Diamond tradeoff: Difference between revisions
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Cmloegcmluin (talk | contribs) work through example for meantone |
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On the other hand, if you go outside these boundaries - for example, if you make 4/3 even flatter than pure - then you're making some intervals in the 5-limit diamond worse without making ''any'' of them better. You're past the realm of compromises and now you're just damaging things for no reason. | On the other hand, if you go outside these boundaries - for example, if you make 4/3 even flatter than pure - then you're making some intervals in the 5-limit diamond worse without making ''any'' of them better. You're past the realm of compromises and now you're just damaging things for no reason. | ||
== Example == | |||
Here we will demonstrate the calculation of the diamond tradeoff tuning range for meantone. | |||
Here is the mapping, <span><math>M</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & 0 & -4 \\ | |||
0 & 1 & 4 \\ | |||
\end{bmatrix} | |||
</math> | |||
This is a 5-limit temperament, so we consider the 5-limit tonality diamond: [1/1, 6/5, 5/4, 4/3, 3/2, 8/5, 5/3]. Of these seven pitches, there are only three we care about. We don't care about the unison, and half of the remaining pitches are octave-complements of the others are thus irrelevant. So, we'll only look at [4/3, 5/4, 6/5]. | |||
For each of these three diamond consonances, we want to find what generator is required in order that this pitch remains pure after tempering, or in other words, that it is an invariant interval (sometimes called an [[eigenmonzo]]). And we want to know this for the situation where octaves are pure. | |||
Let's do it for 4/3 first. So, we prepare a matrix out of these two invariants, 2/1 and 4/3, and call it <span><math>U</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & 2 \\ | |||
0 & -1 \\ | |||
0 & 0 \\ | |||
\end{bmatrix} | |||
</math> | |||
We multiply <span><math>M⋅U</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & 0 & -4 \\ | |||
0 & 1 & 4 \\ | |||
\end{bmatrix} | |||
\begin{bmatrix} | |||
1 & 2 \\ | |||
0 & -1 \\ | |||
0 & 0 \\ | |||
\end{bmatrix}= | |||
\begin{bmatrix} | |||
1 & 1 \\ | |||
0 & -1 \\ | |||
\end{bmatrix} | |||
</math> | |||
We take the inverse <span><math>(M⋅U)^{-1}</math></span> (which in this case is the same): | |||
<math> | |||
\begin{bmatrix} | |||
1 & 1 \\ | |||
0 & -1 \\ | |||
\end{bmatrix} | |||
</math> | |||
Then find <span><math>T</math></span> which is <span><math>U⋅(M⋅U)^{-1}</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & 2 \\ | |||
0 & -1 \\ | |||
0 & 0 \\ | |||
\end{bmatrix} | |||
\begin{bmatrix} | |||
1 & 1 \\ | |||
0 & -1 \\ | |||
\end{bmatrix}= | |||
\begin{bmatrix} | |||
1 & -1 \\ | |||
0 & 1 \\ | |||
0 & 0 \\ | |||
\end{bmatrix} | |||
</math> | |||
Reading the columns from <span><math>T</math></span>, the first one confirms our period of 2/1, and the second column gives our generator 3/2. Which is unsurprising. In cents, that's 1200¢ × log₂(3/2) ≈ 701.955¢. The next invariant interval will give a more interesting result. | |||
So let's do 5/4 now. We prepare a matrix out of these two invariants, 2/1 and 5/4, and call it <span><math>U</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & -2 \\ | |||
0 & 0 \\ | |||
0 & 1 \\ | |||
\end{bmatrix} | |||
</math> | |||
We multiply <span><math>M⋅U</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & 0 & -4 \\ | |||
0 & 1 & 4 \\ | |||
\end{bmatrix} | |||
\begin{bmatrix} | |||
1 & -2 \\ | |||
0 & 0 \\ | |||
0 & 1 \\ | |||
\end{bmatrix}= | |||
\begin{bmatrix} | |||
1 & -2 \\ | |||
0 & 4 \\ | |||
\end{bmatrix} | |||
</math> | |||
We take the inverse <span><math>(M⋅U)^{-1}</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & \frac12 \\ | |||
0 & \frac14 \\ | |||
\end{bmatrix} | |||
</math> | |||
Then find <span><math>T</math></span> which is <span><math>U⋅(M⋅U)^{-1}</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & -2 \\ | |||
0 & 0 \\ | |||
0 & 1 \\ | |||
\end{bmatrix} | |||
\begin{bmatrix} | |||
1 & \frac12 \\ | |||
0 & \frac14 \\ | |||
\end{bmatrix}= | |||
\begin{bmatrix} | |||
1 & 0 \\ | |||
0 & 0 \\ | |||
0 & \frac14 \\ | |||
\end{bmatrix} | |||
</math> | |||
This tells us our generator is 5^(1/4). In cents, that's 1200¢ × log₂(5¹⸍⁴) ≈ 696.578¢. | |||
Okay, one more invariant interval to check: 6/5. We prepare a matrix out of these two invariants, 2/1 and 6/5, and call it <span><math>U</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & 1 \\ | |||
0 & 1 \\ | |||
0 & -1 \\ | |||
\end{bmatrix} | |||
</math> | |||
We multiply <span><math>M⋅U</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & 0 & -4 \\ | |||
0 & 1 & 4 \\ | |||
\end{bmatrix} | |||
\begin{bmatrix} | |||
1 & 1 \\ | |||
0 & 1 \\ | |||
0 & -1 \\ | |||
\end{bmatrix}= | |||
\begin{bmatrix} | |||
1 & 2 \\ | |||
0 & -3 \\ | |||
\end{bmatrix} | |||
</math> | |||
We take the inverse <span><math>(M⋅U)^{-1}</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & \frac23 \\ | |||
0 & -\frac13 \\ | |||
\end{bmatrix} | |||
</math> | |||
Then find <span><math>T</math></span> which is <span><math>U⋅(M⋅U)^{-1}</math></span>: | |||
<math> | |||
\begin{bmatrix} | |||
1 & 1 \\ | |||
0 & 1 \\ | |||
0 & -1 \\ | |||
\end{bmatrix} | |||
\begin{bmatrix} | |||
1 & \frac23 \\ | |||
0 & -\frac13 \\ | |||
\end{bmatrix}= | |||
\begin{bmatrix} | |||
1 & \frac13 \\ | |||
0 & -\frac13 \\ | |||
0 & \frac13 \\ | |||
\end{bmatrix} | |||
</math> | |||
This tells us our generator is (10/3)^(1/3). In cents, that's 1200¢ × log₂((10/3)¹⸍³) ≈ 694.786¢. | |||
We now have our generator sizes that give us pure consonances in the tonality diamond: 701.955¢, 696.578¢, and 694.786¢. The minimum of those is 694.786¢ and the maximum is 701.955¢, so that's our diamond tradeoff range. Anywhere inside that range, we are making at least one of our diamond consonances purer; outside it, we're making them all less pure. | |||
== See also == | == See also == | ||