Wikipedia:Reference desk/Archives/Mathematics/2013 October 25
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October 25
editInverse speed ramp of video
editI've been given a video clip that has been sped up 156%. By what percentage would I need to slow it down to get it back to 100% playback speed? I'm thinking an equality would do it, but I can't think of what to use: 156% over 100 = x over ??? --209.203.125.162 (talk) 17:59, 25 October 2013 (UTC)
- If sped up by 156% it's now at 2.56 times the original speed, so a reduction of 100 X 156/256, ie 60.9% would be needed. If sped up to 156%, ie by 56%, a reduction of 100 X 56/156, ie 35.9% would be needed. In each case "% reduction" = 100 - "% reduced to".86.176.135.38 (talk) 18:59, 25 October 2013 (UTC)
- That's not right. All you need is to take the reciprocal and multiply. For example, a video-tape reel is running at 29 FPS then sped up by a factor of 1.56 (ie: 156%) thus resulting in a new speed of 45.24 FPS. Now if you only knew the current tape speed and the speed-up factor used, you would just multiply 45.24 * (1/1.56) to obtain 29 FPS. Sebastian Garth (talk) 23:40, 27 October 2013 (UTC)
- 1/1.56=0.641 approx, indeed a reduction of 35.9%. So I suggest that instead of saying "not right", you present your approach as an alternative way of restoring the original. If you look carefully, you'll see that the OP asked for the percentage needed, which is what I gave for different interpretations of "sped up 156%", not just for a way to get the answer.86.176.135.38 (talk) 22:34, 28 October 2013 (UTC)
- Interesting, but the FPS is always constant. I know the speedup/slowup factors, but the FPS is locked at a certain speed. Does this change the calculation? --209.203.125.162 (talk) 17:02, 28 October 2013 (UTC)
- That's not right. All you need is to take the reciprocal and multiply. For example, a video-tape reel is running at 29 FPS then sped up by a factor of 1.56 (ie: 156%) thus resulting in a new speed of 45.24 FPS. Now if you only knew the current tape speed and the speed-up factor used, you would just multiply 45.24 * (1/1.56) to obtain 29 FPS. Sebastian Garth (talk) 23:40, 27 October 2013 (UTC)
- Let's say that the true speed is T and the play speed is P. The video is played at 156% of it's true speed, so P = 1.56 T. You can just divide by 1.56 to get T = 25⁄39 P. (Notice 1÷1.56 = 25⁄39.) If you do the sums, you get 25⁄39 ≈ 0.641, so you need to play it back at about 64.1% of the fast speed. In terms of a reduction, that's about at 35.9% reduction. — Fly by Night (talk) 20:57, 30 October 2013 (UTC)