OD: Difference between revisions

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If you want to describe overtones 1-9 with OD you would need to use 8-OD9, because there are only 8 steps from 1 to 9. You could think of it like 9 is the 8th overtone, so you're really dividing 8 by 8. You're dividing the number of overtones. Alternatively, you could describe his as an [[OS|OS, or overtone sequence]], by simply saying 8-OS.

To find the steps for an n-ODp, begin by recognizing that while the multiplicative interval relating your root position to the end position is <math>p</math> (or <math>\frac p1</math>), if you are going to move arithmetically (by repeated addition) from <math>1</math> to <math>p</math>, then the difference in frequency space that you are dividing up is not actually <math>p</math>, but <math>p - 1</math>. And because you are dividing it into <math>n</math> parts, each step will have a size of <math>\frac{p-1}{n}</math>. So, the formula for the frequency of step <math>k</math> of an n-ODp is:

<math>

f(k) = 1 + (\frac kn)(p-1)

</math>

This way, when <math>k</math> is <math>0</math>, <math>f(k)</math> is simply <math>1</math>. And when <math>k</math> is <math>n</math>, <math>f(k)</math> is simply <math>1 + (p-1) = p</math>.

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 If you want to describe overtones 1-9 with OD you would need to use 8-OD9, because there are only 8 steps from 1 to 9. You could think of it like 9 is the 8th overtone, so you're really dividing 8 by 8. You're dividing the number of overtones. Alternatively, you could describe his as an [[OS|OS, or overtone sequence]], by simply saying 8-OS.
+To find the steps for an n-ODp, begin by recognizing that while the multiplicative interval relating your root position to the end position is <span><math>p</math></span> (or <span><math>\frac p1</math></span>), if you are going to move arithmetically (by repeated addition) from <span><math>1</math></span> to <span><math>p</math></span>, then the difference in frequency space that you are dividing up is not actually <span><math>p</math></span>, but <span><math>p - 1</math></span>. And because you are dividing it into <span><math>n</math></span> parts, each step will have a size of <span><math>\frac{p-1}{n}</math></span>. So, the formula for the frequency of step <span><math>k</math></span> of an n-ODp is:
+<math>
+f(k) = 1 + (\frac kn)(p-1)
+</math>
+This way, when <span><math>k</math></span> is <span><math>0</math></span>, <span><math>f(k)</math></span> is simply <span><math>1</math></span>. And when <span><math>k</math></span> is <span><math>n</math></span>, <span><math>f(k)</math></span> is simply <span><math>1 + (p-1) = p</math></span>.
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