8 ms·
I like doing this kind of estimation via binary logs expressed in base 12. 17 is a bit over 5·5·2/3, which is a bit under 2^(1.4 + 1.4 + 1 – 1.7) = 2^4.1 [base
by jacobolus 9y ago
I like doing this kind of estimation via binary logs expressed in base 12.
17 is a bit over 5·5·2/3, which is a bit under 2^(1.4 + 1.4 + 1 – 1.7) = 2^4.1 [base 12], now to take the square root we get 2^2.06 ~ 4·35/34 [base 10], but we want something a bit higher, so for a round number to cancel the 4 we can guess 4·33/32 = 33/8 [base 10].
- btilly 9y agoBase 12 is more efficient for arithmetic, but I never really mastered it. And converting back and forth seems to me to be more trouble than it is worth.
- jacobolus 9y agoIt’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.