Patent val
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- This revision was by author jdfreivald and made on 2011-08-17 00:31:11 UTC.
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Original Wikitext content:
Given N-edo, the equal division of the octave into N parts, we may for any prime p find a corresponding [[p-limit]] [[val]] in a canonical manner by [[http://en.wikipedia.org/wiki/Scalar_multiplication|scalar multiplying]] <1 [[log2]](3) log2(5) ... log(p)| by N and rounding to the nearest integer. In general this is not guaranteed to be the most accurate available val, but if N-edo has enough relative accuracy in the p-limit, it will be. The name //patent// comes from the fact that "patent" in one sense of the word is a synonym for "obvious"; the patent val may or may not be the best choice but it's the obvious choice. ==Example== multiplying 12 times <1 1.585 2.322 2.807 3.459| yields <12 19.020 27.863 33.688 41.513|, rounded to <12 19 28 34 42|, which is the **11-limit patent val for [[12edo]]**. ==Example for 31 EDO== Paraphrased from the Tuning list: The val contains the number of steps it takes to get to a given prime number, in prime number order: < [2/1] [3/1] [5/1] [7/1] [etc.] | By definition, for any EDO, the number of steps to 2/1 is the EDO division: 31 for 31 EDO. The 2-limit patent val is < 31 |. What't the number of steps to 3/1? The step size for 31 EDO is 38.70967742 cents. 3/1 is 1901.96 in cents. 1901.96 cents / 38.70967742 cents/step = 49.13383752 steps. This is an EDO, though -- I can't take .13383752 steps. So I round. This is clearly closer to 49 steps, so that's the "obvious" or "patent" choice. The 3-limit patent val is < 31 49 | Do the same thing up through 17, and you get an 17-limit patent val of < 31 49 72 87 107 115 127 | To do the whole thing one more time, let's do it for the 19-limit. 19/1 = 5097.51 cents, 5097.51 / 38.70967742 cents/step = 131.6857529 steps. Round to get 132. The 19-limit patent val is < 31 49 72 87 107 115 127 132 |
Original HTML content:
<html><head><title>Patent val</title></head><body>Given N-edo, the equal division of the octave into N parts, we may for any prime p find a corresponding <a class="wiki_link" href="/p-limit">p-limit</a> <a class="wiki_link" href="/val">val</a> in a canonical manner by <a class="wiki_link_ext" href="http://en.wikipedia.org/wiki/Scalar_multiplication" rel="nofollow">scalar multiplying</a> <1 <a class="wiki_link" href="/log2">log2</a>(3) log2(5) ... log(p)| by N and rounding to the nearest integer. In general this is not guaranteed to be the most accurate available val, but if N-edo has enough relative accuracy in the p-limit, it will be. The name <em>patent</em> comes from the fact that "patent" in one sense of the word is a synonym for "obvious"; the patent val may or may not be the best choice but it's the obvious choice.<br /> <br /> <!-- ws:start:WikiTextHeadingRule:0:<h2> --><h2 id="toc0"><a name="x-Example"></a><!-- ws:end:WikiTextHeadingRule:0 -->Example</h2> multiplying 12 times <1 1.585 2.322 2.807 3.459|<br /> yields <12 19.020 27.863 33.688 41.513|,<br /> rounded to <12 19 28 34 42|,<br /> which is the <strong>11-limit patent val for <a class="wiki_link" href="/12edo">12edo</a></strong>.<br /> <br /> <!-- ws:start:WikiTextHeadingRule:2:<h2> --><h2 id="toc1"><a name="x-Example for 31 EDO"></a><!-- ws:end:WikiTextHeadingRule:2 -->Example for 31 EDO</h2> Paraphrased from the Tuning list:<br /> <br /> The val contains the number of steps it takes to get to a given prime number, in prime number order:<br /> < [2/1] [3/1] [5/1] [7/1] [etc.] |<br /> <br /> By definition, for any EDO, the number of steps to 2/1 is the EDO division: 31 for 31 EDO. The 2-limit patent val is < 31 |.<br /> <br /> What't the number of steps to 3/1?<br /> The step size for 31 EDO is 38.70967742 cents.<br /> 3/1 is 1901.96 in cents.<br /> 1901.96 cents / 38.70967742 cents/step = 49.13383752 steps.<br /> This is an EDO, though -- I can't take .13383752 steps. So I round. This is clearly closer to 49 steps, so that's the "obvious" or "patent" choice.<br /> The 3-limit patent val is < 31 49 |<br /> <br /> Do the same thing up through 17, and you get an 17-limit patent val of<br /> < 31 49 72 87 107 115 127 |<br /> <br /> To do the whole thing one more time, let's do it for the 19-limit.<br /> 19/1 = 5097.51 cents, 5097.51 / 38.70967742 cents/step = 131.6857529 steps. Round to get 132. <br /> The 19-limit patent val is<br /> < 31 49 72 87 107 115 127 132 |</body></html>