3 ms·
It’s only the binary logs that need to be base 12. This is the approximations inherent in the 12-step-per-octave equal-tempered Western music scale. Memorizing
by jacobolus 9y ago
It’s only the binary logs that need to be base 12. This is the approximations inherent in the 12-step-per-octave equal-tempered Western music scale. Memorizing just two facts gets you a long way:
3 ~ 2^(19/12) -> lg 3 ~ 1;7 or to be more precise 1;703
5 ~ 2^(7/3) = 2^(28/12) -> lg 5 ~ 2;4 or to be more precise 2;3a4
This makes 2^0;7 just about exactly a “perfect fifth” 3/2 (off by about .1%), and 2^0;4 almost a “major third” 5/4 (off by .8%).
If you step up a scale by 2^(1/12) or in log scale 0;1 increments, you can find close fractional approximations to most of the steps. Then I happen to also know that 0;06 is ~lg 35/34, but that one obviously isn’t nearly as important or easy to work with.