Patent val: Difference between revisions

Wikispaces>jdfreivald
**Imported revision 246416533 - Original comment: **
Wikispaces>genewardsmith
**Imported revision 246419827 - Original comment: **
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<h2>IMPORTED REVISION FROM WIKISPACES</h2>
<h2>IMPORTED REVISION FROM WIKISPACES</h2>
This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
: This revision was by author [[User:jdfreivald|jdfreivald]] and made on <tt>2011-08-17 01:00:15 UTC</tt>.<br>
: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2011-08-17 01:30:11 UTC</tt>.<br>
: The original revision id was <tt>246416533</tt>.<br>
: The original revision id was <tt>246419827</tt>.<br>
: The revision comment was: <tt></tt><br>
: The revision comment was: <tt></tt><br>
The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
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<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;white-space: pre-wrap ! important" class="old-revision-html">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]] &lt;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.
<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;white-space: pre-wrap ! important" class="old-revision-html">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]] &lt;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==  
=A 12 EDO Example=
multiplying 12 times &lt;1 1.585 2.322 2.807 3.459|
multiplying 12 times &lt;1 1.585 2.322 2.807 3.459|
yields &lt;12 19.020 27.863 33.688 41.513|,
yields &lt;12 19.020 27.863 33.688 41.513|,
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which is the **11-limit patent val for [[12edo]]**.
which is the **11-limit patent val for [[12edo]]**.


==Example for 31 EDO==  
=An expanded example for 31 EDO==  
Paraphrased from the Tuning list:
The scalar multiplication described above is mathematically correct, but I (Jake) found this method to be more intuitive.


The val contains the number of steps it takes to get to a given prime number, in prime number order:
The val contains the number of steps it takes to get to a given prime number, in prime number order:
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3/1 is 1901.96 in cents.
3/1 is 1901.96 in cents.
1901.96 cents / 38.70967742 cents/step = 49.13383752 steps.
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.
This is an EDO, but we can't take 0.13383752 steps. So we round. This is clearly closer to 49 steps, so that's the "obvious" or "patent" choice. The 3-limit patent val is &lt; 31 49 |
The 3-limit patent val is &lt; 31 49 |


Do the same thing up through 17, and you get an 17-limit patent val of
Doing the same thing up through 17, and we get an 17-limit patent val of &lt; 31 49 72 87 107 115 127 |
&lt; 31 49 72 87 107 115 127 |


To do the whole thing one more time, let's do it for the 19-limit.
To see how to extend from one limit to another, we may look at what to do for 19/1 and use that to go from the 17-limit to 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
19/1 = 5097.51 cents, 5097.51 / 38.70967742 cents/step = 131.6857529 steps. Round to get 132. The 19-limit patent val is
&lt; 31 49 72 87 107 115 127 132 |</pre></div>
&lt; 31 49 72 87 107 115 127 132 |</pre></div>
<h4>Original HTML content:</h4>
<h4>Original HTML content:</h4>
<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html">&lt;html&gt;&lt;head&gt;&lt;title&gt;Patent val&lt;/title&gt;&lt;/head&gt;&lt;body&gt;Given N-edo, the equal division of the octave into N parts, we may for any prime p find a corresponding &lt;a class="wiki_link" href="/p-limit"&gt;p-limit&lt;/a&gt; &lt;a class="wiki_link" href="/val"&gt;val&lt;/a&gt; in a canonical manner by &lt;a class="wiki_link_ext" href="http://en.wikipedia.org/wiki/Scalar_multiplication" rel="nofollow"&gt;scalar multiplying&lt;/a&gt; &amp;lt;1 &lt;a class="wiki_link" href="/log2"&gt;log2&lt;/a&gt;(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 &lt;em&gt;patent&lt;/em&gt; comes from the fact that &amp;quot;patent&amp;quot; in one sense of the word is a synonym for &amp;quot;obvious&amp;quot;; the patent val may or may not be the best choice but it's the obvious choice.&lt;br /&gt;
<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html">&lt;html&gt;&lt;head&gt;&lt;title&gt;Patent val&lt;/title&gt;&lt;/head&gt;&lt;body&gt;Given N-edo, the equal division of the octave into N parts, we may for any prime p find a corresponding &lt;a class="wiki_link" href="/p-limit"&gt;p-limit&lt;/a&gt; &lt;a class="wiki_link" href="/val"&gt;val&lt;/a&gt; in a canonical manner by &lt;a class="wiki_link_ext" href="http://en.wikipedia.org/wiki/Scalar_multiplication" rel="nofollow"&gt;scalar multiplying&lt;/a&gt; &amp;lt;1 &lt;a class="wiki_link" href="/log2"&gt;log2&lt;/a&gt;(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 &lt;em&gt;patent&lt;/em&gt; comes from the fact that &amp;quot;patent&amp;quot; in one sense of the word is a synonym for &amp;quot;obvious&amp;quot;; the patent val may or may not be the best choice but it's the obvious choice.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;!-- ws:start:WikiTextHeadingRule:0:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc0"&gt;&lt;a name="x-Example"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:0 --&gt;Example&lt;/h2&gt;
&lt;!-- ws:start:WikiTextHeadingRule:0:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc0"&gt;&lt;a name="A 12 EDO Example"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:0 --&gt;A 12 EDO Example&lt;/h1&gt;
multiplying 12 times &amp;lt;1 1.585 2.322 2.807 3.459|&lt;br /&gt;
multiplying 12 times &amp;lt;1 1.585 2.322 2.807 3.459|&lt;br /&gt;
yields &amp;lt;12 19.020 27.863 33.688 41.513|,&lt;br /&gt;
yields &amp;lt;12 19.020 27.863 33.688 41.513|,&lt;br /&gt;
rounded to &amp;lt;12 19 28 34 42|,&lt;br /&gt;
rounded to &amp;lt;12 19 28 34 42|,&lt;br /&gt;
which is the &lt;strong&gt;11-limit patent val for &lt;a class="wiki_link" href="/12edo"&gt;12edo&lt;/a&gt;&lt;/strong&gt;.&lt;br /&gt;
which is the &lt;strong&gt;11-limit patent val for &lt;a class="wiki_link" href="/12edo"&gt;12edo&lt;/a&gt;&lt;/strong&gt;.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;!-- ws:start:WikiTextHeadingRule:2:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc1"&gt;&lt;a name="x-Example for 31 EDO"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:2 --&gt;Example for 31 EDO&lt;/h2&gt;
&lt;!-- ws:start:WikiTextHeadingRule:2:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc1"&gt;&lt;a name="xAn expanded example for 31 EDO="&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:2 --&gt;An expanded example for 31 EDO=&lt;/h1&gt;
  Paraphrased from the Tuning list:&lt;br /&gt;
  &lt;br /&gt;
The scalar multiplication described above is mathematically correct, but I (Jake) found this method to be more intuitive.&lt;br /&gt;
&lt;br /&gt;
The val contains the number of steps it takes to get to a given prime number, in prime number order:&lt;br /&gt;
The val contains the number of steps it takes to get to a given prime number, in prime number order:&lt;br /&gt;
&amp;lt; [2/1] [3/1] [5/1] [7/1] [etc.] |&lt;br /&gt;
&amp;lt; [2/1] [3/1] [5/1] [7/1] [etc.] |&lt;br /&gt;
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3/1 is 1901.96 in cents.&lt;br /&gt;
3/1 is 1901.96 in cents.&lt;br /&gt;
1901.96 cents / 38.70967742 cents/step = 49.13383752 steps.&lt;br /&gt;
1901.96 cents / 38.70967742 cents/step = 49.13383752 steps.&lt;br /&gt;
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 &amp;quot;obvious&amp;quot; or &amp;quot;patent&amp;quot; choice.&lt;br /&gt;
This is an EDO, but we can't take 0.13383752 steps. So we round. This is clearly closer to 49 steps, so that's the &amp;quot;obvious&amp;quot; or &amp;quot;patent&amp;quot; choice. The 3-limit patent val is &amp;lt; 31 49 |&lt;br /&gt;
The 3-limit patent val is &amp;lt; 31 49 |&lt;br /&gt;
&lt;br /&gt;
Doing the same thing up through 17, and we get an 17-limit patent val of &amp;lt; 31 49 72 87 107 115 127 |&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Do the same thing up through 17, and you get an 17-limit patent val of&lt;br /&gt;
To see how to extend from one limit to another, we may look at what to do for 19/1 and use that to go from the 17-limit to the 19-limit.&lt;br /&gt;
&amp;lt; 31 49 72 87 107 115 127 |&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
To do the whole thing one more time, let's do it for the 19-limit.&lt;br /&gt;
19/1 = 5097.51 cents, 5097.51 / 38.70967742 cents/step = 131.6857529 steps. Round to get 132. The 19-limit patent val is&lt;br /&gt;
19/1 = 5097.51 cents, 5097.51 / 38.70967742 cents/step = 131.6857529 steps. Round to get 132.&lt;br /&gt;
The 19-limit patent val is&lt;br /&gt;
&amp;lt; 31 49 72 87 107 115 127 132 |&lt;/body&gt;&lt;/html&gt;</pre></div>
&amp;lt; 31 49 72 87 107 115 127 132 |&lt;/body&gt;&lt;/html&gt;</pre></div>