摘要

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)

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