File:BeatTrackMoirePattern.ogg
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BeatTrackMoirePattern.ogg (Ogg Vorbis sound file, length 1 min 30 s, 14 kbps, file size: 154 KB)
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Summary
[edit]DescriptionBeatTrackMoirePattern.ogg |
English: The product of two "beat tracks" of slightly different speeds overlaid. It takes 82.499427 seconds (roughly 1 minute and 22 seconds) for them to permute unevenly (unevenly meaning they have crossed over but have not yet landed directly on each other) and 108 permutations to finally permute evenly. |
Date | |
Source | Own work |
Author | X-Fi6 |
Knowing that this audio wave is sampled at 48kHz,
Key B1: Length in samples of high-pitched beat B2/B3/B4: Length in samples of low-pitched beat S1: Length in samples of silence after high-pitched beat S2/S3/S4: Length in samples of silence after low-pitched beat .====B1====.====S1====.=B2/B3/B4=.=S2/S3/S4=. slow: | 2946 | 18054 | 2836 | 18164 | Total: 84000 samples fast: | 2728 | 17836 | 2618 | 17946 | Total: 82256 samples (Chosen so that exactly 2 Sine cycles were cut off from the high beat of 440Hz and one Sine cycle was cut off from the low beat of 220Hz, with the silence cut off by the same ratio) Calculating an uneven crossover permutation: When the tracks crossed over, the closest-matching beats started at sample #3951015 and ended at sample #3968852, with the slower track starting 70 frames earlier than the faster track and ending 148 frames later. Knowing this, and knowing that the tracks are linearly proportional to each other (one is simply x times faster than the other), to find where they match up evenly, it is just a matter of doing a weighted average. Thus, 3951015 + 3968852 148-70 ----------------- + ------ = Sample number 3959972.5. 2 2 Dividing this number by the sampling rate (48000) gives us the number of seconds (82.49942708333333...), which we can also convert to minutes:seconds. Calculating an even crossover permutation: This is simply a matter of finding the least-common multiple between the two numbers. The first track is 84000 samples long and the second is 82256. When do they align correctly? Calculating (or rather, brute forcing) the LCM of these two numbers gives us a value of sample #431844000 of when they next align perfectly. 431844000 divided by the sampling rate (48000) is 8996.75 seconds. However, it is more useful to give the number of uneven permutations before each even permutation, so it can be related to a polygon with fractions of sides. (A 2.5-sided polygon goes around [180 degrees divided by 2.5 is 72 degrees on each angle, and then find when it reaches a whole-number multiple of 180, which is 360, and dividing by 180 gives us—] twice before coming back to the starting point) So subtract 3959972.5 from 431844000 to exclude the initial even permutation, and then divide the number by 3959972.5, which gives us 108.052..., we can regard as 108. 108 uneven before each even. (Ratio 108:1, Frac. 108/109)
Licensing
[edit]Public domainPublic domainfalsefalse |
I, the copyright holder of this work, release this work into the public domain. This applies worldwide. In some countries this may not be legally possible; if so: I grant anyone the right to use this work for any purpose, without any conditions, unless such conditions are required by law. |
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Date/Time | Thumbnail | Dimensions | User | Comment | |
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current | 05:49, 10 January 2011 | 1 min 30 s (154 KB) | X-Fi6 (talk | contribs) | {{Information |Description={{en|1=The product of two "beat tracks" of slightly different speeds overlaid. It takes 82.499427 seconds (roughly 1 minute and 22 seconds) for them to permute unevenly (unevenly meaning they have crossed over but have not yet l |
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Transcode status
Update transcode statusFormat | Bitrate | Download | Status | Encode time |
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MP3 | 66 kbps | Completed 10:57, 5 December 2017 | 2.0 s |
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