier. The name should be appropriate to the task it performs, and care should be taken that there are no duplicate names.", null, " A parameter list represents the list of variables that will receive data from a calling function. These variables are considered input data to the function, enabling it to carry out its designated task. Because they stand in for actual values, these variables are referred to as formal parameters. It's worth noting that the terms parameters and arguments are often used interchangeably." ], "text_ocr_list": [ null, null, null, "We can see these text from the image: Return Values And Types\n\nReturn Types\n\nmain()\n{\n\tz = function1();\n}\n\nfunction1()\n{\n\treturn(value);\n} integer Type Specifier\n\nExample:\nint mul()\n{\n\treturn(2.5*3.0);\n}\n2.5*3.0=7.5.\n This function returns the value seven as only that part is reserved. Listen up, this is important. Moving on to studying how a function is called. A function is called by simply using the function name followed by a list of actual parameters.", null, "We can see these text from the image: Function Calls\n\nExample: void main()\n{\nint z,x,y;\nx=5;\ny=10;\nadd(x, y);\nprintf(\"ans=%d\", z);\n}\n\nRAM\n\nMemory Of Main()\n\nMemory Of Add().\n If any information is missing, when a compiler encounters a function call, it transfers control from the current statement to the function definition. In the given example, the control is transferred to the 'add' function. The value of 'x' is assigned to 'a', and the value of 'y' is assigned to 'b'.", null, null, "We can see these text from the image: Function Calls\n\nExample: void main()\n```\n{\n int z,x,y;\n x=5;\n y=10;\n z=add(x, y);\n printf(\"ans=%d\", z);\n}\n\nint add(int a, int b)\n{\n int p;\n p=a+b;\n return(p);\n}\n```\n\nRAM\n\nMemory Of Main()\n\nMemory Of Add()\n\n5 10 15.\n The function is then executed till its last statement or until the return statement is encountered and the value is returned to the main function and assigned to Z. There are various ways in which we can call a function.", null, null, "We can see these text from the image: add(10, 5);\nadd(x, 5);\nadd(10, y);\nadd(x, y);\nadd(x + 4, 5);\nadd(10, mul(x, y));\nadd(expression1, expression2);.\n We can place the actual arguments or the constants or a combination of them or another function call or expressions within the function call. When an expression is used in a call, it must be evaluated to a single value in order to pass it in the call as a parameter.", null, "We can see these text from the image: Examples: printf(\"%d\", mul(x,y));\nz=add(x,.\n A function which returns a value can be used in any expression like other variables. However, \"add a b equals 12\" is an invalid statement.", null, null, null, null, "We can see these text from the image: Function Definitions\n\nwww.ezed.in.\n Let's go to the function definition which is also known as function implementation. It includes the function name, type, and parameters combined into the function header. Meanwhile, local variables, function statements, and return statements are grouped into the function body. The general format is: the function type defines the type of value that a function is going to return to the calling function.", null, "We can see these text from the image: Function Definitions\n\nGeneral Format:\nreturn type ? function_type function_name(parameter_list) Function Header\n{\nlocal variable declaration;\nexecutablestatement1;\nexecutablestatement2;\nreturn statement;\n} Function Body.\n If the return type is not mentioned, then C will assume it to be of type integer. If the function does not return anything, then the return type is void. The function name should be a valid identifier. The name should be appropriate to the task it performs, and care should be taken that there are no duplicate names.", null, "We can see these text from the image: Function Definitions\n\nGeneral Format:\n\nfunction_type function_name(parameter_list)\n{\nlocal variable declaration;\nexecutablestatement1;\nexecutablestatement2;\nreturn statement;\n}.\n A parameter list represents the list of variables that will receive data from a calling function. These variables are considered input data to the function, enabling it to carry out its designated task. Because they stand in for actual values, these variables are referred to as formal parameters. It's worth noting that the terms parameters and arguments are often used interchangeably." ], "metadata": [ { "vid": "G9QoX51IDzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/C Programming tutorial on Functions in Imperative Programming Languages_30.json#####audio#####doingASR#####FinishASR/G9QoX51IDzI/537.54_557.4.mp4", "refined_asr": " This function returns the value seven as only that part is reserved. Listen up, this is important. Moving on to studying how a function is called. A function is called by simply using the function name followed by a list of actual parameters.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@537.54_557.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@537.54_557.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@537.54_557.4#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@537.54_557.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@537.54_557.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@537.54_557.4#3.jpg" ], "ocr_qwen2_vl_72b": "Return Values And Types\n\nReturn Types\n\nmain()\n{\n\tz = function1();\n}\n\nfunction1()\n{\n\treturn(value);\n} integer Type Specifier\n\nExample:\nint mul()\n{\n\treturn(2.5*3.0);\n}\n2.5*3.0=7.5" }, { "vid": "G9QoX51IDzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/C Programming tutorial on Functions in Imperative Programming Languages_30.json#####audio#####doingASR#####FinishASR/G9QoX51IDzI/557.4_577.0799999999999.mp4", "refined_asr": " If any information is missing, when a compiler encounters a function call, it transfers control from the current statement to the function definition. In the given example, the control is transferred to the 'add' function. The value of 'x' is assigned to 'a', and the value of 'y' is assigned to 'b'.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@557.4_577.0799999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@557.4_577.0799999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@557.4_577.0799999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@557.4_577.0799999999999#3.jpg" ], "ocr_qwen2_vl_72b": "Function Calls\n\nExample: void main()\n{\nint z,x,y;\nx=5;\ny=10;\nadd(x, y);\nprintf(\"ans=%d\", z);\n}\n\nRAM\n\nMemory Of Main()\n\nMemory Of Add()" }, { "vid": "G9QoX51IDzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/C Programming tutorial on Functions in Imperative Programming Languages_30.json#####audio#####doingASR#####FinishASR/G9QoX51IDzI/577.9_590.52.mp4", "refined_asr": " The function is then executed till its last statement or until the return statement is encountered and the value is returned to the main function and assigned to Z. There are various ways in which we can call a function.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@577.9_590.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@577.9_590.52#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@577.9_590.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@577.9_590.52#2.jpg" ], "ocr_qwen2_vl_72b": "Function Calls\n\nExample: void main()\n```\n{\n int z,x,y;\n x=5;\n y=10;\n z=add(x, y);\n printf(\"ans=%d\", z);\n}\n\nint add(int a, int b)\n{\n int p;\n p=a+b;\n return(p);\n}\n```\n\nRAM\n\nMemory Of Main()\n\nMemory Of Add()\n\n5 10 15" }, { "vid": "G9QoX51IDzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/C Programming tutorial on Functions in Imperative Programming Languages_30.json#####audio#####doingASR#####FinishASR/G9QoX51IDzI/590.52_606.52.mp4", "refined_asr": " We can place the actual arguments or the constants or a combination of them or another function call or expressions within the function call. When an expression is used in a call, it must be evaluated to a single value in order to pass it in the call as a parameter.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@590.52_606.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@590.52_606.52#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@590.52_606.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@590.52_606.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@590.52_606.52#3.jpg" ], "ocr_qwen2_vl_72b": "add(10, 5);\nadd(x, 5);\nadd(10, y);\nadd(x, y);\nadd(x + 4, 5);\nadd(10, mul(x, y));\nadd(expression1, expression2);" }, { "vid": "G9QoX51IDzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/C Programming tutorial on Functions in Imperative Programming Languages_30.json#####audio#####doingASR#####FinishASR/G9QoX51IDzI/606.52_621.52.mp4", "refined_asr": " A function which returns a value can be used in any expression like other variables. However, \"add a b equals 12\" is an invalid statement.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@606.52_621.52#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@606.52_621.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@606.52_621.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@606.52_621.52#3.jpg" ], "ocr_qwen2_vl_72b": "Examples: printf(\"%d\", mul(x,y));\nz=add(x," }, { "vid": "G9QoX51IDzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/C Programming tutorial on Functions in Imperative Programming Languages_30.json#####audio#####doingASR#####FinishASR/G9QoX51IDzI/621.52_647.52.mp4", "refined_asr": " Let's go to the function definition which is also known as function implementation. It includes the function name, type, and parameters combined into the function header. Meanwhile, local variables, function statements, and return statements are grouped into the function body. The general format is: the function type defines the type of value that a function is going to return to the calling function.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@621.52_647.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@621.52_647.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@621.52_647.52#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@621.52_647.52#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@621.52_647.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@621.52_647.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@621.52_647.52#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@621.52_647.52#4.jpg" ], "ocr_qwen2_vl_72b": "Function Definitions\n\nwww.ezed.in" }, { "vid": "G9QoX51IDzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/C Programming tutorial on Functions in Imperative Programming Languages_30.json#####audio#####doingASR#####FinishASR/G9QoX51IDzI/648.52_668.28.mp4", "refined_asr": " If the return type is not mentioned, then C will assume it to be of type integer. If the function does not return anything, then the return type is void. The function name should be a valid identifier. The name should be appropriate to the task it performs, and care should be taken that there are no duplicate names.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@648.52_668.28#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@648.52_668.28#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@648.52_668.28#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@648.52_668.28#3.jpg" ], "ocr_qwen2_vl_72b": "Function Definitions\n\nGeneral Format:\nreturn type ? function_type function_name(parameter_list) Function Header\n{\nlocal variable declaration;\nexecutablestatement1;\nexecutablestatement2;\nreturn statement;\n} Function Body" }, { "vid": "G9QoX51IDzI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/C Programming tutorial on Functions in Imperative Programming Languages_30.json#####audio#####doingASR#####FinishASR/G9QoX51IDzI/668.84_686.1999999999999.mp4", "refined_asr": " A parameter list represents the list of variables that will receive data from a calling function. These variables are considered input data to the function, enabling it to carry out its designated task. Because they stand in for actual values, these variables are referred to as formal parameters. It's worth noting that the terms parameters and arguments are often used interchangeably.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@668.84_686.1999999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@668.84_686.1999999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@668.84_686.1999999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/G9QoX51IDzI/G9QoX51IDzI@668.84_686.1999999999999#3.jpg" ], "ocr_qwen2_vl_72b": "Function Definitions\n\nGeneral Format:\n\nfunction_type function_name(parameter_list)\n{\nlocal variable declaration;\nexecutablestatement1;\nexecutablestatement2;\nreturn statement;\n}" } ], "image_num": 15, "text_num": 470, "token_num": 9110 }, { "images": [ "sample_100_images/7I0Xg92_eA4@1033.24_1053.24#1.jpg", null, "sample_100_images/7I0Xg92_eA4@1053.24_1073.24#1.jpg", null, "sample_100_images/7I0Xg92_eA4@1073.24_1092.24#1.jpg", null, "sample_100_images/7I0Xg92_eA4@1092.24_1111.24#1.jpg", null, "sample_100_images/7I0Xg92_eA4@1111.24_1133.24#1.jpg", null, "sample_100_images/7I0Xg92_eA4@1133.24_1148.24#1.jpg", null, "sample_100_images/7I0Xg92_eA4@1148.24_1170.24#1.jpg", null ], "texts": [ null, " Whenever you increase the numerator of a fraction, the value of the fraction goes up. If you increase the denominator, the value of the fraction goes down. So, because the initial concentration is in the numerator, there is a direct relationship. As we increase the initial concentration.", null, " The half-life is going to increase as well. So there is a direct relationship between the two. So C is the true statement: these two are directly proportional. Now what about answer choice D? The half-life of a zero-order reaction is constant.", null, " Is that true or false? So this is the half-life for a zero-order reaction. Is this value T\u00bd? Is it constant? Yes, T\u00bd depends on the initial concentration and k. The '2' is not going to change; that's constant.", null, " K is the rate constant of a reaction. The fact that it's a rate constant means that K doesn't change, that is, if the temperature doesn't change. So we're assuming that the temperature remains constant. If the temperature is constant, then K is going to be constant.", null, " So at constant temperature K is not going to change. Now, the initial concentration, is that constant? Well, the answer is no. Because the purpose of chemical kinetics is to study how fast reactions are going. And if the reaction is going, as A changes to B.", null, " A is going to decrease. So as the reaction proceeds, the concentration of A will not be constant. It's going to decrease over time. So if A goes down, the half-life is going to go down with it because they are directly related.", null, " So therefore, the half-life of a zero-order reaction is not constant. This makes statement D false. Note that the half-life of a zero-order reaction and a second-order reaction depends on the initial concentration of A. Therefore, D is the correct answer choice." ], "text_ocr_list": [ null, "We can see these text from the image: 3. Which of the following statements is false?\n\nt\u00bd = ln 2 / k\nt\u00bd = 1 / k [A]\u2080\nt\u00bd = [A]\u2080 / 2k\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant..\n Whenever you increase the numerator of a fraction, the value of the fraction goes up. If you increase the denominator, the value of the fraction goes down. So, because the initial concentration is in the numerator, there is a direct relationship. As we increase the initial concentration.", null, "We can see these text from the image: 3. Which of the following statements is false?\n\nt\u00bd = ln 2\nk\n\nt\u00bd = 1\nk[A]\u2080\n\nt\u00bd = [A]\u2080\n2k\n\n[A]\u2080 \u2191 t\u00bd \u2191\n\nk \u2191 t\u00bd \u2193\nd \u2191 f\u2193\n\nThe half life of a first order reaction is independent of the concentration of the reactant.\nThe rate constant K is inversely related to the half life of a second order reaction.\nThe half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nThe half life of a zero order reaction is constant.\nThe half life of a second order reaction is inversely proportional to the initial concentration of the reactant..\n The half-life is going to increase as well. So there is a direct relationship between the two. So C is the true statement: these two are directly proportional. Now what about answer choice D? The half-life of a zero-order reaction is constant.", null, "We can see these text from the image: 3. Which of the following statements is false?\n\nt1/2 = ln 2 / k\nt1/2 = 1 / k [A]0\nt1/2 = [A]0 / 2k\n\n[A]0 \u2191 t1/2 \u2191\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant..\n Is that true or false? So this is the half-life for a zero-order reaction. Is this value T\u00bd? Is it constant? Yes, T\u00bd depends on the initial concentration and k. The '2' is not going to change; that's constant.", null, "We can see these text from the image: 3. Which of the following statements is false?\n\nt\u00bd = ln 2 / k\nt\u00bd = 1 / k[A]\u2080\nt\u00bd = [A]\u2080 / 2k\n\n[A]\u2080 \u2191 t\u00bd \u2191\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant..\n K is the rate constant of a reaction. The fact that it's a rate constant means that K doesn't change, that is, if the temperature doesn't change. So we're assuming that the temperature remains constant. If the temperature is constant, then K is going to be constant.", null, "We can see these text from the image: 3. Which of the following statements is false?\n\nt\u00bd = ln 2 / k\nt\u00bd = 1 / k [A]\u2080\nt\u00bd = [A]\u2080 / 2k\n\n[A]\u2080 \u2191 t\u00bd \u2191\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant..\n So at constant temperature K is not going to change. Now, the initial concentration, is that constant? Well, the answer is no. Because the purpose of chemical kinetics is to study how fast reactions are going. And if the reaction is going, as A changes to B.", null, "We can see these text from the image: 3. Which of the following statements is false?\n\nt\u00bd = ln 2 / k\nt\u00bd = 1 / k [A]\u2080\nt\u00bd = [A]\u2080 / 2k\n\n[A]\u2080 \u2191 t\u00bd \u2191\n\nA \u2192 B\nA \u2193\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant..\n A is going to decrease. So as the reaction proceeds, the concentration of A will not be constant. It's going to decrease over time. So if A goes down, the half-life is going to go down with it because they are directly related.", null, "We can see these text from the image: 3. Which of the following statements is false?\n\nt1/2 = ln 2 / k\nt1/2 = 1 / k [A]0\nt1/2 = [A]0 / 2k\n\n[A]0 \u2191 t1/2 \u2191\n\nA \u2192 B\nA \u2193 t1/2 \u2193\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant..\n So therefore, the half-life of a zero-order reaction is not constant. This makes statement D false. Note that the half-life of a zero-order reaction and a second-order reaction depends on the initial concentration of A. Therefore, D is the correct answer choice." ], "metadata": [ { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/1033.24_1053.24.mp4", "refined_asr": " Whenever you increase the numerator of a fraction, the value of the fraction goes up. If you increase the denominator, the value of the fraction goes down. So, because the initial concentration is in the numerator, there is a direct relationship. As we increase the initial concentration.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1033.24_1053.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1033.24_1053.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1033.24_1053.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1033.24_1053.24#3.jpg" ], "ocr_qwen2_vl_72b": "3. Which of the following statements is false?\n\nt\u00bd = ln 2 / k\nt\u00bd = 1 / k [A]\u2080\nt\u00bd = [A]\u2080 / 2k\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant." }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/1053.24_1073.24.mp4", "refined_asr": " The half-life is going to increase as well. So there is a direct relationship between the two. So C is the true statement: these two are directly proportional. Now what about answer choice D? The half-life of a zero-order reaction is constant.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1053.24_1073.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1053.24_1073.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1053.24_1073.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1053.24_1073.24#3.jpg" ], "ocr_qwen2_vl_72b": "3. Which of the following statements is false?\n\nt\u00bd = ln 2\nk\n\nt\u00bd = 1\nk[A]\u2080\n\nt\u00bd = [A]\u2080\n2k\n\n[A]\u2080 \u2191 t\u00bd \u2191\n\nk \u2191 t\u00bd \u2193\nd \u2191 f\u2193\n\nThe half life of a first order reaction is independent of the concentration of the reactant.\nThe rate constant K is inversely related to the half life of a second order reaction.\nThe half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nThe half life of a zero order reaction is constant.\nThe half life of a second order reaction is inversely proportional to the initial concentration of the reactant." }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/1073.24_1092.24.mp4", "refined_asr": " Is that true or false? So this is the half-life for a zero-order reaction. Is this value T\u00bd? Is it constant? Yes, T\u00bd depends on the initial concentration and k. The '2' is not going to change; that's constant.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1073.24_1092.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1073.24_1092.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1073.24_1092.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1073.24_1092.24#3.jpg" ], "ocr_qwen2_vl_72b": "3. Which of the following statements is false?\n\nt1/2 = ln 2 / k\nt1/2 = 1 / k [A]0\nt1/2 = [A]0 / 2k\n\n[A]0 \u2191 t1/2 \u2191\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant." }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/1092.24_1111.24.mp4", "refined_asr": " K is the rate constant of a reaction. The fact that it's a rate constant means that K doesn't change, that is, if the temperature doesn't change. So we're assuming that the temperature remains constant. If the temperature is constant, then K is going to be constant.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1092.24_1111.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1092.24_1111.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1092.24_1111.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1092.24_1111.24#3.jpg" ], "ocr_qwen2_vl_72b": "3. Which of the following statements is false?\n\nt\u00bd = ln 2 / k\nt\u00bd = 1 / k[A]\u2080\nt\u00bd = [A]\u2080 / 2k\n\n[A]\u2080 \u2191 t\u00bd \u2191\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant." }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/1111.24_1133.24.mp4", "refined_asr": " So at constant temperature K is not going to change. Now, the initial concentration, is that constant? Well, the answer is no. Because the purpose of chemical kinetics is to study how fast reactions are going. And if the reaction is going, as A changes to B.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1111.24_1133.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1111.24_1133.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1111.24_1133.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1111.24_1133.24#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1111.24_1133.24#4.jpg" ], "ocr_qwen2_vl_72b": "3. Which of the following statements is false?\n\nt\u00bd = ln 2 / k\nt\u00bd = 1 / k [A]\u2080\nt\u00bd = [A]\u2080 / 2k\n\n[A]\u2080 \u2191 t\u00bd \u2191\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant." }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/1133.24_1148.24.mp4", "refined_asr": " A is going to decrease. So as the reaction proceeds, the concentration of A will not be constant. It's going to decrease over time. So if A goes down, the half-life is going to go down with it because they are directly related.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1133.24_1148.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1133.24_1148.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1133.24_1148.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1133.24_1148.24#3.jpg" ], "ocr_qwen2_vl_72b": "3. Which of the following statements is false?\n\nt\u00bd = ln 2 / k\nt\u00bd = 1 / k [A]\u2080\nt\u00bd = [A]\u2080 / 2k\n\n[A]\u2080 \u2191 t\u00bd \u2191\n\nA \u2192 B\nA \u2193\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant." }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/1148.24_1170.24.mp4", "refined_asr": " So therefore, the half-life of a zero-order reaction is not constant. This makes statement D false. Note that the half-life of a zero-order reaction and a second-order reaction depends on the initial concentration of A. Therefore, D is the correct answer choice.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1148.24_1170.24#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1148.24_1170.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1148.24_1170.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1148.24_1170.24#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@1148.24_1170.24#4.jpg" ], "ocr_qwen2_vl_72b": "3. Which of the following statements is false?\n\nt1/2 = ln 2 / k\nt1/2 = 1 / k [A]0\nt1/2 = [A]0 / 2k\n\n[A]0 \u2191 t1/2 \u2191\n\nA \u2192 B\nA \u2193 t1/2 \u2193\n\nA. The half life of a first order reaction is independent of the concentration of the reactant.\nB. The rate constant K is inversely related to the half life of a second order reaction.\nC. The half life of a zero order reaction is directly proportional to the initial concentration of the reactant.\nD. The half life of a zero order reaction is constant.\nE. The half life of a second order reaction is inversely proportional to the initial concentration of the reactant." } ], "image_num": 7, "text_num": 410, "token_num": 4442 }, { "images": [ "sample_100_images/7I0Xg92_eA4@2474.2400000000002_2488.2400000000002#1.jpg", null, "sample_100_images/7I0Xg92_eA4@2488.2400000000002_2506.2400000000002#1.jpg", null, "sample_100_images/7I0Xg92_eA4@2506.2400000000002_2526.2400000000002#1.jpg", null, "sample_100_images/7I0Xg92_eA4@2526.2400000000002_2546.2400000000002#1.jpg", null, "sample_100_images/7I0Xg92_eA4@2546.2400000000002_2558.2400000000002#1.jpg", null, "sample_100_images/7I0Xg92_eA4@2558.2400000000002_2576.2400000000002#1.jpg", null, "sample_100_images/7I0Xg92_eA4@2576.2400000000002_2590.2400000000002#1.jpg", null, "sample_100_images/7I0Xg92_eA4@2590.2400000000002_2612.2400000000002#1.jpg", "sample_100_images/7I0Xg92_eA4@2590.2400000000002_2612.2400000000002#2.jpg", null ], "texts": [ null, " N is the order of the reaction times T raised to the minus one. So N is not four, N is five rather, and the time is in units of seconds. So this is going to be seconds to the minus one.", null, " We simply replace t with whatever time unit we see in the reaction rate. So it's m raised to the negative 4 s to the minus 1. Let's save this answer. So if you were to ever forget this formula here's another way in which you can derive it.", null, " Get to this point. So in the rate law expression, isolate k. To get k by itself, we need to divide both sides by this. So we're going to get the rate constant, k, is equal to the rate of the entire reaction divided by A squared.", null, " Times b to the third power. So at this point, what you want to do is you want to plug in the units. The unit for rate is molarity over seconds, or you can say molarity times seconds to the minus one. The unit for a is:", null, " Molarity but it's squared. So you can write that as m squared or simply m times m. So that's for a. For b it's m to the third power. So I'm not going to split it for that one.", null, " The reason why I split this into m times m is because I want to cancel one of the m's. So we can see that we have m to the 4th left on the bottom. 1 plus 3 is 4. Now we can move this to the top.", null, " And then we get the units of K as being meters to the negative four, plus the minus one. So you get the same answer as what we got here, but this is just simply a shortcut technique. But you can always do it this way if you want to.", null, null, " If you ever forget now this does not look like any of the answers that we have here so we need to adjust our units molarity is moles over liters this is moles to the negative one power liters to the negative one power" ], "text_ocr_list": [ null, "We can see these text from the image: 9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nn = 5 + h\n\nM^(-n) t^(-1)\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1.\n N is the order of the reaction times T raised to the minus one. So N is not four, N is five rather, and the time is in units of seconds. So this is going to be seconds to the minus one.", null, "We can see these text from the image: 9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nn = 5+1\nM^(-1) t^(-1)\nM^(-5) s^(-1)\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1.\n We simply replace t with whatever time unit we see in the reaction rate. So it's m raised to the negative 4 s to the minus 1. Let's save this answer. So if you were to ever forget this formula here's another way in which you can derive it.", null, "We can see these text from the image: 9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nn = 5 + h\nm = 1 - n\n1 - 5\nM\n- 4\nM\nS^(-1)\n- 1\nM\nS^(-1)\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1.\n Get to this point. So in the rate law expression, isolate k. To get k by itself, we need to divide both sides by this. So we're going to get the rate constant, k, is equal to the rate of the entire reaction divided by A squared.", null, "We can see these text from the image: 9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nk = Rate / [A]^2[B]^3\n\nn = 5 + h\n\n1 - n t^(-1)\n\n1 - 5 s^(-1)\n\nM^(-5) s^(-1)\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1.\n Times b to the third power. So at this point, what you want to do is you want to plug in the units. The unit for rate is molarity over seconds, or you can say molarity times seconds to the minus one. The unit for a is:", null, "We can see these text from the image: 9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nk = Rate / [A]^2[B]^3\n\nk = M\u00b7s^(-1)\n\nn = 5 + h\n\nM\u00b7t^(-1)\n\n1 - 5\n\nM\u00b7s^(-1)\n\n- 9\n\nM\u00b7s^(-1).\n Molarity but it's squared. So you can write that as m squared or simply m times m. So that's for a. For b it's m to the third power. So I'm not going to split it for that one.", null, "We can see these text from the image: 9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nk = Rate / [A]^2[B]^3\n\nk = M/s / (M^2 * M^3)\n\nk = M/s / (M^5)\n\nk = M^-4 s^-1\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1.\n The reason why I split this into m times m is because I want to cancel one of the m's. So we can see that we have m to the 4th left on the bottom. 1 plus 3 is 4. Now we can move this to the top.", null, "We can see these text from the image: 9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nk = Rate / [A]^2[B]^3\n\nk = (M\u00b7s^-1) / ([M]^2[M]^3)\n\nk = M^-4\u00b7s^-1\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1.\n And then we get the units of K as being meters to the negative four, plus the minus one. So you get the same answer as what we got here, but this is just simply a shortcut technique. But you can always do it this way if you want to.", null, null, "We can see these text from the image: 9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nk = Rate / [A]^2[B]^3\n\nk = M\u00b7s^-1 / (M\u00b7M)^2\u00b7M^3\n\nk = s^-1 / M^4\n\nk = M^-4\u00b7s^-1\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1.\n If you ever forget now this does not look like any of the answers that we have here so we need to adjust our units molarity is moles over liters this is moles to the negative one power liters to the negative one power" ], "metadata": [ { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/2474.2400000000002_2488.2400000000002.mp4", "refined_asr": " N is the order of the reaction times T raised to the minus one. So N is not four, N is five rather, and the time is in units of seconds. So this is going to be seconds to the minus one.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2474.2400000000002_2488.2400000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2474.2400000000002_2488.2400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2474.2400000000002_2488.2400000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2474.2400000000002_2488.2400000000002#3.jpg" ], "ocr_qwen2_vl_72b": "9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nn = 5 + h\n\nM^(-n) t^(-1)\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1" }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/2488.2400000000002_2506.2400000000002.mp4", "refined_asr": " We simply replace t with whatever time unit we see in the reaction rate. So it's m raised to the negative 4 s to the minus 1. Let's save this answer. So if you were to ever forget this formula here's another way in which you can derive it.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2488.2400000000002_2506.2400000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2488.2400000000002_2506.2400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2488.2400000000002_2506.2400000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2488.2400000000002_2506.2400000000002#3.jpg" ], "ocr_qwen2_vl_72b": "9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nn = 5+1\nM^(-1) t^(-1)\nM^(-5) s^(-1)\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1" }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/2506.2400000000002_2526.2400000000002.mp4", "refined_asr": " Get to this point. So in the rate law expression, isolate k. To get k by itself, we need to divide both sides by this. So we're going to get the rate constant, k, is equal to the rate of the entire reaction divided by A squared.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2506.2400000000002_2526.2400000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2506.2400000000002_2526.2400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2506.2400000000002_2526.2400000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2506.2400000000002_2526.2400000000002#3.jpg" ], "ocr_qwen2_vl_72b": "9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nn = 5 + h\nm = 1 - n\n1 - 5\nM\n- 4\nM\nS^(-1)\n- 1\nM\nS^(-1)\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1" }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/2526.2400000000002_2546.2400000000002.mp4", "refined_asr": " Times b to the third power. So at this point, what you want to do is you want to plug in the units. The unit for rate is molarity over seconds, or you can say molarity times seconds to the minus one. The unit for a is:", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2526.2400000000002_2546.2400000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2526.2400000000002_2546.2400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2526.2400000000002_2546.2400000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2526.2400000000002_2546.2400000000002#3.jpg" ], "ocr_qwen2_vl_72b": "9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nk = Rate / [A]^2[B]^3\n\nn = 5 + h\n\n1 - n t^(-1)\n\n1 - 5 s^(-1)\n\nM^(-5) s^(-1)\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1" }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/2546.2400000000002_2558.2400000000002.mp4", "refined_asr": " Molarity but it's squared. So you can write that as m squared or simply m times m. So that's for a. For b it's m to the third power. So I'm not going to split it for that one.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2546.2400000000002_2558.2400000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2546.2400000000002_2558.2400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2546.2400000000002_2558.2400000000002#2.jpg" ], "ocr_qwen2_vl_72b": "9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nk = Rate / [A]^2[B]^3\n\nk = M\u00b7s^(-1)\n\nn = 5 + h\n\nM\u00b7t^(-1)\n\n1 - 5\n\nM\u00b7s^(-1)\n\n- 9\n\nM\u00b7s^(-1)" }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/2558.2400000000002_2576.2400000000002.mp4", "refined_asr": " The reason why I split this into m times m is because I want to cancel one of the m's. So we can see that we have m to the 4th left on the bottom. 1 plus 3 is 4. Now we can move this to the top.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2558.2400000000002_2576.2400000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2558.2400000000002_2576.2400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2558.2400000000002_2576.2400000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2558.2400000000002_2576.2400000000002#3.jpg" ], "ocr_qwen2_vl_72b": "9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nk = Rate / [A]^2[B]^3\n\nk = M/s / (M^2 * M^3)\n\nk = M/s / (M^5)\n\nk = M^-4 s^-1\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1" }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/2576.2400000000002_2590.2400000000002.mp4", "refined_asr": " And then we get the units of K as being meters to the negative four, plus the minus one. So you get the same answer as what we got here, but this is just simply a shortcut technique. But you can always do it this way if you want to.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2576.2400000000002_2590.2400000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2576.2400000000002_2590.2400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2576.2400000000002_2590.2400000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2576.2400000000002_2590.2400000000002#3.jpg" ], "ocr_qwen2_vl_72b": "9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nk = Rate / [A]^2[B]^3\n\nk = (M\u00b7s^-1) / ([M]^2[M]^3)\n\nk = M^-4\u00b7s^-1\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1" }, { "vid": "7I0Xg92_eA4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Chemical Engineering Course on Chemical Kinetics focusing on Catalysis_14.json#####audio#####doingASR#####FinishASR/7I0Xg92_eA4/2590.2400000000002_2612.2400000000002.mp4", "refined_asr": " If you ever forget now this does not look like any of the answers that we have here so we need to adjust our units molarity is moles over liters this is moles to the negative one power liters to the negative one power", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2590.2400000000002_2612.2400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2590.2400000000002_2612.2400000000002#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2590.2400000000002_2612.2400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2590.2400000000002_2612.2400000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2590.2400000000002_2612.2400000000002#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/7I0Xg92_eA4/7I0Xg92_eA4@2590.2400000000002_2612.2400000000002#4.jpg" ], "ocr_qwen2_vl_72b": "9. The rate of a certain reaction with units of M/s increases by a factor of 4 when [A] doubles and increases by a factor of 27 when [B] triples. Which of the following represents the units of the rate constant k for this reaction?\n\nRate = k[A]^2[B]^3\n\nk = Rate / [A]^2[B]^3\n\nk = M\u00b7s^-1 / (M\u00b7M)^2\u00b7M^3\n\nk = s^-1 / M^4\n\nk = M^-4\u00b7s^-1\n\nA. L^3 mol^-3 s^-1\nB. mol^3 L^-3 s^-1\nC. L^4 mol^-4 s^-1\nD. mol^4 L^-4 s^-1\nE. M^-5 s^-1" } ], "image_num": 9, "text_num": 452, "token_num": 5636 }, { "images": [ "sample_100_images/lnj3sN2Tx3Y@412.62000000000006_442.56000000000006#1.jpg", null, "sample_100_images/lnj3sN2Tx3Y@442.56000000000006_471.00000000000006#1.jpg", "sample_100_images/lnj3sN2Tx3Y@442.56000000000006_471.00000000000006#4.jpg", null, "sample_100_images/lnj3sN2Tx3Y@471.00000000000006_495.76000000000005#1.jpg", null, "sample_100_images/lnj3sN2Tx3Y@495.76_519.58#1.jpg", null, "sample_100_images/lnj3sN2Tx3Y@519.58_543.1#1.jpg", null ], "texts": [ null, " The first activity is a financial feasibility study, which is one of the important studies to assess the financial aspects before starting a business. It considers startup capital, expenses, revenues, etc. The next activity is the voice of the customer. This is one of the most important activities because your complete design and other activities will be driven based on it.", null, null, " On the requirements we set based on the voice of the customer. So it's the collection of information about the customer expectations, the preferences, and requirements of the product. And the last activity is patent research. This is a search of published patents. If the product is new and unique, there is always a possibility of applying for one.", null, " A patent related to a new product or a new invention is identified at this stage. Basically, at this phase you research and analyze different possibilities and see what best works for your product. So, we'll get into the next phase; it's a planning phase. So, we'll be doing a number of different things.", null, " Plans in this phase - so the first one is project plan. Project planning is the planning related to project execution and project control. This project plan includes planning related to the timeline, the planning related to the number of resources used for the project, and the activities involved as well.", null, " The identification of project deliverables. Next is design and development planning. This describes the overall development program and includes the steps and activities to follow to reach your final product. So, the next plan is the risk management plan. The risk management plan is to foresee risks." ], "text_ocr_list": [ null, "We can see these text from the image: IDEA/INITIATION\n\n* MARKET ANALYSIS\n- STRENGTH\n- WEAKNESS\n- OPPORTUNITIES\n\n* COMPETITIVE ASSESSMENT - CURRENT COMPETITORS * PRODUCTS\n\n* EARLY RISK ASSESSMENT\n\n* FINANCIAL FEASIBILITY STUDY\n\nPHASE 1.\n The first activity is a financial feasibility study, which is one of the important studies to assess the financial aspects before starting a business. It considers startup capital, expenses, revenues, etc. The next activity is the voice of the customer. This is one of the most important activities because your complete design and other activities will be driven based on it.", null, null, "We can see these text from the image: IDEA/INITIATION\n\n* MARKET ANALYSIS\n- STRENGTH\n- WEAKNESS\n- OPPORTUNITIES\n\n* COMPETITIVE ASSESSMENT - CURRENT COMPETITORS * PRODUCTS\n\n* EARLY RISK ASSESSMENT\n\n* FINANCIAL FEASIBILITY STUDY\n\n* VOICE OF CUSTOMER\n- EXPECTATIONS\n\nPHASE 1.\n On the requirements we set based on the voice of the customer. So it's the collection of information about the customer expectations, the preferences, and requirements of the product. And the last activity is patent research. This is a search of published patents. If the product is new and unique, there is always a possibility of applying for one.", null, "We can see these text from the image: PLANNING.\n A patent related to a new product or a new invention is identified at this stage. Basically, at this phase you research and analyze different possibilities and see what best works for your product. So, we'll get into the next phase; it's a planning phase. So, we'll be doing a number of different things.", null, "We can see these text from the image: PLANNING\n\nPHASE 2.\n Plans in this phase - so the first one is project plan. Project planning is the planning related to project execution and project control. This project plan includes planning related to the timeline, the planning related to the number of resources used for the project, and the activities involved as well.", null, "We can see these text from the image: PLANNING\n\n* PROJECT PLAN\n- EXECUTION\n- CONTROL\n- TIMELINE\n- RESOURCES\n\n+ DESIGN & DEVELOPMENT PLAN\n\nPHASE 2.\n The identification of project deliverables. Next is design and development planning. This describes the overall development program and includes the steps and activities to follow to reach your final product. So, the next plan is the risk management plan. The risk management plan is to foresee risks." ], "metadata": [ { "vid": "lnj3sN2Tx3Y.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering tutorials on Design and Development of Medical Devices in Biomedical Devices and Systems course_30.json#####audio#####doingASR#####FinishASR/lnj3sN2Tx3Y/412.62000000000006_442.56000000000006.mp4", "refined_asr": " The first activity is a financial feasibility study, which is one of the important studies to assess the financial aspects before starting a business. It considers startup capital, expenses, revenues, etc. The next activity is the voice of the customer. This is one of the most important activities because your complete design and other activities will be driven based on it.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@412.62000000000006_442.56000000000006#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@412.62000000000006_442.56000000000006#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@412.62000000000006_442.56000000000006#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@412.62000000000006_442.56000000000006#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@412.62000000000006_442.56000000000006#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@412.62000000000006_442.56000000000006#5.jpg" ], "ocr_qwen2_vl_72b": "IDEA/INITIATION\n\n* MARKET ANALYSIS\n- STRENGTH\n- WEAKNESS\n- OPPORTUNITIES\n\n* COMPETITIVE ASSESSMENT - CURRENT COMPETITORS * PRODUCTS\n\n* EARLY RISK ASSESSMENT\n\n* FINANCIAL FEASIBILITY STUDY\n\nPHASE 1" }, { "vid": "lnj3sN2Tx3Y.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering tutorials on Design and Development of Medical Devices in Biomedical Devices and Systems course_30.json#####audio#####doingASR#####FinishASR/lnj3sN2Tx3Y/442.56000000000006_471.00000000000006.mp4", "refined_asr": " On the requirements we set based on the voice of the customer. So it's the collection of information about the customer expectations, the preferences, and requirements of the product. And the last activity is patent research. This is a search of published patents. If the product is new and unique, there is always a possibility of applying for one.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@442.56000000000006_471.00000000000006#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@442.56000000000006_471.00000000000006#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@442.56000000000006_471.00000000000006#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@442.56000000000006_471.00000000000006#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@442.56000000000006_471.00000000000006#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@442.56000000000006_471.00000000000006#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@442.56000000000006_471.00000000000006#5.jpg" ], "ocr_qwen2_vl_72b": "IDEA/INITIATION\n\n* MARKET ANALYSIS\n- STRENGTH\n- WEAKNESS\n- OPPORTUNITIES\n\n* COMPETITIVE ASSESSMENT - CURRENT COMPETITORS * PRODUCTS\n\n* EARLY RISK ASSESSMENT\n\n* FINANCIAL FEASIBILITY STUDY\n\n* VOICE OF CUSTOMER\n- EXPECTATIONS\n\nPHASE 1" }, { "vid": "lnj3sN2Tx3Y.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering tutorials on Design and Development of Medical Devices in Biomedical Devices and Systems course_30.json#####audio#####doingASR#####FinishASR/lnj3sN2Tx3Y/471.00000000000006_495.76000000000005.mp4", "refined_asr": " A patent related to a new product or a new invention is identified at this stage. Basically, at this phase you research and analyze different possibilities and see what best works for your product. So, we'll get into the next phase; it's a planning phase. So, we'll be doing a number of different things.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@471.00000000000006_495.76000000000005#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@471.00000000000006_495.76000000000005#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@471.00000000000006_495.76000000000005#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@471.00000000000006_495.76000000000005#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@471.00000000000006_495.76000000000005#4.jpg" ], "ocr_qwen2_vl_72b": "PLANNING" }, { "vid": "lnj3sN2Tx3Y.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering tutorials on Design and Development of Medical Devices in Biomedical Devices and Systems course_30.json#####audio#####doingASR#####FinishASR/lnj3sN2Tx3Y/495.76_519.58.mp4", "refined_asr": " Plans in this phase - so the first one is project plan. Project planning is the planning related to project execution and project control. This project plan includes planning related to the timeline, the planning related to the number of resources used for the project, and the activities involved as well.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@495.76_519.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@495.76_519.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@495.76_519.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@495.76_519.58#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@495.76_519.58#4.jpg" ], "ocr_qwen2_vl_72b": "PLANNING\n\nPHASE 2" }, { "vid": "lnj3sN2Tx3Y.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Bioengineering tutorials on Design and Development of Medical Devices in Biomedical Devices and Systems course_30.json#####audio#####doingASR#####FinishASR/lnj3sN2Tx3Y/519.58_543.1.mp4", "refined_asr": " The identification of project deliverables. Next is design and development planning. This describes the overall development program and includes the steps and activities to follow to reach your final product. So, the next plan is the risk management plan. The risk management plan is to foresee risks.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@519.58_543.1#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@519.58_543.1#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@519.58_543.1#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@519.58_543.1#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lnj3sN2Tx3Y/lnj3sN2Tx3Y@519.58_543.1#4.jpg" ], "ocr_qwen2_vl_72b": "PLANNING\n\n* PROJECT PLAN\n- EXECUTION\n- CONTROL\n- TIMELINE\n- RESOURCES\n\n+ DESIGN & DEVELOPMENT PLAN\n\nPHASE 2" } ], "image_num": 6, "text_num": 336, "token_num": 3792 }, { "images": [ "sample_100_images/rA3KEH3B-rM@695.28_711.5400000000001#1.jpg", "sample_100_images/rA3KEH3B-rM@695.28_711.5400000000001#2.jpg", null, "sample_100_images/rA3KEH3B-rM@711.5400000000001_730.0200000000001#1.jpg", "sample_100_images/rA3KEH3B-rM@711.5400000000001_730.0200000000001#2.jpg", "sample_100_images/rA3KEH3B-rM@711.5400000000001_730.0200000000001#3.jpg", null, "sample_100_images/rA3KEH3B-rM@730.0200000000001_746.3000000000001#1.jpg", "sample_100_images/rA3KEH3B-rM@730.0200000000001_746.3000000000001#2.jpg", null, "sample_100_images/rA3KEH3B-rM@746.3000000000001_762.46#1.jpg", "sample_100_images/rA3KEH3B-rM@746.3000000000001_762.46#2.jpg", null, "sample_100_images/rA3KEH3B-rM@762.46_776.32#1.jpg", "sample_100_images/rA3KEH3B-rM@762.46_776.32#2.jpg", null, "sample_100_images/rA3KEH3B-rM@776.4200000000001_791.0600000000001#1.jpg", "sample_100_images/rA3KEH3B-rM@776.4200000000001_791.0600000000001#2.jpg", null, "sample_100_images/rA3KEH3B-rM@791.0600000000001_819.4200000000001#1.jpg", "sample_100_images/rA3KEH3B-rM@791.0600000000001_819.4200000000001#2.jpg", "sample_100_images/rA3KEH3B-rM@791.0600000000001_819.4200000000001#3.jpg", "sample_100_images/rA3KEH3B-rM@791.0600000000001_819.4200000000001#4.jpg", null ], "texts": [ null, null, " In order to do that, we don't want to vary these factors. For our examples here, we have to hold the rate in at constant; you're pouring in at the same rate. We have to hold the concentration in at constant; what you're pouring in all has the same concentration. And then we have to hold the rate out constant as well.", null, null, null, " What's leaving the tank, how fast it's leaving the tank is constant. What's not constant, what we can't hold constant, is the concentration. You see, unless you stir the container and what you're pouring in has exactly the same concentration, what you're pouring out is.", null, null, " Always changing. If you have this pure water and you start pouring in a concentration of salt, it's going to start getting more and more salty until you start draining stuff out. So that concentration out is always changing. Now there is one thing.", null, null, " We're going to assume that in this vat or this lake or whatever, it is very well mixed. So there's this mixer going crazy. As soon as you add something to it, it's immediately perfectly mixed. Realistic? I don't know. But otherwise, remember that trade-off.", null, null, " Often in differential equations, we want to ensure we're not complicating matters unnecessarily. We aim to keep problems practical yet solvable. One assumption we often make for practicality and manageability is that the substances are well-mixed. Without this assumption, dealing with non-uniformly mixed substances can significantly complicate our models.", null, null, " We have different concentrations in the same vat, very hard to deal with. So, rate in held the same, concentration in held the same. I hope you understand why we kind of have to do that. Otherwise, we'd have several different problems. Rate out held the same. So,", null, null, null, null, " You're letting it drain at the same rate. But the concentration out is changing because what you're adding into it and what you're pouring out of it affects how much is actually there. Let's talk a bit more about that. The concentration out: If it's not constant, what's it based on?" ], "text_ocr_list": [ null, null, "We can see these text from the image: Applications for\nLinear 1st Order Diff Eq.\n\nMixture Problems...\nHow a \"Solution\"\nChanges over Time\n\nAmount of solute X(t)\n\nIN - Amount OUT\nConcentration * Rate - Rate * Concentration\nOut Out\n\n- C0 * Co * dt.\n In order to do that, we don't want to vary these factors. For our examples here, we have to hold the rate in at constant; you're pouring in at the same rate. We have to hold the concentration in at constant; what you're pouring in all has the same concentration. And then we have to hold the rate out constant as well.", null, null, null, "We can see these text from the image: Applications for\nLinear 1st Order Diff. Eq.\n\nMixture Problems...\nHow a \"Solution\"\nChanges over Time\n\nAmount of So...\n\nIn - Amount Out\nConcentration * Rate * Conc. Time\nOut * Time.\n What's leaving the tank, how fast it's leaving the tank is constant. What's not constant, what we can't hold constant, is the concentration. You see, unless you stir the container and what you're pouring in has exactly the same concentration, what you're pouring out is.", null, null, "We can see these text from the image: - Applications for Linear 1st Order Diff Eq.\n- Mixture Problems...\n- How a \"Solution\" Changes over Time\n- Amount of Solute\n- IN - Amount OUT\n- CONC. TIME = RATE * CONC. TIME\n- OUT.\n Always changing. If you have this pure water and you start pouring in a concentration of salt, it's going to start getting more and more salty until you start draining stuff out. So that concentration out is always changing. Now there is one thing.", null, null, "We can see these text from the image: - Applications for Linear 1st Order Diff Eq.\n- Mixture Problems...\n- How a \"Solution\" Changes over Time\n- Amount of Solute X(t)\n- IN - AMOUNT OUT\n- FLOW RATE * CONC. TIME\n- OUT = R * C * DT.\n We're going to assume that in this vat or this lake or whatever, it is very well mixed. So there's this mixer going crazy. As soon as you add something to it, it's immediately perfectly mixed. Realistic? I don't know. But otherwise, remember that trade-off.", null, null, "We can see these text from the image: - Applications for Linear 1st Order Diff Eq.\n- Mixture Prob\n- How a \"Solution\" Changes over Time\n- Amount X(t)\n- Amount In - Amount Out\n- Concentration * Rate * Conc * Time Out\n- dX/dt = C_in * C_out * dt.\n Often in differential equations, we want to ensure we're not complicating matters unnecessarily. We aim to keep problems practical yet solvable. One assumption we often make for practicality and manageability is that the substances are well-mixed. Without this assumption, dealing with non-uniformly mixed substances can significantly complicate our models.", null, null, "We can see these text from the image: - Applications for Linear 1st Order Diff Eq.\n- Mixture Problems\n- How a \"Solute\" Changes over\n- Amount of X(t)\n- IN - Amount OUT\n- INFLOW = RATE * CONC * TIME\n- OUTFLOW = C0 * C0 * DT.\n We have different concentrations in the same vat, very hard to deal with. So, rate in held the same, concentration in held the same. I hope you understand why we kind of have to do that. Otherwise, we'd have several different problems. Rate out held the same. So,", null, null, null, null, "We can see these text from the image: - Applications for Linear 1st Order Diff Eq.\n- Mixture Problems\n- How a \"Solution\" Changes over Time\n- Amount of X(t)\n- Amount IN - Amount OUT\n- Concentration * Rate IN - Rate * Concentration OUT\n- dX/dt = C_in * C_out * dt.\n You're letting it drain at the same rate. But the concentration out is changing because what you're adding into it and what you're pouring out of it affects how much is actually there. Let's talk a bit more about that. The concentration out: If it's not constant, what's it based on?" ], "metadata": [ { "vid": "rA3KEH3B-rM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Modeling with First Order Differential Equations_30.json#####audio#####doingASR#####FinishASR/rA3KEH3B-rM/695.28_711.5400000000001.mp4", "refined_asr": " In order to do that, we don't want to vary these factors. For our examples here, we have to hold the rate in at constant; you're pouring in at the same rate. We have to hold the concentration in at constant; what you're pouring in all has the same concentration. And then we have to hold the rate out constant as well.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@695.28_711.5400000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@695.28_711.5400000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@695.28_711.5400000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@695.28_711.5400000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@695.28_711.5400000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Applications for\nLinear 1st Order Diff Eq.\n\nMixture Problems...\nHow a \"Solution\"\nChanges over Time\n\nAmount of solute X(t)\n\nIN - Amount OUT\nConcentration * Rate - Rate * Concentration\nOut Out\n\n- C0 * Co * dt" }, { "vid": "rA3KEH3B-rM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Modeling with First Order Differential Equations_30.json#####audio#####doingASR#####FinishASR/rA3KEH3B-rM/711.5400000000001_730.0200000000001.mp4", "refined_asr": " What's leaving the tank, how fast it's leaving the tank is constant. What's not constant, what we can't hold constant, is the concentration. You see, unless you stir the container and what you're pouring in has exactly the same concentration, what you're pouring out is.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@711.5400000000001_730.0200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@711.5400000000001_730.0200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@711.5400000000001_730.0200000000001#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@711.5400000000001_730.0200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@711.5400000000001_730.0200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@711.5400000000001_730.0200000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Applications for\nLinear 1st Order Diff. Eq.\n\nMixture Problems...\nHow a \"Solution\"\nChanges over Time\n\nAmount of So...\n\nIn - Amount Out\nConcentration * Rate * Conc. Time\nOut * Time" }, { "vid": "rA3KEH3B-rM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Modeling with First Order Differential Equations_30.json#####audio#####doingASR#####FinishASR/rA3KEH3B-rM/730.0200000000001_746.3000000000001.mp4", "refined_asr": " Always changing. If you have this pure water and you start pouring in a concentration of salt, it's going to start getting more and more salty until you start draining stuff out. So that concentration out is always changing. Now there is one thing.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@730.0200000000001_746.3000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@730.0200000000001_746.3000000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@730.0200000000001_746.3000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@730.0200000000001_746.3000000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@730.0200000000001_746.3000000000001#3.jpg" ], "ocr_qwen2_vl_72b": "- Applications for Linear 1st Order Diff Eq.\n- Mixture Problems...\n- How a \"Solution\" Changes over Time\n- Amount of Solute\n- IN - Amount OUT\n- CONC. TIME = RATE * CONC. TIME\n- OUT" }, { "vid": "rA3KEH3B-rM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Modeling with First Order Differential Equations_30.json#####audio#####doingASR#####FinishASR/rA3KEH3B-rM/746.3000000000001_762.46.mp4", "refined_asr": " We're going to assume that in this vat or this lake or whatever, it is very well mixed. So there's this mixer going crazy. As soon as you add something to it, it's immediately perfectly mixed. Realistic? I don't know. But otherwise, remember that trade-off.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@746.3000000000001_762.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@746.3000000000001_762.46#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@746.3000000000001_762.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@746.3000000000001_762.46#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@746.3000000000001_762.46#3.jpg" ], "ocr_qwen2_vl_72b": "- Applications for Linear 1st Order Diff Eq.\n- Mixture Problems...\n- How a \"Solution\" Changes over Time\n- Amount of Solute X(t)\n- IN - AMOUNT OUT\n- FLOW RATE * CONC. TIME\n- OUT = R * C * DT" }, { "vid": "rA3KEH3B-rM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Modeling with First Order Differential Equations_30.json#####audio#####doingASR#####FinishASR/rA3KEH3B-rM/762.46_776.32.mp4", "refined_asr": " Often in differential equations, we want to ensure we're not complicating matters unnecessarily. We aim to keep problems practical yet solvable. One assumption we often make for practicality and manageability is that the substances are well-mixed. Without this assumption, dealing with non-uniformly mixed substances can significantly complicate our models.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@762.46_776.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@762.46_776.32#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@762.46_776.32#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@762.46_776.32#2.jpg" ], "ocr_qwen2_vl_72b": "- Applications for Linear 1st Order Diff Eq.\n- Mixture Prob\n- How a \"Solution\" Changes over Time\n- Amount X(t)\n- Amount In - Amount Out\n- Concentration * Rate * Conc * Time Out\n- dX/dt = C_in * C_out * dt" }, { "vid": "rA3KEH3B-rM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Modeling with First Order Differential Equations_30.json#####audio#####doingASR#####FinishASR/rA3KEH3B-rM/776.4200000000001_791.0600000000001.mp4", "refined_asr": " We have different concentrations in the same vat, very hard to deal with. So, rate in held the same, concentration in held the same. I hope you understand why we kind of have to do that. Otherwise, we'd have several different problems. Rate out held the same. So,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@776.4200000000001_791.0600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@776.4200000000001_791.0600000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@776.4200000000001_791.0600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@776.4200000000001_791.0600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@776.4200000000001_791.0600000000001#3.jpg" ], "ocr_qwen2_vl_72b": "- Applications for Linear 1st Order Diff Eq.\n- Mixture Problems\n- How a \"Solute\" Changes over\n- Amount of X(t)\n- IN - Amount OUT\n- INFLOW = RATE * CONC * TIME\n- OUTFLOW = C0 * C0 * DT" }, { "vid": "rA3KEH3B-rM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Calculus tutorial on Modeling with First Order Differential Equations_30.json#####audio#####doingASR#####FinishASR/rA3KEH3B-rM/791.0600000000001_819.4200000000001.mp4", "refined_asr": " You're letting it drain at the same rate. But the concentration out is changing because what you're adding into it and what you're pouring out of it affects how much is actually there. Let's talk a bit more about that. The concentration out: If it's not constant, what's it based on?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@791.0600000000001_819.4200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@791.0600000000001_819.4200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@791.0600000000001_819.4200000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@791.0600000000001_819.4200000000001#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@791.0600000000001_819.4200000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@791.0600000000001_819.4200000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@791.0600000000001_819.4200000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@791.0600000000001_819.4200000000001#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/rA3KEH3B-rM/rA3KEH3B-rM@791.0600000000001_819.4200000000001#5.jpg" ], "ocr_qwen2_vl_72b": "- Applications for Linear 1st Order Diff Eq.\n- Mixture Problems\n- How a \"Solution\" Changes over Time\n- Amount of X(t)\n- Amount IN - Amount OUT\n- Concentration * Rate IN - Rate * Concentration OUT\n- dX/dt = C_in * C_out * dt" } ], "image_num": 17, "text_num": 483, "token_num": 10275 }, { "images": [ "sample_100_images/81eBix8hDMA@1835.26_1854.54#1.jpg", null, "sample_100_images/81eBix8hDMA@1855.1000000000001_1882.3000000000002#1.jpg", null, "sample_100_images/81eBix8hDMA@1882.3000000000002_1900.8000000000002#1.jpg", null, "sample_100_images/81eBix8hDMA@1900.8000000000002_1928.6000000000001#1.jpg", "sample_100_images/81eBix8hDMA@1900.8000000000002_1928.6000000000001#4.jpg", null, "sample_100_images/81eBix8hDMA@1928.6000000000001_1946.8600000000001#1.jpg", null, "sample_100_images/81eBix8hDMA@1947.7400000000002_1965.2600000000002#1.jpg", null, "sample_100_images/81eBix8hDMA@1965.3400000000001_1983.42#1.jpg", null ], "texts": [ null, " Let's start with the Recovery Appliance providing unique capabilities for ransomware protection. With the Recovery Appliance, you have a Cyber Vault with air gap settings. There's nothing particularly special about the configuration for the Recovery Appliance; it's essentially a basic setup. What is unique, however?", null, " For the secondary world configuration, it's that we do not have a persistent connection between the primary data center and the secondary data center. This means there's just yet another layer of defense, ensuring that if you have some sort of breach or issue with the primary data center's accessibility.", null, " In the backup environment, you're still going to be most likely protected because Cyber Vault is going to be initiating the reverse request for the replication on a random schedule. So most likely, bad actors are not going to be able to identify when actually a connection is made.", null, null, " It's going to be occurring between the Cyber Vault and the primary data center. So why is it important? Because if you take everything we've discussed until now, it's pretty evident that the Recovery Appliance Compliance Mode with immutable backup storage is a complementary solution to the RA Cyber Vault architecture overall. So putting it all together in a sense.", null, " Let's talk about operational availability and management. You have your Oracle databases, and as you know, the recovery appliance supports pretty much all the current releases from 12c to 21c, on any platform. You have real-time transaction protection, of course, and you have point-in-time recovery.", null, " Provided by the recovery appliance. From the management standpoint, if you can imagine a set of policies, this is basically just how it works overall. So you have local backups with a recovery window of 90 days, and you can set your compliance.", null, " Retention is set to 60 days. Then there's going to be a fan out for the replication to the Cyber Vault disaster recovery appliance. You set another policy for the recovery window to 14 days, compliance retention to seven days. Then you have your archiving tier, as in this particular case." ], "text_ocr_list": [ null, "We can see these text from the image: Ransomware Protection with RA Cyber Vault with AirGap\n\nPrimary Data Center\n\nPeriodic Sync to Otherwise Disconnected Vault\n\nCyber Vault.\n Let's start with the Recovery Appliance providing unique capabilities for ransomware protection. With the Recovery Appliance, you have a Cyber Vault with air gap settings. There's nothing particularly special about the configuration for the Recovery Appliance; it's essentially a basic setup. What is unique, however?", null, "We can see these text from the image: Ransomware Protection with RA Cyber Vault with AirGap\n\nPrimary Data Center\n\nPeriodic Sync to Otherwise Disconnected Vault\n\nCyber Vault.\n For the secondary world configuration, it's that we do not have a persistent connection between the primary data center and the secondary data center. This means there's just yet another layer of defense, ensuring that if you have some sort of breach or issue with the primary data center's accessibility.", null, "We can see these text from the image: Ransomware Protection with RA Cyber Vault with AirGap\n\nPrimary Data Center\n\nPeriodic Sync to Otherwise Disconnected Vault\n\nCyber Vault.\n In the backup environment, you're still going to be most likely protected because Cyber Vault is going to be initiating the reverse request for the replication on a random schedule. So most likely, bad actors are not going to be able to identify when actually a connection is made.", null, null, "We can see these text from the image: Ransomware Protection with RA Cyber Vault with AirGap\n\nRA Compliance Mode with Immutable Backup Storage is a perfect complementary solution to the RA Cyber Vault architecture.\n It's going to be occurring between the Cyber Vault and the primary data center. So why is it important? Because if you take everything we've discussed until now, it's pretty evident that the Recovery Appliance Compliance Mode with immutable backup storage is a complementary solution to the RA Cyber Vault architecture overall. So putting it all together in a sense.", null, "We can see these text from the image: End-to-End Compliance with Immutable Backups\n\nPutting It All Together\n\nReal-Time Transaction Protection\n\nOracle DB 12c-21c on Any Platform\n\nPoint-in-time Recovery, Fast Data Restore with Zero Data Loss.\n Let's talk about operational availability and management. You have your Oracle databases, and as you know, the recovery appliance supports pretty much all the current releases from 12c to 21c, on any platform. You have real-time transaction protection, of course, and you have point-in-time recovery.", null, "We can see these text from the image: End-to-End Compliance with Immutable Backups\n\nPutting It All Together\n\nReal-Time Transaction Protection\n\nLocal Backups\nRecovery Window Goal = 90 days\nCompliance Retention = 60 days\n\nOracle DB 12c-21c on Any Platform\n\nPoint-in-time Recovery, Fast Data Restore with Zero Data Loss.\n Provided by the recovery appliance. From the management standpoint, if you can imagine a set of policies, this is basically just how it works overall. So you have local backups with a recovery window of 90 days, and you can set your compliance.", null, "We can see these text from the image: End-to-End Compliance with Immutable Backups\n\nPutting It All Together\n\nReal-Time Transaction Protection\n\nLocal Backups\nRecovery Window Goal = 90 days\nCompliance Retention = 60 days\n\nCyber Vault / DR\nRecovery Window Goal = 14 days\nCompliance Retention = 7 days\n\nOracle DB 12c-21c on Any Platform\n\nPoint-in-time Recovery, Fast Data Restore with Zero Data Loss.\n Retention is set to 60 days. Then there's going to be a fan out for the replication to the Cyber Vault disaster recovery appliance. You set another policy for the recovery window to 14 days, compliance retention to seven days. Then you have your archiving tier, as in this particular case." ], "metadata": [ { "vid": "81eBix8hDMA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Computer Science Databases course on Database Security focusing on Backup and Recovery_30.json#####audio#####doingASR#####FinishASR/81eBix8hDMA/1835.26_1854.54.mp4", "refined_asr": " Let's start with the Recovery Appliance providing unique capabilities for ransomware protection. With the Recovery Appliance, you have a Cyber Vault with air gap settings. There's nothing particularly special about the configuration for the Recovery Appliance; it's essentially a basic setup. What is unique, however?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1835.26_1854.54#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1835.26_1854.54#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1835.26_1854.54#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1835.26_1854.54#3.jpg" ], "ocr_qwen2_vl_72b": "Ransomware Protection with RA Cyber Vault with AirGap\n\nPrimary Data Center\n\nPeriodic Sync to Otherwise Disconnected Vault\n\nCyber Vault" }, { "vid": "81eBix8hDMA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Computer Science Databases course on Database Security focusing on Backup and Recovery_30.json#####audio#####doingASR#####FinishASR/81eBix8hDMA/1855.1000000000001_1882.3000000000002.mp4", "refined_asr": " For the secondary world configuration, it's that we do not have a persistent connection between the primary data center and the secondary data center. This means there's just yet another layer of defense, ensuring that if you have some sort of breach or issue with the primary data center's accessibility.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1855.1000000000001_1882.3000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1855.1000000000001_1882.3000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1855.1000000000001_1882.3000000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1855.1000000000001_1882.3000000000002#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1855.1000000000001_1882.3000000000002#4.jpg" ], "ocr_qwen2_vl_72b": "Ransomware Protection with RA Cyber Vault with AirGap\n\nPrimary Data Center\n\nPeriodic Sync to Otherwise Disconnected Vault\n\nCyber Vault" }, { "vid": "81eBix8hDMA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Computer Science Databases course on Database Security focusing on Backup and Recovery_30.json#####audio#####doingASR#####FinishASR/81eBix8hDMA/1882.3000000000002_1900.8000000000002.mp4", "refined_asr": " In the backup environment, you're still going to be most likely protected because Cyber Vault is going to be initiating the reverse request for the replication on a random schedule. So most likely, bad actors are not going to be able to identify when actually a connection is made.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1882.3000000000002_1900.8000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1882.3000000000002_1900.8000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1882.3000000000002_1900.8000000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1882.3000000000002_1900.8000000000002#3.jpg" ], "ocr_qwen2_vl_72b": "Ransomware Protection with RA Cyber Vault with AirGap\n\nPrimary Data Center\n\nPeriodic Sync to Otherwise Disconnected Vault\n\nCyber Vault" }, { "vid": "81eBix8hDMA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Computer Science Databases course on Database Security focusing on Backup and Recovery_30.json#####audio#####doingASR#####FinishASR/81eBix8hDMA/1900.8000000000002_1928.6000000000001.mp4", "refined_asr": " It's going to be occurring between the Cyber Vault and the primary data center. So why is it important? Because if you take everything we've discussed until now, it's pretty evident that the Recovery Appliance Compliance Mode with immutable backup storage is a complementary solution to the RA Cyber Vault architecture overall. So putting it all together in a sense.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1900.8000000000002_1928.6000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1900.8000000000002_1928.6000000000001#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1900.8000000000002_1928.6000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1900.8000000000002_1928.6000000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1900.8000000000002_1928.6000000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1900.8000000000002_1928.6000000000001#4.jpg" ], "ocr_qwen2_vl_72b": "Ransomware Protection with RA Cyber Vault with AirGap\n\nRA Compliance Mode with Immutable Backup Storage is a perfect complementary solution to the RA Cyber Vault architecture" }, { "vid": "81eBix8hDMA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Computer Science Databases course on Database Security focusing on Backup and Recovery_30.json#####audio#####doingASR#####FinishASR/81eBix8hDMA/1928.6000000000001_1946.8600000000001.mp4", "refined_asr": " Let's talk about operational availability and management. You have your Oracle databases, and as you know, the recovery appliance supports pretty much all the current releases from 12c to 21c, on any platform. You have real-time transaction protection, of course, and you have point-in-time recovery.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1928.6000000000001_1946.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1928.6000000000001_1946.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1928.6000000000001_1946.8600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1928.6000000000001_1946.8600000000001#3.jpg" ], "ocr_qwen2_vl_72b": "End-to-End Compliance with Immutable Backups\n\nPutting It All Together\n\nReal-Time Transaction Protection\n\nOracle DB 12c-21c on Any Platform\n\nPoint-in-time Recovery, Fast Data Restore with Zero Data Loss" }, { "vid": "81eBix8hDMA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Computer Science Databases course on Database Security focusing on Backup and Recovery_30.json#####audio#####doingASR#####FinishASR/81eBix8hDMA/1947.7400000000002_1965.2600000000002.mp4", "refined_asr": " Provided by the recovery appliance. From the management standpoint, if you can imagine a set of policies, this is basically just how it works overall. So you have local backups with a recovery window of 90 days, and you can set your compliance.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1947.7400000000002_1965.2600000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1947.7400000000002_1965.2600000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1947.7400000000002_1965.2600000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1947.7400000000002_1965.2600000000002#3.jpg" ], "ocr_qwen2_vl_72b": "End-to-End Compliance with Immutable Backups\n\nPutting It All Together\n\nReal-Time Transaction Protection\n\nLocal Backups\nRecovery Window Goal = 90 days\nCompliance Retention = 60 days\n\nOracle DB 12c-21c on Any Platform\n\nPoint-in-time Recovery, Fast Data Restore with Zero Data Loss" }, { "vid": "81eBix8hDMA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Computer Science Databases course on Database Security focusing on Backup and Recovery_30.json#####audio#####doingASR#####FinishASR/81eBix8hDMA/1965.3400000000001_1983.42.mp4", "refined_asr": " Retention is set to 60 days. Then there's going to be a fan out for the replication to the Cyber Vault disaster recovery appliance. You set another policy for the recovery window to 14 days, compliance retention to seven days. Then you have your archiving tier, as in this particular case.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1965.3400000000001_1983.42#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1965.3400000000001_1983.42#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1965.3400000000001_1983.42#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/81eBix8hDMA/81eBix8hDMA@1965.3400000000001_1983.42#3.jpg" ], "ocr_qwen2_vl_72b": "End-to-End Compliance with Immutable Backups\n\nPutting It All Together\n\nReal-Time Transaction Protection\n\nLocal Backups\nRecovery Window Goal = 90 days\nCompliance Retention = 60 days\n\nCyber Vault / DR\nRecovery Window Goal = 14 days\nCompliance Retention = 7 days\n\nOracle DB 12c-21c on Any Platform\n\nPoint-in-time Recovery, Fast Data Restore with Zero Data Loss" } ], "image_num": 8, "text_num": 477, "token_num": 5085 }, { "images": [ "sample_100_images/eCkX8XMpTxA@3520.6000000000004_3547.8#1.jpg", null, "sample_100_images/eCkX8XMpTxA@3548.1200000000003_3573.6400000000003#1.jpg", null, "sample_100_images/eCkX8XMpTxA@3574.7400000000002_3603.0800000000004#1.jpg", null, "sample_100_images/eCkX8XMpTxA@3603.32_3631.46#1.jpg", "sample_100_images/eCkX8XMpTxA@3603.32_3631.46#4.jpg", null, "sample_100_images/u3pjVGWecJ0@0.0_34.760000000000005#1.jpg", "sample_100_images/u3pjVGWecJ0@0.0_34.760000000000005#2.jpg", "sample_100_images/u3pjVGWecJ0@0.0_34.760000000000005#3.jpg", null, "sample_100_images/u3pjVGWecJ0@34.760000000000005_54.22#1.jpg", null ], "texts": [ null, " It is always possible to decompose a relation into relations in BCNF and the decomposition is always lossless but may not preserve dependencies. Right. So this is the one that I am omitting. In 3NF the decomposition is always lossless and preserving dependencies but in BCNF", null, " The decomposition is lossless but may not preserve dependencies. Further, 3NF may have redundancy, whereas BCNF does not have redundancy. If a relation is in BCNF, it is always in 3NF.", null, " If a relation is in BCNF, it is always in 3NF. Further, BCNF is more stringent than 3NF. Okay, so these are the important points that you need to remember as far as the comparison of BCNF and 3NF is concerned.", null, null, " So with this, we conclude our discussion with respect to BCNF. Right? And next time, we'll go over a few more higher normal forms. So let's conclude today's session. Thank you.", null, null, null, " In the 1600s, astronomer and mathematician Galileo Galilei first improved upon the telescope and used it to look at the heavens above. Galileo noted all sorts of interesting observations that had never been seen before, including the fact that the planet Venus appeared to undergo phases similar to the Moon.", null, "1600s, when astronomer and mathematician Galileo Galilei first improved upon the telescope and used it to look at the heavens above. Galileo noted all sorts of interesting observations that had never been seen before, including the fact that the planet Venus appeared to." ], "text_ocr_list": [ null, "We can see these text from the image: - It is always possible to decompose a relation into relations in 3NF and\n- the decomposition is lossless\n- the dependencies are preserved\n\n- It is always possible to decompose a relation into relations in BCNF and\n- the decomposition is lossless\n- it may not be possible to preserve dependencies\n\n- 3NF may have redundancy\n\n- If a relation is in BCNF it is in 3NF\n- BCNF is more stringent than 3NF.\n It is always possible to decompose a relation into relations in BCNF and the decomposition is always lossless but may not preserve dependencies. Right. So this is the one that I am omitting. In 3NF the decomposition is always lossless and preserving dependencies but in BCNF", null, "We can see these text from the image: Comparison of BCNF and 3NF\n\n- It is always possible to decompose a relation into relations in 3NF and\n- the decomposition is lossless \u2714\n- the dependencies are preserved \u2714\n\n- It is always possible to decompose a relation into relations in BCNF \u2714 and\n- the decomposition is lossless \u2714\n- it may not be possible to preserve dependencies \u2714\n\n- 3NF may have redundancy\n\n- If a relation is in BCNF it is in 3NF\n- BCNF is more stringent than 3NF.\n The decomposition is lossless but may not preserve dependencies. Further, 3NF may have redundancy, whereas BCNF does not have redundancy. If a relation is in BCNF, it is always in 3NF.", null, "We can see these text from the image: Comparison of BCNF and 3NF\n\n- It is always possible to decompose a relation into relations in 3NF and\n- the decomposition is lossless\n- the dependencies are preserved\n\n- It is always possible to decompose a relation into relations in BCNF and\n- the decomposition is lossless\n- it may not be possible to preserve dependencies\n\n- 3NF may have redundancy\n\n- If a relation is in BCNF it is in 3NF\n- BCNF is more stringent than 3NF.\n If a relation is in BCNF, it is always in 3NF. Further, BCNF is more stringent than 3NF. Okay, so these are the important points that you need to remember as far as the comparison of BCNF and 3NF is concerned.", null, null, "We can see these text from the image: Comparison of BCNF and 3NF\n\n- It is always possible to decompose a relation into relations in 3NF and\n- the decomposition is lossless\n- the dependencies are preserved\n\n- It is always possible to decompose a relation into relations in BCNF and\n- the decomposition is lossless\n- it may not be possible to preserve dependencies\n\n- 3NF may have redundancy\n\n- If a relation is in BCNF it is in 3NF\n- BCNF is more stringent than 3NF.\n So with this, we conclude our discussion with respect to BCNF. Right? And next time, we'll go over a few more higher normal forms. So let's conclude today's session. Thank you.", null, null, null, "We can see these text from the image: EARTH SCIENCE\n\nwith Mr. Sammartano.\n In the 1600s, astronomer and mathematician Galileo Galilei first improved upon the telescope and used it to look at the heavens above. Galileo noted all sorts of interesting observations that had never been seen before, including the fact that the planet Venus appeared to undergo phases similar to the Moon.", null, "We can see these text from the image: ```\n\n```1600s, when astronomer and mathematician Galileo Galilei first improved upon the telescope and used it to look at the heavens above. Galileo noted all sorts of interesting observations that had never been seen before, including the fact that the planet Venus appeared to." ], "metadata": [ { "vid": "eCkX8XMpTxA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Database Design Tutorial on Boyce-Codd Normal Form (BCNF) in Computer Science_30.json#####audio#####doingASR#####FinishASR/eCkX8XMpTxA/3520.6000000000004_3547.8.mp4", "refined_asr": " It is always possible to decompose a relation into relations in BCNF and the decomposition is always lossless but may not preserve dependencies. Right. So this is the one that I am omitting. In 3NF the decomposition is always lossless and preserving dependencies but in BCNF", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3520.6000000000004_3547.8#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3520.6000000000004_3547.8#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3520.6000000000004_3547.8#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3520.6000000000004_3547.8#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3520.6000000000004_3547.8#4.jpg" ], "ocr_qwen2_vl_72b": "- It is always possible to decompose a relation into relations in 3NF and\n- the decomposition is lossless\n- the dependencies are preserved\n\n- It is always possible to decompose a relation into relations in BCNF and\n- the decomposition is lossless\n- it may not be possible to preserve dependencies\n\n- 3NF may have redundancy\n\n- If a relation is in BCNF it is in 3NF\n- BCNF is more stringent than 3NF" }, { "vid": "eCkX8XMpTxA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Database Design Tutorial on Boyce-Codd Normal Form (BCNF) in Computer Science_30.json#####audio#####doingASR#####FinishASR/eCkX8XMpTxA/3548.1200000000003_3573.6400000000003.mp4", "refined_asr": " The decomposition is lossless but may not preserve dependencies. Further, 3NF may have redundancy, whereas BCNF does not have redundancy. If a relation is in BCNF, it is always in 3NF.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3548.1200000000003_3573.6400000000003#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3548.1200000000003_3573.6400000000003#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3548.1200000000003_3573.6400000000003#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3548.1200000000003_3573.6400000000003#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3548.1200000000003_3573.6400000000003#4.jpg" ], "ocr_qwen2_vl_72b": "Comparison of BCNF and 3NF\n\n- It is always possible to decompose a relation into relations in 3NF and\n- the decomposition is lossless \u2714\n- the dependencies are preserved \u2714\n\n- It is always possible to decompose a relation into relations in BCNF \u2714 and\n- the decomposition is lossless \u2714\n- it may not be possible to preserve dependencies \u2714\n\n- 3NF may have redundancy\n\n- If a relation is in BCNF it is in 3NF\n- BCNF is more stringent than 3NF" }, { "vid": "eCkX8XMpTxA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Database Design Tutorial on Boyce-Codd Normal Form (BCNF) in Computer Science_30.json#####audio#####doingASR#####FinishASR/eCkX8XMpTxA/3574.7400000000002_3603.0800000000004.mp4", "refined_asr": " If a relation is in BCNF, it is always in 3NF. Further, BCNF is more stringent than 3NF. Okay, so these are the important points that you need to remember as far as the comparison of BCNF and 3NF is concerned.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3574.7400000000002_3603.0800000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3574.7400000000002_3603.0800000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3574.7400000000002_3603.0800000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3574.7400000000002_3603.0800000000004#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3574.7400000000002_3603.0800000000004#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3574.7400000000002_3603.0800000000004#5.jpg" ], "ocr_qwen2_vl_72b": "Comparison of BCNF and 3NF\n\n- It is always possible to decompose a relation into relations in 3NF and\n- the decomposition is lossless\n- the dependencies are preserved\n\n- It is always possible to decompose a relation into relations in BCNF and\n- the decomposition is lossless\n- it may not be possible to preserve dependencies\n\n- 3NF may have redundancy\n\n- If a relation is in BCNF it is in 3NF\n- BCNF is more stringent than 3NF" }, { "vid": "eCkX8XMpTxA.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Database Design Tutorial on Boyce-Codd Normal Form (BCNF) in Computer Science_30.json#####audio#####doingASR#####FinishASR/eCkX8XMpTxA/3603.32_3631.46.mp4", "refined_asr": " So with this, we conclude our discussion with respect to BCNF. Right? And next time, we'll go over a few more higher normal forms. So let's conclude today's session. Thank you.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3603.32_3631.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3603.32_3631.46#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3603.32_3631.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3603.32_3631.46#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3603.32_3631.46#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3603.32_3631.46#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eCkX8XMpTxA/eCkX8XMpTxA@3603.32_3631.46#5.jpg" ], "ocr_qwen2_vl_72b": "Comparison of BCNF and 3NF\n\n- It is always possible to decompose a relation into relations in 3NF and\n- the decomposition is lossless\n- the dependencies are preserved\n\n- It is always possible to decompose a relation into relations in BCNF and\n- the decomposition is lossless\n- it may not be possible to preserve dependencies\n\n- 3NF may have redundancy\n\n- If a relation is in BCNF it is in 3NF\n- BCNF is more stringent than 3NF" }, { "vid": "u3pjVGWecJ0.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Elementary Physics Tutorial on the Nature of Light focusing on the Electromagnetic Spectrum_30.json#####audio#####doingASR#####FinishASR/u3pjVGWecJ0/0.0_34.760000000000005.mp4", "refined_asr": " In the 1600s, astronomer and mathematician Galileo Galilei first improved upon the telescope and used it to look at the heavens above. Galileo noted all sorts of interesting observations that had never been seen before, including the fact that the planet Venus appeared to undergo phases similar to the Moon.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@0.0_34.760000000000005#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@0.0_34.760000000000005#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@0.0_34.760000000000005#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@0.0_34.760000000000005#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@0.0_34.760000000000005#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@0.0_34.760000000000005#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@0.0_34.760000000000005#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@0.0_34.760000000000005#5.jpg" ], "ocr_qwen2_vl_72b": "EARTH SCIENCE\n\nwith Mr. Sammartano" }, { "vid": "u3pjVGWecJ0.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Elementary Physics Tutorial on the Nature of Light focusing on the Electromagnetic Spectrum_30.json#####audio#####doingASR#####FinishASR/u3pjVGWecJ0/34.760000000000005_54.22.mp4", "refined_asr": null, "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@34.760000000000005_54.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@34.760000000000005_54.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@34.760000000000005_54.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/u3pjVGWecJ0/u3pjVGWecJ0@34.760000000000005_54.22#3.jpg" ], "ocr_qwen2_vl_72b": "```\n\n```" } ], "image_num": 9, "text_num": 355, "token_num": 5539 }, { "images": [ "sample_100_images/eTs7MbyO2tc@1258.6_1284.02#1.jpg", "sample_100_images/eTs7MbyO2tc@1258.6_1284.02#2.jpg", "sample_100_images/eTs7MbyO2tc@1258.6_1284.02#3.jpg", "sample_100_images/eTs7MbyO2tc@1258.6_1284.02#4.jpg", null, "sample_100_images/eTs7MbyO2tc@1284.02_1322.34#1.jpg", "sample_100_images/eTs7MbyO2tc@1284.02_1322.34#2.jpg", "sample_100_images/eTs7MbyO2tc@1284.02_1322.34#4.jpg", "sample_100_images/eTs7MbyO2tc@1284.02_1322.34#5.jpg", null, "sample_100_images/eTs7MbyO2tc@1322.34_1335.76#1.jpg", null, "sample_100_images/eTs7MbyO2tc@1335.76_1349.58#1.jpg", null, "sample_100_images/eTs7MbyO2tc@1349.58_1364.82#1.jpg", null, "sample_100_images/eTs7MbyO2tc@1364.82_1394.52#1.jpg", "sample_100_images/eTs7MbyO2tc@1364.82_1394.52#2.jpg", "sample_100_images/eTs7MbyO2tc@1364.82_1394.52#3.jpg", "sample_100_images/eTs7MbyO2tc@1364.82_1394.52#4.jpg", null, "sample_100_images/eTs7MbyO2tc@1394.52_1409.8799999999999#1.jpg", "sample_100_images/eTs7MbyO2tc@1394.52_1409.8799999999999#2.jpg", null ], "texts": [ null, null, null, null, " So first, we will calculate the voltage across the resistor. V across R is nothing but equal to R into I, and we have just derived what I is. I is nothing but V by R into 1 minus e, to the power minus t by tau. So I will substitute this value of current here.", null, null, null, null, " So it's R into V by R into 1 minus e to the power minus t by tau. Tau is the time constant, equal to L by R. So these R's will cancel out and what we're left with is V into 1 minus e to the power minus T by tau. This is the voltage across the resistor.", null, " How to plot it? Approximately, how to plot it. So, when time t is 0, e raised to the power of 0 will be 1, and 1 minus 1 will be 0 again. So, again, at the start, the voltage across the resistor will be 0.", null, " See again: I put time T equal to 0, e to the power of 0 is 1, 1 minus 1 is 0. So it is going to be 0 again. And when I put T equal to infinity, e raised to minus infinity, it will become:", null, " When I substitute t equals infinity e to the power minus infinity will be zero. 1 minus 0 will be 1. So it is going to be V. The maximum value for it is going to be V, and the graph will reflect this behavior.", null, null, null, null, " Now, what will be the voltage across the inductor? For that, you'll write the voltage across the inductor, which is VL. If you see the original diagram, there is an inductor. The voltage across the inductor is nothing but VL, and VL has a formula: it's equal to L times.", null, null, " DI by DT. So it is going to be L into the value for current I which we have obtained and we will use it. It is this value, V by R, 1 minus e to the power minus T by Tau. So it:" ], "text_ocr_list": [ null, null, null, null, "We can see these text from the image: Voltage across.\n So first, we will calculate the voltage across the resistor. V across R is nothing but equal to R into I, and we have just derived what I is. I is nothing but V by R into 1 minus e, to the power minus t by tau. So I will substitute this value of current here.", null, null, null, null, "We can see these text from the image: Voltage across Resistor\n\nVr = R i = R \u221aV.\n So it's R into V by R into 1 minus e to the power minus t by tau. Tau is the time constant, equal to L by R. So these R's will cancel out and what we're left with is V into 1 minus e to the power minus T by tau. This is the voltage across the resistor.", null, "We can see these text from the image: Voltage across Resistor\n\nV_R = R i = R \\left( \\frac{V}{R} (1 - e^{-t/\\tau}) \\right)\n\nV_R = V (1 - e^{-t/\\tau}).\n How to plot it? Approximately, how to plot it. So, when time t is 0, e raised to the power of 0 will be 1, and 1 minus 1 will be 0 again. So, again, at the start, the voltage across the resistor will be 0.", null, "We can see these text from the image: - Voltage across Resistor\n- \\( V_R = R \\cdot i = R \\left[ \\frac{V}{R} (1 - e^{-\\frac{t}{\\tau}}) \\right] \\)\n- \\( V_R = V (1 - e^{-\\frac{t}{\\tau}}) \\)\n- \\( V_R \\)\n- \\( t \\).\n See again: I put time T equal to 0, e to the power of 0 is 1, 1 minus 1 is 0. So it is going to be 0 again. And when I put T equal to infinity, e raised to minus infinity, it will become:", null, "We can see these text from the image: - Voltage across Resistor\n- \\( V_R = R \\cdot i = R \\left[ \\frac{V}{R} (1 - e^{-t/\\tau}) \\right] \\)\n- \\( V_R = V (1 - e^{-t/\\tau}) \\).\n When I substitute t equals infinity e to the power minus infinity will be zero. 1 minus 0 will be 1. So it is going to be V. The maximum value for it is going to be V, and the graph will reflect this behavior.", null, null, null, null, "We can see these text from the image: Voltage across Resistor\n\nVR = Ri = R [V (1 - e^(-t/\u03c4))]\n\nVR = V (1 - e^(-t/\u03c4))\n\nVR\nV\nt.\n Now, what will be the voltage across the inductor? For that, you'll write the voltage across the inductor, which is VL. If you see the original diagram, there is an inductor. The voltage across the inductor is nothing but VL, and VL has a formula: it's equal to L times.", null, null, "We can see these text from the image: Voltage across Inductor\n\nV_L = L \\frac{di}{dt}.\n DI by DT. So it is going to be L into the value for current I which we have obtained and we will use it. It is this value, V by R, 1 minus e to the power minus T by Tau. So it:" ], "metadata": [ { "vid": "eTs7MbyO2tc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Electrical Engineering course video on Time Domain Analysis in Signal Analysis_30.json#####audio#####doingASR#####FinishASR/eTs7MbyO2tc/1258.6_1284.02.mp4", "refined_asr": " So first, we will calculate the voltage across the resistor. V across R is nothing but equal to R into I, and we have just derived what I is. I is nothing but V by R into 1 minus e, to the power minus t by tau. So I will substitute this value of current here.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1258.6_1284.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1258.6_1284.02#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1258.6_1284.02#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1258.6_1284.02#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1258.6_1284.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1258.6_1284.02#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1258.6_1284.02#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1258.6_1284.02#4.jpg" ], "ocr_qwen2_vl_72b": "Voltage across" }, { "vid": "eTs7MbyO2tc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Electrical Engineering course video on Time Domain Analysis in Signal Analysis_30.json#####audio#####doingASR#####FinishASR/eTs7MbyO2tc/1284.02_1322.34.mp4", "refined_asr": " So it's R into V by R into 1 minus e to the power minus t by tau. Tau is the time constant, equal to L by R. So these R's will cancel out and what we're left with is V into 1 minus e to the power minus T by tau. This is the voltage across the resistor.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1284.02_1322.34#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1284.02_1322.34#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1284.02_1322.34#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1284.02_1322.34#5.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1284.02_1322.34#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1284.02_1322.34#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1284.02_1322.34#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1284.02_1322.34#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1284.02_1322.34#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1284.02_1322.34#6.jpg" ], "ocr_qwen2_vl_72b": "Voltage across Resistor\n\nVr = R i = R \u221aV" }, { "vid": "eTs7MbyO2tc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Electrical Engineering course video on Time Domain Analysis in Signal Analysis_30.json#####audio#####doingASR#####FinishASR/eTs7MbyO2tc/1322.34_1335.76.mp4", "refined_asr": " How to plot it? Approximately, how to plot it. So, when time t is 0, e raised to the power of 0 will be 1, and 1 minus 1 will be 0 again. So, again, at the start, the voltage across the resistor will be 0.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1322.34_1335.76#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1322.34_1335.76#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1322.34_1335.76#2.jpg" ], "ocr_qwen2_vl_72b": "Voltage across Resistor\n\nV_R = R i = R \\left( \\frac{V}{R} (1 - e^{-t/\\tau}) \\right)\n\nV_R = V (1 - e^{-t/\\tau})" }, { "vid": "eTs7MbyO2tc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Electrical Engineering course video on Time Domain Analysis in Signal Analysis_30.json#####audio#####doingASR#####FinishASR/eTs7MbyO2tc/1335.76_1349.58.mp4", "refined_asr": " See again: I put time T equal to 0, e to the power of 0 is 1, 1 minus 1 is 0. So it is going to be 0 again. And when I put T equal to infinity, e raised to minus infinity, it will become:", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1335.76_1349.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1335.76_1349.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1335.76_1349.58#2.jpg" ], "ocr_qwen2_vl_72b": "- Voltage across Resistor\n- \\( V_R = R \\cdot i = R \\left[ \\frac{V}{R} (1 - e^{-\\frac{t}{\\tau}}) \\right] \\)\n- \\( V_R = V (1 - e^{-\\frac{t}{\\tau}}) \\)\n- \\( V_R \\)\n- \\( t \\)" }, { "vid": "eTs7MbyO2tc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Electrical Engineering course video on Time Domain Analysis in Signal Analysis_30.json#####audio#####doingASR#####FinishASR/eTs7MbyO2tc/1349.58_1364.82.mp4", "refined_asr": " When I substitute t equals infinity e to the power minus infinity will be zero. 1 minus 0 will be 1. So it is going to be V. The maximum value for it is going to be V, and the graph will reflect this behavior.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1349.58_1364.82#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1349.58_1364.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1349.58_1364.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1349.58_1364.82#3.jpg" ], "ocr_qwen2_vl_72b": "- Voltage across Resistor\n- \\( V_R = R \\cdot i = R \\left[ \\frac{V}{R} (1 - e^{-t/\\tau}) \\right] \\)\n- \\( V_R = V (1 - e^{-t/\\tau}) \\)" }, { "vid": "eTs7MbyO2tc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Electrical Engineering course video on Time Domain Analysis in Signal Analysis_30.json#####audio#####doingASR#####FinishASR/eTs7MbyO2tc/1364.82_1394.52.mp4", "refined_asr": " Now, what will be the voltage across the inductor? For that, you'll write the voltage across the inductor, which is VL. If you see the original diagram, there is an inductor. The voltage across the inductor is nothing but VL, and VL has a formula: it's equal to L times.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1364.82_1394.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1364.82_1394.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1364.82_1394.52#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1364.82_1394.52#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1364.82_1394.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1364.82_1394.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1364.82_1394.52#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1364.82_1394.52#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1364.82_1394.52#5.jpg" ], "ocr_qwen2_vl_72b": "Voltage across Resistor\n\nVR = Ri = R [V (1 - e^(-t/\u03c4))]\n\nVR = V (1 - e^(-t/\u03c4))\n\nVR\nV\nt" }, { "vid": "eTs7MbyO2tc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Electrical Engineering course video on Time Domain Analysis in Signal Analysis_30.json#####audio#####doingASR#####FinishASR/eTs7MbyO2tc/1394.52_1409.8799999999999.mp4", "refined_asr": " DI by DT. So it is going to be L into the value for current I which we have obtained and we will use it. It is this value, V by R, 1 minus e to the power minus T by Tau. So it:", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1394.52_1409.8799999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1394.52_1409.8799999999999#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1394.52_1409.8799999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1394.52_1409.8799999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/eTs7MbyO2tc/eTs7MbyO2tc@1394.52_1409.8799999999999#3.jpg" ], "ocr_qwen2_vl_72b": "Voltage across Inductor\n\nV_L = L \\frac{di}{dt}" } ], "image_num": 17, "text_num": 454, "token_num": 10246 }, { "images": [ "sample_100_images/JcfWQ1gKL5Q@1581.3000000000002_1596.8600000000001#1.jpg", null, "sample_100_images/JcfWQ1gKL5Q@1596.94_1607.8200000000002#1.jpg", null, "sample_100_images/JcfWQ1gKL5Q@1607.94_1631.96#1.jpg", null, "sample_100_images/JcfWQ1gKL5Q@1631.96_1650.98#1.jpg", null, "sample_100_images/JcfWQ1gKL5Q@1651.6599999999999_1670.02#1.jpg", "sample_100_images/JcfWQ1gKL5Q@1651.6599999999999_1670.02#2.jpg", "sample_100_images/JcfWQ1gKL5Q@1651.6599999999999_1670.02#3.jpg", null, "sample_100_images/JcfWQ1gKL5Q@1670.62_1691.08#1.jpg", null, "sample_100_images/JcfWQ1gKL5Q@1691.08_1709.08#1.jpg", "sample_100_images/JcfWQ1gKL5Q@1691.08_1709.08#2.jpg", null, "sample_100_images/JcfWQ1gKL5Q@1709.08_1726.44#1.jpg", "sample_100_images/JcfWQ1gKL5Q@1709.08_1726.44#2.jpg", null ], "texts": [ null, " Let's say that I had this double bonded carbon and up here I had a CH3. Over here I would have my H and down here I would continue on with a CH2 and so forth. It might take you a while.", null, " You might have to look at this on your own again. I'm not getting anything. I'm not getting into all these little things, but I do want to show you these things. I'm going to draw my X-axis over here. That is the Y-axis.", null, " They're on the same side these are both on the same side with trans it actually jumps sides if I draw this same picture ch two c c and I have this same picture I have this double bonded carbon in the middle", null, " With these branches coming out and instead of my polymer chain both being on the same side it flops. My polymer chain moves over there and over here I am going to have my CH3 let's say and my H over here.", null, null, null, " Or whatever it happens to be. They really could be any type of thing attached. But this one, it goes across and that's trans. Alright, now we have even more things that we want to talk about. I could have blocks of.", null, " Or sorry I could look at my polymers and I could have kind of a random structure I could have let's say that each one of these circles is one of my monomers one of my units of my polymer.", null, null, " Right, one of my examples up here is a triangle-shaped molecule. So that is the circle, and what I could do is have something random where it changes for a while into a slightly different chemical group that's being attached, and then it goes back.", null, null, " And then maybe a little bit more of something else. It just kind of jumps back and forth between some. It kind of has this random structure. I could have alternatives. Let's say, for example, I could have alternating and alternating I think sounds probably pretty standard." ], "text_ocr_list": [ null, "We can see these text from the image: - Atactic\n- Some more (isomerism):\n - cis\n - trans\n- And more (architecture):\n - Random\n - Alternating.\n Let's say that I had this double bonded carbon and up here I had a CH3. Over here I would have my H and down here I would continue on with a CH2 and so forth. It might take you a while.", null, "We can see these text from the image: - Atactic\n- Some more (isomerism):\n - cis\n - trans\n- And more (architecture):\n - Random\n - Alternating.\n You might have to look at this on your own again. I'm not getting anything. I'm not getting into all these little things, but I do want to show you these things. I'm going to draw my X-axis over here. That is the Y-axis.", null, "We can see these text from the image: - Atactic\n- Some more (isomerism):\n - cis\n - trans\n- And more (architecture):\n - Random.\n They're on the same side these are both on the same side with trans it actually jumps sides if I draw this same picture ch two c c and I have this same picture I have this double bonded carbon in the middle", null, "We can see these text from the image: - Atactic\n- Some more (isomerism):\n - cis\n - trans\n- And more (architecture):\n - Random\n - Alternating.\n With these branches coming out and instead of my polymer chain both being on the same side it flops. My polymer chain moves over there and over here I am going to have my CH3 let's say and my H over here.", null, null, null, "We can see these text from the image: - Atactic\n- Some more (isomerism):\n - cis\n - trans\n- And more (architecture):\n - Random\n - Alternating.\n Or whatever it happens to be. They really could be any type of thing attached. But this one, it goes across and that's trans. Alright, now we have even more things that we want to talk about. I could have blocks of.", null, "We can see these text from the image: trans\n\nAnd more (architecture).\n\nRandom\n\nAlternating\n\nBlocky\n\nBranched\n\nA polymer chain showing both crystallization and amorphousness.\n Or sorry I could look at my polymers and I could have kind of a random structure I could have let's say that each one of these circles is one of my monomers one of my units of my polymer.", null, null, "We can see these text from the image: - Syndiotactic\n- Atactic\n- Some more (isomerism):\n - cis: \\(\\text{CH}_3\\)\n - trans: \\(\\text{CH}_2\\)\n- And more (architecture): Random.\n Right, one of my examples up here is a triangle-shaped molecule. So that is the circle, and what I could do is have something random where it changes for a while into a slightly different chemical group that's being attached, and then it goes back.", null, null, "We can see these text from the image: - cis\n- trans\n\nAnd more (architecture).\n\n- Random\n- Alternating\n- Blocky\n- Branched.\n And then maybe a little bit more of something else. It just kind of jumps back and forth between some. It kind of has this random structure. I could have alternatives. Let's say, for example, I could have alternating and alternating I think sounds probably pretty standard." ], "metadata": [ { "vid": "JcfWQ1gKL5Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Engineering Fundamentals - Materials Chemistry Tutorial on Ceramics_30.json#####audio#####doingASR#####FinishASR/JcfWQ1gKL5Q/1581.3000000000002_1596.8600000000001.mp4", "refined_asr": " Let's say that I had this double bonded carbon and up here I had a CH3. Over here I would have my H and down here I would continue on with a CH2 and so forth. It might take you a while.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1581.3000000000002_1596.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1581.3000000000002_1596.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1581.3000000000002_1596.8600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1581.3000000000002_1596.8600000000001#3.jpg" ], "ocr_qwen2_vl_72b": "- Atactic\n- Some more (isomerism):\n - cis\n - trans\n- And more (architecture):\n - Random\n - Alternating" }, { "vid": "JcfWQ1gKL5Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Engineering Fundamentals - Materials Chemistry Tutorial on Ceramics_30.json#####audio#####doingASR#####FinishASR/JcfWQ1gKL5Q/1596.94_1607.8200000000002.mp4", "refined_asr": " You might have to look at this on your own again. I'm not getting anything. I'm not getting into all these little things, but I do want to show you these things. I'm going to draw my X-axis over here. That is the Y-axis.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1596.94_1607.8200000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1596.94_1607.8200000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1596.94_1607.8200000000002#2.jpg" ], "ocr_qwen2_vl_72b": "- Atactic\n- Some more (isomerism):\n - cis\n - trans\n- And more (architecture):\n - Random\n - Alternating" }, { "vid": "JcfWQ1gKL5Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Engineering Fundamentals - Materials Chemistry Tutorial on Ceramics_30.json#####audio#####doingASR#####FinishASR/JcfWQ1gKL5Q/1607.94_1631.96.mp4", "refined_asr": " They're on the same side these are both on the same side with trans it actually jumps sides if I draw this same picture ch two c c and I have this same picture I have this double bonded carbon in the middle", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1607.94_1631.96#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1607.94_1631.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1607.94_1631.96#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1607.94_1631.96#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1607.94_1631.96#4.jpg" ], "ocr_qwen2_vl_72b": "- Atactic\n- Some more (isomerism):\n - cis\n - trans\n- And more (architecture):\n - Random" }, { "vid": "JcfWQ1gKL5Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Engineering Fundamentals - Materials Chemistry Tutorial on Ceramics_30.json#####audio#####doingASR#####FinishASR/JcfWQ1gKL5Q/1631.96_1650.98.mp4", "refined_asr": " With these branches coming out and instead of my polymer chain both being on the same side it flops. My polymer chain moves over there and over here I am going to have my CH3 let's say and my H over here.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1631.96_1650.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1631.96_1650.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1631.96_1650.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1631.96_1650.98#3.jpg" ], "ocr_qwen2_vl_72b": "- Atactic\n- Some more (isomerism):\n - cis\n - trans\n- And more (architecture):\n - Random\n - Alternating" }, { "vid": "JcfWQ1gKL5Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Engineering Fundamentals - Materials Chemistry Tutorial on Ceramics_30.json#####audio#####doingASR#####FinishASR/JcfWQ1gKL5Q/1651.6599999999999_1670.02.mp4", "refined_asr": " Or whatever it happens to be. They really could be any type of thing attached. But this one, it goes across and that's trans. Alright, now we have even more things that we want to talk about. I could have blocks of.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1651.6599999999999_1670.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1651.6599999999999_1670.02#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1651.6599999999999_1670.02#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1651.6599999999999_1670.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1651.6599999999999_1670.02#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1651.6599999999999_1670.02#3.jpg" ], "ocr_qwen2_vl_72b": "- Atactic\n- Some more (isomerism):\n - cis\n - trans\n- And more (architecture):\n - Random\n - Alternating" }, { "vid": "JcfWQ1gKL5Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Engineering Fundamentals - Materials Chemistry Tutorial on Ceramics_30.json#####audio#####doingASR#####FinishASR/JcfWQ1gKL5Q/1670.62_1691.08.mp4", "refined_asr": " Or sorry I could look at my polymers and I could have kind of a random structure I could have let's say that each one of these circles is one of my monomers one of my units of my polymer.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1670.62_1691.08#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1670.62_1691.08#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1670.62_1691.08#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1670.62_1691.08#3.jpg" ], "ocr_qwen2_vl_72b": "trans\n\nAnd more (architecture).\n\nRandom\n\nAlternating\n\nBlocky\n\nBranched\n\nA polymer chain showing both crystallization and amorphousness" }, { "vid": "JcfWQ1gKL5Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Engineering Fundamentals - Materials Chemistry Tutorial on Ceramics_30.json#####audio#####doingASR#####FinishASR/JcfWQ1gKL5Q/1691.08_1709.08.mp4", "refined_asr": " Right, one of my examples up here is a triangle-shaped molecule. So that is the circle, and what I could do is have something random where it changes for a while into a slightly different chemical group that's being attached, and then it goes back.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1691.08_1709.08#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1691.08_1709.08#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1691.08_1709.08#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1691.08_1709.08#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1691.08_1709.08#3.jpg" ], "ocr_qwen2_vl_72b": "- Syndiotactic\n- Atactic\n- Some more (isomerism):\n - cis: \\(\\text{CH}_3\\)\n - trans: \\(\\text{CH}_2\\)\n- And more (architecture): Random" }, { "vid": "JcfWQ1gKL5Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Engineering Fundamentals - Materials Chemistry Tutorial on Ceramics_30.json#####audio#####doingASR#####FinishASR/JcfWQ1gKL5Q/1709.08_1726.44.mp4", "refined_asr": " And then maybe a little bit more of something else. It just kind of jumps back and forth between some. It kind of has this random structure. I could have alternatives. Let's say, for example, I could have alternating and alternating I think sounds probably pretty standard.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1709.08_1726.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1709.08_1726.44#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1709.08_1726.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1709.08_1726.44#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/JcfWQ1gKL5Q/JcfWQ1gKL5Q@1709.08_1726.44#3.jpg" ], "ocr_qwen2_vl_72b": "- cis\n- trans\n\nAnd more (architecture).\n\n- Random\n- Alternating\n- Blocky\n- Branched" } ], "image_num": 12, "text_num": 430, "token_num": 7342 }, { "images": [ "sample_100_images/sJBMOjPVwtE@1241.38_1253.74#1.jpg", null, "sample_100_images/sJBMOjPVwtE@1253.74_1275.24#1.jpg", "sample_100_images/sJBMOjPVwtE@1253.74_1275.24#3.jpg", null, "sample_100_images/sJBMOjPVwtE@1275.24_1295.02#1.jpg", "sample_100_images/sJBMOjPVwtE@1275.24_1295.02#2.jpg", null, "sample_100_images/sJBMOjPVwtE@1295.02_1318.3799999999999#1.jpg", null, "sample_100_images/sJBMOjPVwtE@1318.3799999999999_1342.26#1.jpg", null, "sample_100_images/sJBMOjPVwtE@1342.26_1362.94#1.jpg", "sample_100_images/sJBMOjPVwtE@1342.26_1362.94#2.jpg", "sample_100_images/sJBMOjPVwtE@1342.26_1362.94#3.jpg", null, "sample_100_images/sJBMOjPVwtE@1362.94_1380.7#1.jpg", null ], "texts": [ null, " Right. And again this is similar to our pH and pKa graph. If this were the acid-base relevant graph what would it be? Here you would have HA or the protonated form. Here you would have the deprotonated form.", null, null, " Right. So similarly, here we have the reduced form, which means with the electrons. And here you would have the oxidized form, meaning without the electrons. Right. So that is how you can understand the system. This was the case when we had a simple example of what we mean by oxidized and electrons.", null, null, " Being accepted in the reduced form, I guess, right? Being formed. So now we are going to look at a couple of examples, I believe. So for this case, let us try to understand how to write the relevant aspects. Right. So obviously, P will be equal to P0 minus 1 by N. N is the number of electrons being.", null, " Transferred here it is only 1. So I am going to write minus 1 into log Q, dash, Q, dash. Right. So activity of Fe2+, which is reduced form, by activity of Fe3+. Yes. So that is what you see here again, pe is equal to pe0 here, pardon me, right. pe0 minus log.", null, " Activity of Fe2+ by activity of Fe3+. Right. And obviously again as we looked at it from the graph when are the concentrations of the reduced form and the oxidized form going to be equal? Right. Obviously you see that that is going to be the case when pe is equal to pe0.", null, null, null, " Right. So I think that is something that is easy to understand. Again, this is the simplest case when you have only the oxidized form, which is taking in or accepting electrons, and the reduced form being formed. Obviously, this is your oxidized form, right? It's oxidized, taking in electrons.", null, " Forming the reduced form. Right. The oxidized state here is 3 and because it is accepting 1 electron, it is reducing to 2. Right. I think that is pretty simple to visualize. So let's look at one or a couple more examples when we also have other compounds." ], "text_ocr_list": [ null, "We can see these text from the image: - pe > pe\u00b0\n- A_red + n e^(-) \u2192 A_oxd\n- pe = pe\u00b0 - (1/n) log [A_red] / [A_oxd]\n- pe = pe\u00b0\n- [A_red] = [A_oxd]\n\n- high [e^-] reduced species\n- low [e^-] oxidized species\n\n- pe\u00b0 (mV at pH +1)\n- pe-pe\u00b0.\n Right. And again this is similar to our pH and pKa graph. If this were the acid-base relevant graph what would it be? Here you would have HA or the protonated form. Here you would have the deprotonated form.", null, null, "We can see these text from the image: - pe > pe\u00b0\n- Ared + ne\u207b \u2192 Ared\n- pe = pe\u00b0 - (1/n) log [Ared]/[Aoxd]\n- pe = pe\u00b0\n- [Ared] = [Aoxd]\n\nConc.\n- high [e\u207b]\n- reduced species\n- low [e\u207b]\n- oxidized species\n\npe\n- pe\u00b0 (neutral pH)\n- -1\n- +1.\n Right. So similarly, here we have the reduced form, which means with the electrons. And here you would have the oxidized form, meaning without the electrons. Right. So that is how you can understand the system. This was the case when we had a simple example of what we mean by oxidized and electrons.", null, null, "We can see these text from the image: There is no text or formula present in the image..\n Being accepted in the reduced form, I guess, right? Being formed. So now we are going to look at a couple of examples, I believe. So for this case, let us try to understand how to write the relevant aspects. Right. So obviously, P will be equal to P0 minus 1 by N. N is the number of electrons being.", null, "We can see these text from the image: - Example\n - Fe\u00b3\u207a + e\u207b \u2192 Fe\u00b2\u207a\n\npe = pe - 1/2.\n Transferred here it is only 1. So I am going to write minus 1 into log Q, dash, Q, dash. Right. So activity of Fe2+, which is reduced form, by activity of Fe3+. Yes. So that is what you see here again, pe is equal to pe0 here, pardon me, right. pe0 minus log.", null, "We can see these text from the image: - Fe\u00b3\u207a + e\u207b \u2192 Fe\u00b2\u207a\n\npe = pe\u00b0 - log [Fe\u00b2\u207a]\npe = pe\u00b0 - log [Fe\u00b3\u207a].\n Activity of Fe2+ by activity of Fe3+. Right. And obviously again as we looked at it from the graph when are the concentrations of the reduced form and the oxidized form going to be equal? Right. Obviously you see that that is going to be the case when pe is equal to pe0.", null, null, null, "We can see these text from the image: - Example\n - Fe\u00b3\u207a + e\u207b \u2192 Fe\u00b2\u207a\n\npe = pe\u00b0 - log [Fe\u00b2\u207a]\n [Fe\u00b3\u207a]\n\npe = pe\u00b0 - log [Fe\u00b2\u207a]\n [Fe\u00b3\u207a]\n\n[Fe\u00b2\u207a] = [Fe\u00b3\u207a]\n\npe = pe\u00b0.\n Right. So I think that is something that is easy to understand. Again, this is the simplest case when you have only the oxidized form, which is taking in or accepting electrons, and the reduced form being formed. Obviously, this is your oxidized form, right? It's oxidized, taking in electrons.", null, "We can see these text from the image: - Example\n\nFe\u00b3\u207a + e\u207b \u2192 Fe\u00b2\u207a\n\npe = pe\u00b0 - log [Fe\u00b2\u207a]\npe = pe\u00b0 - log [Fe\u00b3\u207a]\n\n[Fe\u00b2\u207a] = [Fe\u00b3\u207a]\n\npe = pe\u00b0.\n Forming the reduced form. Right. The oxidized state here is 3 and because it is accepting 1 electron, it is reducing to 2. Right. I think that is pretty simple to visualize. So let's look at one or a couple more examples when we also have other compounds." ], "metadata": [ { "vid": "sJBMOjPVwtE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Environmental Engineering Redox Reactions in Environmental Chemistry Tutorial_30.json#####audio#####doingASR#####FinishASR/sJBMOjPVwtE/1241.38_1253.74.mp4", "refined_asr": " Right. And again this is similar to our pH and pKa graph. If this were the acid-base relevant graph what would it be? Here you would have HA or the protonated form. Here you would have the deprotonated form.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1241.38_1253.74#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1241.38_1253.74#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1241.38_1253.74#2.jpg" ], "ocr_qwen2_vl_72b": "- pe > pe\u00b0\n- A_red + n e^(-) \u2192 A_oxd\n- pe = pe\u00b0 - (1/n) log [A_red] / [A_oxd]\n- pe = pe\u00b0\n- [A_red] = [A_oxd]\n\n- high [e^-] reduced species\n- low [e^-] oxidized species\n\n- pe\u00b0 (mV at pH +1)\n- pe-pe\u00b0" }, { "vid": "sJBMOjPVwtE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Environmental Engineering Redox Reactions in Environmental Chemistry Tutorial_30.json#####audio#####doingASR#####FinishASR/sJBMOjPVwtE/1253.74_1275.24.mp4", "refined_asr": " Right. So similarly, here we have the reduced form, which means with the electrons. And here you would have the oxidized form, meaning without the electrons. Right. So that is how you can understand the system. This was the case when we had a simple example of what we mean by oxidized and electrons.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1253.74_1275.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1253.74_1275.24#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1253.74_1275.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1253.74_1275.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1253.74_1275.24#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1253.74_1275.24#4.jpg" ], "ocr_qwen2_vl_72b": "- pe > pe\u00b0\n- Ared + ne\u207b \u2192 Ared\n- pe = pe\u00b0 - (1/n) log [Ared]/[Aoxd]\n- pe = pe\u00b0\n- [Ared] = [Aoxd]\n\nConc.\n- high [e\u207b]\n- reduced species\n- low [e\u207b]\n- oxidized species\n\npe\n- pe\u00b0 (neutral pH)\n- -1\n- +1" }, { "vid": "sJBMOjPVwtE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Environmental Engineering Redox Reactions in Environmental Chemistry Tutorial_30.json#####audio#####doingASR#####FinishASR/sJBMOjPVwtE/1275.24_1295.02.mp4", "refined_asr": " Being accepted in the reduced form, I guess, right? Being formed. So now we are going to look at a couple of examples, I believe. So for this case, let us try to understand how to write the relevant aspects. Right. So obviously, P will be equal to P0 minus 1 by N. N is the number of electrons being.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1275.24_1295.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1275.24_1295.02#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1275.24_1295.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1275.24_1295.02#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1275.24_1295.02#3.jpg" ], "ocr_qwen2_vl_72b": "There is no text or formula present in the image." }, { "vid": "sJBMOjPVwtE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Environmental Engineering Redox Reactions in Environmental Chemistry Tutorial_30.json#####audio#####doingASR#####FinishASR/sJBMOjPVwtE/1295.02_1318.3799999999999.mp4", "refined_asr": " Transferred here it is only 1. So I am going to write minus 1 into log Q, dash, Q, dash. Right. So activity of Fe2+, which is reduced form, by activity of Fe3+. Yes. So that is what you see here again, pe is equal to pe0 here, pardon me, right. pe0 minus log.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1295.02_1318.3799999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1295.02_1318.3799999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1295.02_1318.3799999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1295.02_1318.3799999999999#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1295.02_1318.3799999999999#4.jpg" ], "ocr_qwen2_vl_72b": "- Example\n - Fe\u00b3\u207a + e\u207b \u2192 Fe\u00b2\u207a\n\npe = pe - 1/2" }, { "vid": "sJBMOjPVwtE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Environmental Engineering Redox Reactions in Environmental Chemistry Tutorial_30.json#####audio#####doingASR#####FinishASR/sJBMOjPVwtE/1318.3799999999999_1342.26.mp4", "refined_asr": " Activity of Fe2+ by activity of Fe3+. Right. And obviously again as we looked at it from the graph when are the concentrations of the reduced form and the oxidized form going to be equal? Right. Obviously you see that that is going to be the case when pe is equal to pe0.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1318.3799999999999_1342.26#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1318.3799999999999_1342.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1318.3799999999999_1342.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1318.3799999999999_1342.26#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1318.3799999999999_1342.26#4.jpg" ], "ocr_qwen2_vl_72b": "- Fe\u00b3\u207a + e\u207b \u2192 Fe\u00b2\u207a\n\npe = pe\u00b0 - log [Fe\u00b2\u207a]\npe = pe\u00b0 - log [Fe\u00b3\u207a]" }, { "vid": "sJBMOjPVwtE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Environmental Engineering Redox Reactions in Environmental Chemistry Tutorial_30.json#####audio#####doingASR#####FinishASR/sJBMOjPVwtE/1342.26_1362.94.mp4", "refined_asr": " Right. So I think that is something that is easy to understand. Again, this is the simplest case when you have only the oxidized form, which is taking in or accepting electrons, and the reduced form being formed. Obviously, this is your oxidized form, right? It's oxidized, taking in electrons.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1342.26_1362.94#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1342.26_1362.94#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1342.26_1362.94#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1342.26_1362.94#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1342.26_1362.94#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1342.26_1362.94#3.jpg" ], "ocr_qwen2_vl_72b": "- Example\n - Fe\u00b3\u207a + e\u207b \u2192 Fe\u00b2\u207a\n\npe = pe\u00b0 - log [Fe\u00b2\u207a]\n [Fe\u00b3\u207a]\n\npe = pe\u00b0 - log [Fe\u00b2\u207a]\n [Fe\u00b3\u207a]\n\n[Fe\u00b2\u207a] = [Fe\u00b3\u207a]\n\npe = pe\u00b0" }, { "vid": "sJBMOjPVwtE.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Environmental Engineering Redox Reactions in Environmental Chemistry Tutorial_30.json#####audio#####doingASR#####FinishASR/sJBMOjPVwtE/1362.94_1380.7.mp4", "refined_asr": " Forming the reduced form. Right. The oxidized state here is 3 and because it is accepting 1 electron, it is reducing to 2. Right. I think that is pretty simple to visualize. So let's look at one or a couple more examples when we also have other compounds.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1362.94_1380.7#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1362.94_1380.7#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1362.94_1380.7#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/sJBMOjPVwtE/sJBMOjPVwtE@1362.94_1380.7#3.jpg" ], "ocr_qwen2_vl_72b": "- Example\n\nFe\u00b3\u207a + e\u207b \u2192 Fe\u00b2\u207a\n\npe = pe\u00b0 - log [Fe\u00b2\u207a]\npe = pe\u00b0 - log [Fe\u00b3\u207a]\n\n[Fe\u00b2\u207a] = [Fe\u00b3\u207a]\n\npe = pe\u00b0" } ], "image_num": 11, "text_num": 482, "token_num": 6818 }, { "images": [ "sample_100_images/nZjXjULXk-I@4049.0_4077.0#1.jpg", null, "sample_100_images/nZjXjULXk-I@4077.0_4099.0#1.jpg", null, "sample_100_images/nZjXjULXk-I@4099.0_4121.0#1.jpg", "sample_100_images/nZjXjULXk-I@4099.0_4121.0#3.jpg", null, "sample_100_images/nZjXjULXk-I@4121.0_4139.0#1.jpg", "sample_100_images/nZjXjULXk-I@4121.0_4139.0#2.jpg", null, "sample_100_images/nZjXjULXk-I@4139.0_4169.0#1.jpg", "sample_100_images/nZjXjULXk-I@4139.0_4169.0#4.jpg", null, "sample_100_images/nZjXjULXk-I@4169.0_4191.0#1.jpg", null ], "texts": [ null, " And then we go into hippos and whales. So what we think happened is that dogs go back to form wolves, wolves then come back out of the forest to form hippos. And what's neat about this area is that within the last 20 years there have been so many discoveries. \n\n(Note: The original text has some factual errors that are not corrected here as per the instruction to only correct grammar, repetition, and transcription errors.)", null, " So the Valley of the Whales, that was a discovery in Morocco. At that time, what is today Northern Africa, a desert, was actually a part of the Mediterranean Sea that had flooded that area. Whales would go in there and spawn, and so we have all these whale fossils from a very long period of time.", null, null, " And we've been able to put together - paleontologists have been able to put together - a whole sequence of whales that show a transition from a dog-sized creature to a whale. And that's pretty powerful. Then there's these guys here. And then it starts to get into the little human evolution thing.", null, null, " Which I'm not going to go into. If you want to go deep into human evolution, you can take an anthropology class. That topic gets very emotional for people. When you talk about human evolution, unlike when you talk about animals and plants, it hits closer to home, yeah.", null, null, " So where do humans come from? Well, most anthropologists will tell you that humans, as we know them, appeared on Earth about 3 million years ago. We do have very close genetics to chimpanzees - we're not too far off. So, primates - and this gets to the heart of anthropology.", null, " Which I'm going to avoid. You know, the evolution of the head, the brain, cranial capacity. That's the big thing as it turns out with humans. There are two species of humans. There are actually two lineages, one died off, and that's the whole thing in anthropology right now." ], "text_ocr_list": [ null, "We can see these text from the image: The close relationship between hippopotamuses and whales is indicated on this evolutionary tree of major artiodactyl groups..\n And then we go into hippos and whales. So what we think happened is that dogs go back to form wolves, wolves then come back out of the forest to form hippos. And what's neat about this area is that within the last 20 years there have been so many discoveries. \n\n(Note: The original text has some factual errors that are not corrected here as per the instruction to only correct grammar, repetition, and transcription errors.)", null, "We can see these text from the image: The close relationship between hippopotamuses and whales is indicated on this evolutionary tree of major artiodactyl groups..\n So the Valley of the Whales, that was a discovery in Morocco. At that time, what is today Northern Africa, a desert, was actually a part of the Mediterranean Sea that had flooded that area. Whales would go in there and spawn, and so we have all these whale fossils from a very long period of time.", null, null, "We can see these text from the image: The close relationship between hippopotamuses and whales is indicated on evolutionary tree of four artiodactyl groups..\n And we've been able to put together - paleontologists have been able to put together - a whole sequence of whales that show a transition from a dog-sized creature to a whale. And that's pretty powerful. Then there's these guys here. And then it starts to get into the little human evolution thing.", null, null, "We can see these text from the image: CEBOIDEA: NEW WORLD MONKEYS\n\nCercopithecidae: Old World Monkeys\n\nFigure 15-5 (p. 544)\n\nRepresentative New World and Old World monkeys.\n\nFrom Coonrum, E. L. et al. 1990. Zoology. Philadelphia: W. B. Saunders Company.\n\nLewis, The Earth Through Time, Seventh Edition\n\nCopyright \u00a9 2008 John Wiley and Sons.\n Which I'm not going to go into. If you want to go deep into human evolution, you can take an anthropology class. That topic gets very emotional for people. When you talk about human evolution, unlike when you talk about animals and plants, it hits closer to home, yeah.", null, null, "We can see these text from the image: CEBOIDEA: NEW WORLD MONKEYS\n\nCERCOPITHECOIDEA: OLD WORLD MONKEYS\n\nFigure 15-5 (p. 544)\n\nRepresentative New World and Old World monkeys.\n\nFrom Cockrum, E. L. et al. 1982. Zoology. Philadelphia: W. B. Saunders Company.\n\nLewis, The Earth Through Time, Seventh Edition\n\nCopyright \u00a9 2003 John Wiley and Sons..\n So where do humans come from? Well, most anthropologists will tell you that humans, as we know them, appeared on Earth about 3 million years ago. We do have very close genetics to chimpanzees - we're not too far off. So, primates - and this gets to the heart of anthropology.", null, "We can see these text from the image: Figure 15-10 (p. 548)\n\nSkull of Proconsul (a dryomorph) from Lake Victoria, Kenya.\n\nCopyright \u00a9 2008 John Wiley and Sons..\n Which I'm going to avoid. You know, the evolution of the head, the brain, cranial capacity. That's the big thing as it turns out with humans. There are two species of humans. There are actually two lineages, one died off, and that's the whole thing in anthropology right now." ], "metadata": [ { "vid": "nZjXjULXk-I.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology Course on Mesozoic Era in Historical Geology_30.json#####audio#####doingASR#####FinishASR/nZjXjULXk-I/4049.0_4077.0.mp4", "refined_asr": " And then we go into hippos and whales. So what we think happened is that dogs go back to form wolves, wolves then come back out of the forest to form hippos. And what's neat about this area is that within the last 20 years there have been so many discoveries. \n\n(Note: The original text has some factual errors that are not corrected here as per the instruction to only correct grammar, repetition, and transcription errors.)", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4049.0_4077.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4049.0_4077.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4049.0_4077.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4049.0_4077.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4049.0_4077.0#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4049.0_4077.0#5.jpg" ], "ocr_qwen2_vl_72b": "The close relationship between hippopotamuses and whales is indicated on this evolutionary tree of major artiodactyl groups." }, { "vid": "nZjXjULXk-I.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology Course on Mesozoic Era in Historical Geology_30.json#####audio#####doingASR#####FinishASR/nZjXjULXk-I/4077.0_4099.0.mp4", "refined_asr": " So the Valley of the Whales, that was a discovery in Morocco. At that time, what is today Northern Africa, a desert, was actually a part of the Mediterranean Sea that had flooded that area. Whales would go in there and spawn, and so we have all these whale fossils from a very long period of time.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4077.0_4099.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4077.0_4099.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4077.0_4099.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4077.0_4099.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4077.0_4099.0#4.jpg" ], "ocr_qwen2_vl_72b": "The close relationship between hippopotamuses and whales is indicated on this evolutionary tree of major artiodactyl groups." }, { "vid": "nZjXjULXk-I.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology Course on Mesozoic Era in Historical Geology_30.json#####audio#####doingASR#####FinishASR/nZjXjULXk-I/4099.0_4121.0.mp4", "refined_asr": " And we've been able to put together - paleontologists have been able to put together - a whole sequence of whales that show a transition from a dog-sized creature to a whale. And that's pretty powerful. Then there's these guys here. And then it starts to get into the little human evolution thing.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4099.0_4121.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4099.0_4121.0#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4099.0_4121.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4099.0_4121.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4099.0_4121.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4099.0_4121.0#4.jpg" ], "ocr_qwen2_vl_72b": "The close relationship between hippopotamuses and whales is indicated on evolutionary tree of four artiodactyl groups." }, { "vid": "nZjXjULXk-I.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology Course on Mesozoic Era in Historical Geology_30.json#####audio#####doingASR#####FinishASR/nZjXjULXk-I/4121.0_4139.0.mp4", "refined_asr": " Which I'm not going to go into. If you want to go deep into human evolution, you can take an anthropology class. That topic gets very emotional for people. When you talk about human evolution, unlike when you talk about animals and plants, it hits closer to home, yeah.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4121.0_4139.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4121.0_4139.0#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4121.0_4139.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4121.0_4139.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4121.0_4139.0#3.jpg" ], "ocr_qwen2_vl_72b": "CEBOIDEA: NEW WORLD MONKEYS\n\nCercopithecidae: Old World Monkeys\n\nFigure 15-5 (p. 544)\n\nRepresentative New World and Old World monkeys.\n\nFrom Coonrum, E. L. et al. 1990. Zoology. Philadelphia: W. B. Saunders Company.\n\nLewis, The Earth Through Time, Seventh Edition\n\nCopyright \u00a9 2008 John Wiley and Sons" }, { "vid": "nZjXjULXk-I.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology Course on Mesozoic Era in Historical Geology_30.json#####audio#####doingASR#####FinishASR/nZjXjULXk-I/4139.0_4169.0.mp4", "refined_asr": " So where do humans come from? Well, most anthropologists will tell you that humans, as we know them, appeared on Earth about 3 million years ago. We do have very close genetics to chimpanzees - we're not too far off. So, primates - and this gets to the heart of anthropology.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4139.0_4169.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4139.0_4169.0#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4139.0_4169.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4139.0_4169.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4139.0_4169.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4139.0_4169.0#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4139.0_4169.0#5.jpg" ], "ocr_qwen2_vl_72b": "CEBOIDEA: NEW WORLD MONKEYS\n\nCERCOPITHECOIDEA: OLD WORLD MONKEYS\n\nFigure 15-5 (p. 544)\n\nRepresentative New World and Old World monkeys.\n\nFrom Cockrum, E. L. et al. 1982. Zoology. Philadelphia: W. B. Saunders Company.\n\nLewis, The Earth Through Time, Seventh Edition\n\nCopyright \u00a9 2003 John Wiley and Sons." }, { "vid": "nZjXjULXk-I.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology Course on Mesozoic Era in Historical Geology_30.json#####audio#####doingASR#####FinishASR/nZjXjULXk-I/4169.0_4191.0.mp4", "refined_asr": " Which I'm going to avoid. You know, the evolution of the head, the brain, cranial capacity. That's the big thing as it turns out with humans. There are two species of humans. There are actually two lineages, one died off, and that's the whole thing in anthropology right now.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4169.0_4191.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4169.0_4191.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4169.0_4191.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4169.0_4191.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/nZjXjULXk-I/nZjXjULXk-I@4169.0_4191.0#4.jpg" ], "ocr_qwen2_vl_72b": "Figure 15-10 (p. 548)\n\nSkull of Proconsul (a dryomorph) from Lake Victoria, Kenya.\n\nCopyright \u00a9 2008 John Wiley and Sons." } ], "image_num": 9, "text_num": 459, "token_num": 5643 }, { "images": [ "sample_100_images/ZuUqLj-rmtY@67.92_79.68#1.jpg", null, "sample_100_images/ZuUqLj-rmtY@79.68_95.3#1.jpg", null, "sample_100_images/ZuUqLj-rmtY@95.3_118.66000000000001#1.jpg", "sample_100_images/ZuUqLj-rmtY@95.3_118.66000000000001#3.jpg", null, "sample_100_images/ZuUqLj-rmtY@119.14_139.66#1.jpg", null ], "texts": [ null, " Every drought that we have had in the past 40 to 60 years includes the 1977 drought, the drought of the late 80s and early 90s, and there was a drought in the early 2000s.", null, " In the late 2000s. And then we're currently in a drought since 2011. Predictably, in all of those, groundwater levels in many places go down because groundwater is not replenished at a normal rate and it's being extracted.", null, null, " Rates that far exceed the rate of replenishment in some places the annual drop in water levels may be 10 to 15 feet in other places often not far away the annual drop in water levels from spring to spring may be as much as 40 feet per year in many areas of California", null, " Because of the rapid sequence of droughts that we have seen over the last 15 years, we're now at water levels that are far lower than we have ever recorded in the history of this state. This is showing the change in water levels between the beginning of 2011 when we were just" ], "text_ocr_list": [ null, "We can see these text from the image: - Groundwater Levels during Drought\n- Groundwater Levels for Well 22S25E08N001M\n- 10 ft/y\n- 10 ft/y\n- 15 ft/y.\n Every drought that we have had in the past 40 to 60 years includes the 1977 drought, the drought of the late 80s and early 90s, and there was a drought in the early 2000s.", null, "We can see these text from the image: - Groundwater Levels during Drought\n- Groundwater Levels for Well 22S25E08N001M\n- 10 ft/y\n- 10 ft/y\n- 15 ft/y.\n In the late 2000s. And then we're currently in a drought since 2011. Predictably, in all of those, groundwater levels in many places go down because groundwater is not replenished at a normal rate and it's being extracted.", null, null, "We can see these text from the image: Groundwater Levels during Drought\n\nGroundwater Levels for Well 22S25E08N001M\n\n10 ft/y\n10 ft/y\n15 ft/y.\n Rates that far exceed the rate of replenishment in some places the annual drop in water levels may be 10 to 15 feet in other places often not far away the annual drop in water levels from spring to spring may be as much as 40 feet per year in many areas of California", null, "We can see these text from the image: Groundwater Level Change:\n\nSpring 2011\nto Spring 2016\n\nIncrease > 10 feet\nIncrease 10 to 2.5 feet\nChange +/- 2.5 feet\nDecrease 2.5-10 feet\nDecrease > 10 feet.\n Because of the rapid sequence of droughts that we have seen over the last 15 years, we're now at water levels that are far lower than we have ever recorded in the history of this state. This is showing the change in water levels between the beginning of 2011 when we were just" ], "metadata": [ { "vid": "ZuUqLj-rmtY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology Course Groundwater Management - Understanding Groundwater Overdraft_29.json#####audio#####doingASR#####FinishASR/ZuUqLj-rmtY/67.92_79.68.mp4", "refined_asr": " Every drought that we have had in the past 40 to 60 years includes the 1977 drought, the drought of the late 80s and early 90s, and there was a drought in the early 2000s.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@67.92_79.68#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@67.92_79.68#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@67.92_79.68#2.jpg" ], "ocr_qwen2_vl_72b": "- Groundwater Levels during Drought\n- Groundwater Levels for Well 22S25E08N001M\n- 10 ft/y\n- 10 ft/y\n- 15 ft/y" }, { "vid": "ZuUqLj-rmtY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology Course Groundwater Management - Understanding Groundwater Overdraft_29.json#####audio#####doingASR#####FinishASR/ZuUqLj-rmtY/79.68_95.3.mp4", "refined_asr": " In the late 2000s. And then we're currently in a drought since 2011. Predictably, in all of those, groundwater levels in many places go down because groundwater is not replenished at a normal rate and it's being extracted.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@79.68_95.3#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@79.68_95.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@79.68_95.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@79.68_95.3#3.jpg" ], "ocr_qwen2_vl_72b": "- Groundwater Levels during Drought\n- Groundwater Levels for Well 22S25E08N001M\n- 10 ft/y\n- 10 ft/y\n- 15 ft/y" }, { "vid": "ZuUqLj-rmtY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology Course Groundwater Management - Understanding Groundwater Overdraft_29.json#####audio#####doingASR#####FinishASR/ZuUqLj-rmtY/95.3_118.66000000000001.mp4", "refined_asr": " Rates that far exceed the rate of replenishment in some places the annual drop in water levels may be 10 to 15 feet in other places often not far away the annual drop in water levels from spring to spring may be as much as 40 feet per year in many areas of California", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@95.3_118.66000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@95.3_118.66000000000001#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@95.3_118.66000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@95.3_118.66000000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@95.3_118.66000000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@95.3_118.66000000000001#4.jpg" ], "ocr_qwen2_vl_72b": "Groundwater Levels during Drought\n\nGroundwater Levels for Well 22S25E08N001M\n\n10 ft/y\n10 ft/y\n15 ft/y" }, { "vid": "ZuUqLj-rmtY.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology Course Groundwater Management - Understanding Groundwater Overdraft_29.json#####audio#####doingASR#####FinishASR/ZuUqLj-rmtY/119.14_139.66.mp4", "refined_asr": " Because of the rapid sequence of droughts that we have seen over the last 15 years, we're now at water levels that are far lower than we have ever recorded in the history of this state. This is showing the change in water levels between the beginning of 2011 when we were just", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@119.14_139.66#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@119.14_139.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@119.14_139.66#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ZuUqLj-rmtY/ZuUqLj-rmtY@119.14_139.66#3.jpg" ], "ocr_qwen2_vl_72b": "Groundwater Level Change:\n\nSpring 2011\nto Spring 2016\n\nIncrease > 10 feet\nIncrease 10 to 2.5 feet\nChange +/- 2.5 feet\nDecrease 2.5-10 feet\nDecrease > 10 feet" } ], "image_num": 5, "text_num": 258, "token_num": 3138 }, { "images": [ "sample_100_images/pyZuqrQtLhI@3058.3_3074.84#1.jpg", null, "sample_100_images/pyZuqrQtLhI@3074.84_3091.82#1.jpg", null, "sample_100_images/pyZuqrQtLhI@3091.82_3110.92#1.jpg", null, "sample_100_images/pyZuqrQtLhI@3111.86_3130.56#1.jpg", "sample_100_images/pyZuqrQtLhI@3111.86_3130.56#3.jpg", null, "sample_100_images/pyZuqrQtLhI@3130.56_3147.1#1.jpg", null, "sample_100_images/pyZuqrQtLhI@3147.1_3162.36#1.jpg", null, "sample_100_images/pyZuqrQtLhI@3162.36_3184.04#1.jpg", "sample_100_images/pyZuqrQtLhI@3162.36_3184.04#2.jpg", "sample_100_images/pyZuqrQtLhI@3162.36_3184.04#3.jpg", null, "sample_100_images/pyZuqrQtLhI@3184.04_3203.9#1.jpg", "sample_100_images/pyZuqrQtLhI@3184.04_3203.9#3.jpg", null ], "texts": [ null, " Because that's where the plates have merged and have become too stuck together. Each of the suture zones represents an orogeny as well because that's where they collided and that's where they deformed. We can start filling them in here as well.", null, " And if you look closely enough at your screen, you will see the gray hatched areas appearing here. They're scattered a little bit all across Europe. We only did the ones from the past 600 million years. That's why Baltica is not really popping up.", null, " But basically, you've got four big, three to four big groups of these ranges that have been formed. These have shaped Europe and are quite important to see. Some, especially the Alps and the various shields, are the ones that are going to pop up later.", null, null, " Now importantly what we see here in the east are the Urals or the Ural Mountains as they're called geographically because these help us actually to get the missing boundary that we had to define Europe. These are the three boundaries.", null, " You've got two active plate tectonic margins to the west and to the south. To the south, the African plate margin. And to the east, you have the suture zone which is the Uralides, and everything in between.", null, " Geologically speaking this is the definition of Europe. It's a scientific definition and one that I like because I'm a geologist and geology should come first. Now let's explore. Having done this definition let's start with the claim of Iceland.", null, null, null, " That there you can walk between the continents. And yeah let's zoom in a little bit to Iceland and actually the yellow star will show you where the bridge is that features in Game of Thrones. Let's check here. Yeah there it is.", null, null, " And you see indeed that it is sitting on this metoceanic ridge because Iceland is a very interesting and very unique place in the world. It's actually where the metoceanic ridge pops up. And indeed since we have proven that Europe is a continent" ], "text_ocr_list": [ null, "We can see these text from the image: GeoERA - GEO-ENERGY\n\nBuilding up Europe\n\nKris Piessens\n\nLaurentia\n\nBaltica\n\nLaurentia\n\nAvalonia\n\nALCAPA\n\nTisza\n\nAdria\n\nDacia\n\n(South) Armorica\n\nIberia.\n Because that's where the plates have merged and have become too stuck together. Each of the suture zones represents an orogeny as well because that's where they collided and that's where they deformed. We can start filling them in here as well.", null, "We can see these text from the image: Building up Europe\n\nKris Piessens.\n And if you look closely enough at your screen, you will see the gray hatched areas appearing here. They're scattered a little bit all across Europe. We only did the ones from the past 600 million years. That's why Baltica is not really popping up.", null, "We can see these text from the image: - GeoERA - GEO-ENERGY\n- Building up Europe\n- Kris Piessens\n- Caledonian orogen\n- Variscan orogen\n- Alpine orogen\n- Uralides.\n But basically, you've got four big, three to four big groups of these ranges that have been formed. These have shaped Europe and are quite important to see. Some, especially the Alps and the various shields, are the ones that are going to pop up later.", null, null, "We can see these text from the image: - GeoERA - GEO-ENERGY\n- Building up Europe\n- Kris Piessens\n- Caledonian orogen\n- Variscan orogen\n- Alpine orogen\n- Uralides.\n Now importantly what we see here in the east are the Urals or the Ural Mountains as they're called geographically because these help us actually to get the missing boundary that we had to define Europe. These are the three boundaries.", null, "We can see these text from the image: GeoERA - GEO-ENERGY\n\nBuilding up Europe\n\nKris Piessens.\n You've got two active plate tectonic margins to the west and to the south. To the south, the African plate margin. And to the east, you have the suture zone which is the Uralides, and everything in between.", null, "We can see these text from the image: - GeoERA - GEO-ENERGY\n- Building up Europe\n- This is Europe!\n- Kris Piessens\n- ZOOM\n- GEO CONNECT3D.\n Geologically speaking this is the definition of Europe. It's a scientific definition and one that I like because I'm a geologist and geology should come first. Now let's explore. Having done this definition let's start with the claim of Iceland.", null, null, null, "We can see these text from the image: Fact-checking an Icelandic Claim\n\nKris Piessens.\n That there you can walk between the continents. And yeah let's zoom in a little bit to Iceland and actually the yellow star will show you where the bridge is that features in Game of Thrones. Let's check here. Yeah there it is.", null, null, "We can see these text from the image: - GeoERA - GEO-ENERGY\n- FACT-CHECKING AN ICELANDIC CLAIM\n- Kris Piessens\n- GEOSITE\n- ZOOM\n- GEO CONNECT3D.\n And you see indeed that it is sitting on this metoceanic ridge because Iceland is a very interesting and very unique place in the world. It's actually where the metoceanic ridge pops up. And indeed since we have proven that Europe is a continent" ], "metadata": [ { "vid": "pyZuqrQtLhI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology and Society Geotourism tutorial_30.json#####audio#####doingASR#####FinishASR/pyZuqrQtLhI/3058.3_3074.84.mp4", "refined_asr": " Because that's where the plates have merged and have become too stuck together. Each of the suture zones represents an orogeny as well because that's where they collided and that's where they deformed. We can start filling them in here as well.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3058.3_3074.84#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3058.3_3074.84#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3058.3_3074.84#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3058.3_3074.84#3.jpg" ], "ocr_qwen2_vl_72b": "GeoERA - GEO-ENERGY\n\nBuilding up Europe\n\nKris Piessens\n\nLaurentia\n\nBaltica\n\nLaurentia\n\nAvalonia\n\nALCAPA\n\nTisza\n\nAdria\n\nDacia\n\n(South) Armorica\n\nIberia" }, { "vid": "pyZuqrQtLhI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology and Society Geotourism tutorial_30.json#####audio#####doingASR#####FinishASR/pyZuqrQtLhI/3074.84_3091.82.mp4", "refined_asr": " And if you look closely enough at your screen, you will see the gray hatched areas appearing here. They're scattered a little bit all across Europe. We only did the ones from the past 600 million years. That's why Baltica is not really popping up.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3074.84_3091.82#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3074.84_3091.82#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3074.84_3091.82#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3074.84_3091.82#3.jpg" ], "ocr_qwen2_vl_72b": "Building up Europe\n\nKris Piessens" }, { "vid": "pyZuqrQtLhI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology and Society Geotourism tutorial_30.json#####audio#####doingASR#####FinishASR/pyZuqrQtLhI/3091.82_3110.92.mp4", "refined_asr": " But basically, you've got four big, three to four big groups of these ranges that have been formed. These have shaped Europe and are quite important to see. Some, especially the Alps and the various shields, are the ones that are going to pop up later.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3091.82_3110.92#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3091.82_3110.92#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3091.82_3110.92#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3091.82_3110.92#3.jpg" ], "ocr_qwen2_vl_72b": "- GeoERA - GEO-ENERGY\n- Building up Europe\n- Kris Piessens\n- Caledonian orogen\n- Variscan orogen\n- Alpine orogen\n- Uralides" }, { "vid": "pyZuqrQtLhI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology and Society Geotourism tutorial_30.json#####audio#####doingASR#####FinishASR/pyZuqrQtLhI/3111.86_3130.56.mp4", "refined_asr": " Now importantly what we see here in the east are the Urals or the Ural Mountains as they're called geographically because these help us actually to get the missing boundary that we had to define Europe. These are the three boundaries.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3111.86_3130.56#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3111.86_3130.56#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3111.86_3130.56#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3111.86_3130.56#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3111.86_3130.56#3.jpg" ], "ocr_qwen2_vl_72b": "- GeoERA - GEO-ENERGY\n- Building up Europe\n- Kris Piessens\n- Caledonian orogen\n- Variscan orogen\n- Alpine orogen\n- Uralides" }, { "vid": "pyZuqrQtLhI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology and Society Geotourism tutorial_30.json#####audio#####doingASR#####FinishASR/pyZuqrQtLhI/3130.56_3147.1.mp4", "refined_asr": " You've got two active plate tectonic margins to the west and to the south. To the south, the African plate margin. And to the east, you have the suture zone which is the Uralides, and everything in between.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3130.56_3147.1#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3130.56_3147.1#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3130.56_3147.1#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3130.56_3147.1#3.jpg" ], "ocr_qwen2_vl_72b": "GeoERA - GEO-ENERGY\n\nBuilding up Europe\n\nKris Piessens" }, { "vid": "pyZuqrQtLhI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology and Society Geotourism tutorial_30.json#####audio#####doingASR#####FinishASR/pyZuqrQtLhI/3147.1_3162.36.mp4", "refined_asr": " Geologically speaking this is the definition of Europe. It's a scientific definition and one that I like because I'm a geologist and geology should come first. Now let's explore. Having done this definition let's start with the claim of Iceland.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3147.1_3162.36#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3147.1_3162.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3147.1_3162.36#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3147.1_3162.36#3.jpg" ], "ocr_qwen2_vl_72b": "- GeoERA - GEO-ENERGY\n- Building up Europe\n- This is Europe!\n- Kris Piessens\n- ZOOM\n- GEO CONNECT3D" }, { "vid": "pyZuqrQtLhI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology and Society Geotourism tutorial_30.json#####audio#####doingASR#####FinishASR/pyZuqrQtLhI/3162.36_3184.04.mp4", "refined_asr": " That there you can walk between the continents. And yeah let's zoom in a little bit to Iceland and actually the yellow star will show you where the bridge is that features in Game of Thrones. Let's check here. Yeah there it is.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3162.36_3184.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3162.36_3184.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3162.36_3184.04#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3162.36_3184.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3162.36_3184.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3162.36_3184.04#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3162.36_3184.04#4.jpg" ], "ocr_qwen2_vl_72b": "Fact-checking an Icelandic Claim\n\nKris Piessens" }, { "vid": "pyZuqrQtLhI.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Geology and Society Geotourism tutorial_30.json#####audio#####doingASR#####FinishASR/pyZuqrQtLhI/3184.04_3203.9.mp4", "refined_asr": " And you see indeed that it is sitting on this metoceanic ridge because Iceland is a very interesting and very unique place in the world. It's actually where the metoceanic ridge pops up. And indeed since we have proven that Europe is a continent", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3184.04_3203.9#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3184.04_3203.9#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3184.04_3203.9#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3184.04_3203.9#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/pyZuqrQtLhI/pyZuqrQtLhI@3184.04_3203.9#3.jpg" ], "ocr_qwen2_vl_72b": "- GeoERA - GEO-ENERGY\n- FACT-CHECKING AN ICELANDIC CLAIM\n- Kris Piessens\n- GEOSITE\n- ZOOM\n- GEO CONNECT3D" } ], "image_num": 12, "text_num": 459, "token_num": 7371 }, { "images": [ "sample_100_images/w2wYPjlDxQQ@176.0_204.0#1.jpg", null, "sample_100_images/w2wYPjlDxQQ@204.0_238.0#1.jpg", null, "sample_100_images/w2wYPjlDxQQ@238.0_260.0#1.jpg", null, "sample_100_images/w2wYPjlDxQQ@260.0_276.0#1.jpg", null, "sample_100_images/w2wYPjlDxQQ@276.0_298.0#1.jpg", null, "sample_100_images/w2wYPjlDxQQ@298.0_322.0#1.jpg", null, null ], "texts": [ null, " To the reservoir should equal the total flux out. So in the first three questions, show that that's the case for the ocean reservoir, for the atmosphere reservoir, and for the groundwater reservoir. Show that the fluxes going into it add together to equal.", null, " The total fluxes going out of the reservoir added together. We can also think about the water cycle in terms of residence times. The average amount of time water remains in a given reservoir. The residence time equals the volume of that reservoir divided by the total flux leaving it.", null, " Of the fluxes in or out of that reservoir but not both. You can add up the total fluxes in or you can add up the total fluxes out. You're dividing the volume by that number. Why does this give a residence time?", null, " Time will have units of years, days, months, or any other unit of time. What we're doing is we have the volume - in this case, it's in cubic kilometers. We're dividing by fluxes, which.", null, " The units are in cubic kilometers per year. By dividing volume by flux, our cubic kilometers cancel out. We're left with units of time in years, which would be an appropriate unit for a residence time. For example, to find the residence time of water in the atmosphere.", null, " Take the volume of water in the atmosphere, divide it by the total fluxes, either out of the atmosphere - in other words, the total of those two numbers - or the total flux into the atmosphere, the sum of those two numbers. You could do it either way.", " And you should get the same answer." ], "text_ocr_list": [ null, "We can see these text from the image: - Atmosphere: 13,000 km\u00b3\n- Precipitation onto Continents: 111,000 km\u00b3/yr\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Plants and Animals: 10,000 km\u00b3\n- Glaciers: 29,200,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Infiltration: 16,000 km\u00b3/yr\n- Runoff: 24,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n- Oceans: 1,320,000,000 km\u00b3\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n- Precipitation into Oceans: 385,000 km\u00b3/yr.\n To the reservoir should equal the total flux out. So in the first three questions, show that that's the case for the ocean reservoir, for the atmosphere reservoir, and for the groundwater reservoir. Show that the fluxes going into it add together to equal.", null, "We can see these text from the image: - Atmosphere: 13,000 km\u00b3\n- Precipitation onto Continents: 111,000 km\u00b3/yr\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Plants and Animals: 10,000 km\u00b3\n- Glaciers: 29,200,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Infiltration: 16,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Runoff: 24,000 km\u00b3/yr\n- Oceans: 1,320,000,000 km\u00b3\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n- Precipitation into Oceans: 385,000 km\u00b3/yr\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n\nTotal flux in = total flux out.\n The total fluxes going out of the reservoir added together. We can also think about the water cycle in terms of residence times. The average amount of time water remains in a given reservoir. The residence time equals the volume of that reservoir divided by the total flux leaving it.", null, "We can see these text from the image: - Precipitation onto Continents: 111,000 km\u00b3/yr\n- Atmosphere: 13,000 km\u00b3\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Glaciers: 29,200,000 km\u00b3\n- Plants and Animals: 10,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Infiltration: 16,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Runoff: 24,000 km\u00b3/yr\n- Precipitation into Oceans: 385,000 km\u00b3/yr\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n- Oceans: 1,320,000,000 km\u00b3\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n\nResidence time = Volume / Total flux.\n Of the fluxes in or out of that reservoir but not both. You can add up the total fluxes in or you can add up the total fluxes out. You're dividing the volume by that number. Why does this give a residence time?", null, "We can see these text from the image: - Precipitation onto Continents: 111,000 km\u00b3/yr\n- Atmosphere: 13,000 km\u00b3\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Glaciers: 29,200,000 km\u00b3\n- Plants and Animals: 10,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Infiltration: 16,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Runoff: 24,000 km\u00b3/yr\n- Oceans: 1,320,000,000 km\u00b3\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n- Precipitation into Oceans: 385,000 km\u00b3/yr\n\nResidence time = Volume / total flux (in or out).\n Time will have units of years, days, months, or any other unit of time. What we're doing is we have the volume - in this case, it's in cubic kilometers. We're dividing by fluxes, which.", null, "We can see these text from the image: - Precipitation onto Continents: 111,000 km\u00b3/yr\n- Atmosphere: 13,000 km\u00b3\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Glaciers: 29,200,000 km\u00b3\n- Plants and Animals: 10,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Infiltration: 16,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Runoff: 24,000 km\u00b3/yr\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n- Oceans: 1,320,000,000 km\u00b3\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n- Precipitation into Oceans: 385,000 km\u00b3/yr\n\nResidence time = Volume (km\u00b3) / Total fluxes in (or out) (km\u00b3/yr).\n The units are in cubic kilometers per year. By dividing volume by flux, our cubic kilometers cancel out. We're left with units of time in years, which would be an appropriate unit for a residence time. For example, to find the residence time of water in the atmosphere.", null, "We can see these text from the image: - Precipitation onto Continents: 111,000 km\u00b3/yr\n- Atmosphere: 13,000 km\u00b3\n- Precipitation into Oceans: 385,000 km\u00b3/yr\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Glaciers: 29,200,000 km\u00b3\n- Plants and Animals: 10,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Runoff: 24,000 km\u00b3/yr\n- Infiltration: 16,000 km\u00b3/yr\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Oceans: 1,320,000,000 km\u00b3\n\nResidence time = Volume (km\u00b3) / Total fluxes in (or out) (km\u00b3/yr).\n Take the volume of water in the atmosphere, divide it by the total fluxes, either out of the atmosphere - in other words, the total of those two numbers - or the total flux into the atmosphere, the sum of those two numbers. You could do it either way.", " And you should get the same answer." ], "metadata": [ { "vid": "w2wYPjlDxQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology course on Thermal Stratification in Lakes and Reservoirs_30.json#####audio#####doingASR#####FinishASR/w2wYPjlDxQQ/176.0_204.0.mp4", "refined_asr": " To the reservoir should equal the total flux out. So in the first three questions, show that that's the case for the ocean reservoir, for the atmosphere reservoir, and for the groundwater reservoir. Show that the fluxes going into it add together to equal.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@176.0_204.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@176.0_204.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@176.0_204.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@176.0_204.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@176.0_204.0#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@176.0_204.0#5.jpg" ], "ocr_qwen2_vl_72b": "- Atmosphere: 13,000 km\u00b3\n- Precipitation onto Continents: 111,000 km\u00b3/yr\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Plants and Animals: 10,000 km\u00b3\n- Glaciers: 29,200,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Infiltration: 16,000 km\u00b3/yr\n- Runoff: 24,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n- Oceans: 1,320,000,000 km\u00b3\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n- Precipitation into Oceans: 385,000 km\u00b3/yr" }, { "vid": "w2wYPjlDxQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology course on Thermal Stratification in Lakes and Reservoirs_30.json#####audio#####doingASR#####FinishASR/w2wYPjlDxQQ/204.0_238.0.mp4", "refined_asr": " The total fluxes going out of the reservoir added together. We can also think about the water cycle in terms of residence times. The average amount of time water remains in a given reservoir. The residence time equals the volume of that reservoir divided by the total flux leaving it.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@204.0_238.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@204.0_238.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@204.0_238.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@204.0_238.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@204.0_238.0#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@204.0_238.0#5.jpg" ], "ocr_qwen2_vl_72b": "- Atmosphere: 13,000 km\u00b3\n- Precipitation onto Continents: 111,000 km\u00b3/yr\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Plants and Animals: 10,000 km\u00b3\n- Glaciers: 29,200,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Infiltration: 16,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Runoff: 24,000 km\u00b3/yr\n- Oceans: 1,320,000,000 km\u00b3\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n- Precipitation into Oceans: 385,000 km\u00b3/yr\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n\nTotal flux in = total flux out" }, { "vid": "w2wYPjlDxQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology course on Thermal Stratification in Lakes and Reservoirs_30.json#####audio#####doingASR#####FinishASR/w2wYPjlDxQQ/238.0_260.0.mp4", "refined_asr": " Of the fluxes in or out of that reservoir but not both. You can add up the total fluxes in or you can add up the total fluxes out. You're dividing the volume by that number. Why does this give a residence time?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@238.0_260.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@238.0_260.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@238.0_260.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@238.0_260.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@238.0_260.0#4.jpg" ], "ocr_qwen2_vl_72b": "- Precipitation onto Continents: 111,000 km\u00b3/yr\n- Atmosphere: 13,000 km\u00b3\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Glaciers: 29,200,000 km\u00b3\n- Plants and Animals: 10,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Infiltration: 16,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Runoff: 24,000 km\u00b3/yr\n- Precipitation into Oceans: 385,000 km\u00b3/yr\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n- Oceans: 1,320,000,000 km\u00b3\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n\nResidence time = Volume / Total flux" }, { "vid": "w2wYPjlDxQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology course on Thermal Stratification in Lakes and Reservoirs_30.json#####audio#####doingASR#####FinishASR/w2wYPjlDxQQ/260.0_276.0.mp4", "refined_asr": " Time will have units of years, days, months, or any other unit of time. What we're doing is we have the volume - in this case, it's in cubic kilometers. We're dividing by fluxes, which.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@260.0_276.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@260.0_276.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@260.0_276.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@260.0_276.0#3.jpg" ], "ocr_qwen2_vl_72b": "- Precipitation onto Continents: 111,000 km\u00b3/yr\n- Atmosphere: 13,000 km\u00b3\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Glaciers: 29,200,000 km\u00b3\n- Plants and Animals: 10,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Infiltration: 16,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Runoff: 24,000 km\u00b3/yr\n- Oceans: 1,320,000,000 km\u00b3\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n- Precipitation into Oceans: 385,000 km\u00b3/yr\n\nResidence time = Volume / total flux (in or out)" }, { "vid": "w2wYPjlDxQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology course on Thermal Stratification in Lakes and Reservoirs_30.json#####audio#####doingASR#####FinishASR/w2wYPjlDxQQ/276.0_298.0.mp4", "refined_asr": " The units are in cubic kilometers per year. By dividing volume by flux, our cubic kilometers cancel out. We're left with units of time in years, which would be an appropriate unit for a residence time. For example, to find the residence time of water in the atmosphere.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@276.0_298.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@276.0_298.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@276.0_298.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@276.0_298.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@276.0_298.0#4.jpg" ], "ocr_qwen2_vl_72b": "- Precipitation onto Continents: 111,000 km\u00b3/yr\n- Atmosphere: 13,000 km\u00b3\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Glaciers: 29,200,000 km\u00b3\n- Plants and Animals: 10,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Infiltration: 16,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Runoff: 24,000 km\u00b3/yr\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n- Oceans: 1,320,000,000 km\u00b3\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n- Precipitation into Oceans: 385,000 km\u00b3/yr\n\nResidence time = Volume (km\u00b3) / Total fluxes in (or out) (km\u00b3/yr)" }, { "vid": "w2wYPjlDxQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology course on Thermal Stratification in Lakes and Reservoirs_30.json#####audio#####doingASR#####FinishASR/w2wYPjlDxQQ/298.0_322.0.mp4", "refined_asr": " Take the volume of water in the atmosphere, divide it by the total fluxes, either out of the atmosphere - in other words, the total of those two numbers - or the total flux into the atmosphere, the sum of those two numbers. You could do it either way.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@298.0_322.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@298.0_322.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@298.0_322.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@298.0_322.0#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@298.0_322.0#4.jpg" ], "ocr_qwen2_vl_72b": "- Precipitation onto Continents: 111,000 km\u00b3/yr\n- Atmosphere: 13,000 km\u00b3\n- Precipitation into Oceans: 385,000 km\u00b3/yr\n- Evaporation from Ocean: 425,000 km\u00b3/yr\n- Evapotranspiration from Continents: 71,000 km\u00b3/yr\n- Glaciers: 29,200,000 km\u00b3\n- Plants and Animals: 10,000 km\u00b3\n- Lakes and Rivers: 230,000 km\u00b3\n- Runoff: 24,000 km\u00b3/yr\n- Infiltration: 16,000 km\u00b3/yr\n- Groundwater Discharge to Ocean: 16,000 km\u00b3/yr\n- Groundwater: 8,417,000 km\u00b3\n- Oceans: 1,320,000,000 km\u00b3\n\nResidence time = Volume (km\u00b3) / Total fluxes in (or out) (km\u00b3/yr)" }, { "vid": "w2wYPjlDxQQ.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Hydrology course on Thermal Stratification in Lakes and Reservoirs_30.json#####audio#####doingASR#####FinishASR/w2wYPjlDxQQ/322.0_324.0.mp4", "refined_asr": " And you should get the same answer.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/w2wYPjlDxQQ/w2wYPjlDxQQ@322.0_324.0#1.jpg" ], "ocr_qwen2_vl_72b": null } ], "image_num": 6, "text_num": 364, "token_num": 3820 }, { "images": [ "sample_100_images/lUUte2o2Sn8@2587.8_2608.04#1.jpg", "sample_100_images/lUUte2o2Sn8@2587.8_2608.04#3.jpg", null, "sample_100_images/lUUte2o2Sn8@2608.04_2635.52#1.jpg", "sample_100_images/lUUte2o2Sn8@2608.04_2635.52#2.jpg", "sample_100_images/lUUte2o2Sn8@2608.04_2635.52#3.jpg", "sample_100_images/lUUte2o2Sn8@2608.04_2635.52#4.jpg", null, "sample_100_images/lUUte2o2Sn8@2635.52_2652.7#1.jpg", null, "sample_100_images/lUUte2o2Sn8@2652.7_2676.72#1.jpg", "sample_100_images/lUUte2o2Sn8@2652.7_2676.72#3.jpg", null, "sample_100_images/lUUte2o2Sn8@2676.72_2688.44#1.jpg", "sample_100_images/lUUte2o2Sn8@2676.72_2688.44#2.jpg", null, "sample_100_images/lUUte2o2Sn8@2688.44_2712.3#1.jpg", "sample_100_images/lUUte2o2Sn8@2688.44_2712.3#2.jpg", null, "sample_100_images/lUUte2o2Sn8@2712.3_2729.0600000000004#1.jpg", "sample_100_images/lUUte2o2Sn8@2712.3_2729.0600000000004#2.jpg", null ], "texts": [ null, null, " So that tells me if I find one set of Xs that works then two times that set of Xs will work five times that set of Xs will work because it'll just multiply all through because I'll have zero equals zero and I can safely multiply by five.", null, null, null, null, " So I'm expecting two solutions, two separate solutions. And I'll write those maybe up above: x1, x2, x3, x4, yeah. So give me. So I have two equations. And for these four unknowns, let me.", null, " Start out with x3 equal to 1 and x4 equal to 0. So I'm just sort of freely setting x3 and x4 to 1 and 0. And then, either I or you will figure out the rest.", null, null, " What do x1 and x2 have to be? So what's the answer then? If x3 is 1 and x4 is 0, what is x1? This is obviously 0. So what is x1? That leads to x1 equals what?", null, null, " Negative three Chief Justice Gonzalez negative three thank you and what's X squared that'll come from this equation what will X squared be", null, null, " Negative four. So that's a solution. We've solved this original problem. Oh, I erased it. It was much worse at the beginning, but we found a solution. But again, we have four unknowns and we have only two real equations.", null, null, " Two independent equations because this third equation turned out to be a fraud. It was just a combination of one and two. So we really have only two equations for unknowns. So we'd expect two parameters in the solution and I found one of them." ], "text_ocr_list": [ null, null, "We can see these text from the image: 1 2 11 17\n\n10 4 6\n\n0 0 0\n\n10 3 5\n\n0 1 4\n\n0 0 0\n\nx\u2081 + 0x\u2082 + 3x\u2083 + 5x\u2084 = 0\n\n0x\u2081 + 1x\u2082 + 4x\u2083 + 6x\u2084 = 0.\n So that tells me if I find one set of Xs that works then two times that set of Xs will work five times that set of Xs will work because it'll just multiply all through because I'll have zero equals zero and I can safely multiply by five.", null, null, null, null, "We can see these text from the image: 1 2 11 17\n\n0 1 4 6\n\n5 0\n\nx\u2081 + 0x\u2082 + 3x\u2083 + 5x\u2084 = 0\n\n0x\u2081 + 1x\u2082 + 4x\u2083 + 6x\u2084 = 0.\n So I'm expecting two solutions, two separate solutions. And I'll write those maybe up above: x1, x2, x3, x4, yeah. So give me. So I have two equations. And for these four unknowns, let me.", null, "We can see these text from the image: 1 2 11 17\n3 7 37 57\n0 0 0 0\n\nA x = 0\n\nm=3 eqns n=4 unknowns RHS=0\none soln: x1=x2=x3=x4=0\nmore solns:\nx1, x2, x3, x4\n\n1 2 11 17\n10 1 4 6\n0 0 0 0\n\n10 5\n0 6\n\nx1 + 0x2 + 3x3 + 5x4 = 0\n0x1 + 1x2 + 4x3 + 6x4 = 0.\n Start out with x3 equal to 1 and x4 equal to 0. So I'm just sort of freely setting x3 and x4 to 1 and 0. And then, either I or you will figure out the rest.", null, null, "We can see these text from the image: 1 2 11 17\n3 7 37 57\n0 0 0 0\n\nA x = 0\n\nm=3 eqns n=4 unknowns RHS=0\none soln: x1=x2=x3=x4=0\nmore solns:\nx1,x2,x3,x4\n\n1 2 11 17\n10 1 4 0\n0 0 0 0\n\n[10 35\n0 1 4 6\n0 0 0 0]\n\nx1 + 0x2 + 3x3 + 5x4 = 0\n0x1 + 1x2 + 4x3 + 6x4 = 0.\n What do x1 and x2 have to be? So what's the answer then? If x3 is 1 and x4 is 0, what is x1? This is obviously 0. So what is x1? That leads to x1 equals what?", null, null, "We can see these text from the image: 1 2 11 17\n3 7 37 57\n\nm=3 eqns n=4 unknowns RHS=0\none soln: x=x2=x3=x4=0\nmore solns:\nx1,x2,x3,x4\n\nA x=0\n\n1 2 11 17 10\n0 1 4 6 0\n0 0 0 0 0\n\nx1+0x2+3x3+5x4=0\n0x1+1x2+4x3+6x4=0.\n Negative three Chief Justice Gonzalez negative three thank you and what's X squared that'll come from this equation what will X squared be", null, null, "We can see these text from the image: - \\( m = 3 \\)\n- \\( n = 4 \\)\n- \\( \\text{unknowns} \\)\n- \\( \\text{RHS} = 0 \\)\n- \\( A \\mathbf{x} = \\mathbf{0} \\)\n- \\( x_1 + 3x_2 - x_3 + x_4 = 0 \\)\n- \\( x_1 + 4x_2 + 5x_3 + 6x_4 = 0 \\).\n Negative four. So that's a solution. We've solved this original problem. Oh, I erased it. It was much worse at the beginning, but we found a solution. But again, we have four unknowns and we have only two real equations.", null, null, "We can see these text from the image: 1 2 11 17 x1 0\n3 7 37 57 x2 0\n0 0 0 0 x3 0\nA x=0 x4 0\n\nm=3 eqns n=4 unknowns RHS=0\none soln: x=x2=x3=x4=0\nmore solns:\nx1 -3 -4 1 0\nx2 x3 x4\n\n1 2 11 17 10 35\n4 6 0 0 0 46\n0 0 0 0 0 0\nx1+0x2+3x3+5x4=0\n0x1+1x2+4x3+6x4=0.\n Two independent equations because this third equation turned out to be a fraud. It was just a combination of one and two. So we really have only two equations for unknowns. So we'd expect two parameters in the solution and I found one of them." ], "metadata": [ { "vid": "lUUte2o2Sn8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Introduction to Matrices in Advanced Algebra Course_30.json#####audio#####doingASR#####FinishASR/lUUte2o2Sn8/2587.8_2608.04.mp4", "refined_asr": " So that tells me if I find one set of Xs that works then two times that set of Xs will work five times that set of Xs will work because it'll just multiply all through because I'll have zero equals zero and I can safely multiply by five.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2587.8_2608.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2587.8_2608.04#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2587.8_2608.04#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2587.8_2608.04#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2587.8_2608.04#3.jpg" ], "ocr_qwen2_vl_72b": "1 2 11 17\n\n10 4 6\n\n0 0 0\n\n10 3 5\n\n0 1 4\n\n0 0 0\n\nx\u2081 + 0x\u2082 + 3x\u2083 + 5x\u2084 = 0\n\n0x\u2081 + 1x\u2082 + 4x\u2083 + 6x\u2084 = 0" }, { "vid": "lUUte2o2Sn8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Introduction to Matrices in Advanced Algebra Course_30.json#####audio#####doingASR#####FinishASR/lUUte2o2Sn8/2608.04_2635.52.mp4", "refined_asr": " So I'm expecting two solutions, two separate solutions. And I'll write those maybe up above: x1, x2, x3, x4, yeah. So give me. So I have two equations. And for these four unknowns, let me.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2608.04_2635.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2608.04_2635.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2608.04_2635.52#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2608.04_2635.52#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2608.04_2635.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2608.04_2635.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2608.04_2635.52#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2608.04_2635.52#4.jpg" ], "ocr_qwen2_vl_72b": "1 2 11 17\n\n0 1 4 6\n\n5 0\n\nx\u2081 + 0x\u2082 + 3x\u2083 + 5x\u2084 = 0\n\n0x\u2081 + 1x\u2082 + 4x\u2083 + 6x\u2084 = 0" }, { "vid": "lUUte2o2Sn8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Introduction to Matrices in Advanced Algebra Course_30.json#####audio#####doingASR#####FinishASR/lUUte2o2Sn8/2635.52_2652.7.mp4", "refined_asr": " Start out with x3 equal to 1 and x4 equal to 0. So I'm just sort of freely setting x3 and x4 to 1 and 0. And then, either I or you will figure out the rest.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2635.52_2652.7#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2635.52_2652.7#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2635.52_2652.7#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2635.52_2652.7#3.jpg" ], "ocr_qwen2_vl_72b": "1 2 11 17\n3 7 37 57\n0 0 0 0\n\nA x = 0\n\nm=3 eqns n=4 unknowns RHS=0\none soln: x1=x2=x3=x4=0\nmore solns:\nx1, x2, x3, x4\n\n1 2 11 17\n10 1 4 6\n0 0 0 0\n\n10 5\n0 6\n\nx1 + 0x2 + 3x3 + 5x4 = 0\n0x1 + 1x2 + 4x3 + 6x4 = 0" }, { "vid": "lUUte2o2Sn8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Introduction to Matrices in Advanced Algebra Course_30.json#####audio#####doingASR#####FinishASR/lUUte2o2Sn8/2652.7_2676.72.mp4", "refined_asr": " What do x1 and x2 have to be? So what's the answer then? If x3 is 1 and x4 is 0, what is x1? This is obviously 0. So what is x1? That leads to x1 equals what?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2652.7_2676.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2652.7_2676.72#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2652.7_2676.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2652.7_2676.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2652.7_2676.72#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2652.7_2676.72#4.jpg" ], "ocr_qwen2_vl_72b": "1 2 11 17\n3 7 37 57\n0 0 0 0\n\nA x = 0\n\nm=3 eqns n=4 unknowns RHS=0\none soln: x1=x2=x3=x4=0\nmore solns:\nx1,x2,x3,x4\n\n1 2 11 17\n10 1 4 0\n0 0 0 0\n\n[10 35\n0 1 4 6\n0 0 0 0]\n\nx1 + 0x2 + 3x3 + 5x4 = 0\n0x1 + 1x2 + 4x3 + 6x4 = 0" }, { "vid": "lUUte2o2Sn8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Introduction to Matrices in Advanced Algebra Course_30.json#####audio#####doingASR#####FinishASR/lUUte2o2Sn8/2676.72_2688.44.mp4", "refined_asr": " Negative three Chief Justice Gonzalez negative three thank you and what's X squared that'll come from this equation what will X squared be", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2676.72_2688.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2676.72_2688.44#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2676.72_2688.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2676.72_2688.44#2.jpg" ], "ocr_qwen2_vl_72b": "1 2 11 17\n3 7 37 57\n\nm=3 eqns n=4 unknowns RHS=0\none soln: x=x2=x3=x4=0\nmore solns:\nx1,x2,x3,x4\n\nA x=0\n\n1 2 11 17 10\n0 1 4 6 0\n0 0 0 0 0\n\nx1+0x2+3x3+5x4=0\n0x1+1x2+4x3+6x4=0" }, { "vid": "lUUte2o2Sn8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Introduction to Matrices in Advanced Algebra Course_30.json#####audio#####doingASR#####FinishASR/lUUte2o2Sn8/2688.44_2712.3.mp4", "refined_asr": " Negative four. So that's a solution. We've solved this original problem. Oh, I erased it. It was much worse at the beginning, but we found a solution. But again, we have four unknowns and we have only two real equations.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2688.44_2712.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2688.44_2712.3#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2688.44_2712.3#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2688.44_2712.3#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2688.44_2712.3#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2688.44_2712.3#4.jpg" ], "ocr_qwen2_vl_72b": "- \\( m = 3 \\)\n- \\( n = 4 \\)\n- \\( \\text{unknowns} \\)\n- \\( \\text{RHS} = 0 \\)\n- \\( A \\mathbf{x} = \\mathbf{0} \\)\n- \\( x_1 + 3x_2 - x_3 + x_4 = 0 \\)\n- \\( x_1 + 4x_2 + 5x_3 + 6x_4 = 0 \\)" }, { "vid": "lUUte2o2Sn8.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Introduction to Matrices in Advanced Algebra Course_30.json#####audio#####doingASR#####FinishASR/lUUte2o2Sn8/2712.3_2729.0600000000004.mp4", "refined_asr": " Two independent equations because this third equation turned out to be a fraud. It was just a combination of one and two. So we really have only two equations for unknowns. So we'd expect two parameters in the solution and I found one of them.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2712.3_2729.0600000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2712.3_2729.0600000000004#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2712.3_2729.0600000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2712.3_2729.0600000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/lUUte2o2Sn8/lUUte2o2Sn8@2712.3_2729.0600000000004#3.jpg" ], "ocr_qwen2_vl_72b": "1 2 11 17 x1 0\n3 7 37 57 x2 0\n0 0 0 0 x3 0\nA x=0 x4 0\n\nm=3 eqns n=4 unknowns RHS=0\none soln: x=x2=x3=x4=0\nmore solns:\nx1 -3 -4 1 0\nx2 x3 x4\n\n1 2 11 17 10 35\n4 6 0 0 0 46\n0 0 0 0 0 0\nx1+0x2+3x3+5x4=0\n0x1+1x2+4x3+6x4=0" } ], "image_num": 15, "text_num": 369, "token_num": 9009 }, { "images": [ "sample_100_images/v2VuM0AIZVU@615.3399999999999_630.5799999999999#1.jpg", "sample_100_images/v2VuM0AIZVU@615.3399999999999_630.5799999999999#2.jpg", null, "sample_100_images/v2VuM0AIZVU@630.5799999999999_649.42#1.jpg", "sample_100_images/v2VuM0AIZVU@630.5799999999999_649.42#2.jpg", "sample_100_images/v2VuM0AIZVU@630.5799999999999_649.42#3.jpg", null, "sample_100_images/v2VuM0AIZVU@650.04_663.52#1.jpg", "sample_100_images/v2VuM0AIZVU@650.04_663.52#2.jpg", null, "sample_100_images/v2VuM0AIZVU@663.52_691.8199999999999#1.jpg", "sample_100_images/v2VuM0AIZVU@663.52_691.8199999999999#2.jpg", "sample_100_images/v2VuM0AIZVU@663.52_691.8199999999999#3.jpg", "sample_100_images/v2VuM0AIZVU@663.52_691.8199999999999#4.jpg", null, "sample_100_images/v2VuM0AIZVU@691.8199999999999_705.4599999999999#1.jpg", "sample_100_images/v2VuM0AIZVU@691.8199999999999_705.4599999999999#2.jpg", null ], "texts": [ null, null, " Number of input symbols: 2. What are they? 0, 1. Yes or no? We check it out. a0, b1. In that way, we can uniquely encode it. That means a Turing machine can be encoded. Now, one more thing is, how do we represent, or rather, how do we encode the transition?", null, null, null, "\nFunction. That is the final thing, right? Example: You have a description like this transition function. Let me take one state: q, okay? Q on something. Otherwise, I take start state. Start state on seeing input: a. Generally, we write like this: S on a.", null, null, " Go to Q and make it B, and go to the right side. How to encode it for that? What I do is, I am going to use 0s and 1s, and I am going to do the same thing. First of all, blindly write like this: 0 power S.", null, null, null, null, " Start a, 0 power a, start q, b r. We know that any transition function will have five components, right: start, a, b, state, symbol, state, symbol, direction. That is why 0 power s, 1, 0 power a, 0 power q. 1, 0 power b, 1, 0 power r. Okay, so let me use:", null, null, " L means 0 R means 1 This is the encoding for directions okay let's do that otherwise I use L means 1 R means 2 Anything we can use not a problem now you see" ], "text_ocr_list": [ null, null, "We can see these text from the image: Universal Turing machine\n\nTuring machine encoding\n\n010101...\n\na, b, x, y, z\n\ns, t, q\n\ninput states tape\n\ns = 0 a = 1 t = 2 q = 3 a = 0 b = 1 x = 2 y = 3\n\n01010...\n\n0, 1, 2, 3 | 0, 1, 2, 3 | 01.\n Number of input symbols: 2. What are they? 0, 1. Yes or no? We check it out. a0, b1. In that way, we can uniquely encode it. That means a Turing machine can be encoded. Now, one more thing is, how do we represent, or rather, how do we encode the transition?", null, null, null, "We can see these text from the image: Universal Turing machine\n\nTuring machine encoding\n\n010101...\n\ns, b, s, t, b, z\ninput states tape\ns = 0, a = 1, t = 2, b = 3, a = 0, b = 1, x = 2\n\n01010...\n\nC.\n\nFunction. That is the final thing, right? Example: You have a description like this transition function. Let me take one state: q, okay? Q on something. Otherwise, I take start state. Start state on seeing input: a. Generally, we write like this: S on a.", null, null, "We can see these text from the image: Universal Turing machine\n\nTuring machine encoding\n\n010101...\n\na, b, x, y, z\n\ninput states tape\n\ns = 0 a = 1 t = 2 b = 1 x = 2 y = 3\n\n010/0...\n\n(S, a) = (q, b, L).\n Go to Q and make it B, and go to the right side. How to encode it for that? What I do is, I am going to use 0s and 1s, and I am going to do the same thing. First of all, blindly write like this: 0 power S.", null, null, null, null, "We can see these text from the image: Universal Turing machine\n\nTuring machine encoding\n\n010101...\n\ns, b2, s, t, b, z\ninput states tape\n\ns = 0, b = 1, t = 2, b = 3\na = 0, b = 1, x = 2, y = 3\n\n01010...\n\n(S, a) = (q, b, R) => 01.\n Start a, 0 power a, start q, b r. We know that any transition function will have five components, right: start, a, b, state, symbol, state, symbol, direction. That is why 0 power s, 1, 0 power a, 0 power q. 1, 0 power b, 1, 0 power r. Okay, so let me use:", null, null, "We can see these text from the image: Universal Turing machine\n\nTuring machine encoding\n\n010101...\n\n{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z}\n\n{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}\n\nL = 0\n\nR = 1\n\nS = 0\n\na = 1\n\nb = 2\n\nc = 3\n\nd = 4\n\ne = 5\n\nf = 6\n\ng = 7\n\nh = 8\n\ni = 9\n\nj = 10\n\nk = 11\n\nl = 12\n\nm = 13\n\nn = 14\n\no = 15\n\np = 16\n\nq = 17\n\nr = 18\n\ns = 19\n\nt = 20\n\nu = 21\n\nv = 22\n\nw = 23\n\nx = 24\n\ny = 25\n\nz = 26\n\n(0, 1) = (2, 3)\n\n(0, 2) = (4, 5)\n\n(0, 3) = (6, 7)\n\n(0, 4) = (8, 9)\n\n(0, 5) = (10, 11)\n\n(0, 6) = (12, 13)\n\n(0, 7) = (14, 15)\n\n(0, 8) = (16, 17)\n\n(0, 9) = (18, 19)\n\n(0, 10) = (20, 21)\n\n(0, 11) = (22, 23)\n\n(0, 12) = (24, 25)\n\n(0, 13) = (26, 27)\n\n(0, 14) = (28, 29)\n\n(0, 15) = (30, 31)\n\n(0, 16) = (32, 33)\n\n(0, 17) = (34, 35)\n\n(0, 18) = (36, 37)\n\n(0, 19) = (38, 39)\n\n(0, 20) = (40, 41)\n\n(0, 21) = (42, 43)\n\n(0, 22) = (44, 45)\n\n(0, 23) = (46, 47)\n\n(0, 24) = (48, 49)\n\n(0, 25) = (50, 51)\n\n(0, 26) = (52, 53)\n\n(0, 27) = (54, 55)\n\n(0, 28) = (56, 57)\n\n(0, 29) = (58, 59)\n\n(0, 30) = (60, 61)\n\n(0, 31) = (62, 63)\n\n(0, 32) = (64, 65)\n\n(0, 33) = (66, 67)\n\n(0, 34) = (68, 69)\n\n(0, 35) = (70, 71)\n\n(0, 36) = (72, 73)\n\n(0, 37) = (74, 75)\n\n(0, 38) = (76, 77)\n\n(0, 39) = (78, 79)\n\n(0, 40) = (80, 81)\n\n(0, 41) = (82, 83)\n\n(0, 42) = (84, 85)\n\n(0, 43) = (86, 87)\n\n(0, 44) = (88, 89)\n\n(0, 45) = (90, 91)\n\n(0, 46) = (92, 93)\n\n(0, 47) = (94, 95)\n\n(0, 48) = (96, 97)\n\n(0, 49) = (98, 99)\n\n(0, 50) = (100, 101)\n\n(0, 51) = (102, 103)\n\n(0, 52) = (104, 105)\n\n(0, 53) = (106, 107)\n\n(0, 54) = (108, 109)\n\n(0, 55) = (110, 111)\n\n(0, 56) = (112, 113)\n\n(0, 57) = (114, 115)\n\n(0, 58) = (116, 117)\n\n(0, 59) = (118, 119)\n\n(0, 60) = (120, 121)\n\n(0, 61) = (122, 123)\n\n(0, 62) = (124, 125)\n\n(0, 63) = (126, 127)\n\n(0, 64) = (128, 129)\n\n(0, 65) = (130, 131)\n\n(0, 66) = (132, 133)\n\n(0, 67) = (134, 135)\n\n(0, 68) = (136, 137)\n\n(0, 69) = (138, 139)\n\n(0, 70) = (140, 141)\n\n(0, 71) = (142, 143)\n\n(0, 72) = (144, 145)\n\n(0, 73) = (146, 147)\n\n(0, 74) = (148, 149)\n\n(0, 75) = (150, 151)\n\n(0, 76) = (152, 153)\n\n(0, 77) = (154, 155)\n\n(0, 78) = (156, 157)\n\n(0, 79) = (158, 159)\n\n(0, 80) = (160, 161)\n\n(0, 81) = (162, 163)\n\n(0, 82) = (164, 165)\n\n(0, 83) = (166, 167)\n\n(0, 84) = (168, 169)\n\n(0, 85) = (170, 171)\n\n(0, 86) = (172, 173)\n\n(0, 87) = (174, 175)\n\n(0, 88) = (176, 177)\n\n(0, 89) = (178, 179)\n\n(0, 90) = (180, 181)\n\n(0, 91) = (182, 183)\n\n(0, 92) = (184, 185)\n\n(0, 93) = (186, 187)\n\n(0, 94) = (188, 189)\n\n(0, 95) = (190, 191)\n\n(0, 96) = (192, 193)\n\n(0, 97) = (194, 195)\n\n(0, 98) = (196, 197)\n\n(0, 99) = (198, 199)\n\n(0, 100) = (200, 201)\n\n(0, 101) = (202, 203)\n\n(0, 102) = (204, 205)\n\n(0, 103) = (206, 207)\n\n(0, 104) = (208, 209)\n\n(0, 105) = (210, 211)\n\n(0, 106) = (212, 213)\n\n(0, 107) = (214, 215)\n\n(0, 108) = (216, 2.\n L means 0 R means 1 This is the encoding for directions okay let's do that otherwise I use L means 1 R means 2 Anything we can use not a problem now you see" ], "metadata": [ { "vid": "v2VuM0AIZVU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Mathematical Logic Tutorial on Decidable and Undecidable Problems in Computability Theory_30.json#####audio#####doingASR#####FinishASR/v2VuM0AIZVU/615.3399999999999_630.5799999999999.mp4", "refined_asr": " Number of input symbols: 2. What are they? 0, 1. Yes or no? We check it out. a0, b1. In that way, we can uniquely encode it. That means a Turing machine can be encoded. Now, one more thing is, how do we represent, or rather, how do we encode the transition?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@615.3399999999999_630.5799999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@615.3399999999999_630.5799999999999#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@615.3399999999999_630.5799999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@615.3399999999999_630.5799999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@615.3399999999999_630.5799999999999#3.jpg" ], "ocr_qwen2_vl_72b": "Universal Turing machine\n\nTuring machine encoding\n\n010101...\n\na, b, x, y, z\n\ns, t, q\n\ninput states tape\n\ns = 0 a = 1 t = 2 q = 3 a = 0 b = 1 x = 2 y = 3\n\n01010...\n\n0, 1, 2, 3 | 0, 1, 2, 3 | 01" }, { "vid": "v2VuM0AIZVU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Mathematical Logic Tutorial on Decidable and Undecidable Problems in Computability Theory_30.json#####audio#####doingASR#####FinishASR/v2VuM0AIZVU/630.5799999999999_649.42.mp4", "refined_asr": "\nFunction. That is the final thing, right? Example: You have a description like this transition function. Let me take one state: q, okay? Q on something. Otherwise, I take start state. Start state on seeing input: a. Generally, we write like this: S on a.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@630.5799999999999_649.42#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@630.5799999999999_649.42#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@630.5799999999999_649.42#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@630.5799999999999_649.42#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@630.5799999999999_649.42#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@630.5799999999999_649.42#3.jpg" ], "ocr_qwen2_vl_72b": "Universal Turing machine\n\nTuring machine encoding\n\n010101...\n\ns, b, s, t, b, z\ninput states tape\ns = 0, a = 1, t = 2, b = 3, a = 0, b = 1, x = 2\n\n01010...\n\nC" }, { "vid": "v2VuM0AIZVU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Mathematical Logic Tutorial on Decidable and Undecidable Problems in Computability Theory_30.json#####audio#####doingASR#####FinishASR/v2VuM0AIZVU/650.04_663.52.mp4", "refined_asr": " Go to Q and make it B, and go to the right side. How to encode it for that? What I do is, I am going to use 0s and 1s, and I am going to do the same thing. First of all, blindly write like this: 0 power S.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@650.04_663.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@650.04_663.52#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@650.04_663.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@650.04_663.52#2.jpg" ], "ocr_qwen2_vl_72b": "Universal Turing machine\n\nTuring machine encoding\n\n010101...\n\na, b, x, y, z\n\ninput states tape\n\ns = 0 a = 1 t = 2 b = 1 x = 2 y = 3\n\n010/0...\n\n(S, a) = (q, b, L)" }, { "vid": "v2VuM0AIZVU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Mathematical Logic Tutorial on Decidable and Undecidable Problems in Computability Theory_30.json#####audio#####doingASR#####FinishASR/v2VuM0AIZVU/663.52_691.8199999999999.mp4", "refined_asr": " Start a, 0 power a, start q, b r. We know that any transition function will have five components, right: start, a, b, state, symbol, state, symbol, direction. That is why 0 power s, 1, 0 power a, 0 power q. 1, 0 power b, 1, 0 power r. Okay, so let me use:", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@663.52_691.8199999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@663.52_691.8199999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@663.52_691.8199999999999#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@663.52_691.8199999999999#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@663.52_691.8199999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@663.52_691.8199999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@663.52_691.8199999999999#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@663.52_691.8199999999999#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@663.52_691.8199999999999#5.jpg" ], "ocr_qwen2_vl_72b": "Universal Turing machine\n\nTuring machine encoding\n\n010101...\n\ns, b2, s, t, b, z\ninput states tape\n\ns = 0, b = 1, t = 2, b = 3\na = 0, b = 1, x = 2, y = 3\n\n01010...\n\n(S, a) = (q, b, R) => 01" }, { "vid": "v2VuM0AIZVU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Mathematical Logic Tutorial on Decidable and Undecidable Problems in Computability Theory_30.json#####audio#####doingASR#####FinishASR/v2VuM0AIZVU/691.8199999999999_705.4599999999999.mp4", "refined_asr": " L means 0 R means 1 This is the encoding for directions okay let's do that otherwise I use L means 1 R means 2 Anything we can use not a problem now you see", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@691.8199999999999_705.4599999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@691.8199999999999_705.4599999999999#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@691.8199999999999_705.4599999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/v2VuM0AIZVU/v2VuM0AIZVU@691.8199999999999_705.4599999999999#2.jpg" ], "ocr_qwen2_vl_72b": "Universal Turing machine\n\nTuring machine encoding\n\n010101...\n\n{a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z}\n\n{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}\n\nL = 0\n\nR = 1\n\nS = 0\n\na = 1\n\nb = 2\n\nc = 3\n\nd = 4\n\ne = 5\n\nf = 6\n\ng = 7\n\nh = 8\n\ni = 9\n\nj = 10\n\nk = 11\n\nl = 12\n\nm = 13\n\nn = 14\n\no = 15\n\np = 16\n\nq = 17\n\nr = 18\n\ns = 19\n\nt = 20\n\nu = 21\n\nv = 22\n\nw = 23\n\nx = 24\n\ny = 25\n\nz = 26\n\n(0, 1) = (2, 3)\n\n(0, 2) = (4, 5)\n\n(0, 3) = (6, 7)\n\n(0, 4) = (8, 9)\n\n(0, 5) = (10, 11)\n\n(0, 6) = (12, 13)\n\n(0, 7) = (14, 15)\n\n(0, 8) = (16, 17)\n\n(0, 9) = (18, 19)\n\n(0, 10) = (20, 21)\n\n(0, 11) = (22, 23)\n\n(0, 12) = (24, 25)\n\n(0, 13) = (26, 27)\n\n(0, 14) = (28, 29)\n\n(0, 15) = (30, 31)\n\n(0, 16) = (32, 33)\n\n(0, 17) = (34, 35)\n\n(0, 18) = (36, 37)\n\n(0, 19) = (38, 39)\n\n(0, 20) = (40, 41)\n\n(0, 21) = (42, 43)\n\n(0, 22) = (44, 45)\n\n(0, 23) = (46, 47)\n\n(0, 24) = (48, 49)\n\n(0, 25) = (50, 51)\n\n(0, 26) = (52, 53)\n\n(0, 27) = (54, 55)\n\n(0, 28) = (56, 57)\n\n(0, 29) = (58, 59)\n\n(0, 30) = (60, 61)\n\n(0, 31) = (62, 63)\n\n(0, 32) = (64, 65)\n\n(0, 33) = (66, 67)\n\n(0, 34) = (68, 69)\n\n(0, 35) = (70, 71)\n\n(0, 36) = (72, 73)\n\n(0, 37) = (74, 75)\n\n(0, 38) = (76, 77)\n\n(0, 39) = (78, 79)\n\n(0, 40) = (80, 81)\n\n(0, 41) = (82, 83)\n\n(0, 42) = (84, 85)\n\n(0, 43) = (86, 87)\n\n(0, 44) = (88, 89)\n\n(0, 45) = (90, 91)\n\n(0, 46) = (92, 93)\n\n(0, 47) = (94, 95)\n\n(0, 48) = (96, 97)\n\n(0, 49) = (98, 99)\n\n(0, 50) = (100, 101)\n\n(0, 51) = (102, 103)\n\n(0, 52) = (104, 105)\n\n(0, 53) = (106, 107)\n\n(0, 54) = (108, 109)\n\n(0, 55) = (110, 111)\n\n(0, 56) = (112, 113)\n\n(0, 57) = (114, 115)\n\n(0, 58) = (116, 117)\n\n(0, 59) = (118, 119)\n\n(0, 60) = (120, 121)\n\n(0, 61) = (122, 123)\n\n(0, 62) = (124, 125)\n\n(0, 63) = (126, 127)\n\n(0, 64) = (128, 129)\n\n(0, 65) = (130, 131)\n\n(0, 66) = (132, 133)\n\n(0, 67) = (134, 135)\n\n(0, 68) = (136, 137)\n\n(0, 69) = (138, 139)\n\n(0, 70) = (140, 141)\n\n(0, 71) = (142, 143)\n\n(0, 72) = (144, 145)\n\n(0, 73) = (146, 147)\n\n(0, 74) = (148, 149)\n\n(0, 75) = (150, 151)\n\n(0, 76) = (152, 153)\n\n(0, 77) = (154, 155)\n\n(0, 78) = (156, 157)\n\n(0, 79) = (158, 159)\n\n(0, 80) = (160, 161)\n\n(0, 81) = (162, 163)\n\n(0, 82) = (164, 165)\n\n(0, 83) = (166, 167)\n\n(0, 84) = (168, 169)\n\n(0, 85) = (170, 171)\n\n(0, 86) = (172, 173)\n\n(0, 87) = (174, 175)\n\n(0, 88) = (176, 177)\n\n(0, 89) = (178, 179)\n\n(0, 90) = (180, 181)\n\n(0, 91) = (182, 183)\n\n(0, 92) = (184, 185)\n\n(0, 93) = (186, 187)\n\n(0, 94) = (188, 189)\n\n(0, 95) = (190, 191)\n\n(0, 96) = (192, 193)\n\n(0, 97) = (194, 195)\n\n(0, 98) = (196, 197)\n\n(0, 99) = (198, 199)\n\n(0, 100) = (200, 201)\n\n(0, 101) = (202, 203)\n\n(0, 102) = (204, 205)\n\n(0, 103) = (206, 207)\n\n(0, 104) = (208, 209)\n\n(0, 105) = (210, 211)\n\n(0, 106) = (212, 213)\n\n(0, 107) = (214, 215)\n\n(0, 108) = (216, 2" } ], "image_num": 13, "text_num": 339, "token_num": 7827 }, { "images": [ "sample_100_images/4k-_KSx6-N4@1310.34_1327.3999999999999#1.jpg", null, "sample_100_images/4k-_KSx6-N4@1327.76_1345.94#1.jpg", null, "sample_100_images/4k-_KSx6-N4@1345.94_1362.42#1.jpg", null, "sample_100_images/4k-_KSx6-N4@1362.54_1379.24#1.jpg", "sample_100_images/4k-_KSx6-N4@1362.54_1379.24#2.jpg", null, "sample_100_images/4k-_KSx6-N4@1379.24_1392.26#1.jpg", null, "sample_100_images/4k-_KSx6-N4@1392.54_1407.22#1.jpg", null ], "texts": [ null, " Fine. So, you can't say that this number is greater than because it depends on the value of alpha. Using the max norm, it will be represented here. And similarly, for this case, you will obtain this number. Now, if I were to consider the minimum, I can substitute both these values into this equation.", null, " You can see from here that this number will definitely be at a maximum because each quantity has a similar value. Now I can substitute equation number one and equation number two into this equation. Alright then your target is to prove that this will be the minimum. How can you do that?", null, " Prove because I can simply first multiply this quantity inside. So, I can return this number as max. I can multiply 0.2 alpha; this alpha will be canceled. So, it is my 0.4 of this plus it will be.", null, null, " When you multiply this it will be my point. It's a 10 it's a 6 comma. Then I can multiply 2.5 with this quantity you will get as this expression fine. Now the question arises when it will be minimum.", null, " So for what value of Alpha can you choose so that this will be at a minimum? This will be at a minimum only because we all know that if I say this is my X and this is my Y, it will be minimum when X is equal to Y.", null, " Because whenever x is less than y then you can say the minimum is my x but when you take x is greater than y the minimum will be y so this will be minimum only when both the numbers are the same so can you" ], "text_ocr_list": [ null, "We can see these text from the image: Example: Let \\( A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\). Determine \\(\\alpha\\) such that \\( K(A(\\alpha)) \\) is minimized. Use the maximum norm.\n\nSolution: For a matrix \\(A\\),\n\n\\[ K(A) = \\|A\\| \\|A^{-1}\\| \\]\n\nNow,\n\n\\[ A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\]\n\n\\[ \\Rightarrow A^{-1}(\\alpha) = \\frac{1}{0.05\\alpha} \\begin{bmatrix} 1.5 & -0.1\\alpha \\\\ -1.0 & 0.1\\alpha \\end{bmatrix} \\]\n\n\\[ \\|A\\|_{\\infty} = \\max(0.2|\\alpha|, 2.5) \\].\n Fine. So, you can't say that this number is greater than because it depends on the value of alpha. Using the max norm, it will be represented here. And similarly, for this case, you will obtain this number. Now, if I were to consider the minimum, I can substitute both these values into this equation.", null, "We can see these text from the image: Example: Let \\( A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\). Determine \\(\\alpha\\) such that \\( K(A(\\alpha)) \\) is minimized. Use the maximum norm.\n\nSolution: For a matrix \\( A \\),\n\\[ K(A) = \\|A\\| \\|A^{-1}\\| \\]\n\nNow,\n\\[ A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\]\n\\[ \\Rightarrow A^{-1}(\\alpha) = \\frac{1}{0.05\\alpha} \\begin{bmatrix} 1.5 & -0.1\\alpha \\\\ -1.0 & 0.1\\alpha \\end{bmatrix} \\]\n\nUsing maximum norm, we get\n\\[ \\|A(\\alpha)\\| = \\max\\{0.2|\\alpha|, 2.5\\} \\]\n\\[ \\|A^{-1}(\\alpha)\\| = \\max\\left\\{\\frac{2|\\alpha| + 30}{|\\alpha|}, \\frac{2|\\alpha| + 20}{|\\alpha|}\\right\\} \\]\n\\[ = \\frac{2|\\alpha| + 30}{|\\alpha|} \\].\n You can see from here that this number will definitely be at a maximum because each quantity has a similar value. Now I can substitute equation number one and equation number two into this equation. Alright then your target is to prove that this will be the minimum. How can you do that?", null, "We can see these text from the image: Example: Let \\( A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\). Determine \\(\\alpha\\) such that \\( K(A(\\alpha)) \\) is minimized. Use the maximum norm.\n\nSolution: For a matrix \\(A\\),\n\\[ K(A) = \\|A\\| \\|A^{-1}\\| \\]\n\nNow,\n\\[ A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\]\n\\[ \\Rightarrow A^{-1}(\\alpha) = \\frac{1}{0.05\\alpha} \\begin{bmatrix} 1.5 & -0.1\\alpha \\\\ -1.0 & 0.1\\alpha \\end{bmatrix} \\]\n\nUsing maximum norm, we get\n\\[ \\|A(\\alpha)\\| = \\max\\{0.2|\\alpha|, 2.5\\} \\]\n\\[ \\|A^{-1}(\\alpha)\\| = \\max\\left\\{\\frac{2|\\alpha| + 30}{|\\alpha|}, \\frac{2|\\alpha| + 20}{|\\alpha|}\\right\\} \\]\n\\[ = \\frac{2|\\alpha| + 30}{|\\alpha|} \\]\n\nHence,\n\\[ K(A(\\alpha)) = \\left( \\frac{2|\\alpha| + 30}{|\\alpha|} \\right) \\max\\{0.2|\\alpha|, 2.5\\} \\].\n Prove because I can simply first multiply this quantity inside. So, I can return this number as max. I can multiply 0.2 alpha; this alpha will be canceled. So, it is my 0.4 of this plus it will be.", null, null, "We can see these text from the image: Example: Let \\( A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\). Determine \\(\\alpha\\) such that \\( K(A(\\alpha)) \\) is minimized. Use the maximum norm.\n\nSolution: For a matrix \\(A\\),\n\n\\[ K(A) = \\|A\\| \\|A^{-1}\\| \\]\n\nNow,\n\n\\[ A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\]\n\n\\[ \\Rightarrow A^{-1}(\\alpha) = \\frac{1}{0.05\\alpha} \\begin{bmatrix} 1.5 & -0.1\\alpha \\\\ -1.0 & 0.1\\alpha \\end{bmatrix} \\]\n\nUsing maximum norm, we get\n\n\\[ \\|A(\\alpha)\\| = \\max\\{0.2|\\alpha|, 2.5\\} \\]\n\n\\[ \\|A^{-1}(\\alpha)\\| = \\max\\left\\{\\frac{2|\\alpha| + 30}{|\\alpha|}, \\frac{2|\\alpha| + 20}{|\\alpha|}\\right\\} \\]\n\n\\[ = \\frac{2|\\alpha| + 30}{|\\alpha|} \\]\n\nHence,\n\n\\[ K(A(\\alpha)) = \\left( \\frac{2|\\alpha| + 30}{|\\alpha|} \\right) \\max\\{0.2|\\alpha|, 2.5\\} \\]\n\n\\[ = \\max\\left[0.4|\\alpha| + 6, 12\\right] \\].\n When you multiply this it will be my point. It's a 10 it's a 6 comma. Then I can multiply 2.5 with this quantity you will get as this expression fine. Now the question arises when it will be minimum.", null, "We can see these text from the image: Using maximum norm, we get\n\n\\[\n\\|A(\\alpha)\\| = \\max\\{0.2|\\alpha|, 2.5\\}\n\\]\n\n\\[\n\\|A^{-1}(\\alpha)\\| = \\max\\left\\{\\frac{2|\\alpha| + 30}{|\\alpha|}, \\frac{2|\\alpha| + 20}{|\\alpha|}\\right\\}\n\\]\n\n\\[\n= \\frac{2|\\alpha| + 30}{|\\alpha|}\n\\]\n\nHence,\n\n\\[\nK(A(\\alpha)) = \\left(\\frac{2|\\alpha| + 30}{|\\alpha|}\\right) \\max\\{0.2|\\alpha|, 2.5\\}\n\\]\n\nTo find \\(\\alpha\\), such that \\(K(A(\\alpha))\\) is minimum, we have\n\n\\[\nK(A(\\alpha)) = \\max\\left[0.4|\\alpha| + 6, 5 + \\frac{75}{|\\alpha|}\\right]\n\\]\n\nFor minimum \\(K(A)\\),\n\nwe can choose \\(\\alpha\\) such that.\n So for what value of Alpha can you choose so that this will be at a minimum? This will be at a minimum only because we all know that if I say this is my X and this is my Y, it will be minimum when X is equal to Y.", null, "We can see these text from the image: Using maximum norm, we get\n\n\\[\n\\|A(\\alpha)\\| = \\max\\{0.2|\\alpha|, 2.5\\}\n\\]\n\n\\[\n\\|A^{-1}(\\alpha)\\| = \\max\\left\\{\\frac{2|\\alpha| + 30}{|\\alpha|}, \\frac{2|\\alpha| + 20}{|\\alpha|}\\right\\}\n\\]\n\n\\[\n= \\frac{2|\\alpha| + 30}{|\\alpha|}\n\\]\n\nHence,\n\n\\[\nK(A(\\alpha)) = \\left(\\frac{2|\\alpha| + 30}{|\\alpha|}\\right) \\max\\{0.2|\\alpha|, 2.5\\}\n\\]\n\nTo find \\(\\alpha\\), such that \\(K(A(\\alpha))\\) is minimum, we have\n\n\\[\nK(A(\\alpha)) = \\max\\left[0.4|\\alpha| + 6, 5 + \\frac{75}{|\\alpha|}\\right]\n\\]\n\nFor minimum \\(K(A)\\),\n\nwe can choose \\(\\alpha\\) such that.\n Because whenever x is less than y then you can say the minimum is my x but when you take x is greater than y the minimum will be y so this will be minimum only when both the numbers are the same so can you" ], "metadata": [ { "vid": "4k-_KSx6-N4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Linear algebra tutorial on Condition Number in computational aspects of numerical linear algebra_30.json#####audio#####doingASR#####FinishASR/4k-_KSx6-N4/1310.34_1327.3999999999999.mp4", "refined_asr": " Fine. So, you can't say that this number is greater than because it depends on the value of alpha. Using the max norm, it will be represented here. And similarly, for this case, you will obtain this number. Now, if I were to consider the minimum, I can substitute both these values into this equation.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1310.34_1327.3999999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1310.34_1327.3999999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1310.34_1327.3999999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1310.34_1327.3999999999999#3.jpg" ], "ocr_qwen2_vl_72b": "Example: Let \\( A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\). Determine \\(\\alpha\\) such that \\( K(A(\\alpha)) \\) is minimized. Use the maximum norm.\n\nSolution: For a matrix \\(A\\),\n\n\\[ K(A) = \\|A\\| \\|A^{-1}\\| \\]\n\nNow,\n\n\\[ A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\]\n\n\\[ \\Rightarrow A^{-1}(\\alpha) = \\frac{1}{0.05\\alpha} \\begin{bmatrix} 1.5 & -0.1\\alpha \\\\ -1.0 & 0.1\\alpha \\end{bmatrix} \\]\n\n\\[ \\|A\\|_{\\infty} = \\max(0.2|\\alpha|, 2.5) \\]" }, { "vid": "4k-_KSx6-N4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Linear algebra tutorial on Condition Number in computational aspects of numerical linear algebra_30.json#####audio#####doingASR#####FinishASR/4k-_KSx6-N4/1327.76_1345.94.mp4", "refined_asr": " You can see from here that this number will definitely be at a maximum because each quantity has a similar value. Now I can substitute equation number one and equation number two into this equation. Alright then your target is to prove that this will be the minimum. How can you do that?", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1327.76_1345.94#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1327.76_1345.94#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1327.76_1345.94#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1327.76_1345.94#3.jpg" ], "ocr_qwen2_vl_72b": "Example: Let \\( A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\). Determine \\(\\alpha\\) such that \\( K(A(\\alpha)) \\) is minimized. Use the maximum norm.\n\nSolution: For a matrix \\( A \\),\n\\[ K(A) = \\|A\\| \\|A^{-1}\\| \\]\n\nNow,\n\\[ A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\]\n\\[ \\Rightarrow A^{-1}(\\alpha) = \\frac{1}{0.05\\alpha} \\begin{bmatrix} 1.5 & -0.1\\alpha \\\\ -1.0 & 0.1\\alpha \\end{bmatrix} \\]\n\nUsing maximum norm, we get\n\\[ \\|A(\\alpha)\\| = \\max\\{0.2|\\alpha|, 2.5\\} \\]\n\\[ \\|A^{-1}(\\alpha)\\| = \\max\\left\\{\\frac{2|\\alpha| + 30}{|\\alpha|}, \\frac{2|\\alpha| + 20}{|\\alpha|}\\right\\} \\]\n\\[ = \\frac{2|\\alpha| + 30}{|\\alpha|} \\]" }, { "vid": "4k-_KSx6-N4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Linear algebra tutorial on Condition Number in computational aspects of numerical linear algebra_30.json#####audio#####doingASR#####FinishASR/4k-_KSx6-N4/1345.94_1362.42.mp4", "refined_asr": " Prove because I can simply first multiply this quantity inside. So, I can return this number as max. I can multiply 0.2 alpha; this alpha will be canceled. So, it is my 0.4 of this plus it will be.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1345.94_1362.42#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1345.94_1362.42#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1345.94_1362.42#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1345.94_1362.42#3.jpg" ], "ocr_qwen2_vl_72b": "Example: Let \\( A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\). Determine \\(\\alpha\\) such that \\( K(A(\\alpha)) \\) is minimized. Use the maximum norm.\n\nSolution: For a matrix \\(A\\),\n\\[ K(A) = \\|A\\| \\|A^{-1}\\| \\]\n\nNow,\n\\[ A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\]\n\\[ \\Rightarrow A^{-1}(\\alpha) = \\frac{1}{0.05\\alpha} \\begin{bmatrix} 1.5 & -0.1\\alpha \\\\ -1.0 & 0.1\\alpha \\end{bmatrix} \\]\n\nUsing maximum norm, we get\n\\[ \\|A(\\alpha)\\| = \\max\\{0.2|\\alpha|, 2.5\\} \\]\n\\[ \\|A^{-1}(\\alpha)\\| = \\max\\left\\{\\frac{2|\\alpha| + 30}{|\\alpha|}, \\frac{2|\\alpha| + 20}{|\\alpha|}\\right\\} \\]\n\\[ = \\frac{2|\\alpha| + 30}{|\\alpha|} \\]\n\nHence,\n\\[ K(A(\\alpha)) = \\left( \\frac{2|\\alpha| + 30}{|\\alpha|} \\right) \\max\\{0.2|\\alpha|, 2.5\\} \\]" }, { "vid": "4k-_KSx6-N4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Linear algebra tutorial on Condition Number in computational aspects of numerical linear algebra_30.json#####audio#####doingASR#####FinishASR/4k-_KSx6-N4/1362.54_1379.24.mp4", "refined_asr": " When you multiply this it will be my point. It's a 10 it's a 6 comma. Then I can multiply 2.5 with this quantity you will get as this expression fine. Now the question arises when it will be minimum.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1362.54_1379.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1362.54_1379.24#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1362.54_1379.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1362.54_1379.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1362.54_1379.24#3.jpg" ], "ocr_qwen2_vl_72b": "Example: Let \\( A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\). Determine \\(\\alpha\\) such that \\( K(A(\\alpha)) \\) is minimized. Use the maximum norm.\n\nSolution: For a matrix \\(A\\),\n\n\\[ K(A) = \\|A\\| \\|A^{-1}\\| \\]\n\nNow,\n\n\\[ A(\\alpha) = \\begin{bmatrix} 0.1\\alpha & 0.1\\alpha \\\\ 1.0 & 1.5 \\end{bmatrix} \\]\n\n\\[ \\Rightarrow A^{-1}(\\alpha) = \\frac{1}{0.05\\alpha} \\begin{bmatrix} 1.5 & -0.1\\alpha \\\\ -1.0 & 0.1\\alpha \\end{bmatrix} \\]\n\nUsing maximum norm, we get\n\n\\[ \\|A(\\alpha)\\| = \\max\\{0.2|\\alpha|, 2.5\\} \\]\n\n\\[ \\|A^{-1}(\\alpha)\\| = \\max\\left\\{\\frac{2|\\alpha| + 30}{|\\alpha|}, \\frac{2|\\alpha| + 20}{|\\alpha|}\\right\\} \\]\n\n\\[ = \\frac{2|\\alpha| + 30}{|\\alpha|} \\]\n\nHence,\n\n\\[ K(A(\\alpha)) = \\left( \\frac{2|\\alpha| + 30}{|\\alpha|} \\right) \\max\\{0.2|\\alpha|, 2.5\\} \\]\n\n\\[ = \\max\\left[0.4|\\alpha| + 6, 12\\right] \\]" }, { "vid": "4k-_KSx6-N4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Linear algebra tutorial on Condition Number in computational aspects of numerical linear algebra_30.json#####audio#####doingASR#####FinishASR/4k-_KSx6-N4/1379.24_1392.26.mp4", "refined_asr": " So for what value of Alpha can you choose so that this will be at a minimum? This will be at a minimum only because we all know that if I say this is my X and this is my Y, it will be minimum when X is equal to Y.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1379.24_1392.26#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1379.24_1392.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1379.24_1392.26#2.jpg" ], "ocr_qwen2_vl_72b": "Using maximum norm, we get\n\n\\[\n\\|A(\\alpha)\\| = \\max\\{0.2|\\alpha|, 2.5\\}\n\\]\n\n\\[\n\\|A^{-1}(\\alpha)\\| = \\max\\left\\{\\frac{2|\\alpha| + 30}{|\\alpha|}, \\frac{2|\\alpha| + 20}{|\\alpha|}\\right\\}\n\\]\n\n\\[\n= \\frac{2|\\alpha| + 30}{|\\alpha|}\n\\]\n\nHence,\n\n\\[\nK(A(\\alpha)) = \\left(\\frac{2|\\alpha| + 30}{|\\alpha|}\\right) \\max\\{0.2|\\alpha|, 2.5\\}\n\\]\n\nTo find \\(\\alpha\\), such that \\(K(A(\\alpha))\\) is minimum, we have\n\n\\[\nK(A(\\alpha)) = \\max\\left[0.4|\\alpha| + 6, 5 + \\frac{75}{|\\alpha|}\\right]\n\\]\n\nFor minimum \\(K(A)\\),\n\nwe can choose \\(\\alpha\\) such that" }, { "vid": "4k-_KSx6-N4.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Linear algebra tutorial on Condition Number in computational aspects of numerical linear algebra_30.json#####audio#####doingASR#####FinishASR/4k-_KSx6-N4/1392.54_1407.22.mp4", "refined_asr": " Because whenever x is less than y then you can say the minimum is my x but when you take x is greater than y the minimum will be y so this will be minimum only when both the numbers are the same so can you", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1392.54_1407.22#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1392.54_1407.22#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1392.54_1407.22#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/4k-_KSx6-N4/4k-_KSx6-N4@1392.54_1407.22#3.jpg" ], "ocr_qwen2_vl_72b": "Using maximum norm, we get\n\n\\[\n\\|A(\\alpha)\\| = \\max\\{0.2|\\alpha|, 2.5\\}\n\\]\n\n\\[\n\\|A^{-1}(\\alpha)\\| = \\max\\left\\{\\frac{2|\\alpha| + 30}{|\\alpha|}, \\frac{2|\\alpha| + 20}{|\\alpha|}\\right\\}\n\\]\n\n\\[\n= \\frac{2|\\alpha| + 30}{|\\alpha|}\n\\]\n\nHence,\n\n\\[\nK(A(\\alpha)) = \\left(\\frac{2|\\alpha| + 30}{|\\alpha|}\\right) \\max\\{0.2|\\alpha|, 2.5\\}\n\\]\n\nTo find \\(\\alpha\\), such that \\(K(A(\\alpha))\\) is minimum, we have\n\n\\[\nK(A(\\alpha)) = \\max\\left[0.4|\\alpha| + 6, 5 + \\frac{75}{|\\alpha|}\\right]\n\\]\n\nFor minimum \\(K(A)\\),\n\nwe can choose \\(\\alpha\\) such that" } ], "image_num": 7, "text_num": 343, "token_num": 4375 }, { "images": [ "sample_100_images/ecn8bPDV6Sc@25.48_81.34#1.jpg", "sample_100_images/ecn8bPDV6Sc@25.48_81.34#2.jpg", "sample_100_images/ecn8bPDV6Sc@25.48_81.34#3.jpg", "sample_100_images/ecn8bPDV6Sc@25.48_81.34#5.jpg", "sample_100_images/ecn8bPDV6Sc@25.48_81.34#6.jpg", "sample_100_images/ecn8bPDV6Sc@25.48_81.34#7.jpg", null, "sample_100_images/ecn8bPDV6Sc@81.34_115.02000000000001#1.jpg", null, "sample_100_images/ecn8bPDV6Sc@115.02_169.1#1.jpg", "sample_100_images/ecn8bPDV6Sc@115.02_169.1#3.jpg", "sample_100_images/ecn8bPDV6Sc@115.02_169.1#4.jpg", "sample_100_images/ecn8bPDV6Sc@115.02_169.1#8.jpg", null, "sample_100_images/ecn8bPDV6Sc@169.1_205.54#1.jpg", "sample_100_images/ecn8bPDV6Sc@169.1_205.54#3.jpg", null ], "texts": [ null, null, null, null, null, null, " Till now, the ideal crystal we discussed was in terms of a lattice and a motif. We said that the lattice was a 3D periodic arrangement of points - a three-dimensional periodic arrangement of points. And the motif, we then placed an atom or a group of atoms on each of these points.", null, " With each lattice point. So this was the theme, and we developed this theme in great detail with the several examples we have seen. All those examples till now were of an ideal crystal. These crystals were not supposed to have any defects. But then, real crystals will show deviations from ideality. Real crystals will", null, null, null, null, " Deviations from ideality are what we term as defects or imperfections. They are a very important topic in materials science because these defects or imperfections control the properties of materials. Therefore, they are crucial in determining both the physical and mechanical properties.", null, null, " So it is important to have a serious study of defects in crystals. Of course, to study defects, we spent considerable time on ideal crystals. This is because defects can always be defined as deviations from ideality. So we have to." ], "text_ocr_list": [ null, null, null, null, null, null, "We can see these text from the image: Ideal Crystal.\n Till now, the ideal crystal we discussed was in terms of a lattice and a motif. We said that the lattice was a 3D periodic arrangement of points - a three-dimensional periodic arrangement of points. And the motif, we then placed an atom or a group of atoms on each of these points.", null, "We can see these text from the image: Ideal Crystal\n\n= Lattice + Motif\n\n3D Periodic arrangement of points\n\nAn atom or group of atoms.\n With each lattice point. So this was the theme, and we developed this theme in great detail with the several examples we have seen. All those examples till now were of an ideal crystal. These crystals were not supposed to have any defects. But then, real crystals will show deviations from ideality. Real crystals will", null, null, null, null, "We can see these text from the image: - Ideal Crystal\n - Lattice\n - 3D Periodic arrangement of points\n - Motif\n - An atom or group of atoms\n\n- Real Crystals: Deviation.\n Deviations from ideality are what we term as defects or imperfections. They are a very important topic in materials science because these defects or imperfections control the properties of materials. Therefore, they are crucial in determining both the physical and mechanical properties.", null, null, "We can see these text from the image: - Ideal Crystal\n - Lattice + Motif\n - 3D Periodic arrangement of points\n - An atom or group of atoms\n\n- Real Crystals: Deviations from ideality\n - Defects or Imperfections\n - Controls the physical and.\n So it is important to have a serious study of defects in crystals. Of course, to study defects, we spent considerable time on ideal crystals. This is because defects can always be defined as deviations from ideality. So we have to." ], "metadata": [ { "vid": "ecn8bPDV6Sc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Materials Science and Engineering course introduction to imperfections in solids focusing on surface defects tutorial_30.json#####audio#####doingASR#####FinishASR/ecn8bPDV6Sc/25.48_81.34.mp4", "refined_asr": " Till now, the ideal crystal we discussed was in terms of a lattice and a motif. We said that the lattice was a 3D periodic arrangement of points - a three-dimensional periodic arrangement of points. And the motif, we then placed an atom or a group of atoms on each of these points.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#7.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#7.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@25.48_81.34#8.jpg" ], "ocr_qwen2_vl_72b": "Ideal Crystal" }, { "vid": "ecn8bPDV6Sc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Materials Science and Engineering course introduction to imperfections in solids focusing on surface defects tutorial_30.json#####audio#####doingASR#####FinishASR/ecn8bPDV6Sc/81.34_115.02000000000001.mp4", "refined_asr": " With each lattice point. So this was the theme, and we developed this theme in great detail with the several examples we have seen. All those examples till now were of an ideal crystal. These crystals were not supposed to have any defects. But then, real crystals will show deviations from ideality. Real crystals will", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@81.34_115.02000000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@81.34_115.02000000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@81.34_115.02000000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@81.34_115.02000000000001#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@81.34_115.02000000000001#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@81.34_115.02000000000001#5.jpg" ], "ocr_qwen2_vl_72b": "Ideal Crystal\n\n= Lattice + Motif\n\n3D Periodic arrangement of points\n\nAn atom or group of atoms" }, { "vid": "ecn8bPDV6Sc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Materials Science and Engineering course introduction to imperfections in solids focusing on surface defects tutorial_30.json#####audio#####doingASR#####FinishASR/ecn8bPDV6Sc/115.02_169.1.mp4", "refined_asr": " Deviations from ideality are what we term as defects or imperfections. They are a very important topic in materials science because these defects or imperfections control the properties of materials. Therefore, they are crucial in determining both the physical and mechanical properties.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#8.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#6.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#7.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@115.02_169.1#8.jpg" ], "ocr_qwen2_vl_72b": "- Ideal Crystal\n - Lattice\n - 3D Periodic arrangement of points\n - Motif\n - An atom or group of atoms\n\n- Real Crystals: Deviation" }, { "vid": "ecn8bPDV6Sc.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Materials Science and Engineering course introduction to imperfections in solids focusing on surface defects tutorial_30.json#####audio#####doingASR#####FinishASR/ecn8bPDV6Sc/169.1_205.54.mp4", "refined_asr": " So it is important to have a serious study of defects in crystals. Of course, to study defects, we spent considerable time on ideal crystals. This is because defects can always be defined as deviations from ideality. So we have to.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@169.1_205.54#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@169.1_205.54#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@169.1_205.54#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@169.1_205.54#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@169.1_205.54#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@169.1_205.54#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@169.1_205.54#5.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ecn8bPDV6Sc/ecn8bPDV6Sc@169.1_205.54#6.jpg" ], "ocr_qwen2_vl_72b": "- Ideal Crystal\n - Lattice + Motif\n - 3D Periodic arrangement of points\n - An atom or group of atoms\n\n- Real Crystals: Deviations from ideality\n - Defects or Imperfections\n - Controls the physical and" } ], "image_num": 13, "text_num": 254, "token_num": 7742 }, { "images": [ "sample_100_images/PTNfRk0v6YM@642.26_661.52#1.jpg", "sample_100_images/PTNfRk0v6YM@642.26_661.52#2.jpg", "sample_100_images/PTNfRk0v6YM@642.26_661.52#3.jpg", null, "sample_100_images/PTNfRk0v6YM@661.52_677.84#1.jpg", null, "sample_100_images/PTNfRk0v6YM@677.84_695.12#1.jpg", null, "sample_100_images/PTNfRk0v6YM@695.12_714.86#1.jpg", null, "sample_100_images/PTNfRk0v6YM@714.86_738.4399999999999#1.jpg", null, "sample_100_images/PTNfRk0v6YM@738.4399999999999_758.42#1.jpg", null, "sample_100_images/PTNfRk0v6YM@758.42_776.6#1.jpg", null ], "texts": [ null, null, null, " A row whose distances I will try to preserve when I do a dimension reduction. Okay, let's go through examples now. This is the most interesting part. So our first example is this: Let me change the color here, just in the interest of variety. Okay, I'm taking dark green here. Suppose that", null, " I have orange points that are already there so I don't want to mess it up. So imagine originally I have my dataset like this: I have a bunch of blue points and a bunch of orange points. My step is 5000, which means I'm running it for 5000 iterations. This is very very important.", null, " Okay now the first thing that you'll notice is if I make my perplexity very low If I keep my perplexity to two which means I am trying to preserve only distances to two nearest neighbors or two points which are the closest to a given point Okay if I make my", null, " Perplexity at 5, 30, 50, and 100. Remember, I have 50 blue points here and 50 orange points here. Okay, that's how I constructed this data. When perplexity is very small, say 5, we want to get this structure right, which you can see. And when perplexity equals 30,", null, " Perplexity equals to 30 and 50, you're getting roughly this shape - roughly, not necessarily exactly, but roughly this shape. Now comes the interesting part: when perplexity is very small, you are getting a structure like this. If these are your blue points and if these are your green or orange points, this structure is completely absurd. So when perplexity", null, " If it's small the results you get can be crazy. Right as perplexity increases your shape will start to get much more sensible again. This is not sensible. This seems more sensible that you have two groups of points. So for example if this is some high", null, " Dimensional data. As soon as you visualize this, I say, yes, there are a bunch of points here, there are a bunch of points here, and they're well separated. Okay, now there is a catch here. So the lesson from this, the lesson from this, is number one: always run your t-SNE with multiple" ], "text_ocr_list": [ null, null, null, "We can see these text from the image: - https://distill.pub/2016/misread-tsne/\n- Secure\n- Step 5,000\n- Number Of Points 50\n- Perplexity 10\n- Epsilon 1.\n A row whose distances I will try to preserve when I do a dimension reduction. Okay, let's go through examples now. This is the most interesting part. So our first example is this: Let me change the color here, just in the interest of variety. Okay, I'm taking dark green here. Suppose that", null, "We can see these text from the image: 1. Those hyperparameters really matter\n\nLet's start with the \"hello world\" of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we'll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\nPerplexity: 2 Step: 5,000\nPerplexity: 5 Step: 5,000\nPerplexity: 30 Step: 5,000\nPerplexity: 50 Step: 5,000\nPerplexity: 100 Step: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very.\n I have orange points that are already there so I don't want to mess it up. So imagine originally I have my dataset like this: I have a bunch of blue points and a bunch of orange points. My step is 5000, which means I'm running it for 5000 iterations. This is very very important.", null, "We can see these text from the image: 1. Those hyperparameters really matter\n\nLet\u2019s start with the \u201chello world\u201d of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we\u2019ll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\n\nPerplexity: 2 Step: 5,000\n\nPerplexity: 5 Step: 5,000\n\nPerplexity: 30 Step: 5,000\n\nPerplexity: 50 Step: 5,000\n\nPerplexity: 100 Step: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very.\n Okay now the first thing that you'll notice is if I make my perplexity very low If I keep my perplexity to two which means I am trying to preserve only distances to two nearest neighbors or two points which are the closest to a given point Okay if I make my", null, "We can see these text from the image: 1. Those hyperparameters really matter\n\nLet\u2019s start with the \u201chello world\u201d of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we\u2019ll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\n\nPerplexity: 2 Step: 5,000\n\nPerplexity: 5 Step: 5,000\n\nPerplexity: 30 Step: 5,000\n\nPerplexity: 50 Step: 5,000\n\nPerplexity: 70 Step: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very.\n Perplexity at 5, 30, 50, and 100. Remember, I have 50 blue points here and 50 orange points here. Okay, that's how I constructed this data. When perplexity is very small, say 5, we want to get this structure right, which you can see. And when perplexity equals 30,", null, "We can see these text from the image: 1. Those hyperparameters really matter\n\nLet's start with the \"hello world\" of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we'll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\nPerplexity: 2\nStep: 5,000\nPerplexity: 5\nStep: 5,000\nPerplexity: 30\nStep: 5,000\nPerplexity: 50\nStep: 5,000\nPerplexity: 100\nStep: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very.\n Perplexity equals to 30 and 50, you're getting roughly this shape - roughly, not necessarily exactly, but roughly this shape. Now comes the interesting part: when perplexity is very small, you are getting a structure like this. If these are your blue points and if these are your green or orange points, this structure is completely absurd. So when perplexity", null, "We can see these text from the image: 1. Those hyperparameters really matter\n\nLet\u2019s start with the \u201chello world\u201d of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we\u2019ll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\n\nPerplexity: 2 Step: 5,000\n\nPerplexity: 5 Step: 5,000\n\nPerplexity: 30 Step: 5,000\n\nPerplexity: 50 Step: 5,000\n\nPerplexity: 100 Step: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very.\n If it's small the results you get can be crazy. Right as perplexity increases your shape will start to get much more sensible again. This is not sensible. This seems more sensible that you have two groups of points. So for example if this is some high", null, "We can see these text from the image: 1. Those hyperparameters really matter\n\nLet's start with the \"hello world\" of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we'll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\n\nPerplexity: 2 Step: 5,000\n\nPerplexity: 5 Step: 5,000\n\nPerplexity: 30 Step: 5,000\n\nPerplexity: 50 Step: 5,000\n\nPerplexity: 100 Step: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very.\n Dimensional data. As soon as you visualize this, I say, yes, there are a bunch of points here, there are a bunch of points here, and they're well separated. Okay, now there is a catch here. So the lesson from this, the lesson from this, is number one: always run your t-SNE with multiple" ], "metadata": [ { "vid": "PTNfRk0v6YM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Machine Learning tutorial on t-SNE for Dimensionality Reduction in Unsupervised Learning_30.json#####audio#####doingASR#####FinishASR/PTNfRk0v6YM/642.26_661.52.mp4", "refined_asr": " A row whose distances I will try to preserve when I do a dimension reduction. Okay, let's go through examples now. This is the most interesting part. So our first example is this: Let me change the color here, just in the interest of variety. Okay, I'm taking dark green here. Suppose that", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@642.26_661.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@642.26_661.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@642.26_661.52#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@642.26_661.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@642.26_661.52#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@642.26_661.52#3.jpg" ], "ocr_qwen2_vl_72b": "- https://distill.pub/2016/misread-tsne/\n- Secure\n- Step 5,000\n- Number Of Points 50\n- Perplexity 10\n- Epsilon 1" }, { "vid": "PTNfRk0v6YM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Machine Learning tutorial on t-SNE for Dimensionality Reduction in Unsupervised Learning_30.json#####audio#####doingASR#####FinishASR/PTNfRk0v6YM/661.52_677.84.mp4", "refined_asr": " I have orange points that are already there so I don't want to mess it up. So imagine originally I have my dataset like this: I have a bunch of blue points and a bunch of orange points. My step is 5000, which means I'm running it for 5000 iterations. This is very very important.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@661.52_677.84#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@661.52_677.84#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@661.52_677.84#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@661.52_677.84#3.jpg" ], "ocr_qwen2_vl_72b": "1. Those hyperparameters really matter\n\nLet's start with the \"hello world\" of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we'll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\nPerplexity: 2 Step: 5,000\nPerplexity: 5 Step: 5,000\nPerplexity: 30 Step: 5,000\nPerplexity: 50 Step: 5,000\nPerplexity: 100 Step: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very" }, { "vid": "PTNfRk0v6YM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Machine Learning tutorial on t-SNE for Dimensionality Reduction in Unsupervised Learning_30.json#####audio#####doingASR#####FinishASR/PTNfRk0v6YM/677.84_695.12.mp4", "refined_asr": " Okay now the first thing that you'll notice is if I make my perplexity very low If I keep my perplexity to two which means I am trying to preserve only distances to two nearest neighbors or two points which are the closest to a given point Okay if I make my", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@677.84_695.12#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@677.84_695.12#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@677.84_695.12#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@677.84_695.12#3.jpg" ], "ocr_qwen2_vl_72b": "1. Those hyperparameters really matter\n\nLet\u2019s start with the \u201chello world\u201d of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we\u2019ll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\n\nPerplexity: 2 Step: 5,000\n\nPerplexity: 5 Step: 5,000\n\nPerplexity: 30 Step: 5,000\n\nPerplexity: 50 Step: 5,000\n\nPerplexity: 100 Step: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very" }, { "vid": "PTNfRk0v6YM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Machine Learning tutorial on t-SNE for Dimensionality Reduction in Unsupervised Learning_30.json#####audio#####doingASR#####FinishASR/PTNfRk0v6YM/695.12_714.86.mp4", "refined_asr": " Perplexity at 5, 30, 50, and 100. Remember, I have 50 blue points here and 50 orange points here. Okay, that's how I constructed this data. When perplexity is very small, say 5, we want to get this structure right, which you can see. And when perplexity equals 30,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@695.12_714.86#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@695.12_714.86#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@695.12_714.86#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@695.12_714.86#3.jpg" ], "ocr_qwen2_vl_72b": "1. Those hyperparameters really matter\n\nLet\u2019s start with the \u201chello world\u201d of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we\u2019ll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\n\nPerplexity: 2 Step: 5,000\n\nPerplexity: 5 Step: 5,000\n\nPerplexity: 30 Step: 5,000\n\nPerplexity: 50 Step: 5,000\n\nPerplexity: 70 Step: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very" }, { "vid": "PTNfRk0v6YM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Machine Learning tutorial on t-SNE for Dimensionality Reduction in Unsupervised Learning_30.json#####audio#####doingASR#####FinishASR/PTNfRk0v6YM/714.86_738.4399999999999.mp4", "refined_asr": " Perplexity equals to 30 and 50, you're getting roughly this shape - roughly, not necessarily exactly, but roughly this shape. Now comes the interesting part: when perplexity is very small, you are getting a structure like this. If these are your blue points and if these are your green or orange points, this structure is completely absurd. So when perplexity", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@714.86_738.4399999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@714.86_738.4399999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@714.86_738.4399999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@714.86_738.4399999999999#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@714.86_738.4399999999999#4.jpg" ], "ocr_qwen2_vl_72b": "1. Those hyperparameters really matter\n\nLet's start with the \"hello world\" of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we'll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\nPerplexity: 2\nStep: 5,000\nPerplexity: 5\nStep: 5,000\nPerplexity: 30\nStep: 5,000\nPerplexity: 50\nStep: 5,000\nPerplexity: 100\nStep: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very" }, { "vid": "PTNfRk0v6YM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Machine Learning tutorial on t-SNE for Dimensionality Reduction in Unsupervised Learning_30.json#####audio#####doingASR#####FinishASR/PTNfRk0v6YM/738.4399999999999_758.42.mp4", "refined_asr": " If it's small the results you get can be crazy. Right as perplexity increases your shape will start to get much more sensible again. This is not sensible. This seems more sensible that you have two groups of points. So for example if this is some high", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@738.4399999999999_758.42#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@738.4399999999999_758.42#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@738.4399999999999_758.42#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@738.4399999999999_758.42#3.jpg" ], "ocr_qwen2_vl_72b": "1. Those hyperparameters really matter\n\nLet\u2019s start with the \u201chello world\u201d of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we\u2019ll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\n\nPerplexity: 2 Step: 5,000\n\nPerplexity: 5 Step: 5,000\n\nPerplexity: 30 Step: 5,000\n\nPerplexity: 50 Step: 5,000\n\nPerplexity: 100 Step: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very" }, { "vid": "PTNfRk0v6YM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Machine Learning tutorial on t-SNE for Dimensionality Reduction in Unsupervised Learning_30.json#####audio#####doingASR#####FinishASR/PTNfRk0v6YM/758.42_776.6.mp4", "refined_asr": " Dimensional data. As soon as you visualize this, I say, yes, there are a bunch of points here, there are a bunch of points here, and they're well separated. Okay, now there is a catch here. So the lesson from this, the lesson from this, is number one: always run your t-SNE with multiple", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@758.42_776.6#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@758.42_776.6#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@758.42_776.6#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/PTNfRk0v6YM/PTNfRk0v6YM@758.42_776.6#3.jpg" ], "ocr_qwen2_vl_72b": "1. Those hyperparameters really matter\n\nLet's start with the \"hello world\" of t-SNE: a data set of two widely separated clusters. To make things as simple as possible, we'll consider clusters in a 2D plane, as shown in the lefthand diagram. (For clarity, the two clusters are color coded.) The diagrams at right show t-SNE plots for five different perplexity values.\n\nOriginal\n\nPerplexity: 2 Step: 5,000\n\nPerplexity: 5 Step: 5,000\n\nPerplexity: 30 Step: 5,000\n\nPerplexity: 50 Step: 5,000\n\nPerplexity: 100 Step: 5,000\n\nWith perplexity values in the range (5 - 50) suggested by van der Maaten & Hinton, the diagrams do show these clusters, although with very" } ], "image_num": 9, "text_num": 503, "token_num": 5687 }, { "images": [ "sample_100_images/d4ME6kCFr1A@1147.68_1163.0400000000002#1.jpg", null, "sample_100_images/d4ME6kCFr1A@1163.0400000000002_1177.8600000000001#1.jpg", null, "sample_100_images/d4ME6kCFr1A@1177.8600000000001_1190.7600000000002#1.jpg", null, "sample_100_images/d4ME6kCFr1A@1190.7600000000002_1203.48#1.jpg", "sample_100_images/d4ME6kCFr1A@1190.7600000000002_1203.48#2.jpg", null, "sample_100_images/d4ME6kCFr1A@1203.48_1215.5#1.jpg", null, "sample_100_images/d4ME6kCFr1A@1215.5_1227.08#1.jpg", null, "sample_100_images/d4ME6kCFr1A@1227.08_1243.58#1.jpg", null, "sample_100_images/d4ME6kCFr1A@1243.58_1259.18#1.jpg", null, "sample_100_images/d4ME6kCFr1A@1259.18_1271.5000000000002#1.jpg", null ], "texts": [ null, " So that's not so bad. And here is just a plot that shows a comparison between the effective prescale as a function of the object energy. This is the effective prescale, so a higher number means you're throwing away more events randomly. A prescale of 1 means you save everything.", null, " The blue is if your associated object is a photon. And the red is if your associated object is a quark or a gluon. And, as Katarina said, we have higher pre-scales for quarks and gluons. The pre-scale for those are like 400 in these units, GeV.", null, " And for photons it's a bit lower but in both cases. So the black solid line here represents what you get if you do this zero bias trigger offline and then the dashed line shows that you can actually get a significant improvement over these standard techniques by using the data that you have available.", null, null, " Already took but not the data you would think you should look at. So this is the upshot: We have this huge data set that we're currently ignoring. New physics might be hiding in these data. And it's most powerful when combined with a trigger level.", null, " Analysis actually. So at level one in the hardware you trigger on say electrons and then in software you look at say jets that Katrina mentioned but this requires designing the algorithms now because they have to be if they're not in the software", null, " They're not going to be able to do that. They're going to be more than lost forever. And I think the key takeaway message is that the bottom line for any analysis at the LHC is not zero. So if you don't trigger on it, you're not totally hosed.", null, " You still have a budget line which is much bigger than zero so that's good news. And I think the key point here is that the sky and not the trigger is really the limit when it comes to doing analysis; you're not limited by triggers.", null, " Okay so now I want to move on and go a little bit further down on this timescale ladder to focusing now on improvements that one could use in the software-based trigger decision and also event reconstruction. So here the machine learning will start coming in.", null, " And once again I'm going to talk about pileup. So this pileup is really a key issue at the LHC because the pileup is a source of noise. Sometimes the pileup has interesting events, but most of the time the pileup is just soft energy." ], "text_ocr_list": [ null, "We can see these text from the image: - How does the ZBT compare to the other paradigms?\n- Cross-sections with Leptophobic Z' in MG5\n- Effective Prescale\n- Associated photon production\n- Associated quark/gluon production\n- ZBT offline\n- ZBT @ HLT\n- Unprescaled photon trigger\n- Unprescaled quark/gluon trigger\n- Object p_T [GeV].\n So that's not so bad. And here is just a plot that shows a comparison between the effective prescale as a function of the object energy. This is the effective prescale, so a higher number means you're throwing away more events randomly. A prescale of 1 means you save everything.", null, "We can see these text from the image: - How does the ZBT compare to the other paradigms?\n- Cross-sections with Leptophobic Z' in MG5\n- Effective Prescale\n- Associated photon production\n- Associated quark/gluon production\n- Object p_T [GeV].\n The blue is if your associated object is a photon. And the red is if your associated object is a quark or a gluon. And, as Katarina said, we have higher pre-scales for quarks and gluons. The pre-scale for those are like 400 in these units, GeV.", null, "We can see these text from the image: - How does the ZBT compare to the other paradigms?\n- Cross-sections with Leptophobic Z' in MG5\n- Effective Prescale\n- Associated photon production\n- Associated quark/gluon production\n- ZBT offline\n- ZBT @ HLT\n- Object p_T [GeV].\n And for photons it's a bit lower but in both cases. So the black solid line here represents what you get if you do this zero bias trigger offline and then the dashed line shows that you can actually get a significant improvement over these standard techniques by using the data that you have available.", null, null, "We can see these text from the image: - How does the ZBT compare to the other paradigms?\n- Cross-sections with Leptophobic Z' in MG5\n- Effective Prescale\n- Associated photon production\n- Associated quark/gluon production\n- ZBT offline\n- ZBT @ HLT\n- Object p_T [GeV].\n Already took but not the data you would think you should look at. So this is the upshot: We have this huge data set that we're currently ignoring. New physics might be hiding in these data. And it's most powerful when combined with a trigger level.", null, "We can see these text from the image: Upshot for the zero bias trigger\n\nThere is a huge dataset that we are currently ignoring.\n\nNew physics may be hiding in these data and we are collecting them anyway\n\nMost powerful when combined with trigger-level analysis (so need to design ASAP!)\n\nTakeaway message: the baseline is the ZBT >> 0!\n\nThink creatively about new possibilities... the sky, and not the trigger, is the limit!\n\n(also, remember ZBT offline has ~infinite time for processing).\n Analysis actually. So at level one in the hardware you trigger on say electrons and then in software you look at say jets that Katrina mentioned but this requires designing the algorithms now because they have to be if they're not in the software", null, "We can see these text from the image: Upshot for the zero bias trigger\n\nThere is a huge dataset that we are currently ignoring.\n\nNew physics may be hiding in these data and we are collecting them anyway\n\nMost powerful when combined with trigger-level analysis (so need to design ASAP!)\n\nTakeaway message: the baseline is the ZBT >> 0!\n\nThink creatively about new possibilities... the sky, and not the trigger, is the limit!\n\n(also, remember ZBT offline has ~infinite time for processing).\n They're not going to be able to do that. They're going to be more than lost forever. And I think the key takeaway message is that the bottom line for any analysis at the LHC is not zero. So if you don't trigger on it, you're not totally hosed.", null, "We can see these text from the image: Upshot for the zero bias trigger\n\nThere is a huge dataset that we are currently ignoring.\n\nNew physics may be hiding in these data\nand we are collecting them anyway\n\nMost powerful when combined with trigger-\nlevel analysis (so need to design ASAP!)\n\nTakeaway message: the baseline is the ZBT >> 0!\n\nThink creatively about new possibilities...\nthe sky, and not the trigger, is the limit!\n\n(also, remember ZBT offline has ~infinite time for processing).\n You still have a budget line which is much bigger than zero so that's good news. And I think the key point here is that the sky and not the trigger is really the limit when it comes to doing analysis; you're not limited by triggers.", null, "We can see these text from the image: Data pipeline at the LHC\n\n25 ns\n10^-19-10^-15 s ~ms\n0.01-20 ns ~min\n1-100 ns ~100 ms\n2.5 \u03bcs\n200 ms\n~100 ms\n~months\n\nO(100) pp collisions data / simulation\n(sub-)nuclear physics\nout-going particles interact with detector\ndetector response (signal formation + digitization)\nhardware-based trigger decision\nsoftware-based trigger decision\nevent reconstruction\nevent processing (skim, thin, augment)\nfinal data analysis (uses millions of events).\n Okay so now I want to move on and go a little bit further down on this timescale ladder to focusing now on improvements that one could use in the software-based trigger decision and also event reconstruction. So here the machine learning will start coming in.", null, "We can see these text from the image: The extra pileup collisions add unwanted soft radiation on top of the event\n\nThis degrades trigger and offline performance\n\nakin to image de-noising ... we can use ML for that!.\n And once again I'm going to talk about pileup. So this pileup is really a key issue at the LHC because the pileup is a source of noise. Sometimes the pileup has interesting events, but most of the time the pileup is just soft energy." ], "metadata": [ { "vid": "d4ME6kCFr1A.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Data Analysis Tutorial on Event Reconstruction_30.json#####audio#####doingASR#####FinishASR/d4ME6kCFr1A/1147.68_1163.0400000000002.mp4", "refined_asr": " So that's not so bad. And here is just a plot that shows a comparison between the effective prescale as a function of the object energy. This is the effective prescale, so a higher number means you're throwing away more events randomly. A prescale of 1 means you save everything.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1147.68_1163.0400000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1147.68_1163.0400000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1147.68_1163.0400000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1147.68_1163.0400000000002#3.jpg" ], "ocr_qwen2_vl_72b": "- How does the ZBT compare to the other paradigms?\n- Cross-sections with Leptophobic Z' in MG5\n- Effective Prescale\n- Associated photon production\n- Associated quark/gluon production\n- ZBT offline\n- ZBT @ HLT\n- Unprescaled photon trigger\n- Unprescaled quark/gluon trigger\n- Object p_T [GeV]" }, { "vid": "d4ME6kCFr1A.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Data Analysis Tutorial on Event Reconstruction_30.json#####audio#####doingASR#####FinishASR/d4ME6kCFr1A/1163.0400000000002_1177.8600000000001.mp4", "refined_asr": " The blue is if your associated object is a photon. And the red is if your associated object is a quark or a gluon. And, as Katarina said, we have higher pre-scales for quarks and gluons. The pre-scale for those are like 400 in these units, GeV.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1163.0400000000002_1177.8600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1163.0400000000002_1177.8600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1163.0400000000002_1177.8600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1163.0400000000002_1177.8600000000001#3.jpg" ], "ocr_qwen2_vl_72b": "- How does the ZBT compare to the other paradigms?\n- Cross-sections with Leptophobic Z' in MG5\n- Effective Prescale\n- Associated photon production\n- Associated quark/gluon production\n- Object p_T [GeV]" }, { "vid": "d4ME6kCFr1A.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Data Analysis Tutorial on Event Reconstruction_30.json#####audio#####doingASR#####FinishASR/d4ME6kCFr1A/1177.8600000000001_1190.7600000000002.mp4", "refined_asr": " And for photons it's a bit lower but in both cases. So the black solid line here represents what you get if you do this zero bias trigger offline and then the dashed line shows that you can actually get a significant improvement over these standard techniques by using the data that you have available.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1177.8600000000001_1190.7600000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1177.8600000000001_1190.7600000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1177.8600000000001_1190.7600000000002#2.jpg" ], "ocr_qwen2_vl_72b": "- How does the ZBT compare to the other paradigms?\n- Cross-sections with Leptophobic Z' in MG5\n- Effective Prescale\n- Associated photon production\n- Associated quark/gluon production\n- ZBT offline\n- ZBT @ HLT\n- Object p_T [GeV]" }, { "vid": "d4ME6kCFr1A.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Data Analysis Tutorial on Event Reconstruction_30.json#####audio#####doingASR#####FinishASR/d4ME6kCFr1A/1190.7600000000002_1203.48.mp4", "refined_asr": " Already took but not the data you would think you should look at. So this is the upshot: We have this huge data set that we're currently ignoring. New physics might be hiding in these data. And it's most powerful when combined with a trigger level.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1190.7600000000002_1203.48#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1190.7600000000002_1203.48#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1190.7600000000002_1203.48#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1190.7600000000002_1203.48#2.jpg" ], "ocr_qwen2_vl_72b": "- How does the ZBT compare to the other paradigms?\n- Cross-sections with Leptophobic Z' in MG5\n- Effective Prescale\n- Associated photon production\n- Associated quark/gluon production\n- ZBT offline\n- ZBT @ HLT\n- Object p_T [GeV]" }, { "vid": "d4ME6kCFr1A.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Data Analysis Tutorial on Event Reconstruction_30.json#####audio#####doingASR#####FinishASR/d4ME6kCFr1A/1203.48_1215.5.mp4", "refined_asr": " Analysis actually. So at level one in the hardware you trigger on say electrons and then in software you look at say jets that Katrina mentioned but this requires designing the algorithms now because they have to be if they're not in the software", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1203.48_1215.5#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1203.48_1215.5#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1203.48_1215.5#2.jpg" ], "ocr_qwen2_vl_72b": "Upshot for the zero bias trigger\n\nThere is a huge dataset that we are currently ignoring.\n\nNew physics may be hiding in these data and we are collecting them anyway\n\nMost powerful when combined with trigger-level analysis (so need to design ASAP!)\n\nTakeaway message: the baseline is the ZBT >> 0!\n\nThink creatively about new possibilities... the sky, and not the trigger, is the limit!\n\n(also, remember ZBT offline has ~infinite time for processing)" }, { "vid": "d4ME6kCFr1A.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Data Analysis Tutorial on Event Reconstruction_30.json#####audio#####doingASR#####FinishASR/d4ME6kCFr1A/1215.5_1227.08.mp4", "refined_asr": " They're not going to be able to do that. They're going to be more than lost forever. And I think the key takeaway message is that the bottom line for any analysis at the LHC is not zero. So if you don't trigger on it, you're not totally hosed.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1215.5_1227.08#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1215.5_1227.08#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1215.5_1227.08#2.jpg" ], "ocr_qwen2_vl_72b": "Upshot for the zero bias trigger\n\nThere is a huge dataset that we are currently ignoring.\n\nNew physics may be hiding in these data and we are collecting them anyway\n\nMost powerful when combined with trigger-level analysis (so need to design ASAP!)\n\nTakeaway message: the baseline is the ZBT >> 0!\n\nThink creatively about new possibilities... the sky, and not the trigger, is the limit!\n\n(also, remember ZBT offline has ~infinite time for processing)" }, { "vid": "d4ME6kCFr1A.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Data Analysis Tutorial on Event Reconstruction_30.json#####audio#####doingASR#####FinishASR/d4ME6kCFr1A/1227.08_1243.58.mp4", "refined_asr": " You still have a budget line which is much bigger than zero so that's good news. And I think the key point here is that the sky and not the trigger is really the limit when it comes to doing analysis; you're not limited by triggers.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1227.08_1243.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1227.08_1243.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1227.08_1243.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1227.08_1243.58#3.jpg" ], "ocr_qwen2_vl_72b": "Upshot for the zero bias trigger\n\nThere is a huge dataset that we are currently ignoring.\n\nNew physics may be hiding in these data\nand we are collecting them anyway\n\nMost powerful when combined with trigger-\nlevel analysis (so need to design ASAP!)\n\nTakeaway message: the baseline is the ZBT >> 0!\n\nThink creatively about new possibilities...\nthe sky, and not the trigger, is the limit!\n\n(also, remember ZBT offline has ~infinite time for processing)" }, { "vid": "d4ME6kCFr1A.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Data Analysis Tutorial on Event Reconstruction_30.json#####audio#####doingASR#####FinishASR/d4ME6kCFr1A/1243.58_1259.18.mp4", "refined_asr": " Okay so now I want to move on and go a little bit further down on this timescale ladder to focusing now on improvements that one could use in the software-based trigger decision and also event reconstruction. So here the machine learning will start coming in.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1243.58_1259.18#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1243.58_1259.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1243.58_1259.18#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1243.58_1259.18#3.jpg" ], "ocr_qwen2_vl_72b": "Data pipeline at the LHC\n\n25 ns\n10^-19-10^-15 s ~ms\n0.01-20 ns ~min\n1-100 ns ~100 ms\n2.5 \u03bcs\n200 ms\n~100 ms\n~months\n\nO(100) pp collisions data / simulation\n(sub-)nuclear physics\nout-going particles interact with detector\ndetector response (signal formation + digitization)\nhardware-based trigger decision\nsoftware-based trigger decision\nevent reconstruction\nevent processing (skim, thin, augment)\nfinal data analysis (uses millions of events)" }, { "vid": "d4ME6kCFr1A.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Data Analysis Tutorial on Event Reconstruction_30.json#####audio#####doingASR#####FinishASR/d4ME6kCFr1A/1259.18_1271.5000000000002.mp4", "refined_asr": " And once again I'm going to talk about pileup. So this pileup is really a key issue at the LHC because the pileup is a source of noise. Sometimes the pileup has interesting events, but most of the time the pileup is just soft energy.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1259.18_1271.5000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1259.18_1271.5000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/d4ME6kCFr1A/d4ME6kCFr1A@1259.18_1271.5000000000002#2.jpg" ], "ocr_qwen2_vl_72b": "The extra pileup collisions add unwanted soft radiation on top of the event\n\nThis degrades trigger and offline performance\n\nakin to image de-noising ... we can use ML for that!" } ], "image_num": 10, "text_num": 551, "token_num": 6311 }, { "images": [ "sample_100_images/D2wLMkU3CCo@407.0_423.0#1.jpg", null, "sample_100_images/D2wLMkU3CCo@423.0_437.0#1.jpg", "sample_100_images/D2wLMkU3CCo@423.0_437.0#2.jpg", null, "sample_100_images/D2wLMkU3CCo@437.0_447.0#1.jpg", null, "sample_100_images/D2wLMkU3CCo@447.0_463.6#1.jpg", null, "sample_100_images/D2wLMkU3CCo@463.6_475.0#1.jpg", null, "sample_100_images/D2wLMkU3CCo@475.0_495.0#1.jpg", null, "sample_100_images/D2wLMkU3CCo@495.0_510.14#1.jpg", "sample_100_images/D2wLMkU3CCo@495.0_510.14#2.jpg", null, "sample_100_images/D2wLMkU3CCo@510.14_522.34#1.jpg", null, "sample_100_images/D2wLMkU3CCo@522.5600000000001_532.94#1.jpg", null ], "texts": [ null, " From the illustrious collaboration the left-hand side is the real observation the right-hand side is the mock observation and this looks identical and you don't just do it by seeing by the way you do various statistical measurements and you can show that this actually matches extremely well", null, null, " So this is a theoretical proof that dark matter exists right? Given the fact that we have been making these observations for if you include Zwicky's observations it's almost 90 years what do we know of dark matter? Well we actually know a lot about dark matter.", null, " Right. And also, we do not know a lot about dark matter. So what do we know? First thing we know is that dark matter must be non-relativistic. Currently, that is something that we know.", null, " So any candidate that has to be dark matter must be non-relativistic. Now, secondly, dark matter must have weak interactions with standard model particles. Alright, so here I put 'weak' in quotation marks.", null, " What does that mean? That means it doesn't have to be the strong weak force. The weak force just has to be weak in an English way. It can also be the weak force that we know, but this is where weak is also used as an English term.", null, " Right. So, we know that the lifetime of whatever is the dark matter candidate must be larger than the age of the universe. We see dark matter around us; the smallest length scale in which we see dark matter is roughly a kiloparsec. So, if you make observations around the Sun at a distance of kiloparsecs,", null, null, " You will find dark matter. So that tells you that whatever the dark matter candidate is, it must exist now. These are profound things that we know. But we also do not know many different things. First, what is the mass of the dark matter candidate? Dark matter must be.", null, " So what is the dark matter made up of? If it's made up of something, what is the mass of that? We don't know. We don't know what the lifetime of the dark matter candidate is; I told you it must exist now.", null, " But what is that number? When I say it must exist, that tells you the lifetime is greater than some number. But what is that number? It's an inequality. Is there an equality? What is the lifetime of the dark matter candidate? We do not know that answer." ], "text_ocr_list": [ null, "We can see these text from the image: - Gravitational detection of dark matter\n- NGC 6503\n- Radius (kpc)\n- Vc (km s^-1)\n- Begeman, Broeils & Sanders MNRAS 249 (1991) 523\n- ESA website: http://sci.esa.int/plank/51557-plank-cosmo-recipe/\n- Dark Matter: 26.8%\n- Baryons: 4.9%\n- Dark Energy: 68.3%\n- Bullet cluster\n- https://en.wikipedia.org/wiki/File:1e0657_xray.jpg\n- Real observation from Hubble eXtreme Deep Field Observations: left side\n- Mock observation from Illustris: right side\n- Illustris website.\n From the illustrious collaboration the left-hand side is the real observation the right-hand side is the mock observation and this looks identical and you don't just do it by seeing by the way you do various statistical measurements and you can show that this actually matches extremely well", null, null, "We can see these text from the image: - Gravitational detection of dark matter\n- NGC 6503\n- Radius (kpc)\n- Dark gas\n- Dark Matter: 26.8%\n- Baryonic matter: 4.9%\n- Dark Energy: 68.3%\n- ESA website: http://sci.esa.int/planck/51557-planck-cosmo-recipe/\n- Bullet cluster\n- Real observation from Hubble eXtreme Deep Field Observations: left side\n- Mock observation from Illustris: right side\n- Illustris website.\n So this is a theoretical proof that dark matter exists right? Given the fact that we have been making these observations for if you include Zwicky's observations it's almost 90 years what do we know of dark matter? Well we actually know a lot about dark matter.", null, "We can see these text from the image: What do we know?\n\n- Structure formation tells us that the candidate must be non-relativistic\n- It must have \"weak\" interactions with other Standard Model particles\n- The lifetime of the candidate must be larger than the age of the Universe.\n Right. And also, we do not know a lot about dark matter. So what do we know? First thing we know is that dark matter must be non-relativistic. Currently, that is something that we know.", null, "We can see these text from the image: - Structure formation tells us that the candidate must be non-relativistic\n- It must have \"weak\" interactions with other Standard Model particles\n- The lifetime of the candidate must be larger than the age of the Universe.\n So any candidate that has to be dark matter must be non-relativistic. Now, secondly, dark matter must have weak interactions with standard model particles. Alright, so here I put 'weak' in quotation marks.", null, "We can see these text from the image: - Structure formation tells us that the candidate must be non-relativistic\n- It must have \"weak\" interactions with other Standard Model particles\n- The lifetime of the candidate must be larger than the age of the Universe.\n What does that mean? That means it doesn't have to be the strong weak force. The weak force just has to be weak in an English way. It can also be the weak force that we know, but this is where weak is also used as an English term.", null, "We can see these text from the image: - Structure formation tells us that the candidate must be non-relativistic\n\n- It must have \"weak\" interactions with other Standard Model particles\n\n- The lifetime of the candidate must be larger than the age of the Universe.\n Right. So, we know that the lifetime of whatever is the dark matter candidate must be larger than the age of the universe. We see dark matter around us; the smallest length scale in which we see dark matter is roughly a kiloparsec. So, if you make observations around the Sun at a distance of kiloparsecs,", null, null, "We can see these text from the image: - Structure formation tells us that the candidate must be non-relativistic\n\n- It must have \"weak\" interactions with other Standard Model particles\n\n- The lifetime of the candidate must be larger than the age of the Universe.\n You will find dark matter. So that tells you that whatever the dark matter candidate is, it must exist now. These are profound things that we know. But we also do not know many different things. First, what is the mass of the dark matter candidate? Dark matter must be.", null, "We can see these text from the image: What do we want to know?\n\n\u2022 Mass of the dark matter candidate\n\n\u2022 Lifetime of the dark matter candidate\n\n\u2022 Interaction strength of the dark matter candidate with itself and other Standard Model particles.\n So what is the dark matter made up of? If it's made up of something, what is the mass of that? We don't know. We don't know what the lifetime of the dark matter candidate is; I told you it must exist now.", null, "We can see these text from the image: What do we want to know?\n\n\u2022 Mass of the dark matter candidate\n\n\u2022 Lifetime of the dark matter candidate\n\n\u2022 Interaction strength of the dark matter candidate with itself and other Standard Model particles.\n But what is that number? When I say it must exist, that tells you the lifetime is greater than some number. But what is that number? It's an inequality. Is there an equality? What is the lifetime of the dark matter candidate? We do not know that answer." ], "metadata": [ { "vid": "D2wLMkU3CCo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Direct and Indirect Detection of Dark Matter Tutorial_30.json#####audio#####doingASR#####FinishASR/D2wLMkU3CCo/407.0_423.0.mp4", "refined_asr": " From the illustrious collaboration the left-hand side is the real observation the right-hand side is the mock observation and this looks identical and you don't just do it by seeing by the way you do various statistical measurements and you can show that this actually matches extremely well", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@407.0_423.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@407.0_423.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@407.0_423.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@407.0_423.0#3.jpg" ], "ocr_qwen2_vl_72b": "- Gravitational detection of dark matter\n- NGC 6503\n- Radius (kpc)\n- Vc (km s^-1)\n- Begeman, Broeils & Sanders MNRAS 249 (1991) 523\n- ESA website: http://sci.esa.int/plank/51557-plank-cosmo-recipe/\n- Dark Matter: 26.8%\n- Baryons: 4.9%\n- Dark Energy: 68.3%\n- Bullet cluster\n- https://en.wikipedia.org/wiki/File:1e0657_xray.jpg\n- Real observation from Hubble eXtreme Deep Field Observations: left side\n- Mock observation from Illustris: right side\n- Illustris website" }, { "vid": "D2wLMkU3CCo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Direct and Indirect Detection of Dark Matter Tutorial_30.json#####audio#####doingASR#####FinishASR/D2wLMkU3CCo/423.0_437.0.mp4", "refined_asr": " So this is a theoretical proof that dark matter exists right? Given the fact that we have been making these observations for if you include Zwicky's observations it's almost 90 years what do we know of dark matter? Well we actually know a lot about dark matter.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@423.0_437.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@423.0_437.0#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@423.0_437.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@423.0_437.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@423.0_437.0#3.jpg" ], "ocr_qwen2_vl_72b": "- Gravitational detection of dark matter\n- NGC 6503\n- Radius (kpc)\n- Dark gas\n- Dark Matter: 26.8%\n- Baryonic matter: 4.9%\n- Dark Energy: 68.3%\n- ESA website: http://sci.esa.int/planck/51557-planck-cosmo-recipe/\n- Bullet cluster\n- Real observation from Hubble eXtreme Deep Field Observations: left side\n- Mock observation from Illustris: right side\n- Illustris website" }, { "vid": "D2wLMkU3CCo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Direct and Indirect Detection of Dark Matter Tutorial_30.json#####audio#####doingASR#####FinishASR/D2wLMkU3CCo/437.0_447.0.mp4", "refined_asr": " Right. And also, we do not know a lot about dark matter. So what do we know? First thing we know is that dark matter must be non-relativistic. Currently, that is something that we know.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@437.0_447.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@437.0_447.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@437.0_447.0#2.jpg" ], "ocr_qwen2_vl_72b": "What do we know?\n\n- Structure formation tells us that the candidate must be non-relativistic\n- It must have \"weak\" interactions with other Standard Model particles\n- The lifetime of the candidate must be larger than the age of the Universe" }, { "vid": "D2wLMkU3CCo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Direct and Indirect Detection of Dark Matter Tutorial_30.json#####audio#####doingASR#####FinishASR/D2wLMkU3CCo/447.0_463.6.mp4", "refined_asr": " So any candidate that has to be dark matter must be non-relativistic. Now, secondly, dark matter must have weak interactions with standard model particles. Alright, so here I put 'weak' in quotation marks.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@447.0_463.6#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@447.0_463.6#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@447.0_463.6#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@447.0_463.6#3.jpg" ], "ocr_qwen2_vl_72b": "- Structure formation tells us that the candidate must be non-relativistic\n- It must have \"weak\" interactions with other Standard Model particles\n- The lifetime of the candidate must be larger than the age of the Universe" }, { "vid": "D2wLMkU3CCo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Direct and Indirect Detection of Dark Matter Tutorial_30.json#####audio#####doingASR#####FinishASR/D2wLMkU3CCo/463.6_475.0.mp4", "refined_asr": " What does that mean? That means it doesn't have to be the strong weak force. The weak force just has to be weak in an English way. It can also be the weak force that we know, but this is where weak is also used as an English term.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@463.6_475.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@463.6_475.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@463.6_475.0#2.jpg" ], "ocr_qwen2_vl_72b": "- Structure formation tells us that the candidate must be non-relativistic\n- It must have \"weak\" interactions with other Standard Model particles\n- The lifetime of the candidate must be larger than the age of the Universe" }, { "vid": "D2wLMkU3CCo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Direct and Indirect Detection of Dark Matter Tutorial_30.json#####audio#####doingASR#####FinishASR/D2wLMkU3CCo/475.0_495.0.mp4", "refined_asr": " Right. So, we know that the lifetime of whatever is the dark matter candidate must be larger than the age of the universe. We see dark matter around us; the smallest length scale in which we see dark matter is roughly a kiloparsec. So, if you make observations around the Sun at a distance of kiloparsecs,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@475.0_495.0#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@475.0_495.0#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@475.0_495.0#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@475.0_495.0#3.jpg" ], "ocr_qwen2_vl_72b": "- Structure formation tells us that the candidate must be non-relativistic\n\n- It must have \"weak\" interactions with other Standard Model particles\n\n- The lifetime of the candidate must be larger than the age of the Universe" }, { "vid": "D2wLMkU3CCo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Direct and Indirect Detection of Dark Matter Tutorial_30.json#####audio#####doingASR#####FinishASR/D2wLMkU3CCo/495.0_510.14.mp4", "refined_asr": " You will find dark matter. So that tells you that whatever the dark matter candidate is, it must exist now. These are profound things that we know. But we also do not know many different things. First, what is the mass of the dark matter candidate? Dark matter must be.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@495.0_510.14#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@495.0_510.14#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@495.0_510.14#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@495.0_510.14#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@495.0_510.14#3.jpg" ], "ocr_qwen2_vl_72b": "- Structure formation tells us that the candidate must be non-relativistic\n\n- It must have \"weak\" interactions with other Standard Model particles\n\n- The lifetime of the candidate must be larger than the age of the Universe" }, { "vid": "D2wLMkU3CCo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Direct and Indirect Detection of Dark Matter Tutorial_30.json#####audio#####doingASR#####FinishASR/D2wLMkU3CCo/510.14_522.34.mp4", "refined_asr": " So what is the dark matter made up of? If it's made up of something, what is the mass of that? We don't know. We don't know what the lifetime of the dark matter candidate is; I told you it must exist now.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@510.14_522.34#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@510.14_522.34#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@510.14_522.34#2.jpg" ], "ocr_qwen2_vl_72b": "What do we want to know?\n\n\u2022 Mass of the dark matter candidate\n\n\u2022 Lifetime of the dark matter candidate\n\n\u2022 Interaction strength of the dark matter candidate with itself and other Standard Model particles" }, { "vid": "D2wLMkU3CCo.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Direct and Indirect Detection of Dark Matter Tutorial_30.json#####audio#####doingASR#####FinishASR/D2wLMkU3CCo/522.5600000000001_532.94.mp4", "refined_asr": " But what is that number? When I say it must exist, that tells you the lifetime is greater than some number. But what is that number? It's an inequality. Is there an equality? What is the lifetime of the dark matter candidate? We do not know that answer.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@522.5600000000001_532.94#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@522.5600000000001_532.94#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/D2wLMkU3CCo/D2wLMkU3CCo@522.5600000000001_532.94#2.jpg" ], "ocr_qwen2_vl_72b": "What do we want to know?\n\n\u2022 Mass of the dark matter candidate\n\n\u2022 Lifetime of the dark matter candidate\n\n\u2022 Interaction strength of the dark matter candidate with itself and other Standard Model particles" } ], "image_num": 11, "text_num": 515, "token_num": 6851 }, { "images": [ "sample_100_images/q7rc4J4t-pM@1221.1_1237.6399999999999#1.jpg", null, "sample_100_images/q7rc4J4t-pM@1237.6399999999999_1256.1999999999998#1.jpg", null, "sample_100_images/q7rc4J4t-pM@1256.8799999999999_1280.36#1.jpg", null, "sample_100_images/q7rc4J4t-pM@1280.36_1307.74#1.jpg", null, "sample_100_images/q7rc4J4t-pM@1307.74_1330.72#1.jpg", null, "sample_100_images/q7rc4J4t-pM@1330.72_1350.64#1.jpg", null, "sample_100_images/q7rc4J4t-pM@1350.64_1366.4#1.jpg", null ], "texts": [ null, " To study CP violation there, as I'll explain, we look at how we visualize it in the standard model through something called the unitarity triangle. So, we use what's known as the Wolfenstein parameterization. This is an unitary parameterization where there is this parameter lambda, which is just sine.", null, " Of the Cabibbo angle. So you write this, the first two generation mixing in terms of sine of the Cabibbo angle. And then you introduce these other parameters, a, rho, and eta, to describe these off-diagonal elements. And these parameters are all of order one. The expansion parameter lambda is.", null, " Making the off-diagonal elements smaller. So lambda is 0.22 in the standard model. You then have higher order terms. But you essentially look at these to this triangle, to order of these, the CKM matrix, to order lambda cubed. And then you exploit the fact that it must.", null, " Be a unitary matrix. And if you look at the first and third columns, you can construct this relationship between the various elements of the CKM matrix which have to add up to zero. And because you have these complex elements due to the CP violating phase, you can also visualize this in the Argand plane. Each of these terms in this expression.", null, " It can be seen as the sides of a triangle that has to close if you're satisfying this unitarity constraint. What you should notice about this triangle is that each side depends on a coupling with the b quark: VUB, VTB, and VCB. Hence, we're very interested in looking at studying CP.", null, " Violation in beauty quarks And this is the only triangle where you get the sides all to be roughly the same length because each side is of order lambda cubed whereas if you look at other triangles you'll have one side that's very small and the other side of order lambda or something So this", null, " It isn't the only triangle in which this happens. So you can see that the two sides are the same length and so you get non-trivial values of the angles which Bell calls Phi one, Phi two, Phi three, and the rest of the world calls Alpha, Beta, and Gamma. So these are just the internal angles of this." ], "text_ocr_list": [ null, "We can see these text from the image: Visualising CP violation: the unitarity triangle\n\nINO online series 5/8/2020.\n To study CP violation there, as I'll explain, we look at how we visualize it in the standard model through something called the unitarity triangle. So, we use what's known as the Wolfenstein parameterization. This is an unitary parameterization where there is this parameter lambda, which is just sine.", null, "We can see these text from the image: Visualising CP violation: the unitarity triangle\n\n1) \\(\\begin{pmatrix}\n1 - \\lambda^2 / 2 & \\lambda & A\\lambda^3 (\\rho - i\\eta) \\\\\n-\\lambda & 1 - \\lambda^2 / 2 & A\\lambda^2 \\\\\nA\\lambda^3 [1 - (\\rho - i\\eta)] & -A\\lambda^2 & 1\n\\end{pmatrix} + O(\\lambda^4)\\).\n Of the Cabibbo angle. So you write this, the first two generation mixing in terms of sine of the Cabibbo angle. And then you introduce these other parameters, a, rho, and eta, to describe these off-diagonal elements. And these parameters are all of order one. The expansion parameter lambda is.", null, "We can see these text from the image: Visualising CP violation: the unitarity triangle\n\n1) \\(\\begin{pmatrix}\n1 - \\lambda^2 / 2 & \\lambda & A\\lambda^3 (\\rho - i\\eta) \\\\\n-\\lambda & 1 - \\lambda^2 / 2 & A\\lambda^2 \\\\\nA\\lambda^3 [1 - (\\rho - i\\eta)] & -A\\lambda^2 & 1\n\\end{pmatrix} + O(\\lambda^4)\\).\n Making the off-diagonal elements smaller. So lambda is 0.22 in the standard model. You then have higher order terms. But you essentially look at these to this triangle, to order of these, the CKM matrix, to order lambda cubed. And then you exploit the fact that it must.", null, "We can see these text from the image: Visualising CP violation: the unitarity triangle\n\n1) \\(\\begin{pmatrix} 1 - \\lambda^2 / 2 & \\lambda & A\\lambda^3 (\\rho - i\\eta) \\\\ -\\lambda & 1 - \\lambda^2 / 2 & A\\lambda^2 \\\\ A\\lambda^3 (1 - (\\rho - i\\eta)) & -A\\lambda^2 & 1 \\end{pmatrix} + O(\\lambda^4)\\)\n\n2) Exploit unitarity (1st and 3rd col.)\n\n\\(V_{ud} V_{ub}^* + V_{cd} V_{cb}^* + V_{td} V_{tb}^* = 0\\).\n Be a unitary matrix. And if you look at the first and third columns, you can construct this relationship between the various elements of the CKM matrix which have to add up to zero. And because you have these complex elements due to the CP violating phase, you can also visualize this in the Argand plane. Each of these terms in this expression.", null, "We can see these text from the image: 1) \\[\n\\begin{pmatrix}\n1 - \\lambda^2 / 2 & \\lambda \\\\\n-\\lambda & 1 - \\lambda^2 / 2 \\\\\nA \\lambda^3 \\left[ 1 - (\\rho - i \\eta) \\right] & -A \\lambda^2\n\\end{pmatrix}\n\\]\n\n2) Exploit unitarity (1st and 3rd col.)\n\n3) \\[\nV_{ud} V_{ub}^* + V_{cd} V_{cb}^* + V_{td} V_{tb}^* = 0\n\\]\n\n\\[\n\\phi_1\n\\]\n\\[\n\\phi_2\n\\]\n\\[\n\\phi_3\n\\].\n It can be seen as the sides of a triangle that has to close if you're satisfying this unitarity constraint. What you should notice about this triangle is that each side depends on a coupling with the b quark: VUB, VTB, and VCB. Hence, we're very interested in looking at studying CP.", null, "We can see these text from the image: 1) \\[\n\\begin{pmatrix}\n1 - \\lambda^2 / 2 & \\lambda & A \\lambda^3 (\\rho - i \\eta) \\\\\n-\\lambda & 1 - \\lambda^2 / 2 & A \\lambda^2 \\\\\nA \\lambda^3 [1 - (\\rho - i \\eta)] & -A \\lambda^2 & 1\n\\end{pmatrix} + O(\\lambda^4)\n\\]\n\n2) Exploit unitarity (1st and 3rd col.)\n\n\\[ V_{ud} V_{ub}^* + V_{cd} V_{cb}^* + V_{td} V_{tb}^* = 0 \\]\n\n3) \\[\n\\phi_1 \\quad V_{ud} V_{ub}^* \\quad \\phi_2 \\quad V_{td} V_{tb}^* \\quad \\phi_3 \\quad V_{cd} V_{cb}^*\n\\].\n Violation in beauty quarks And this is the only triangle where you get the sides all to be roughly the same length because each side is of order lambda cubed whereas if you look at other triangles you'll have one side that's very small and the other side of order lambda or something So this", null, "We can see these text from the image: 1) \\[\n\\begin{pmatrix}\n1 - \\lambda^2 / 2 & \\lambda \\\\\n-\\lambda & 1 - \\lambda^2 / 2 \\\\\nA \\lambda^3 \\left[ 1 - (\\rho - i \\eta) \\right] & -A \\lambda^2\n\\end{pmatrix}\n\\]\n\n2) Exploit unitarity (1st and 3rd col.)\n\n3) \\[\nV_{ud} V_{ub}^* + V_{cd} V_{cb}^* + V_{td} V_{tb}^* = 0\n\\]\n\n\\[\n\\phi_1\n\\]\n\n\\[\n\\phi_2\n\\]\n\n\\[\n\\phi_3\n\\].\n It isn't the only triangle in which this happens. So you can see that the two sides are the same length and so you get non-trivial values of the angles which Bell calls Phi one, Phi two, Phi three, and the rest of the world calls Alpha, Beta, and Gamma. So these are just the internal angles of this." ], "metadata": [ { "vid": "q7rc4J4t-pM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Major Experiments: BaBar and Belle Tutorial_30.json#####audio#####doingASR#####FinishASR/q7rc4J4t-pM/1221.1_1237.6399999999999.mp4", "refined_asr": " To study CP violation there, as I'll explain, we look at how we visualize it in the standard model through something called the unitarity triangle. So, we use what's known as the Wolfenstein parameterization. This is an unitary parameterization where there is this parameter lambda, which is just sine.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1221.1_1237.6399999999999#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1221.1_1237.6399999999999#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1221.1_1237.6399999999999#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1221.1_1237.6399999999999#3.jpg" ], "ocr_qwen2_vl_72b": "Visualising CP violation: the unitarity triangle\n\nINO online series 5/8/2020" }, { "vid": "q7rc4J4t-pM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Major Experiments: BaBar and Belle Tutorial_30.json#####audio#####doingASR#####FinishASR/q7rc4J4t-pM/1237.6399999999999_1256.1999999999998.mp4", "refined_asr": " Of the Cabibbo angle. So you write this, the first two generation mixing in terms of sine of the Cabibbo angle. And then you introduce these other parameters, a, rho, and eta, to describe these off-diagonal elements. And these parameters are all of order one. The expansion parameter lambda is.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1237.6399999999999_1256.1999999999998#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1237.6399999999999_1256.1999999999998#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1237.6399999999999_1256.1999999999998#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1237.6399999999999_1256.1999999999998#3.jpg" ], "ocr_qwen2_vl_72b": "Visualising CP violation: the unitarity triangle\n\n1) \\(\\begin{pmatrix}\n1 - \\lambda^2 / 2 & \\lambda & A\\lambda^3 (\\rho - i\\eta) \\\\\n-\\lambda & 1 - \\lambda^2 / 2 & A\\lambda^2 \\\\\nA\\lambda^3 [1 - (\\rho - i\\eta)] & -A\\lambda^2 & 1\n\\end{pmatrix} + O(\\lambda^4)\\)" }, { "vid": "q7rc4J4t-pM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Major Experiments: BaBar and Belle Tutorial_30.json#####audio#####doingASR#####FinishASR/q7rc4J4t-pM/1256.8799999999999_1280.36.mp4", "refined_asr": " Making the off-diagonal elements smaller. So lambda is 0.22 in the standard model. You then have higher order terms. But you essentially look at these to this triangle, to order of these, the CKM matrix, to order lambda cubed. And then you exploit the fact that it must.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1256.8799999999999_1280.36#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1256.8799999999999_1280.36#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1256.8799999999999_1280.36#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1256.8799999999999_1280.36#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1256.8799999999999_1280.36#4.jpg" ], "ocr_qwen2_vl_72b": "Visualising CP violation: the unitarity triangle\n\n1) \\(\\begin{pmatrix}\n1 - \\lambda^2 / 2 & \\lambda & A\\lambda^3 (\\rho - i\\eta) \\\\\n-\\lambda & 1 - \\lambda^2 / 2 & A\\lambda^2 \\\\\nA\\lambda^3 [1 - (\\rho - i\\eta)] & -A\\lambda^2 & 1\n\\end{pmatrix} + O(\\lambda^4)\\)" }, { "vid": "q7rc4J4t-pM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Major Experiments: BaBar and Belle Tutorial_30.json#####audio#####doingASR#####FinishASR/q7rc4J4t-pM/1280.36_1307.74.mp4", "refined_asr": " Be a unitary matrix. And if you look at the first and third columns, you can construct this relationship between the various elements of the CKM matrix which have to add up to zero. And because you have these complex elements due to the CP violating phase, you can also visualize this in the Argand plane. Each of these terms in this expression.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1280.36_1307.74#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1280.36_1307.74#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1280.36_1307.74#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1280.36_1307.74#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1280.36_1307.74#4.jpg" ], "ocr_qwen2_vl_72b": "Visualising CP violation: the unitarity triangle\n\n1) \\(\\begin{pmatrix} 1 - \\lambda^2 / 2 & \\lambda & A\\lambda^3 (\\rho - i\\eta) \\\\ -\\lambda & 1 - \\lambda^2 / 2 & A\\lambda^2 \\\\ A\\lambda^3 (1 - (\\rho - i\\eta)) & -A\\lambda^2 & 1 \\end{pmatrix} + O(\\lambda^4)\\)\n\n2) Exploit unitarity (1st and 3rd col.)\n\n\\(V_{ud} V_{ub}^* + V_{cd} V_{cb}^* + V_{td} V_{tb}^* = 0\\)" }, { "vid": "q7rc4J4t-pM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Major Experiments: BaBar and Belle Tutorial_30.json#####audio#####doingASR#####FinishASR/q7rc4J4t-pM/1307.74_1330.72.mp4", "refined_asr": " It can be seen as the sides of a triangle that has to close if you're satisfying this unitarity constraint. What you should notice about this triangle is that each side depends on a coupling with the b quark: VUB, VTB, and VCB. Hence, we're very interested in looking at studying CP.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1307.74_1330.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1307.74_1330.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1307.74_1330.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1307.74_1330.72#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1307.74_1330.72#4.jpg" ], "ocr_qwen2_vl_72b": "1) \\[\n\\begin{pmatrix}\n1 - \\lambda^2 / 2 & \\lambda \\\\\n-\\lambda & 1 - \\lambda^2 / 2 \\\\\nA \\lambda^3 \\left[ 1 - (\\rho - i \\eta) \\right] & -A \\lambda^2\n\\end{pmatrix}\n\\]\n\n2) Exploit unitarity (1st and 3rd col.)\n\n3) \\[\nV_{ud} V_{ub}^* + V_{cd} V_{cb}^* + V_{td} V_{tb}^* = 0\n\\]\n\n\\[\n\\phi_1\n\\]\n\\[\n\\phi_2\n\\]\n\\[\n\\phi_3\n\\]" }, { "vid": "q7rc4J4t-pM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Major Experiments: BaBar and Belle Tutorial_30.json#####audio#####doingASR#####FinishASR/q7rc4J4t-pM/1330.72_1350.64.mp4", "refined_asr": " Violation in beauty quarks And this is the only triangle where you get the sides all to be roughly the same length because each side is of order lambda cubed whereas if you look at other triangles you'll have one side that's very small and the other side of order lambda or something So this", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1330.72_1350.64#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1330.72_1350.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1330.72_1350.64#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1330.72_1350.64#3.jpg" ], "ocr_qwen2_vl_72b": "1) \\[\n\\begin{pmatrix}\n1 - \\lambda^2 / 2 & \\lambda & A \\lambda^3 (\\rho - i \\eta) \\\\\n-\\lambda & 1 - \\lambda^2 / 2 & A \\lambda^2 \\\\\nA \\lambda^3 [1 - (\\rho - i \\eta)] & -A \\lambda^2 & 1\n\\end{pmatrix} + O(\\lambda^4)\n\\]\n\n2) Exploit unitarity (1st and 3rd col.)\n\n\\[ V_{ud} V_{ub}^* + V_{cd} V_{cb}^* + V_{td} V_{tb}^* = 0 \\]\n\n3) \\[\n\\phi_1 \\quad V_{ud} V_{ub}^* \\quad \\phi_2 \\quad V_{td} V_{tb}^* \\quad \\phi_3 \\quad V_{cd} V_{cb}^*\n\\]" }, { "vid": "q7rc4J4t-pM.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Particle Physics Major Experiments: BaBar and Belle Tutorial_30.json#####audio#####doingASR#####FinishASR/q7rc4J4t-pM/1350.64_1366.4.mp4", "refined_asr": " It isn't the only triangle in which this happens. So you can see that the two sides are the same length and so you get non-trivial values of the angles which Bell calls Phi one, Phi two, Phi three, and the rest of the world calls Alpha, Beta, and Gamma. So these are just the internal angles of this.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1350.64_1366.4#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1350.64_1366.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1350.64_1366.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/q7rc4J4t-pM/q7rc4J4t-pM@1350.64_1366.4#3.jpg" ], "ocr_qwen2_vl_72b": "1) \\[\n\\begin{pmatrix}\n1 - \\lambda^2 / 2 & \\lambda \\\\\n-\\lambda & 1 - \\lambda^2 / 2 \\\\\nA \\lambda^3 \\left[ 1 - (\\rho - i \\eta) \\right] & -A \\lambda^2\n\\end{pmatrix}\n\\]\n\n2) Exploit unitarity (1st and 3rd col.)\n\n3) \\[\nV_{ud} V_{ub}^* + V_{cd} V_{cb}^* + V_{td} V_{tb}^* = 0\n\\]\n\n\\[\n\\phi_1\n\\]\n\n\\[\n\\phi_2\n\\]\n\n\\[\n\\phi_3\n\\]" } ], "image_num": 7, "text_num": 502, "token_num": 4534 }, { "images": [ null, null, "sample_100_images/dX8396ZmSPk@16234.78_16261.78#1.jpg", "sample_100_images/dX8396ZmSPk@16234.78_16261.78#2.jpg", "sample_100_images/dX8396ZmSPk@16234.78_16261.78#3.jpg", "sample_100_images/dX8396ZmSPk@16234.78_16261.78#4.jpg", null, "sample_100_images/dX8396ZmSPk@16261.78_16275.92#1.jpg", "sample_100_images/dX8396ZmSPk@16261.78_16275.92#2.jpg", null, "sample_100_images/dX8396ZmSPk@16275.92_16291.66#1.jpg", "sample_100_images/dX8396ZmSPk@16275.92_16291.66#2.jpg", null, "sample_100_images/dX8396ZmSPk@16291.66_16307.78#1.jpg", "sample_100_images/dX8396ZmSPk@16291.66_16307.78#2.jpg", null, "sample_100_images/dX8396ZmSPk@16307.78_16322.96#1.jpg", "sample_100_images/dX8396ZmSPk@16307.78_16322.96#2.jpg", null, "sample_100_images/dX8396ZmSPk@16322.96_16340.78#1.jpg", "sample_100_images/dX8396ZmSPk@16322.96_16340.78#2.jpg", "sample_100_images/dX8396ZmSPk@16322.96_16340.78#3.jpg", null, "sample_100_images/dX8396ZmSPk@16340.78_16353.96#1.jpg", "sample_100_images/dX8396ZmSPk@16340.78_16353.96#2.jpg", null ], "texts": [ " Okay.", " Okay.", null, null, null, null, " Okay. So we had this initial wireframe here which looked a bit like let's see yeah we had", null, null, " This initial wireframe, which looked like that, we wanted to improve. So we made it look like this. But we had this table and we had 'submit your application' form, so that gave us a visual reference for what we wanted to create.", null, null, " And of course, this is the iterative method of software development. You build something, deploy it, and let people use it. Then you come back to improve it. And then we came in and created a table. So, we created tables using the 'table', 'tr', and 'td' tags, as well as the 'th' tags.", null, null, " So first, create a table tag. Then create the rows using the TR tag. Next, create heading cells using TH and data cells using TD. And that's exactly what we did. We looked at some data on a spreadsheet and simply transferred it.", null, null, " We translated it into HTML code which allowed us to create a table that looked like this. Of course this is not a very pretty looking table so the next step was to style the HTML table to make it look a little nicer. The basic styling that we did with CSS was adding borders around the table and the", null, null, null, " Table cells. So you can see here we've added borders to the jobs table. Also, we've set 'border-collapse' equal to 'collapse'. Otherwise, we get these double borders that we were looking at earlier. So to avoid the double borders like this, you have to set the 'border-collapse' property.", null, null, " Just something to keep in mind: you can always look it up. Then we also wanted to make the table full width, so that it occupies the entire width of the parent element, which itself is limited to about 800 pixels at the center of the screen." ], "text_ocr_list": [ " Okay.", " Okay.", null, null, null, null, "We can see these text from the image: **Problem Statement**\n\nWe'll explore abovementioned topics in HTML & CSS by attempting to solve this problem statement:\n\nImprove the Jovian Careers website created in [the previous tutorial](https://jovian.ai/learn/html-and-css-for-data-science) by making the following changes:\n- Show the list of jobs in a tabular format with columns for job title, location, salary, and posted date\n- Show an application form below the jobs table where a user can fill out and submit an application\n- Improve the aesthetics of the page using appropriate fonts, text sizes and colors\n- Deploy the page to the cloud and ensure it previews properly via a link.\n Okay. So we had this initial wireframe here which looked a bit like let's see yeah we had", null, null, "We can see these text from the image: - About Jovian\n- Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua.\n- Job Opportunities\n- Developer\n- Apply\n- Developer\n- Apply\n- Test\n- Apply\n- Courses\n- Programs\n- YouTube\n- (c) 2023, Jovian\n- Jovian\n- About Jovian\n- Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua.\n- Job Opportunities\n- Job Title\n- Location.\n This initial wireframe, which looked like that, we wanted to improve. So we made it look like this. But we had this table and we had 'submit your application' form, so that gave us a visual reference for what we wanted to create.", null, null, "We can see these text from the image: - Improved readability: Tables make it easier for users to scan and compare information.\n- Better organization: By organizing the job openings in a table, we can sort them by department, location, or other relevant criteria..\n And of course, this is the iterative method of software development. You build something, deploy it, and let people use it. Then you come back to improve it. And then we came in and created a table. So, we created tables using the 'table', 'tr', and 'td' tags, as well as the 'th' tags.", null, null, "We can see these text from the image: Here's an example:\n\n```html\n\n \n \n \n \n \n \n \n \n \n \n \n \n
Heading 1Heading 2
Data 1Data 2
Data 3Data 4
\n```\n\nThe `border` attribute is used to provide a border width for the table..\n So first, create a table tag. Then create the rows using the TR tag. Next, create heading cells using TH and data cells using TD. And that's exactly what we did. We looked at some data on a spreadsheet and simply transferred it.", null, null, "We can see these text from the image: We've given the table tag an id and added classes for the tr tags for styling with CSS.\n\nHere's what the table looks like:\n\nJob Opportunities\n\n| Job Title | Location | Salary | Posted On |\n| --- | --- | --- | --- |\n| Frontend Developer | Bengaluru, India | \u20b912,00,000 | Mar 3, 2023 |\n| Full Stack Developer | New Delhi, India | \u20b915,00,000 | Feb 1, 2023 |\n| Data Scientist | San Francisco, USA | $175,000 | Dec 22, 2022 |\n| ML Engineer | Remote | $80,000 | Sep 19, 2022 |\n\nEXERCISE: Add more rows and columns to the jobs table e.g. responsibilities etc. Try including paragraphs and lists within so, you might need to set vary the widths of different columns https://www.w3schools.com/html/html_table_sizes.asp.\n We translated it into HTML code which allowed us to create a table that looked like this. Of course this is not a very pretty looking table so the next step was to style the HTML table to make it look a little nicer. The basic styling that we did with CSS was adding borders around the table and the", null, null, null, "We can see these text from the image: - styles.css\n- .form-group input[type=\"checkbox\"]\n- #team-image\n- border-radius: 5px;\n- width: 100%;\n- #jobs-table\n- width: 100%;\n- border: 1px solid gray;\n- border-collapse: collapse;\n- .jobs-header-row th\n- border: 1px solid gray;\n- padding: 8px;\n- background-color: #BFDBFE;\n- text-align: center;\n- color: #2E3440;.\n Table cells. So you can see here we've added borders to the jobs table. Also, we've set 'border-collapse' equal to 'collapse'. Otherwise, we get these double borders that we were looking at earlier. So to avoid the double borders like this, you have to set the 'border-collapse' property.", null, null, "We can see these text from the image: We can use simple CSS properties to make the following changes to our table:\n\n1. We'll add borders around the table and table cells\n2. We'll make the table full width on the page\n3. We'll left align the headers so that they're in line with values\n4. We'll show a different background color for the headers\n5. We'll add alternate colors to the rows using nth-child (even) and odd\n\nHere are the CSS properties we need to apply in the styles.css file:\n\n```css\n#jobs-table {\n width: 100%;\n border: 1px solid grey;\n border-collapse: collapse;\n}\n```.\n Just something to keep in mind: you can always look it up. Then we also wanted to make the table full width, so that it occupies the entire width of the parent element, which itself is limited to about 800 pixels at the center of the screen." ], "metadata": [ { "vid": "dX8396ZmSPk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Web Development tutorial on CSS Syntax Basics in HTML and CSS course_30.json#####audio#####doingASR#####FinishASR/dX8396ZmSPk/16232.640000000001_16233.82.mp4", "refined_asr": " Okay.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16232.640000000001_16233.82#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "dX8396ZmSPk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Web Development tutorial on CSS Syntax Basics in HTML and CSS course_30.json#####audio#####doingASR#####FinishASR/dX8396ZmSPk/16233.82_16234.78.mp4", "refined_asr": " Okay.", "keyframe_ssim": null, "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16233.82_16234.78#1.jpg" ], "ocr_qwen2_vl_72b": null }, { "vid": "dX8396ZmSPk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Web Development tutorial on CSS Syntax Basics in HTML and CSS course_30.json#####audio#####doingASR#####FinishASR/dX8396ZmSPk/16234.78_16261.78.mp4", "refined_asr": " Okay. So we had this initial wireframe here which looked a bit like let's see yeah we had", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16234.78_16261.78#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16234.78_16261.78#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16234.78_16261.78#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16234.78_16261.78#4.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16234.78_16261.78#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16234.78_16261.78#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16234.78_16261.78#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16234.78_16261.78#4.jpg" ], "ocr_qwen2_vl_72b": "**Problem Statement**\n\nWe'll explore abovementioned topics in HTML & CSS by attempting to solve this problem statement:\n\nImprove the Jovian Careers website created in [the previous tutorial](https://jovian.ai/learn/html-and-css-for-data-science) by making the following changes:\n- Show the list of jobs in a tabular format with columns for job title, location, salary, and posted date\n- Show an application form below the jobs table where a user can fill out and submit an application\n- Improve the aesthetics of the page using appropriate fonts, text sizes and colors\n- Deploy the page to the cloud and ensure it previews properly via a link" }, { "vid": "dX8396ZmSPk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Web Development tutorial on CSS Syntax Basics in HTML and CSS course_30.json#####audio#####doingASR#####FinishASR/dX8396ZmSPk/16261.78_16275.92.mp4", "refined_asr": " This initial wireframe, which looked like that, we wanted to improve. So we made it look like this. But we had this table and we had 'submit your application' form, so that gave us a visual reference for what we wanted to create.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16261.78_16275.92#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16261.78_16275.92#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16261.78_16275.92#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16261.78_16275.92#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16261.78_16275.92#3.jpg" ], "ocr_qwen2_vl_72b": "- About Jovian\n- Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua.\n- Job Opportunities\n- Developer\n- Apply\n- Developer\n- Apply\n- Test\n- Apply\n- Courses\n- Programs\n- YouTube\n- (c) 2023, Jovian\n- Jovian\n- About Jovian\n- Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua.\n- Job Opportunities\n- Job Title\n- Location" }, { "vid": "dX8396ZmSPk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Web Development tutorial on CSS Syntax Basics in HTML and CSS course_30.json#####audio#####doingASR#####FinishASR/dX8396ZmSPk/16275.92_16291.66.mp4", "refined_asr": " And of course, this is the iterative method of software development. You build something, deploy it, and let people use it. Then you come back to improve it. And then we came in and created a table. So, we created tables using the 'table', 'tr', and 'td' tags, as well as the 'th' tags.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16275.92_16291.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16275.92_16291.66#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16275.92_16291.66#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16275.92_16291.66#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16275.92_16291.66#3.jpg" ], "ocr_qwen2_vl_72b": "- Improved readability: Tables make it easier for users to scan and compare information.\n- Better organization: By organizing the job openings in a table, we can sort them by department, location, or other relevant criteria." }, { "vid": "dX8396ZmSPk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Web Development tutorial on CSS Syntax Basics in HTML and CSS course_30.json#####audio#####doingASR#####FinishASR/dX8396ZmSPk/16291.66_16307.78.mp4", "refined_asr": " So first, create a table tag. Then create the rows using the TR tag. Next, create heading cells using TH and data cells using TD. And that's exactly what we did. We looked at some data on a spreadsheet and simply transferred it.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16291.66_16307.78#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16291.66_16307.78#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16291.66_16307.78#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16291.66_16307.78#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16291.66_16307.78#3.jpg" ], "ocr_qwen2_vl_72b": "Here's an example:\n\n```html\n\n \n \n \n \n \n \n \n \n \n \n \n \n
Heading 1Heading 2
Data 1Data 2
Data 3Data 4
\n```\n\nThe `border` attribute is used to provide a border width for the table." }, { "vid": "dX8396ZmSPk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Web Development tutorial on CSS Syntax Basics in HTML and CSS course_30.json#####audio#####doingASR#####FinishASR/dX8396ZmSPk/16307.78_16322.96.mp4", "refined_asr": " We translated it into HTML code which allowed us to create a table that looked like this. Of course this is not a very pretty looking table so the next step was to style the HTML table to make it look a little nicer. The basic styling that we did with CSS was adding borders around the table and the", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16307.78_16322.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16307.78_16322.96#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16307.78_16322.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16307.78_16322.96#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16307.78_16322.96#3.jpg" ], "ocr_qwen2_vl_72b": "We've given the table tag an id and added classes for the tr tags for styling with CSS.\n\nHere's what the table looks like:\n\nJob Opportunities\n\n| Job Title | Location | Salary | Posted On |\n| --- | --- | --- | --- |\n| Frontend Developer | Bengaluru, India | \u20b912,00,000 | Mar 3, 2023 |\n| Full Stack Developer | New Delhi, India | \u20b915,00,000 | Feb 1, 2023 |\n| Data Scientist | San Francisco, USA | $175,000 | Dec 22, 2022 |\n| ML Engineer | Remote | $80,000 | Sep 19, 2022 |\n\nEXERCISE: Add more rows and columns to the jobs table e.g. responsibilities etc. Try including paragraphs and lists within so, you might need to set vary the widths of different columns https://www.w3schools.com/html/html_table_sizes.asp" }, { "vid": "dX8396ZmSPk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Web Development tutorial on CSS Syntax Basics in HTML and CSS course_30.json#####audio#####doingASR#####FinishASR/dX8396ZmSPk/16322.96_16340.78.mp4", "refined_asr": " Table cells. So you can see here we've added borders to the jobs table. Also, we've set 'border-collapse' equal to 'collapse'. Otherwise, we get these double borders that we were looking at earlier. So to avoid the double borders like this, you have to set the 'border-collapse' property.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16322.96_16340.78#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16322.96_16340.78#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16322.96_16340.78#3.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16322.96_16340.78#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16322.96_16340.78#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16322.96_16340.78#3.jpg" ], "ocr_qwen2_vl_72b": "- styles.css\n- .form-group input[type=\"checkbox\"]\n- #team-image\n- border-radius: 5px;\n- width: 100%;\n- #jobs-table\n- width: 100%;\n- border: 1px solid gray;\n- border-collapse: collapse;\n- .jobs-header-row th\n- border: 1px solid gray;\n- padding: 8px;\n- background-color: #BFDBFE;\n- text-align: center;\n- color: #2E3440;" }, { "vid": "dX8396ZmSPk.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/Web Development tutorial on CSS Syntax Basics in HTML and CSS course_30.json#####audio#####doingASR#####FinishASR/dX8396ZmSPk/16340.78_16353.96.mp4", "refined_asr": " Just something to keep in mind: you can always look it up. Then we also wanted to make the table full width, so that it occupies the entire width of the parent element, which itself is limited to about 800 pixels at the center of the screen.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16340.78_16353.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16340.78_16353.96#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16340.78_16353.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/dX8396ZmSPk/dX8396ZmSPk@16340.78_16353.96#2.jpg" ], "ocr_qwen2_vl_72b": "We can use simple CSS properties to make the following changes to our table:\n\n1. We'll add borders around the table and table cells\n2. We'll make the table full width on the page\n3. We'll left align the headers so that they're in line with values\n4. We'll show a different background color for the headers\n5. We'll add alternate colors to the rows using nth-child (even) and odd\n\nHere are the CSS properties we need to apply in the styles.css file:\n\n```css\n#jobs-table {\n width: 100%;\n border: 1px solid grey;\n border-collapse: collapse;\n}\n```" } ], "image_num": 17, "text_num": 403, "token_num": 10195 }, { "images": [ "sample_100_images/SEh3yhEFK1w@684.46_704.02#1.jpg", null, "sample_100_images/SEh3yhEFK1w@704.26_717.72#1.jpg", null, "sample_100_images/SEh3yhEFK1w@719.72_735.2#1.jpg", null, "sample_100_images/SEh3yhEFK1w@735.22_749.26#1.jpg", null, "sample_100_images/SEh3yhEFK1w@749.4_765.0600000000001#1.jpg", null, "sample_100_images/SEh3yhEFK1w@765.22_784.4#1.jpg", null, "sample_100_images/SEh3yhEFK1w@784.4_801.98#1.jpg", null ], "texts": [ null, " Row operations were necessary to turn these values into zero. And that's the way that I prefer to do it. But again for the sake of this video not being 37 minutes long let's not. If I do it that way what happens here is I would end up with my three solutions and that's why I like it.", null, " So I might as well just keep going. I'm going to keep doing row operations until I've actually solved the system. But for now, we can see that we have solved the system again using an augmented matrix and row operations, and in this case, also back substitution.", null, " You're going to see the question of existence and uniqueness pop up quite often throughout the course. When we're talking about existence, we're essentially asking: is the system consistent or does a solution exist? Uniqueness means that if a solution exists, it is the only one.", null, " So whether a solution does exist is it unique meaning is there one solution or are there many solutions So let's take a look at this one together And again we're trying to determine So far up to this point we have only looked at consistent systems with one solution", null, " So let's take a look at maybe one that isn't like that. The first thing I'm going to do, as always, is write my matrix: 0, 1, negative 4, 8; 2, negative 3;", null, " Two, one, four, negative eight, twelve, one. And just as we talked about before, I really want this column to end up being ones. But I've got this glaring issue right here with this zero, so I can't have a zero there.", null, " In the next section, we'll talk about all the terminology. But essentially, right now, I'm just telling you that we have to have something there other than a zero. So the easiest thing I can do is essentially swap R2 and R3. With one? I'm sorry, nope, just kidding." ], "text_ocr_list": [ null, "We can see these text from the image: Solve the system (again)\n\nThis time use an augmented matrix and row operations\n\nx1 - 2x2 + x3 = 0\n2x2 - 8x3 = 8\n-4x1 + x2 + x3 = -9\n\nBack Subst.\n\nx1 - 2x2 + x3 = 0\nx2 - 4x3 = 4\nx3 = 3\n\nx2 - 4(3) = 4\nx2 - 12 = 4\nx2 = 16\n\nx1 - 2(16) + 3 = 0\nx1 - 32 + 3 = 0\nx1 - 29 = 0\nx1 = 29\n\n(29, 16, 3).\n Row operations were necessary to turn these values into zero. And that's the way that I prefer to do it. But again for the sake of this video not being 37 minutes long let's not. If I do it that way what happens here is I would end up with my three solutions and that's why I like it.", null, "We can see these text from the image: Solve the system (again)\n\nThis time use an augmented matrix and row operations\n\nx\u2081 - 2x\u2082 + x\u2083 = 0\n2x\u2082 - 8x\u2083 = 8\n-4x\u2081 + x\u2082 + x\u2083 = -9\n\nBack substitute:\n\nx\u2081 - 2(16) + 3 = 0\nx\u2081 - 32 + 3 = 0\nx\u2081 - 29 = 0\nx\u2081 = 29\n\n(29, 16, 3).\n So I might as well just keep going. I'm going to keep doing row operations until I've actually solved the system. But for now, we can see that we have solved the system again using an augmented matrix and row operations, and in this case, also back substitution.", null, "We can see these text from the image: EXISTENCE AND UNIQUENESS\n\nDETERMINE IF THE SYSTEM IS CONSISTENT (DOES A SOLUTION EXIST?)\nIF SO, DETERMINE IF THE SOLUTION IS UNIQUE (JUST ONE SOLUTION?)\n\nx\u2082 - 4x\u2083 = 8\n2x\u2081 - 3x\u2082 + 2x\u2083 = 1\n4x\u2081 - 8x\u2082 + 12x\u2083 = 1.\n You're going to see the question of existence and uniqueness pop up quite often throughout the course. When we're talking about existence, we're essentially asking: is the system consistent or does a solution exist? Uniqueness means that if a solution exists, it is the only one.", null, "We can see these text from the image: EXISTENCE AND UNIQUENESS\n\nDETERMINE IF THE SYSTEM IS CONSISTENT (DOES A SOLUTION EXIST?)\n\nIF SO, DETERMINE IF THE SOLUTION IS UNIQUE (JUST ONE SOLUTION?)\n\nx\u2082 - 4x\u2083 = 8\n2x\u2081 - 3x\u2082 + 2x\u2083 = 1\n4x\u2081 - 8x\u2082 + 12x\u2083 = 1.\n So whether a solution does exist is it unique meaning is there one solution or are there many solutions So let's take a look at this one together And again we're trying to determine So far up to this point we have only looked at consistent systems with one solution", null, "We can see these text from the image: EXISTENCE AND UNIQUENESS\n\nDETERMINE IF THE SYSTEM IS CONSISTENT (DOES A SOLUTION EXIST?)\nIF SO, DETERMINE IF THE SOLUTION IS UNIQUE (JUST ONE SOLUTION?)\n\nx\u2082 - 4x\u2083 = 8\n2x\u2081 - 3x\u2082 + 2x\u2083 = 1\n4x\u2081 - 8x\u2082 + 12x\u2083 = 1.\n So let's take a look at maybe one that isn't like that. The first thing I'm going to do, as always, is write my matrix: 0, 1, negative 4, 8; 2, negative 3;", null, "We can see these text from the image: EXISTENCE AND UNIQUENESS\n\nDETERMINE IF THE SYSTEM IS CONSISTENT (DOES A SOLUTION EXIST?)\nIF SO, DETERMINE IF THE SOLUTION IS UNIQUE (JUST ONE SOLUTION?)\n\nx\u2082 - 4x\u2083 = 8\n2x\u2081 - 3x\u2082 + 2x\u2083 = 1\n4x\u2081 - 8x\u2082 + 12x\u2083 = 1\n\n[0 1 -4 | 8]\n[2 -3 2 | 1]\n[4 -8].\n Two, one, four, negative eight, twelve, one. And just as we talked about before, I really want this column to end up being ones. But I've got this glaring issue right here with this zero, so I can't have a zero there.", null, "We can see these text from the image: EXISTENCE AND UNIQUENESS\n\nDETERMINE IF THE SYSTEM IS CONSISTENT (DOES A SOLUTION EXIST?)\nIF SO, DETERMINE IF THE SOLUTION IS UNIQUE (JUST ONE SOLUTION?)\n\nx2 - 4x3 = 8\n2x1 - 3x2 + 2x3 = 1\n4x1 - 8x2 + 12x3 = 1\n\n[0 1 -4 | 8]\n[2 -3 2 | 1]\n[4 -8 12 | 1].\n In the next section, we'll talk about all the terminology. But essentially, right now, I'm just telling you that we have to have something there other than a zero. So the easiest thing I can do is essentially swap R2 and R3. With one? I'm sorry, nope, just kidding." ], "metadata": [ { "vid": "SEh3yhEFK1w.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/linear algebra_500.json#####audio#####doingASR#####FinishASR/SEh3yhEFK1w/684.46_704.02.mp4", "refined_asr": " Row operations were necessary to turn these values into zero. And that's the way that I prefer to do it. But again for the sake of this video not being 37 minutes long let's not. If I do it that way what happens here is I would end up with my three solutions and that's why I like it.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@684.46_704.02#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@684.46_704.02#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@684.46_704.02#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@684.46_704.02#3.jpg" ], "ocr_qwen2_vl_72b": "Solve the system (again)\n\nThis time use an augmented matrix and row operations\n\nx1 - 2x2 + x3 = 0\n2x2 - 8x3 = 8\n-4x1 + x2 + x3 = -9\n\nBack Subst.\n\nx1 - 2x2 + x3 = 0\nx2 - 4x3 = 4\nx3 = 3\n\nx2 - 4(3) = 4\nx2 - 12 = 4\nx2 = 16\n\nx1 - 2(16) + 3 = 0\nx1 - 32 + 3 = 0\nx1 - 29 = 0\nx1 = 29\n\n(29, 16, 3)" }, { "vid": "SEh3yhEFK1w.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/linear algebra_500.json#####audio#####doingASR#####FinishASR/SEh3yhEFK1w/704.26_717.72.mp4", "refined_asr": " So I might as well just keep going. I'm going to keep doing row operations until I've actually solved the system. But for now, we can see that we have solved the system again using an augmented matrix and row operations, and in this case, also back substitution.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@704.26_717.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@704.26_717.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@704.26_717.72#2.jpg" ], "ocr_qwen2_vl_72b": "Solve the system (again)\n\nThis time use an augmented matrix and row operations\n\nx\u2081 - 2x\u2082 + x\u2083 = 0\n2x\u2082 - 8x\u2083 = 8\n-4x\u2081 + x\u2082 + x\u2083 = -9\n\nBack substitute:\n\nx\u2081 - 2(16) + 3 = 0\nx\u2081 - 32 + 3 = 0\nx\u2081 - 29 = 0\nx\u2081 = 29\n\n(29, 16, 3)" }, { "vid": "SEh3yhEFK1w.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/linear algebra_500.json#####audio#####doingASR#####FinishASR/SEh3yhEFK1w/719.72_735.2.mp4", "refined_asr": " You're going to see the question of existence and uniqueness pop up quite often throughout the course. When we're talking about existence, we're essentially asking: is the system consistent or does a solution exist? Uniqueness means that if a solution exists, it is the only one.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@719.72_735.2#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@719.72_735.2#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@719.72_735.2#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@719.72_735.2#3.jpg" ], "ocr_qwen2_vl_72b": "EXISTENCE AND UNIQUENESS\n\nDETERMINE IF THE SYSTEM IS CONSISTENT (DOES A SOLUTION EXIST?)\nIF SO, DETERMINE IF THE SOLUTION IS UNIQUE (JUST ONE SOLUTION?)\n\nx\u2082 - 4x\u2083 = 8\n2x\u2081 - 3x\u2082 + 2x\u2083 = 1\n4x\u2081 - 8x\u2082 + 12x\u2083 = 1" }, { "vid": "SEh3yhEFK1w.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/linear algebra_500.json#####audio#####doingASR#####FinishASR/SEh3yhEFK1w/735.22_749.26.mp4", "refined_asr": " So whether a solution does exist is it unique meaning is there one solution or are there many solutions So let's take a look at this one together And again we're trying to determine So far up to this point we have only looked at consistent systems with one solution", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@735.22_749.26#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@735.22_749.26#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@735.22_749.26#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@735.22_749.26#3.jpg" ], "ocr_qwen2_vl_72b": "EXISTENCE AND UNIQUENESS\n\nDETERMINE IF THE SYSTEM IS CONSISTENT (DOES A SOLUTION EXIST?)\n\nIF SO, DETERMINE IF THE SOLUTION IS UNIQUE (JUST ONE SOLUTION?)\n\nx\u2082 - 4x\u2083 = 8\n2x\u2081 - 3x\u2082 + 2x\u2083 = 1\n4x\u2081 - 8x\u2082 + 12x\u2083 = 1" }, { "vid": "SEh3yhEFK1w.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/linear algebra_500.json#####audio#####doingASR#####FinishASR/SEh3yhEFK1w/749.4_765.0600000000001.mp4", "refined_asr": " So let's take a look at maybe one that isn't like that. The first thing I'm going to do, as always, is write my matrix: 0, 1, negative 4, 8; 2, negative 3;", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@749.4_765.0600000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@749.4_765.0600000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@749.4_765.0600000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@749.4_765.0600000000001#3.jpg" ], "ocr_qwen2_vl_72b": "EXISTENCE AND UNIQUENESS\n\nDETERMINE IF THE SYSTEM IS CONSISTENT (DOES A SOLUTION EXIST?)\nIF SO, DETERMINE IF THE SOLUTION IS UNIQUE (JUST ONE SOLUTION?)\n\nx\u2082 - 4x\u2083 = 8\n2x\u2081 - 3x\u2082 + 2x\u2083 = 1\n4x\u2081 - 8x\u2082 + 12x\u2083 = 1" }, { "vid": "SEh3yhEFK1w.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/linear algebra_500.json#####audio#####doingASR#####FinishASR/SEh3yhEFK1w/765.22_784.4.mp4", "refined_asr": " Two, one, four, negative eight, twelve, one. And just as we talked about before, I really want this column to end up being ones. But I've got this glaring issue right here with this zero, so I can't have a zero there.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@765.22_784.4#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@765.22_784.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@765.22_784.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@765.22_784.4#3.jpg" ], "ocr_qwen2_vl_72b": "EXISTENCE AND UNIQUENESS\n\nDETERMINE IF THE SYSTEM IS CONSISTENT (DOES A SOLUTION EXIST?)\nIF SO, DETERMINE IF THE SOLUTION IS UNIQUE (JUST ONE SOLUTION?)\n\nx\u2082 - 4x\u2083 = 8\n2x\u2081 - 3x\u2082 + 2x\u2083 = 1\n4x\u2081 - 8x\u2082 + 12x\u2083 = 1\n\n[0 1 -4 | 8]\n[2 -3 2 | 1]\n[4 -8]" }, { "vid": "SEh3yhEFK1w.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave/video_clip/linear algebra_500.json#####audio#####doingASR#####FinishASR/SEh3yhEFK1w/784.4_801.98.mp4", "refined_asr": " In the next section, we'll talk about all the terminology. But essentially, right now, I'm just telling you that we have to have something there other than a zero. So the easiest thing I can do is essentially swap R2 and R3. With one? I'm sorry, nope, just kidding.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@784.4_801.98#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@784.4_801.98#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@784.4_801.98#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/SEh3yhEFK1w/SEh3yhEFK1w@784.4_801.98#3.jpg" ], "ocr_qwen2_vl_72b": "EXISTENCE AND UNIQUENESS\n\nDETERMINE IF THE SYSTEM IS CONSISTENT (DOES A SOLUTION EXIST?)\nIF SO, DETERMINE IF THE SOLUTION IS UNIQUE (JUST ONE SOLUTION?)\n\nx2 - 4x3 = 8\n2x1 - 3x2 + 2x3 = 1\n4x1 - 8x2 + 12x3 = 1\n\n[0 1 -4 | 8]\n[2 -3 2 | 1]\n[4 -8 12 | 1]" } ], "image_num": 7, "text_num": 435, "token_num": 4467 }, { "images": [ "sample_100_images/RyesLifeUBw@543.5400000000001_558.2#1.jpg", null, "sample_100_images/RyesLifeUBw@558.2_573.6400000000001#1.jpg", "sample_100_images/RyesLifeUBw@558.2_573.6400000000001#2.jpg", null, "sample_100_images/RyesLifeUBw@573.6400000000001_588.96#1.jpg", "sample_100_images/RyesLifeUBw@573.6400000000001_588.96#2.jpg", null, "sample_100_images/RyesLifeUBw@588.96_605.58#1.jpg", "sample_100_images/RyesLifeUBw@588.96_605.58#2.jpg", null, "sample_100_images/RyesLifeUBw@605.58_621.08#1.jpg", "sample_100_images/RyesLifeUBw@605.58_621.08#2.jpg", null, "sample_100_images/RyesLifeUBw@621.08_634.46#1.jpg", "sample_100_images/RyesLifeUBw@621.08_634.46#2.jpg", null, "sample_100_images/RyesLifeUBw@634.46_651.76#1.jpg", "sample_100_images/RyesLifeUBw@634.46_651.76#2.jpg", null, "sample_100_images/RyesLifeUBw@651.76_668.5400000000001#1.jpg", "sample_100_images/RyesLifeUBw@651.76_668.5400000000001#2.jpg", null ], "texts": [ null, " Therefore, 0, 0 must be on the side of the boundary line where none of the points are included in the solution set. Which means that all of the points on the other side will be included. So that brings us to the last step. We just need to shade the correct side.", null, null, " Let me illustrate it this way. To summarize, if the test point you select makes the inequality true, then you've chosen a point within the solution set. This means you'll shade the region of the graph where that test point lies. But if the test point yields a false statement,", null, null, " Then you know that it's not in the answer set, which means that you'll leave that side blank and shade the other side. So that's the basics of graphing inequalities. But there's a couple more things that you'll need to know to be successful at it. After all, not every inequality that you encounter", null, null, " It will be in a nice, easy-to-work-with form like y is greater than mx plus b. You may have a few things that need to be simplified. Fortunately, you can solve inequalities almost the same way that you do with equations. All of the same principles apply, like combining.", null, null, " Like terms and the idea that anything you do to one side needs to be done to the other side too. However, there are a couple situations where you need to flip the inequality sign. You never had to worry about that with equations because order doesn't matter when both sides", null, null, " Have the exact same value. But inequalities tell us that one side is not equal to the other side. One side has a greater value than the other and the open end of the inequality sign always faces that greater side. That means that if you need to switch the left and right sides,", null, null, " When solving an inequality, you also need to switch the inequality sign. For example, if you have the inequality 2x > y and you want to rearrange it so that y comes first, you'd need to flip the inequality sign when you do that. So just remember, if you switch the sides, switch the inequality sign as well.", null, null, " The inequality sign. Another way to switch the inequality sign is whenever you multiply or divide both sides by a negative number or term. You don't need to do this for any other operations. You can add or subtract positive or negative terms from both sides, and you can multiply." ], "text_ocr_list": [ null, "We can see these text from the image: Test\ny < 2x - 3\n0 < 2\u00b70 - 3\n0 < -3\nFalse\n\n(0, 0).\n Therefore, 0, 0 must be on the side of the boundary line where none of the points are included in the solution set. Which means that all of the points on the other side will be included. So that brings us to the last step. We just need to shade the correct side.", null, null, "We can see these text from the image: Test Point (x, y).\n Let me illustrate it this way. To summarize, if the test point you select makes the inequality true, then you've chosen a point within the solution set. This means you'll shade the region of the graph where that test point lies. But if the test point yields a false statement,", null, null, "We can see these text from the image: Test Point (x, y)\nInequality True\nShade the same side\n\nTest Point (x, y)\nInequality False.\n Then you know that it's not in the answer set, which means that you'll leave that side blank and shade the other side. So that's the basics of graphing inequalities. But there's a couple more things that you'll need to know to be successful at it. After all, not every inequality that you encounter", null, null, "We can see these text from the image: Easy\ny >.\n It will be in a nice, easy-to-work-with form like y is greater than mx plus b. You may have a few things that need to be simplified. Fortunately, you can solve inequalities almost the same way that you do with equations. All of the same principles apply, like combining.", null, null, "We can see these text from the image: Solve or Simplify\n\nCombining 'like' terms\n\nmath Antics.\n Like terms and the idea that anything you do to one side needs to be done to the other side too. However, there are a couple situations where you need to flip the inequality sign. You never had to worry about that with equations because order doesn't matter when both sides", null, null, "We can see these text from the image: Same = Same.\n Have the exact same value. But inequalities tell us that one side is not equal to the other side. One side has a greater value than the other and the open end of the inequality sign always faces that greater side. That means that if you need to switch the left and right sides,", null, null, "We can see these text from the image: Lesser < Greater\n\nGreater > Lesser.\n When solving an inequality, you also need to switch the inequality sign. For example, if you have the inequality 2x > y and you want to rearrange it so that y comes first, you'd need to flip the inequality sign when you do that. So just remember, if you switch the sides, switch the inequality sign as well.", null, null, "We can see these text from the image: Another Case\n\nSwitch >.\n The inequality sign. Another way to switch the inequality sign is whenever you multiply or divide both sides by a negative number or term. You don't need to do this for any other operations. You can add or subtract positive or negative terms from both sides, and you can multiply." ], "metadata": [ { "vid": "RyesLifeUBw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Algebraic inequality in coordinate plane_20.json#####audio#####doingASR#####FinishASR/RyesLifeUBw/543.5400000000001_558.2.mp4", "refined_asr": " Therefore, 0, 0 must be on the side of the boundary line where none of the points are included in the solution set. Which means that all of the points on the other side will be included. So that brings us to the last step. We just need to shade the correct side.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@543.5400000000001_558.2#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@543.5400000000001_558.2#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@543.5400000000001_558.2#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@543.5400000000001_558.2#3.jpg" ], "ocr_qwen2_vl_72b": "Test\ny < 2x - 3\n0 < 2\u00b70 - 3\n0 < -3\nFalse\n\n(0, 0)" }, { "vid": "RyesLifeUBw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Algebraic inequality in coordinate plane_20.json#####audio#####doingASR#####FinishASR/RyesLifeUBw/558.2_573.6400000000001.mp4", "refined_asr": " Let me illustrate it this way. To summarize, if the test point you select makes the inequality true, then you've chosen a point within the solution set. This means you'll shade the region of the graph where that test point lies. But if the test point yields a false statement,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@558.2_573.6400000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@558.2_573.6400000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@558.2_573.6400000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@558.2_573.6400000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@558.2_573.6400000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Test Point (x, y)" }, { "vid": "RyesLifeUBw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Algebraic inequality in coordinate plane_20.json#####audio#####doingASR#####FinishASR/RyesLifeUBw/573.6400000000001_588.96.mp4", "refined_asr": " Then you know that it's not in the answer set, which means that you'll leave that side blank and shade the other side. So that's the basics of graphing inequalities. But there's a couple more things that you'll need to know to be successful at it. After all, not every inequality that you encounter", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@573.6400000000001_588.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@573.6400000000001_588.96#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@573.6400000000001_588.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@573.6400000000001_588.96#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@573.6400000000001_588.96#3.jpg" ], "ocr_qwen2_vl_72b": "Test Point (x, y)\nInequality True\nShade the same side\n\nTest Point (x, y)\nInequality False" }, { "vid": "RyesLifeUBw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Algebraic inequality in coordinate plane_20.json#####audio#####doingASR#####FinishASR/RyesLifeUBw/588.96_605.58.mp4", "refined_asr": " It will be in a nice, easy-to-work-with form like y is greater than mx plus b. You may have a few things that need to be simplified. Fortunately, you can solve inequalities almost the same way that you do with equations. All of the same principles apply, like combining.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@588.96_605.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@588.96_605.58#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@588.96_605.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@588.96_605.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@588.96_605.58#3.jpg" ], "ocr_qwen2_vl_72b": "Easy\ny >" }, { "vid": "RyesLifeUBw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Algebraic inequality in coordinate plane_20.json#####audio#####doingASR#####FinishASR/RyesLifeUBw/605.58_621.08.mp4", "refined_asr": " Like terms and the idea that anything you do to one side needs to be done to the other side too. However, there are a couple situations where you need to flip the inequality sign. You never had to worry about that with equations because order doesn't matter when both sides", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@605.58_621.08#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@605.58_621.08#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@605.58_621.08#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@605.58_621.08#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@605.58_621.08#3.jpg" ], "ocr_qwen2_vl_72b": "Solve or Simplify\n\nCombining 'like' terms\n\nmath Antics" }, { "vid": "RyesLifeUBw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Algebraic inequality in coordinate plane_20.json#####audio#####doingASR#####FinishASR/RyesLifeUBw/621.08_634.46.mp4", "refined_asr": " Have the exact same value. But inequalities tell us that one side is not equal to the other side. One side has a greater value than the other and the open end of the inequality sign always faces that greater side. That means that if you need to switch the left and right sides,", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@621.08_634.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@621.08_634.46#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@621.08_634.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@621.08_634.46#2.jpg" ], "ocr_qwen2_vl_72b": "Same = Same" }, { "vid": "RyesLifeUBw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Algebraic inequality in coordinate plane_20.json#####audio#####doingASR#####FinishASR/RyesLifeUBw/634.46_651.76.mp4", "refined_asr": " When solving an inequality, you also need to switch the inequality sign. For example, if you have the inequality 2x > y and you want to rearrange it so that y comes first, you'd need to flip the inequality sign when you do that. So just remember, if you switch the sides, switch the inequality sign as well.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@634.46_651.76#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@634.46_651.76#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@634.46_651.76#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@634.46_651.76#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@634.46_651.76#3.jpg" ], "ocr_qwen2_vl_72b": "Lesser < Greater\n\nGreater > Lesser" }, { "vid": "RyesLifeUBw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Algebraic inequality in coordinate plane_20.json#####audio#####doingASR#####FinishASR/RyesLifeUBw/651.76_668.5400000000001.mp4", "refined_asr": " The inequality sign. Another way to switch the inequality sign is whenever you multiply or divide both sides by a negative number or term. You don't need to do this for any other operations. You can add or subtract positive or negative terms from both sides, and you can multiply.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@651.76_668.5400000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@651.76_668.5400000000001#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@651.76_668.5400000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@651.76_668.5400000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/RyesLifeUBw/RyesLifeUBw@651.76_668.5400000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Another Case\n\nSwitch >" } ], "image_num": 15, "text_num": 519, "token_num": 9159 }, { "images": [ "sample_100_images/ehjFf-mMq3Q@427.44_445.44#1.jpg", "sample_100_images/ehjFf-mMq3Q@427.44_445.44#2.jpg", null, "sample_100_images/ehjFf-mMq3Q@445.44_457.44#1.jpg", null, "sample_100_images/ehjFf-mMq3Q@457.44_469.44#1.jpg", null, "sample_100_images/ehjFf-mMq3Q@469.44_485.44#1.jpg", null, "sample_100_images/ehjFf-mMq3Q@485.44_497.44#1.jpg", null, "sample_100_images/ehjFf-mMq3Q@497.44_517.44#1.jpg", null, "sample_100_images/ehjFf-mMq3Q@517.44_535.44#1.jpg", null, "sample_100_images/ehjFf-mMq3Q@535.44_555.44#1.jpg", null, "sample_100_images/ehjFf-mMq3Q@555.44_567.44#1.jpg", null ], "texts": [ null, null, " See what I'm saying, gang? We usually use situational irony to allow writers and artists to create contrasts between expectations and reality. Once again, find your own irony. Sorry, I'm drinking some tea. Find your own example and post it in the class Padlet.", null, " But let's take a look at Kaczynski's work and see if we can find all of these different techniques in there. We're looking for hyperbole, understatement, caricature, visual distortion, allegory, and situational irony.", null, " If we look at our piece right here on the right, it's called Pokemon Go. I think this is a good piece because it really does contain every single one of these things. So let's take a look at these techniques and let's talk about how.", null, " So hyperbole. There's an obvious exaggeration here of people staring down at their phones. His neck is cranked really long. He's staring directly down. This is a clear exaggeration of someone whose neck is cranked over all the time.", null, " But there's also understatement. If we look at the subject's face right here, we see that it has a very placid expression, as though this is no joke.", null, " This is no big deal. This is completely normal. Nothing happening here. Despite the fact that he has a, I don't know, magical little creature on his back. I would call that an understatement given the circumstances, trying to show us how normal it is that we are completely subject to our phones.", null, " We have caricature. We absolutely have caricature. That exaggerated stooped neck is not going to be seen in real people. It's going to be a caricature of this person to exaggerate the fact that they are glued to their phone. And that visual distortion that goes with the neck as well.", null, " We are exaggerating the size and the characteristics of it. But we're also making this entire image visually distorted by doing something really unrealistic. Little creatures don't ride people with saddles. So depicting that within this is going to be visual distortion. It's making something new.", null, " It's making something new. We have the characteristics of a normal picture of somebody leaning over and looking at their phone. We have situational irony. You better believe it. The situational irony here: we are supposed to be controlling the Pokemon." ], "text_ocr_list": [ null, null, "We can see these text from the image: Situational Irony\n\n- When the OPPOSITE of what is expected in a given situation occurs\n- SI allows writers and artists to create contrasts between expectations and reality.\n See what I'm saying, gang? We usually use situational irony to allow writers and artists to create contrasts between expectations and reality. Once again, find your own irony. Sorry, I'm drinking some tea. Find your own example and post it in the class Padlet.", null, "We can see these text from the image: Kuczynski\n\n1. Hyperbole?\n2. Understatement?\n3. Caricature?\n4. Visual Distortion?\n5. Allegory?\n6. Situational Irony?.\n But let's take a look at Kaczynski's work and see if we can find all of these different techniques in there. We're looking for hyperbole, understatement, caricature, visual distortion, allegory, and situational irony.", null, "We can see these text from the image: Kuczynski\n\n1. Hyperbole?\n2. Understatement?\n3. Caricature?\n4. Visual Distortion?\n5. Allegory?\n6. Situational Irony?.\n If we look at our piece right here on the right, it's called Pokemon Go. I think this is a good piece because it really does contain every single one of these things. So let's take a look at these techniques and let's talk about how.", null, "We can see these text from the image: Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n\n2. Understatement?\n a. The characters placid expression\n\n3. Caricature?\n a. The long stooped neck\n\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n\n6. Allegory?\n a. \"We should stop being controlled by our phones.\".\n So hyperbole. There's an obvious exaggeration here of people staring down at their phones. His neck is cranked really long. He's staring directly down. This is a clear exaggeration of someone whose neck is cranked over all the time.", null, "We can see these text from the image: Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n2. Understatement?\n a. The characters placid expression\n3. Caricature?\n a. The long stooped neck\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n6. Allegory?\n a. \"We should stop being controlled by our phones.\".\n But there's also understatement. If we look at the subject's face right here, we see that it has a very placid expression, as though this is no joke.", null, "We can see these text from the image: Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n2. Understatement?\n a. The characters placid expression\n3. Caricature?\n a. The long stooped neck\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n6. Allegory?\n a. \"We should stop being controlled by our phones.\".\n This is no big deal. This is completely normal. Nothing happening here. Despite the fact that he has a, I don't know, magical little creature on his back. I would call that an understatement given the circumstances, trying to show us how normal it is that we are completely subject to our phones.", null, "We can see these text from the image: Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n\n2. Understatement?\n a. The characters placid expression\n\n3. Caricature?\n a. The long stooped neck\n\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n\n6. Allegory?\n a. \"We should stop being controlled by our phones.\".\n We have caricature. We absolutely have caricature. That exaggerated stooped neck is not going to be seen in real people. It's going to be a caricature of this person to exaggerate the fact that they are glued to their phone. And that visual distortion that goes with the neck as well.", null, "We can see these text from the image: Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n\n2. Understatement?\n a. The characters placid expression\n\n3. Caricature?\n a. The long stooped neck\n\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n\n6. Allegory?\n a. \"We should stop being controlled by our phones.\".\n We are exaggerating the size and the characteristics of it. But we're also making this entire image visually distorted by doing something really unrealistic. Little creatures don't ride people with saddles. So depicting that within this is going to be visual distortion. It's making something new.", null, "We can see these text from the image: Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n2. Understatement?\n a. The characters placid expression\n3. Caricature?\n a. The long stooped neck\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n6. Allegory?\n a. \"We should stop being controlled by our phones.\".\n It's making something new. We have the characteristics of a normal picture of somebody leaning over and looking at their phone. We have situational irony. You better believe it. The situational irony here: we are supposed to be controlling the Pokemon." ], "metadata": [ { "vid": "ehjFf-mMq3Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Analyzing literary terms satire and crack._20.json#####audio#####doingASR#####FinishASR/ehjFf-mMq3Q/427.44_445.44.mp4", "refined_asr": " See what I'm saying, gang? We usually use situational irony to allow writers and artists to create contrasts between expectations and reality. Once again, find your own irony. Sorry, I'm drinking some tea. Find your own example and post it in the class Padlet.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@427.44_445.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@427.44_445.44#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@427.44_445.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@427.44_445.44#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@427.44_445.44#3.jpg" ], "ocr_qwen2_vl_72b": "Situational Irony\n\n- When the OPPOSITE of what is expected in a given situation occurs\n- SI allows writers and artists to create contrasts between expectations and reality" }, { "vid": "ehjFf-mMq3Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Analyzing literary terms satire and crack._20.json#####audio#####doingASR#####FinishASR/ehjFf-mMq3Q/445.44_457.44.mp4", "refined_asr": " But let's take a look at Kaczynski's work and see if we can find all of these different techniques in there. We're looking for hyperbole, understatement, caricature, visual distortion, allegory, and situational irony.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@445.44_457.44#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@445.44_457.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@445.44_457.44#2.jpg" ], "ocr_qwen2_vl_72b": "Kuczynski\n\n1. Hyperbole?\n2. Understatement?\n3. Caricature?\n4. Visual Distortion?\n5. Allegory?\n6. Situational Irony?" }, { "vid": "ehjFf-mMq3Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Analyzing literary terms satire and crack._20.json#####audio#####doingASR#####FinishASR/ehjFf-mMq3Q/457.44_469.44.mp4", "refined_asr": " If we look at our piece right here on the right, it's called Pokemon Go. I think this is a good piece because it really does contain every single one of these things. So let's take a look at these techniques and let's talk about how.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@457.44_469.44#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@457.44_469.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@457.44_469.44#2.jpg" ], "ocr_qwen2_vl_72b": "Kuczynski\n\n1. Hyperbole?\n2. Understatement?\n3. Caricature?\n4. Visual Distortion?\n5. Allegory?\n6. Situational Irony?" }, { "vid": "ehjFf-mMq3Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Analyzing literary terms satire and crack._20.json#####audio#####doingASR#####FinishASR/ehjFf-mMq3Q/469.44_485.44.mp4", "refined_asr": " So hyperbole. There's an obvious exaggeration here of people staring down at their phones. His neck is cranked really long. He's staring directly down. This is a clear exaggeration of someone whose neck is cranked over all the time.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@469.44_485.44#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@469.44_485.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@469.44_485.44#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@469.44_485.44#3.jpg" ], "ocr_qwen2_vl_72b": "Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n\n2. Understatement?\n a. The characters placid expression\n\n3. Caricature?\n a. The long stooped neck\n\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n\n6. Allegory?\n a. \"We should stop being controlled by our phones.\"" }, { "vid": "ehjFf-mMq3Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Analyzing literary terms satire and crack._20.json#####audio#####doingASR#####FinishASR/ehjFf-mMq3Q/485.44_497.44.mp4", "refined_asr": " But there's also understatement. If we look at the subject's face right here, we see that it has a very placid expression, as though this is no joke.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@485.44_497.44#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@485.44_497.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@485.44_497.44#2.jpg" ], "ocr_qwen2_vl_72b": "Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n2. Understatement?\n a. The characters placid expression\n3. Caricature?\n a. The long stooped neck\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n6. Allegory?\n a. \"We should stop being controlled by our phones.\"" }, { "vid": "ehjFf-mMq3Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Analyzing literary terms satire and crack._20.json#####audio#####doingASR#####FinishASR/ehjFf-mMq3Q/497.44_517.44.mp4", "refined_asr": " This is no big deal. This is completely normal. Nothing happening here. Despite the fact that he has a, I don't know, magical little creature on his back. I would call that an understatement given the circumstances, trying to show us how normal it is that we are completely subject to our phones.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@497.44_517.44#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@497.44_517.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@497.44_517.44#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@497.44_517.44#3.jpg" ], "ocr_qwen2_vl_72b": "Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n2. Understatement?\n a. The characters placid expression\n3. Caricature?\n a. The long stooped neck\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n6. Allegory?\n a. \"We should stop being controlled by our phones.\"" }, { "vid": "ehjFf-mMq3Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Analyzing literary terms satire and crack._20.json#####audio#####doingASR#####FinishASR/ehjFf-mMq3Q/517.44_535.44.mp4", "refined_asr": " We have caricature. We absolutely have caricature. That exaggerated stooped neck is not going to be seen in real people. It's going to be a caricature of this person to exaggerate the fact that they are glued to their phone. And that visual distortion that goes with the neck as well.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@517.44_535.44#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@517.44_535.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@517.44_535.44#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@517.44_535.44#3.jpg" ], "ocr_qwen2_vl_72b": "Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n\n2. Understatement?\n a. The characters placid expression\n\n3. Caricature?\n a. The long stooped neck\n\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n\n6. Allegory?\n a. \"We should stop being controlled by our phones.\"" }, { "vid": "ehjFf-mMq3Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Analyzing literary terms satire and crack._20.json#####audio#####doingASR#####FinishASR/ehjFf-mMq3Q/535.44_555.44.mp4", "refined_asr": " We are exaggerating the size and the characteristics of it. But we're also making this entire image visually distorted by doing something really unrealistic. Little creatures don't ride people with saddles. So depicting that within this is going to be visual distortion. It's making something new.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@535.44_555.44#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@535.44_555.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@535.44_555.44#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@535.44_555.44#3.jpg" ], "ocr_qwen2_vl_72b": "Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n\n2. Understatement?\n a. The characters placid expression\n\n3. Caricature?\n a. The long stooped neck\n\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n\n6. Allegory?\n a. \"We should stop being controlled by our phones.\"" }, { "vid": "ehjFf-mMq3Q.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Analyzing literary terms satire and crack._20.json#####audio#####doingASR#####FinishASR/ehjFf-mMq3Q/555.44_567.44.mp4", "refined_asr": " It's making something new. We have the characteristics of a normal picture of somebody leaning over and looking at their phone. We have situational irony. You better believe it. The situational irony here: we are supposed to be controlling the Pokemon.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@555.44_567.44#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@555.44_567.44#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/ehjFf-mMq3Q/ehjFf-mMq3Q@555.44_567.44#2.jpg" ], "ocr_qwen2_vl_72b": "Kuczynski\n\n1. Hyperbole?\n a. An obvious exaggeration of people staring down at their phone\n2. Understatement?\n a. The characters placid expression\n3. Caricature?\n a. The long stooped neck\n4. Visual Distortion?\n a. The neck and the critter riding the neck\n5. Situational Irony?\n a. We're supposed to be controlling the Pokemon- not the other way around\n6. Allegory?\n a. \"We should stop being controlled by our phones.\"" } ], "image_num": 10, "text_num": 552, "token_num": 6312 }, { "images": [ "sample_100_images/VkY_lDAIODw@183.98_200.72#1.jpg", null, "sample_100_images/VkY_lDAIODw@200.72_226.14#1.jpg", null, "sample_100_images/VkY_lDAIODw@226.14_239.96#1.jpg", null, "sample_100_images/VkY_lDAIODw@239.96_259.8#1.jpg", null, "sample_100_images/VkY_lDAIODw@259.8_268.52#1.jpg", null, "sample_100_images/VkY_lDAIODw@268.52_286.24#1.jpg", "sample_100_images/VkY_lDAIODw@268.52_286.24#2.jpg", null, "sample_100_images/VkY_lDAIODw@286.24_310.46#1.jpg", null ], "texts": [ null, " So the y-intercept is the y-intercept. For this equation, what would be the value of the slope and the value of the y-intercept? Well, remember, the slope is the value in front of the x when y is isolated. So that would be 5.", null, " And the y-intercept would be negative one. Good, now that was pretty easy, right? Well then, let's imagine that we've got a linear equation like this one. Two different students give their answers to what they think the slope and y-intercept for this equation would be. It's now your turn to determine who has the right answer.", null, " The wrong answer. Well in this case Student B has the right answer. Now it's important to be careful here. Some people might assume that the slope and y-intercept would be negative 1 and the y-intercept would be negative 2. So we shouldn't assume right away that the", null, " The coefficient, the number right in front of the x variable, is automatically the slope. But this is only the case when we've already isolated the equation to have the y variable alone on the left side. In this situation, we've got to first divide both sides by 4 to isolate for y.", null, " Following of course. Now, our slope would simplify down to just one over two. And our y-intercept would be one over two. The slope would be negative one and the y-intercept would be negative one. The slope would be negative one and the y-intercept would be negative two. The slope would be negative one and the", null, null, " The slope would be negative 2, and the slope would be negative 3. Minus 2 over 4. Now we know that the y-intercept can be identified by the b variable here, and that this is the point where the line intersects the y-axis when x is equal to zero.", null, " If we wanted to actually test this out we would substitute zero for x and notice how regardless of what the slope would be we would get the y-intercept. What this means is that if we were given a table of coordinate values and one of the points had an x value of zero then we can expect that the y value to automatically be the y-intercept." ], "text_ocr_list": [ null, "We can see these text from the image: y = 5x - 1.\n So the y-intercept is the y-intercept. For this equation, what would be the value of the slope and the value of the y-intercept? Well, remember, the slope is the value in front of the x when y is isolated. So that would be 5.", null, "We can see these text from the image: y = 5x - 1\n\nSlope = 5\n\nY-intercept = -1.\n And the y-intercept would be negative one. Good, now that was pretty easy, right? Well then, let's imagine that we've got a linear equation like this one. Two different students give their answers to what they think the slope and y-intercept for this equation would be. It's now your turn to determine who has the right answer.", null, "We can see these text from the image: 4y = 2x + 1\n\nStudent A\nSlope = 2\nY-intercept = 1\n\nStudent B\nSlope = 1/2\nY-intercept = 1/4.\n The wrong answer. Well in this case Student B has the right answer. Now it's important to be careful here. Some people might assume that the slope and y-intercept would be negative 1 and the y-intercept would be negative 2. So we shouldn't assume right away that the", null, "We can see these text from the image: 4y = 2x + 1\n\nStudent B\n\nSlope = 1/2\n\nY-intercept = 1/4.\n The coefficient, the number right in front of the x variable, is automatically the slope. But this is only the case when we've already isolated the equation to have the y variable alone on the left side. In this situation, we've got to first divide both sides by 4 to isolate for y.", null, "We can see these text from the image: 4y = 2x + 1\n\ny = \\frac{2x}{4} + \\frac{1}{4}\n\nStudent B\n\nSlope = \\frac{1}{2}\n\nY-intercept = \\frac{1}{4}.\n Following of course. Now, our slope would simplify down to just one over two. And our y-intercept would be one over two. The slope would be negative one and the y-intercept would be negative one. The slope would be negative one and the y-intercept would be negative two. The slope would be negative one and the", null, null, "We can see these text from the image: Y-INTERCEPT\n\ny = mx + b.\n The slope would be negative 2, and the slope would be negative 3. Minus 2 over 4. Now we know that the y-intercept can be identified by the b variable here, and that this is the point where the line intersects the y-axis when x is equal to zero.", null, "We can see these text from the image: Y-INTERCEPT\n\ny = mx + b\n\ny = m(0) + b\n\nY-intercept\n\nb.\n If we wanted to actually test this out we would substitute zero for x and notice how regardless of what the slope would be we would get the y-intercept. What this means is that if we were given a table of coordinate values and one of the points had an x value of zero then we can expect that the y value to automatically be the y-intercept." ], "metadata": [ { "vid": "VkY_lDAIODw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Linear equation x-intercept to y-intercept_20.json#####audio#####doingASR#####FinishASR/VkY_lDAIODw/183.98_200.72.mp4", "refined_asr": " So the y-intercept is the y-intercept. For this equation, what would be the value of the slope and the value of the y-intercept? Well, remember, the slope is the value in front of the x when y is isolated. So that would be 5.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@183.98_200.72#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@183.98_200.72#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@183.98_200.72#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@183.98_200.72#3.jpg" ], "ocr_qwen2_vl_72b": "y = 5x - 1" }, { "vid": "VkY_lDAIODw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Linear equation x-intercept to y-intercept_20.json#####audio#####doingASR#####FinishASR/VkY_lDAIODw/200.72_226.14.mp4", "refined_asr": " And the y-intercept would be negative one. Good, now that was pretty easy, right? Well then, let's imagine that we've got a linear equation like this one. Two different students give their answers to what they think the slope and y-intercept for this equation would be. It's now your turn to determine who has the right answer.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@200.72_226.14#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@200.72_226.14#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@200.72_226.14#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@200.72_226.14#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@200.72_226.14#4.jpg" ], "ocr_qwen2_vl_72b": "y = 5x - 1\n\nSlope = 5\n\nY-intercept = -1" }, { "vid": "VkY_lDAIODw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Linear equation x-intercept to y-intercept_20.json#####audio#####doingASR#####FinishASR/VkY_lDAIODw/226.14_239.96.mp4", "refined_asr": " The wrong answer. Well in this case Student B has the right answer. Now it's important to be careful here. Some people might assume that the slope and y-intercept would be negative 1 and the y-intercept would be negative 2. So we shouldn't assume right away that the", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@226.14_239.96#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@226.14_239.96#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@226.14_239.96#2.jpg" ], "ocr_qwen2_vl_72b": "4y = 2x + 1\n\nStudent A\nSlope = 2\nY-intercept = 1\n\nStudent B\nSlope = 1/2\nY-intercept = 1/4" }, { "vid": "VkY_lDAIODw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Linear equation x-intercept to y-intercept_20.json#####audio#####doingASR#####FinishASR/VkY_lDAIODw/239.96_259.8.mp4", "refined_asr": " The coefficient, the number right in front of the x variable, is automatically the slope. But this is only the case when we've already isolated the equation to have the y variable alone on the left side. In this situation, we've got to first divide both sides by 4 to isolate for y.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@239.96_259.8#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@239.96_259.8#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@239.96_259.8#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@239.96_259.8#3.jpg" ], "ocr_qwen2_vl_72b": "4y = 2x + 1\n\nStudent B\n\nSlope = 1/2\n\nY-intercept = 1/4" }, { "vid": "VkY_lDAIODw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Linear equation x-intercept to y-intercept_20.json#####audio#####doingASR#####FinishASR/VkY_lDAIODw/259.8_268.52.mp4", "refined_asr": " Following of course. Now, our slope would simplify down to just one over two. And our y-intercept would be one over two. The slope would be negative one and the y-intercept would be negative one. The slope would be negative one and the y-intercept would be negative two. The slope would be negative one and the", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@259.8_268.52#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@259.8_268.52#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@259.8_268.52#2.jpg" ], "ocr_qwen2_vl_72b": "4y = 2x + 1\n\ny = \\frac{2x}{4} + \\frac{1}{4}\n\nStudent B\n\nSlope = \\frac{1}{2}\n\nY-intercept = \\frac{1}{4}" }, { "vid": "VkY_lDAIODw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Linear equation x-intercept to y-intercept_20.json#####audio#####doingASR#####FinishASR/VkY_lDAIODw/268.52_286.24.mp4", "refined_asr": " The slope would be negative 2, and the slope would be negative 3. Minus 2 over 4. Now we know that the y-intercept can be identified by the b variable here, and that this is the point where the line intersects the y-axis when x is equal to zero.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@268.52_286.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@268.52_286.24#2.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@268.52_286.24#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@268.52_286.24#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@268.52_286.24#3.jpg" ], "ocr_qwen2_vl_72b": "Y-INTERCEPT\n\ny = mx + b" }, { "vid": "VkY_lDAIODw.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Linear equation x-intercept to y-intercept_20.json#####audio#####doingASR#####FinishASR/VkY_lDAIODw/286.24_310.46.mp4", "refined_asr": " If we wanted to actually test this out we would substitute zero for x and notice how regardless of what the slope would be we would get the y-intercept. What this means is that if we were given a table of coordinate values and one of the points had an x value of zero then we can expect that the y value to automatically be the y-intercept.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@286.24_310.46#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@286.24_310.46#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@286.24_310.46#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@286.24_310.46#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/VkY_lDAIODw/VkY_lDAIODw@286.24_310.46#4.jpg" ], "ocr_qwen2_vl_72b": "Y-INTERCEPT\n\ny = mx + b\n\ny = m(0) + b\n\nY-intercept\n\nb" } ], "image_num": 8, "text_num": 485, "token_num": 5093 }, { "images": [ "sample_100_images/5JGjELKJYD0@158.76000000000002_193.18#1.jpg", null, "sample_100_images/5JGjELKJYD0@193.18_219.4#1.jpg", null, "sample_100_images/5JGjELKJYD0@219.4_233.60000000000002#1.jpg", null, "sample_100_images/5JGjELKJYD0@233.70000000000002_256.08000000000004#1.jpg", null, "sample_100_images/5JGjELKJYD0@256.08000000000004_284.58#1.jpg", null, "sample_100_images/5JGjELKJYD0@284.58_305.53999999999996#1.jpg", null ], "texts": [ null, " We know that or assume it's divisible by 8. Okay. Divisible by 8. Now, on this basis, we are going to show that this is going to be divisible by 8. Now, I said this was slightly different to what we've done in the past. When we're doing these divisibility tests or multiple tests, what we", null, " Always look at is f of k plus 1 minus f of k. Alright. And what I'm going to do now is substitute for what f of k plus 1 would be. So that would be replacing the n here with k plus 1. So we're going to get three to the power of two lots of k plus 1.", null, " And then that's minus one and then we've got minus f of k. So that's going to be minus three to the two K minus one. So I'll put that in brackets though: minus three to the power two K.", null, " Minus 1. Now let's just tidy this up. What we've got here, if we were to expand the brackets, is 3 to the power 2k plus 2. And then we've got minus 1 here and then minus 3 to the power 2k.", null, " And then plus one. And then if we continue down here let's just write this out again. We therefore have F of K plus one minus F of K equals right now the minus one and the one that cancels out. But here when we're adding the powers it's the same.", null, " As doing 3 squared multiplied by 3 to the power 2k. 3 squared, then multiplied by 3 to the power 2k. And then we've got this term here: minus 3 to the power 2k. Well, 3 squared is 9." ], "text_ocr_list": [ null, "We can see these text from the image: Show that \\(3^{2n} - 1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\n\u2234 true when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\u2234 \\(f(k) = 3^{2k} - 1\\) is.\n We know that or assume it's divisible by 8. Okay. Divisible by 8. Now, on this basis, we are going to show that this is going to be divisible by 8. Now, I said this was slightly different to what we've done in the past. When we're doing these divisibility tests or multiple tests, what we", null, "We can see these text from the image: Show that \\(3^{2n} - 1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\n\u2234 true when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\u2234 \\(f(k) = 3^{2k} - 1\\) is divisible by 8\n\nWhen \\(n = k + 1\\)\n\n\\(f(k)\\).\n Always look at is f of k plus 1 minus f of k. Alright. And what I'm going to do now is substitute for what f of k plus 1 would be. So that would be replacing the n here with k plus 1. So we're going to get three to the power of two lots of k plus 1.", null, "We can see these text from the image: Show that \\(3^{2n} - 1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\n\u2234 true when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\u2234 \\(f(k) = 3^{2k} - 1\\) is divisible by 8\n\nWhen \\(n = k + 1\\)\n\n\\(f(k + 1) = 3^{2(k + 1)} - 1\\).\n And then that's minus one and then we've got minus f of k. So that's going to be minus three to the two K minus one. So I'll put that in brackets though: minus three to the power two K.", null, "We can see these text from the image: Show that \\(3^{2n}-1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\n\u2234 true when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\u2234 \\(f(k) = 3^{2k} - 1\\) is divisible by 8\n\nWhen \\(n = k + 1\\)\n\n\\(f(k+1) - f(k) = 3^{2(k+1)} - 1 - (3^{2k} - 1)\\).\n Minus 1. Now let's just tidy this up. What we've got here, if we were to expand the brackets, is 3 to the power 2k plus 2. And then we've got minus 1 here and then minus 3 to the power 2k.", null, "We can see these text from the image: Show that \\(3^{2n} - 1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\n\u2234 true when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\u2234 \\(f(k) = 3^{2k} - 1\\) is divisible by 8\n\nWhen \\(n = k + 1\\)\n\n\\(f(k + 1) - f(k) = 3^{2(k+1)} - 1 - (3^{2k} - 1)\\)\n\n\\(= 3^{2k+2} - 1 - 3^{2k} + 1\\).\n And then plus one. And then if we continue down here let's just write this out again. We therefore have F of K plus one minus F of K equals right now the minus one and the one that cancels out. But here when we're adding the powers it's the same.", null, "We can see these text from the image: Show that \\(3^{2n} - 1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\nTrue when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\\(f(k) = 3^{2k} - 1\\) is divisible by 8\n\nWhen \\(n = k + 1\\)\n\n\\(f(k + 1) - f(k) = 3^{2(k+1)} - 1 - (3^{2k} - 1)\\)\n\n\\(= 3^{2k+2} - 1 - 3^{2k} + 1\\).\n As doing 3 squared multiplied by 3 to the power 2k. 3 squared, then multiplied by 3 to the power 2k. And then we've got this term here: minus 3 to the power 2k. Well, 3 squared is 9." ], "metadata": [ { "vid": "5JGjELKJYD0.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical induction for powers of integers_20.json#####audio#####doingASR#####FinishASR/5JGjELKJYD0/158.76000000000002_193.18.mp4", "refined_asr": " We know that or assume it's divisible by 8. Okay. Divisible by 8. Now, on this basis, we are going to show that this is going to be divisible by 8. Now, I said this was slightly different to what we've done in the past. When we're doing these divisibility tests or multiple tests, what we", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@158.76000000000002_193.18#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@158.76000000000002_193.18#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@158.76000000000002_193.18#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@158.76000000000002_193.18#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@158.76000000000002_193.18#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@158.76000000000002_193.18#5.jpg" ], "ocr_qwen2_vl_72b": "Show that \\(3^{2n} - 1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\n\u2234 true when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\u2234 \\(f(k) = 3^{2k} - 1\\) is" }, { "vid": "5JGjELKJYD0.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical induction for powers of integers_20.json#####audio#####doingASR#####FinishASR/5JGjELKJYD0/193.18_219.4.mp4", "refined_asr": " Always look at is f of k plus 1 minus f of k. Alright. And what I'm going to do now is substitute for what f of k plus 1 would be. So that would be replacing the n here with k plus 1. So we're going to get three to the power of two lots of k plus 1.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@193.18_219.4#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@193.18_219.4#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@193.18_219.4#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@193.18_219.4#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@193.18_219.4#4.jpg" ], "ocr_qwen2_vl_72b": "Show that \\(3^{2n} - 1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\n\u2234 true when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\u2234 \\(f(k) = 3^{2k} - 1\\) is divisible by 8\n\nWhen \\(n = k + 1\\)\n\n\\(f(k)\\)" }, { "vid": "5JGjELKJYD0.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical induction for powers of integers_20.json#####audio#####doingASR#####FinishASR/5JGjELKJYD0/219.4_233.60000000000002.mp4", "refined_asr": " And then that's minus one and then we've got minus f of k. So that's going to be minus three to the two K minus one. So I'll put that in brackets though: minus three to the power two K.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@219.4_233.60000000000002#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@219.4_233.60000000000002#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@219.4_233.60000000000002#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@219.4_233.60000000000002#3.jpg" ], "ocr_qwen2_vl_72b": "Show that \\(3^{2n} - 1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\n\u2234 true when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\u2234 \\(f(k) = 3^{2k} - 1\\) is divisible by 8\n\nWhen \\(n = k + 1\\)\n\n\\(f(k + 1) = 3^{2(k + 1)} - 1\\)" }, { "vid": "5JGjELKJYD0.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical induction for powers of integers_20.json#####audio#####doingASR#####FinishASR/5JGjELKJYD0/233.70000000000002_256.08000000000004.mp4", "refined_asr": " Minus 1. Now let's just tidy this up. What we've got here, if we were to expand the brackets, is 3 to the power 2k plus 2. And then we've got minus 1 here and then minus 3 to the power 2k.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@233.70000000000002_256.08000000000004#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@233.70000000000002_256.08000000000004#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@233.70000000000002_256.08000000000004#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@233.70000000000002_256.08000000000004#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@233.70000000000002_256.08000000000004#4.jpg" ], "ocr_qwen2_vl_72b": "Show that \\(3^{2n}-1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\n\u2234 true when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\u2234 \\(f(k) = 3^{2k} - 1\\) is divisible by 8\n\nWhen \\(n = k + 1\\)\n\n\\(f(k+1) - f(k) = 3^{2(k+1)} - 1 - (3^{2k} - 1)\\)" }, { "vid": "5JGjELKJYD0.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical induction for powers of integers_20.json#####audio#####doingASR#####FinishASR/5JGjELKJYD0/256.08000000000004_284.58.mp4", "refined_asr": " And then plus one. And then if we continue down here let's just write this out again. We therefore have F of K plus one minus F of K equals right now the minus one and the one that cancels out. But here when we're adding the powers it's the same.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@256.08000000000004_284.58#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@256.08000000000004_284.58#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@256.08000000000004_284.58#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@256.08000000000004_284.58#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@256.08000000000004_284.58#4.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@256.08000000000004_284.58#5.jpg" ], "ocr_qwen2_vl_72b": "Show that \\(3^{2n} - 1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\n\u2234 true when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\u2234 \\(f(k) = 3^{2k} - 1\\) is divisible by 8\n\nWhen \\(n = k + 1\\)\n\n\\(f(k + 1) - f(k) = 3^{2(k+1)} - 1 - (3^{2k} - 1)\\)\n\n\\(= 3^{2k+2} - 1 - 3^{2k} + 1\\)" }, { "vid": "5JGjELKJYD0.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical induction for powers of integers_20.json#####audio#####doingASR#####FinishASR/5JGjELKJYD0/284.58_305.53999999999996.mp4", "refined_asr": " As doing 3 squared multiplied by 3 to the power 2k. 3 squared, then multiplied by 3 to the power 2k. And then we've got this term here: minus 3 to the power 2k. Well, 3 squared is 9.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@284.58_305.53999999999996#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@284.58_305.53999999999996#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@284.58_305.53999999999996#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@284.58_305.53999999999996#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/5JGjELKJYD0/5JGjELKJYD0@284.58_305.53999999999996#4.jpg" ], "ocr_qwen2_vl_72b": "Show that \\(3^{2n} - 1\\) is divisible by 8 for all positive integers.\n\nLet \\(f(n) = 3^{2n} - 1\\)\n\n\\(f(1) = 3^2 - 1 = 8\\)\n\nTrue when \\(n = 1\\)\n\nAssume true for \\(n = k\\)\n\n\\(f(k) = 3^{2k} - 1\\) is divisible by 8\n\nWhen \\(n = k + 1\\)\n\n\\(f(k + 1) - f(k) = 3^{2(k+1)} - 1 - (3^{2k} - 1)\\)\n\n\\(= 3^{2k+2} - 1 - 3^{2k} + 1\\)" } ], "image_num": 6, "text_num": 405, "token_num": 3861 }, { "images": [ "sample_100_images/AD-KCP2yolU@7804.4_7820.320000000001#1.jpg", null, "sample_100_images/AD-KCP2yolU@7820.320000000001_7836.38#1.jpg", null, "sample_100_images/AD-KCP2yolU@7836.38_7849.9400000000005#1.jpg", null, "sample_100_images/AD-KCP2yolU@7849.9400000000005_7862.6#1.jpg", null, "sample_100_images/AD-KCP2yolU@7862.6_7882.5599999999995#1.jpg", null, "sample_100_images/AD-KCP2yolU@7882.5599999999995_7907.4400000000005#1.jpg", null, "sample_100_images/AD-KCP2yolU@7908.72_7928.64#1.jpg", null, "sample_100_images/AD-KCP2yolU@7929.4400000000005_7942.56#1.jpg", null ], "texts": [ null, " So 0 delta 0 will be equal to 0 times 0 plus 2. That's the first number times the second number plus 2. 0 delta 0 is equal to 0 times 0 plus 2, which is 0.", null, " 0 plus 2 is 2. And 2 mod 5 is 2. The next one is 0 delta 1. 0 delta 1 is equal to 0 times 1 plus 2. This will give us 0 plus 2, and 0 plus 2 equals 2.", null, " Modulo 5 of 2 is 2. The next one is 0 delta 2. 0 delta 2 is equal to 0 times 1 plus 2. This will give us 0 plus 2. 0 plus 2 is 2.", null, " 2 mod 5 is 2. The next one is 0 delta 3. 0 delta 3 is equal to 0 times 3 plus 2. 0 times 3 is 0. 0 plus 2 is 2.", null, " 2 mod 5 is 2. The next one is 0 delta 4. 0 delta 4 is equal to 0 times 4 plus 2. 0 times 4 is 0. Zero plus 2 will give us 2. 2 mod 5 is 2. Let's move on to the next rule. You have 1 delta.", null, " Zero. One delta zero is equal to one times zero plus two. This will give us zero plus two, and zero plus two is two. Two mod five is two. The next one is one delta one. One delta one is equal to one times one plus two. This will give us one plus two. One plus two is three. Three mod five is three.", null, " The next one is one delta two. One delta two is equal to one times two plus two. This will give us two plus two. Two plus two is four, and four mod five is four. The next one is one delta three. One delta three is equal to one times three plus two.", null, " This will give us 3 plus 2. 3 plus 2 is 5. 5 multiplied by 5 is 25. The next one is 1 delta 4. 1 delta 4 is equal to 1 times 4 plus 2." ], "text_ocr_list": [ null, "We can see these text from the image: Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(0 \\Delta 0 = 0 \\times 0 + 2\\).\n So 0 delta 0 will be equal to 0 times 0 plus 2. That's the first number times the second number plus 2. 0 delta 0 is equal to 0 times 0 plus 2, which is 0.", null, "We can see these text from the image: Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(0 \\Delta 0 = 0 \\times 0 + 2\\)\n\n\\(= 0 + 2\\)\n\n\\(= 2\\)\n\n\\(2 \\mod 5 = 2\\).\n 0 plus 2 is 2. And 2 mod 5 is 2. The next one is 0 delta 1. 0 delta 1 is equal to 0 times 1 plus 2. This will give us 0 plus 2, and 0 plus 2 equals 2.", null, "We can see these text from the image: Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\(a \\Delta b = ab + 2\\).\n Modulo 5 of 2 is 2. The next one is 0 delta 2. 0 delta 2 is equal to 0 times 1 plus 2. This will give us 0 plus 2. 0 plus 2 is 2.", null, "We can see these text from the image: Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\[\n\\begin{array}{c|ccccc}\n\\Delta & 0 & 1 & 2 & 3 & 4 \\\\\n\\hline\n0 & 2 & 2 & 2 & 2 & 2 \\\\\n1 & & & & & \\\\\n2 & & & & & \\\\\n3 & & & & & \\\\\n4 & & & & & \n\\end{array}\n\\]\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(0 \\Delta 2 = 0 \\times 2 + 2\\)\n\n\\(= 0 + 2\\)\n\n\\(= 2\\)\n\n\\(2 \\mod 5 = 2\\)\n\n\\(0 \\Delta 3 = 0 \\times 3 + 2\\).\n 2 mod 5 is 2. The next one is 0 delta 3. 0 delta 3 is equal to 0 times 3 plus 2. 0 times 3 is 0. 0 plus 2 is 2.", null, "We can see these text from the image: Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\[\n\\begin{array}{c|ccccc}\n\\Delta & 0 & 1 & 2 & 3 & 4 \\\\\n\\hline\n0 & 2 & 2 & 2 & 2 & 2 \\\\\n1 & & & & & \\\\\n2 & & & & & \\\\\n3 & & & & & \\\\\n4 & & & & &\n\\end{array}\n\\]\n\n\\(a \\Delta b = ab + 2\\).\n 2 mod 5 is 2. The next one is 0 delta 4. 0 delta 4 is equal to 0 times 4 plus 2. 0 times 4 is 0. Zero plus 2 will give us 2. 2 mod 5 is 2. Let's move on to the next rule. You have 1 delta.", null, "We can see these text from the image: Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\[\n\\begin{array}{c|ccccc}\n\\Delta & 0 & 1 & 2 & 3 & 4 \\\\\n\\hline\n0 & 2 & 2 & 2 & 2 & 2 \\\\\n1 & & & & & \\\\\n2 & & & & & \\\\\n3 & & & & & \\\\\n4 & & & & &\n\\end{array}\n\\]\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(0 \\Delta 4 = 0 \\times 4 + 2\\)\n\n\\(= 0 + 2\\)\n\n\\(= 2\\)\n\n\\(2 \\mod 5 = 2\\)\n\n\\(1 \\Delta 0 = 1 \\times 0 + 2\\).\n Zero. One delta zero is equal to one times zero plus two. This will give us zero plus two, and zero plus two is two. Two mod five is two. The next one is one delta one. One delta one is equal to one times one plus two. This will give us one plus two. One plus two is three. Three mod five is three.", null, "We can see these text from the image: Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\[\n\\begin{array}{c|ccccc}\n\\Delta & 0 & 1 & 2 & 3 & 4 \\\\\n\\hline\n0 & 2 & 2 & 2 & 2 & 2 \\\\\n1 & 2 & 3 & & & \\\\\n2 & & & & & \\\\\n3 & & & & & \\\\\n4 & & & & &\n\\end{array}\n\\]\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(1 \\Delta 1 = 1 \\times 1 + 2\\)\n\n\\(= 1 + 2\\)\n\n\\(= 3\\)\n\n\\(3 \\mod 5 = 3\\)\n\n\\(1 \\Delta 2 = 1 \\times 2 + 2\\).\n The next one is one delta two. One delta two is equal to one times two plus two. This will give us two plus two. Two plus two is four, and four mod five is four. The next one is one delta three. One delta three is equal to one times three plus two.", null, "We can see these text from the image: Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set \\(\\{0, 1, 2, 3, 4\\}\\)\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(1 \\Delta 3 = 1 \\times 3 + 2\\)\n\n\\(= 3 + 2\\)\n\n\\(= 5\\).\n This will give us 3 plus 2. 3 plus 2 is 5. 5 multiplied by 5 is 25. The next one is 1 delta 4. 1 delta 4 is equal to 1 times 4 plus 2." ], "metadata": [ { "vid": "AD-KCP2yolU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical reasoning with modular arithmetic_20.json#####audio#####doingASR#####FinishASR/AD-KCP2yolU/7804.4_7820.320000000001.mp4", "refined_asr": " So 0 delta 0 will be equal to 0 times 0 plus 2. That's the first number times the second number plus 2. 0 delta 0 is equal to 0 times 0 plus 2, which is 0.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7804.4_7820.320000000001#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7804.4_7820.320000000001#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7804.4_7820.320000000001#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7804.4_7820.320000000001#3.jpg" ], "ocr_qwen2_vl_72b": "Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(0 \\Delta 0 = 0 \\times 0 + 2\\)" }, { "vid": "AD-KCP2yolU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical reasoning with modular arithmetic_20.json#####audio#####doingASR#####FinishASR/AD-KCP2yolU/7820.320000000001_7836.38.mp4", "refined_asr": " 0 plus 2 is 2. And 2 mod 5 is 2. The next one is 0 delta 1. 0 delta 1 is equal to 0 times 1 plus 2. This will give us 0 plus 2, and 0 plus 2 equals 2.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7820.320000000001_7836.38#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7820.320000000001_7836.38#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7820.320000000001_7836.38#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7820.320000000001_7836.38#3.jpg" ], "ocr_qwen2_vl_72b": "Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(0 \\Delta 0 = 0 \\times 0 + 2\\)\n\n\\(= 0 + 2\\)\n\n\\(= 2\\)\n\n\\(2 \\mod 5 = 2\\)" }, { "vid": "AD-KCP2yolU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical reasoning with modular arithmetic_20.json#####audio#####doingASR#####FinishASR/AD-KCP2yolU/7836.38_7849.9400000000005.mp4", "refined_asr": " Modulo 5 of 2 is 2. The next one is 0 delta 2. 0 delta 2 is equal to 0 times 1 plus 2. This will give us 0 plus 2. 0 plus 2 is 2.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7836.38_7849.9400000000005#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7836.38_7849.9400000000005#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7836.38_7849.9400000000005#2.jpg" ], "ocr_qwen2_vl_72b": "Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\(a \\Delta b = ab + 2\\)" }, { "vid": "AD-KCP2yolU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical reasoning with modular arithmetic_20.json#####audio#####doingASR#####FinishASR/AD-KCP2yolU/7849.9400000000005_7862.6.mp4", "refined_asr": " 2 mod 5 is 2. The next one is 0 delta 3. 0 delta 3 is equal to 0 times 3 plus 2. 0 times 3 is 0. 0 plus 2 is 2.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7849.9400000000005_7862.6#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7849.9400000000005_7862.6#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7849.9400000000005_7862.6#2.jpg" ], "ocr_qwen2_vl_72b": "Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\[\n\\begin{array}{c|ccccc}\n\\Delta & 0 & 1 & 2 & 3 & 4 \\\\\n\\hline\n0 & 2 & 2 & 2 & 2 & 2 \\\\\n1 & & & & & \\\\\n2 & & & & & \\\\\n3 & & & & & \\\\\n4 & & & & & \n\\end{array}\n\\]\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(0 \\Delta 2 = 0 \\times 2 + 2\\)\n\n\\(= 0 + 2\\)\n\n\\(= 2\\)\n\n\\(2 \\mod 5 = 2\\)\n\n\\(0 \\Delta 3 = 0 \\times 3 + 2\\)" }, { "vid": "AD-KCP2yolU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical reasoning with modular arithmetic_20.json#####audio#####doingASR#####FinishASR/AD-KCP2yolU/7862.6_7882.5599999999995.mp4", "refined_asr": " 2 mod 5 is 2. The next one is 0 delta 4. 0 delta 4 is equal to 0 times 4 plus 2. 0 times 4 is 0. Zero plus 2 will give us 2. 2 mod 5 is 2. Let's move on to the next rule. You have 1 delta.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7862.6_7882.5599999999995#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7862.6_7882.5599999999995#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7862.6_7882.5599999999995#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7862.6_7882.5599999999995#3.jpg" ], "ocr_qwen2_vl_72b": "Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\[\n\\begin{array}{c|ccccc}\n\\Delta & 0 & 1 & 2 & 3 & 4 \\\\\n\\hline\n0 & 2 & 2 & 2 & 2 & 2 \\\\\n1 & & & & & \\\\\n2 & & & & & \\\\\n3 & & & & & \\\\\n4 & & & & &\n\\end{array}\n\\]\n\n\\(a \\Delta b = ab + 2\\)" }, { "vid": "AD-KCP2yolU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical reasoning with modular arithmetic_20.json#####audio#####doingASR#####FinishASR/AD-KCP2yolU/7882.5599999999995_7907.4400000000005.mp4", "refined_asr": " Zero. One delta zero is equal to one times zero plus two. This will give us zero plus two, and zero plus two is two. Two mod five is two. The next one is one delta one. One delta one is equal to one times one plus two. This will give us one plus two. One plus two is three. Three mod five is three.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7882.5599999999995_7907.4400000000005#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7882.5599999999995_7907.4400000000005#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7882.5599999999995_7907.4400000000005#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7882.5599999999995_7907.4400000000005#3.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7882.5599999999995_7907.4400000000005#4.jpg" ], "ocr_qwen2_vl_72b": "Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\[\n\\begin{array}{c|ccccc}\n\\Delta & 0 & 1 & 2 & 3 & 4 \\\\\n\\hline\n0 & 2 & 2 & 2 & 2 & 2 \\\\\n1 & & & & & \\\\\n2 & & & & & \\\\\n3 & & & & & \\\\\n4 & & & & &\n\\end{array}\n\\]\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(0 \\Delta 4 = 0 \\times 4 + 2\\)\n\n\\(= 0 + 2\\)\n\n\\(= 2\\)\n\n\\(2 \\mod 5 = 2\\)\n\n\\(1 \\Delta 0 = 1 \\times 0 + 2\\)" }, { "vid": "AD-KCP2yolU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical reasoning with modular arithmetic_20.json#####audio#####doingASR#####FinishASR/AD-KCP2yolU/7908.72_7928.64.mp4", "refined_asr": " The next one is one delta two. One delta two is equal to one times two plus two. This will give us two plus two. Two plus two is four, and four mod five is four. The next one is one delta three. One delta three is equal to one times three plus two.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7908.72_7928.64#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7908.72_7928.64#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7908.72_7928.64#2.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7908.72_7928.64#3.jpg" ], "ocr_qwen2_vl_72b": "Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set {0, 1, 2, 3, 4}\n\n\\[\n\\begin{array}{c|ccccc}\n\\Delta & 0 & 1 & 2 & 3 & 4 \\\\\n\\hline\n0 & 2 & 2 & 2 & 2 & 2 \\\\\n1 & 2 & 3 & & & \\\\\n2 & & & & & \\\\\n3 & & & & & \\\\\n4 & & & & &\n\\end{array}\n\\]\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(1 \\Delta 1 = 1 \\times 1 + 2\\)\n\n\\(= 1 + 2\\)\n\n\\(= 3\\)\n\n\\(3 \\mod 5 = 3\\)\n\n\\(1 \\Delta 2 = 1 \\times 2 + 2\\)" }, { "vid": "AD-KCP2yolU.mp4", "clip_path": "/mnt/workspace/zwq_data/youtube_interleave2/video_clip/Mathematical reasoning with modular arithmetic_20.json#####audio#####doingASR#####FinishASR/AD-KCP2yolU/7929.4400000000005_7942.56.mp4", "refined_asr": " This will give us 3 plus 2. 3 plus 2 is 5. 5 multiplied by 5 is 25. The next one is 1 delta 4. 1 delta 4 is equal to 1 times 4 plus 2.", "keyframe_ssim": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7929.4400000000005_7942.56#1.jpg" ], "extracted_frames": [ "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7929.4400000000005_7942.56#1.jpg", "/mnt/workspace/zwq_data/interleaved_dataset/dataset_images_interval_7/AD-KCP2yolU/AD-KCP2yolU@7929.4400000000005_7942.56#2.jpg" ], "ocr_qwen2_vl_72b": "Table for \\(a \\Delta b = ab + 2\\) in mod 5 on the set \\(\\{0, 1, 2, 3, 4\\}\\)\n\n\\(a \\Delta b = ab + 2\\)\n\n\\(1 \\Delta 3 = 1 \\times 3 + 2\\)\n\n\\(= 3 + 2\\)\n\n\\(= 5\\)" } ], "image_num": 8, "text_num": 511, "token_num": 5119 } ]