Douglas Blumeyer's RTT How-To: Difference between revisions

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If you’ve seen one map before, it’s probably {{val|12 19 28}}. That’s because this map is the foundation of conventional Western tuning: [[12edo|12 equal temperament]]. A major reason it stuck is because — for its low complexity — it can closely approximate all three of the 5 prime-limit harmonics 2, 3, and 5 at the same time.

One way to think of this is that 12:19:28 is an excellent low integer approximation of log(2:3:5). ~~Why~~ take the logarithm? Because a) 2, 3, and 5 are not exponents, b) 12, 19, and 28 are exponents, and c) logarithms give exponents.

One way to think of this is that 12:19:28 is an excellent low integer approximation of log(2:3:5). That's a really compact way of saying that each of these sets of three numbers has the same ratio between each pair of them:

* <math>\frac{19}{12} = 1.583 ≈ \frac{log(3)}{log(2)} = 1.585</math>

* <math>\frac{28}{12} = 2.333 ≈ \frac{log(5)}{log(2)} = 2.322</math>

* <math>\frac{28}{19} = 1.474 ≈ \frac{log(5)}{log(3)} = 1.465</math>

You may be more familiar with seeing the base specified for a logarithm, but in this case the base is irrelevant as long as you use the same base for both numbers. If you don't see why, try experimenting with different bases and see that the ratio comes out the same<ref>[https://en.wikipedia.org/wiki/List_of_logarithmic_identities This list of logarithmic identities] has been an excellent resource for me in getting my head around logarithmic thinking. As you can see there, <math>\frac{log_{10}{a}}{log_{10}{b}} = log_{b}{a}</math>, so the base doesn't matter; you could put anything in there — 10, 2, e — and it still reduces to <math>log_{b}{a}</math>.</ref>.

But why take the logarithm at all? Because a) 2, 3, and 5 are not exponents, b) 12, 19, and 28 are exponents, and c) logarithms give exponents.

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 If you’ve seen one map before, it’s probably {{val|12 19 28}}. That’s because this map is the foundation of conventional Western tuning: [[12edo|12 equal temperament]]. A major reason it stuck is because — for its low complexity — it can closely approximate all three of the 5 prime-limit harmonics 2, 3, and 5 at the same time.
-One way to think of this is that 12:19:28 is an excellent low integer approximation of log(2:3:5). Why take the logarithm? Because a) 2, 3, and 5 are not exponents, b) 12, 19, and 28 are exponents, and c) logarithms give exponents.
+One way to think of this is that 12:19:28 is an excellent low integer approximation of log(2:3:5). That's a really compact way of saying that each of these sets of three numbers has the same ratio between each pair of them:
+* <span><math>\frac{19}{12} = 1.583 ≈ \frac{log(3)}{log(2)} = 1.585</math></span>
+* <span><math>\frac{28}{12} = 2.333 ≈ \frac{log(5)}{log(2)} = 2.322</math></span>
+* <span><math>\frac{28}{19} = 1.474 ≈ \frac{log(5)}{log(3)} = 1.465</math></span>
+You may be more familiar with seeing the base specified for a logarithm, but in this case the base is irrelevant as long as you use the same base for both numbers. If you don't see why, try experimenting with different bases and see that the ratio comes out the same<ref>[https://en.wikipedia.org/wiki/List_of_logarithmic_identities This list of logarithmic identities] has been an excellent resource for me in getting my head around logarithmic thinking. As you can see there, <span><math>\frac{log_{10}{a}}{log_{10}{b}} = log_{b}{a}</math></span>, so the base doesn't matter; you could put anything in there — 10, 2, e — and it still reduces to <span><math>log_{b}{a}</math></span>.</ref>.
+But why take the logarithm at all? Because a) 2, 3, and 5 are not exponents, b) 12, 19, and 28 are exponents, and c) logarithms give exponents.
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