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| <h2>IMPORTED REVISION FROM WIKISPACES</h2> | | <span style="display: block; text-align: right;">[[特徴的なヴァル|日本語]]</span> |
| This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
| | __FORCETOC__ |
| : This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2016-05-24 19:03:32 UTC</tt>.<br>
| | ----- |
| : The original revision id was <tt>583993521</tt>.<br>
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| : The revision comment was: <tt></tt><br>
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| 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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| <h4>Original Wikitext content:</h4>
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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"><span style="display: block; text-align: right;">[[特徴的なヴァル|日本語]]</span>
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| [[toc|flat]]
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| ----
| | =Introduction= |
| =Introduction= | |
| The patent val for some EDO is the val that you obtain by simply finding the closest rounded-off approximation to each prime in the tuning. For example, the patent val for 17-EDO is <17 27 39|, indicating that the closest mapping for 2/1 is 17 steps, the closest mapping for 3/1 is 27 steps, and the closest mapping for 5/1 is 39 steps. This means, if octaves are pure, that 3/2 is 706 cents, which is what you get if you round off 3/2 to the closest location in 17-equal, and that 5/4 is 353 cents, which is what you get is you round off 5/4 to the closest location in 17-equal. This val can be extended to the case where the number of steps in an octave is a real number rather than an integer; this is called a generalized patent val, or GPV. For instance the 7-limit generalized patent val for 16.9 is <17 27 39 47|, since 16.9 * log2(7) = 47.444, which rounds down to 47. | | The patent val for some EDO is the val that you obtain by simply finding the closest rounded-off approximation to each prime in the tuning. For example, the patent val for 17-EDO is <17 27 39|, indicating that the closest mapping for 2/1 is 17 steps, the closest mapping for 3/1 is 27 steps, and the closest mapping for 5/1 is 39 steps. This means, if octaves are pure, that 3/2 is 706 cents, which is what you get if you round off 3/2 to the closest location in 17-equal, and that 5/4 is 353 cents, which is what you get is you round off 5/4 to the closest location in 17-equal. This val can be extended to the case where the number of steps in an octave is a real number rather than an integer; this is called a generalized patent val, or GPV. For instance the 7-limit generalized patent val for 16.9 is <17 27 39 47|, since 16.9 * log2(7) = 47.444, which rounds down to 47. |
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| You may prefer to use the <17 27 40| val as the 5-limit 17-equal val instead, which rather than <17 27 39| treats 424 cents as 5/4. This val has lower Tenney-Euclidean error than the 17-EDO patent val. However, while <17 27 39| may not necessarily be the "best" val for 17-equal for all purposes, it is the obvious, or "patent" val, that you get by naively rounding primes off within the EDO and taking no further considerations into account. However, <17 27 40| is the generalized patent val for 17.1, since 17.1 * log2(5) = 39.705, which rounds up to 40. | | You may prefer to use the <17 27 40| val as the 5-limit 17-equal val instead, which rather than <17 27 39| treats 424 cents as 5/4. This val has lower Tenney-Euclidean error than the 17-EDO patent val. However, while <17 27 39| may not necessarily be the "best" val for 17-equal for all purposes, it is the obvious, or "patent" val, that you get by naively rounding primes off within the EDO and taking no further considerations into account. However, <17 27 40| is the generalized patent val for 17.1, since 17.1 * log2(5) = 39.705, which rounds up to 40. |
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| =Further explanation= | | =Further explanation= |
| A [[p-limit]] [[Vals and Tuning Space|val]] contains the number of steps it takes to get to each prime number up to p, in prime number order: | | A [[p-limit|p-limit]] [[Vals_and_Tuning_Space|val]] contains the number of steps it takes to get to each prime number up to p, in prime number order: |
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| < [2/1] [3/1] [5/1] [7/1] ... [p/1] | | | < [2/1] [3/1] [5/1] [7/1] ... [p/1] | |
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| Given N-EDO, the equal division of the octave into N parts, we may define vals that map a specific number of N-EDO steps to these primes. | | Given N-EDO, the equal division of the octave into N parts, we may define vals that map a specific number of N-EDO steps to these primes. |
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| For any prime p we can 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. | | For any prime p we can find a corresponding p-limit val in a canonical manner by [http://en.wikipedia.org/wiki/Scalar_multiplication scalar multiplying] <1 [[log2|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. |
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| One way to think of this process is to first ask, "How many 1200-cent steps (octaves) does it take to get to each prime?" It takes one full-octave step to get to 2/1, log2(3) steps to get to 3/1, log2(5) to get to 5/1, and so on. This gives us <1 1.585 2.322 2.807 3.459 ... log2(p) |. | | One way to think of this process is to first ask, "How many 1200-cent steps (octaves) does it take to get to each prime?" It takes one full-octave step to get to 2/1, log2(3) steps to get to 3/1, log2(5) to get to 5/1, and so on. This gives us <1 1.585 2.322 2.807 3.459 ... log2(p) |. |
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| Thus, the way to get the p-limit patent val for N-EDO is to multiply <1 1.585 2.322 2.807 ... log2(p) | by N. Then, since you can't take fractional steps in an EDO, you round the results to the nearest integers. | | Thus, the way to get the p-limit patent val for N-EDO is to multiply <1 1.585 2.322 2.807 ... log2(p) | by N. Then, since you can't take fractional steps in an EDO, you round the results to the nearest integers. |
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| =A 12 EDO Example= | | =A 12 EDO Example= |
| Multiplying 12 times <1 1.585 2.322 2.807 3.459| | | Multiplying 12 times <1 1.585 2.322 2.807 3.459| |
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| yields <12 19.020 27.863 33.688 41.513|, | | yields <12 19.020 27.863 33.688 41.513|, |
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| rounded to <12 19 28 34 42|, | | rounded to <12 19 28 34 42|, |
| which is the **11-limit patent val for [[12edo]]**.
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| =An alternate and expanded example for 31 EDO= | | which is the '''11-limit patent val for [[12edo|12edo]]'''. |
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| | =An alternate and expanded example for 31 EDO= |
| As stated above, the val contains the number of steps it takes to get to a given prime number, in prime number order: | | As stated above, the val contains the number of steps it takes to get to a given prime number, in prime number order: |
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| < [2/1] [3/1] [5/1] [7/1] [etc.] | | | < [2/1] [3/1] [5/1] [7/1] [etc.] | |
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| 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 |. | | 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 |. |
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| What's the number of steps to 3/1? | | What's the number of steps to 3/1? |
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| The step size for 31 EDO is 38.70967742 cents. | | The step size for 31 EDO is 38.70967742 cents. |
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| 3/1 is 1901.96 in cents. | | 3/1 is 1901.96 in cents. |
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| 1901.96 cents / 38.70967742 cents/step = 49.13383752 steps. | | 1901.96 cents / 38.70967742 cents/step = 49.13383752 steps. |
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| This is an EDO, so we can't take 0.13383752 steps. Instead, we round. This is clearly closer to 49 steps, so that's the "obvious" or "patent" choice. The 3-limit patent val is | | This is an EDO, so we can't take 0.13383752 steps. Instead, we round. This is clearly closer to 49 steps, so that's the "obvious" or "patent" choice. The 3-limit patent val is |
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| < 31 49 |. Doing the same thing up through 17, and we get an 17-limit patent val of | | < 31 49 |. Doing the same thing up through 17, and we get an 17-limit patent val of |
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| < 31 49 72 87 107 115 127 | | | < 31 49 72 87 107 115 127 | |
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| 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 |
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| < 31 49 72 87 107 115 127 132 | | | < 31 49 72 87 107 115 127 132 | |
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| Note that these are the same answers you would get if you multiplied 31 times <1 1.585 2.322 2.807 3.459 3.700 4.087 4.248 | and rounded the result. | | Note that these are the same answers you would get if you multiplied 31 times <1 1.585 2.322 2.807 3.459 3.700 4.087 4.248 | and rounded the result. |
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| =How this defines a rank 1 temperament= | | =How this defines a rank 1 temperament= |
| A val defines a rank 1 temperament by defining the deliberate introduction of an error into one or more primes. In 12 EDO, for instance, the perfect fifth (ratio 3/2, or exactly 1.5) is mapped to 700 cents, which is actually just barely flat: a ratio of 2^(700/1200), or 1.4983070769. | | A val defines a rank 1 temperament by defining the deliberate introduction of an error into one or more primes. In 12 EDO, for instance, the perfect fifth (ratio 3/2, or exactly 1.5) is mapped to 700 cents, which is actually just barely flat: a ratio of 2^(700/1200), or 1.4983070769. |
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| As stated above, the 2/1 in the patent val is perfect. The patent val for 12 EDO, <12 19 28 34 42 (etc) |, implies that it takes 12 steps to get the octave, which is does: 12 steps * 100 cents / step = 1200 cents = 1 octave. | | As stated above, the 2/1 in the patent val is perfect. The patent val for 12 EDO, <12 19 28 34 42 (etc) |, implies that it takes 12 steps to get the octave, which is does: 12 steps * 100 cents / step = 1200 cents = 1 octave. |
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| In the patent val for 12 EDO, the number 19 is in the second spot -- the place reserved for 3/1. That implies that it takes 19 steps to get to 3/1. We know that's not true: We rounded numbers off when we created the val in the first place, so essentially we're just //acting as if// 19 steps gets you to 3/1. To put it differently, we're deliberately introducing an error into 3/1. More precisely, any time we would have used 3/1 (e.g., in 9/8, 3/2, etc.), we're going to use a slightly different number instead. | | In the patent val for 12 EDO, the number 19 is in the second spot -- the place reserved for 3/1. That implies that it takes 19 steps to get to 3/1. We know that's not true: We rounded numbers off when we created the val in the first place, so essentially we're just ''acting as if'' 19 steps gets you to 3/1. To put it differently, we're deliberately introducing an error into 3/1. More precisely, any time we would have used 3/1 (e.g., in 9/8, 3/2, etc.), we're going to use a slightly different number instead. |
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| We can calculate the error we're introducing into 3/1 as follows for 12 EDO: | | We can calculate the error we're introducing into 3/1 as follows for 12 EDO: |
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| 12 EDO steps are 100.0 cents each. | | 12 EDO steps are 100.0 cents each. |
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| 19 steps of 12 EDO = 19 steps * 100.0 cents/step = 1900.0 cents. | | 19 steps of 12 EDO = 19 steps * 100.0 cents/step = 1900.0 cents. |
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| 1900.0 cents => 2^(1900/1200), or 2.9966141538. This is the value that 12 EDO uses in place of prime 3. | | 1900.0 cents => 2^(1900/1200), or 2.9966141538. This is the value that 12 EDO uses in place of prime 3. |
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| The patent val for 31 EDO is <31 49 72 87 107 (etc) |. The 49 implies that it takes 49 steps to get to 3/1. | | The patent val for 31 EDO is <31 49 72 87 107 (etc) |. The 49 implies that it takes 49 steps to get to 3/1. |
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| 31 EDO steps are 38.70967742 cents each. | | 31 EDO steps are 38.70967742 cents each. |
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| 49 steps of 31 EDO = 49 steps * 38.70967742 cents/step = 1896.774194 cents. | | 49 steps of 31 EDO = 49 steps * 38.70967742 cents/step = 1896.774194 cents. |
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| 1896.774194 cents => 2^(1896.774194/1200), or 2.991035765. This is what 31 EDO uses in place of prime 3. | | 1896.774194 cents => 2^(1896.774194/1200), or 2.991035765. This is what 31 EDO uses in place of prime 3. |
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| That doesn't make 31 EDO better or worse than 12; it just means there's more error in the 3/1 ratio in 31 EDO than in 12 EDO. If you run these calculations for 5/1 using the patent vals for 12 EDO and 31 EDO, you'll find that 5/1 has more error in 12 EDO than in 31 EDO: 5.0396842 vs. 5.002262078, respectively. 31 EDO may therefore be preferred by people who like sweeter thirds (5/4 ratios) and are willing to have flatter fifths (3/2 ratios). | | That doesn't make 31 EDO better or worse than 12; it just means there's more error in the 3/1 ratio in 31 EDO than in 12 EDO. If you run these calculations for 5/1 using the patent vals for 12 EDO and 31 EDO, you'll find that 5/1 has more error in 12 EDO than in 31 EDO: 5.0396842 vs. 5.002262078, respectively. 31 EDO may therefore be preferred by people who like sweeter thirds (5/4 ratios) and are willing to have flatter fifths (3/2 ratios). |
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| =How this relates to commas= | | =How this relates to commas= |
| These deliberate errors ensure that certain commas get tempered out. The patent vals for both 12 EDO and 31 EDO temper out 81/80. Here are the calculations: | | These deliberate errors ensure that certain commas get tempered out. The patent vals for both 12 EDO and 31 EDO temper out 81/80. Here are the calculations: |
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| 81 = 3*3*3*3. This can also be written as a power of a prime -- 3^4 -- or as a monzo -- | 0 4 >. | | 81 = 3*3*3*3. This can also be written as a power of a prime -- 3^4 -- or as a monzo -- | 0 4 >. |
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| 80 = 2*2*2*2*5. This can also be written as a product of powers of primes -- (2^4)*(5^1) -- or as a monzo -- | 4 0 1 >. | | 80 = 2*2*2*2*5. This can also be written as a product of powers of primes -- (2^4)*(5^1) -- or as a monzo -- | 4 0 1 >. |
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| Substitute in the values for 81/80 in 12 EDO and get this: (2.9966141538^4) / (2^4)*(5.0396842) = 80.6349472 / 80.6349472 = 1/1. | | Substitute in the values for 81/80 in 12 EDO and get this: (2.9966141538^4) / (2^4)*(5.0396842) = 80.6349472 / 80.6349472 = 1/1. |
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| Substitute in the values for 81/80 in 31 EDO and get this: (2.991035765^4) / (2^4)*(5.002262078) = 80.036193 / 80.036193 = 1/1. | | Substitute in the values for 81/80 in 31 EDO and get this: (2.991035765^4) / (2^4)*(5.002262078) = 80.036193 / 80.036193 = 1/1. |
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| You're dividing 81 by 80, so (assuming we're starting at zero, though it works no matter where you start) you add the steps for 81 (+196) and subtract the steps for 80 (-196). 196-196 = 0. This means that it takes zero steps to reach 81/80 -- in other words, 81/80 "vanishes". | | You're dividing 81 by 80, so (assuming we're starting at zero, though it works no matter where you start) you add the steps for 81 (+196) and subtract the steps for 80 (-196). 196-196 = 0. This means that it takes zero steps to reach 81/80 -- in other words, 81/80 "vanishes". |
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| .</pre></div> | | . |
| <h4>Original HTML content:</h4>
| | [[Category:definition]] |
| <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"><html><head><title>Patent val</title></head><body><span style="display: block; text-align: right;"><a class="wiki_link" href="/%E7%89%B9%E5%BE%B4%E7%9A%84%E3%81%AA%E3%83%B4%E3%82%A1%E3%83%AB">日本語</a></span><br />
| | [[Category:math]] |
| <!-- ws:start:WikiTextTocRule:12:&lt;img id=&quot;wikitext@@toc@@flat&quot; class=&quot;WikiMedia WikiMediaTocFlat&quot; title=&quot;Table of Contents&quot; src=&quot;/site/embedthumbnail/toc/flat?w=100&amp;h=16&quot;/&gt; --><!-- ws:end:WikiTextTocRule:12 --><!-- ws:start:WikiTextTocRule:13: --><a href="#Introduction">Introduction</a><!-- ws:end:WikiTextTocRule:13 --><!-- ws:start:WikiTextTocRule:14: --> | <a href="#Further explanation">Further explanation</a><!-- ws:end:WikiTextTocRule:14 --><!-- ws:start:WikiTextTocRule:15: --> | <a href="#A 12 EDO Example">A 12 EDO Example</a><!-- ws:end:WikiTextTocRule:15 --><!-- ws:start:WikiTextTocRule:16: --> | <a href="#An alternate and expanded example for 31 EDO">An alternate and expanded example for 31 EDO</a><!-- ws:end:WikiTextTocRule:16 --><!-- ws:start:WikiTextTocRule:17: --> | <a href="#How this defines a rank 1 temperament">How this defines a rank 1 temperament</a><!-- ws:end:WikiTextTocRule:17 --><!-- ws:start:WikiTextTocRule:18: --> | <a href="#How this relates to commas">How this relates to commas</a><!-- ws:end:WikiTextTocRule:18 --><!-- ws:start:WikiTextTocRule:19: -->
| | [[Category:term]] |
| <!-- ws:end:WikiTextTocRule:19 --><br />
| | [[Category:theory]] |
| <hr />
| | [[Category:val]] |
| <!-- ws:start:WikiTextHeadingRule:0:&lt;h1&gt; --><h1 id="toc0"><a name="Introduction"></a><!-- ws:end:WikiTextHeadingRule:0 -->Introduction</h1>
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| The patent val for some EDO is the val that you obtain by simply finding the closest rounded-off approximation to each prime in the tuning. For example, the patent val for 17-EDO is &lt;17 27 39|, indicating that the closest mapping for 2/1 is 17 steps, the closest mapping for 3/1 is 27 steps, and the closest mapping for 5/1 is 39 steps. This means, if octaves are pure, that 3/2 is 706 cents, which is what you get if you round off 3/2 to the closest location in 17-equal, and that 5/4 is 353 cents, which is what you get is you round off 5/4 to the closest location in 17-equal. This val can be extended to the case where the number of steps in an octave is a real number rather than an integer; this is called a generalized patent val, or GPV. For instance the 7-limit generalized patent val for 16.9 is &lt;17 27 39 47|, since 16.9 * log2(7) = 47.444, which rounds down to 47.<br />
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| <br />
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| You may prefer to use the &lt;17 27 40| val as the 5-limit 17-equal val instead, which rather than &lt;17 27 39| treats 424 cents as 5/4. This val has lower Tenney-Euclidean error than the 17-EDO patent val. However, while &lt;17 27 39| may not necessarily be the &quot;best&quot; val for 17-equal for all purposes, it is the obvious, or &quot;patent&quot; val, that you get by naively rounding primes off within the EDO and taking no further considerations into account. However, &lt;17 27 40| is the generalized patent val for 17.1, since 17.1 * log2(5) = 39.705, which rounds up to 40.<br />
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| <br />
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| <!-- ws:start:WikiTextHeadingRule:2:&lt;h1&gt; --><h1 id="toc1"><a name="Further explanation"></a><!-- ws:end:WikiTextHeadingRule:2 -->Further explanation</h1>
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| A <a class="wiki_link" href="/p-limit">p-limit</a> <a class="wiki_link" href="/Vals%20and%20Tuning%20Space">val</a> contains the number of steps it takes to get to each prime number up to p, in prime number order:<br />
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| &lt; [2/1] [3/1] [5/1] [7/1] ... [p/1] |<br />
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| Given N-EDO, the equal division of the octave into N parts, we may define vals that map a specific number of N-EDO steps to these primes.<br />
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| <br />
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| For any prime p we can find a corresponding p-limit val in a canonical manner by <a class="wiki_link_ext" href="http://en.wikipedia.org/wiki/Scalar_multiplication" rel="nofollow">scalar multiplying</a> &lt;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 &quot;patent&quot; in one sense of the word is a synonym for &quot;obvious&quot;; the patent val may or may not be the best choice but it's the obvious choice.<br />
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| <br />
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| One way to think of this process is to first ask, &quot;How many 1200-cent steps (octaves) does it take to get to each prime?&quot; It takes one full-octave step to get to 2/1, log2(3) steps to get to 3/1, log2(5) to get to 5/1, and so on. This gives us &lt;1 1.585 2.322 2.807 3.459 ... log2(p) |.<br />
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| Then ask, &quot;How many more N-EDO steps does it take to get to the same places?&quot; One 12-EDO step is 1/12th of an octave, by definition; therefore, you need 12 times as many steps to reach 2/1, 3/1, 5/1... Similarly, one 31-EDO step is 1/31st of an octave, so you need 31 times as many steps to reach 2/1, 3/1, 5/1...<br />
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| <br />
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| Thus, the way to get the p-limit patent val for N-EDO is to multiply &lt;1 1.585 2.322 2.807 ... log2(p) | by N. Then, since you can't take fractional steps in an EDO, you round the results to the nearest integers.<br />
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| <br />
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| <!-- ws:start:WikiTextHeadingRule:4:&lt;h1&gt; --><h1 id="toc2"><a name="A 12 EDO Example"></a><!-- ws:end:WikiTextHeadingRule:4 -->A 12 EDO Example</h1>
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| Multiplying 12 times &lt;1 1.585 2.322 2.807 3.459|<br />
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| yields &lt;12 19.020 27.863 33.688 41.513|,<br />
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| rounded to &lt;12 19 28 34 42|,<br />
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| which is the <strong>11-limit patent val for <a class="wiki_link" href="/12edo">12edo</a></strong>.<br />
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| <br />
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| <!-- ws:start:WikiTextHeadingRule:6:&lt;h1&gt; --><h1 id="toc3"><a name="An alternate and expanded example for 31 EDO"></a><!-- ws:end:WikiTextHeadingRule:6 -->An alternate and expanded example for 31 EDO</h1>
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| As stated above, the val contains the number of steps it takes to get to a given prime number, in prime number order:<br />
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| &lt; [2/1] [3/1] [5/1] [7/1] [etc.] |<br />
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| 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 &lt; 31 |.<br />
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| <br />
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| What's the number of steps to 3/1?<br />
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| The step size for 31 EDO is 38.70967742 cents.<br />
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| 3/1 is 1901.96 in cents.<br />
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| 1901.96 cents / 38.70967742 cents/step = 49.13383752 steps.<br />
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| This is an EDO, so we can't take 0.13383752 steps. Instead, we round. This is clearly closer to 49 steps, so that's the &quot;obvious&quot; or &quot;patent&quot; choice. The 3-limit patent val is<br />
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| &lt; 31 49 |. Doing the same thing up through 17, and we get an 17-limit patent val of<br />
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| &lt; 31 49 72 87 107 115 127 |<br />
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| <br />
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| 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.<br />
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| <br />
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| 19/1 = 5097.51 cents, 5097.51 / 38.70967742 cents/step = 131.6857529 steps. Round to get 132. The 19-limit patent val is<br />
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| &lt; 31 49 72 87 107 115 127 132 |<br />
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| <br />
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| Note that these are the same answers you would get if you multiplied 31 times &lt;1 1.585 2.322 2.807 3.459 3.700 4.087 4.248 | and rounded the result.<br />
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| <br />
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| <!-- ws:start:WikiTextHeadingRule:8:&lt;h1&gt; --><h1 id="toc4"><a name="How this defines a rank 1 temperament"></a><!-- ws:end:WikiTextHeadingRule:8 -->How this defines a rank 1 temperament</h1>
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| A val defines a rank 1 temperament by defining the deliberate introduction of an error into one or more primes. In 12 EDO, for instance, the perfect fifth (ratio 3/2, or exactly 1.5) is mapped to 700 cents, which is actually just barely flat: a ratio of 2^(700/1200), or 1.4983070769.<br />
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| <br />
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| As stated above, the 2/1 in the patent val is perfect. The patent val for 12 EDO, &lt;12 19 28 34 42 (etc) |, implies that it takes 12 steps to get the octave, which is does: 12 steps * 100 cents / step = 1200 cents = 1 octave.<br />
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| <br />
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| In the patent val for 12 EDO, the number 19 is in the second spot -- the place reserved for 3/1. That implies that it takes 19 steps to get to 3/1. We know that's not true: We rounded numbers off when we created the val in the first place, so essentially we're just <em>acting as if</em> 19 steps gets you to 3/1. To put it differently, we're deliberately introducing an error into 3/1. More precisely, any time we would have used 3/1 (e.g., in 9/8, 3/2, etc.), we're going to use a slightly different number instead.<br />
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| <br />
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| We can calculate the error we're introducing into 3/1 as follows for 12 EDO:<br />
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| <br />
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| 12 EDO steps are 100.0 cents each.<br />
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| 19 steps of 12 EDO = 19 steps * 100.0 cents/step = 1900.0 cents.<br />
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| 1900.0 cents =&gt; 2^(1900/1200), or 2.9966141538. This is the value that 12 EDO uses in place of prime 3.<br />
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| <br />
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| That means that the 12 EDO patent val substitutes 2.9966141538 any time you would have had prime 3 in a ratio. 9/8 and 3/2 will be somewhat flat. 4/3 will be somewhat sharp. (Note that, for now, the example intervals (3/2, 4/3, 9/8) deliberately avoid any primes other than 3 and 2, and 2 is pure, so the sharpness and flatness comes only from the impure 3/1 value.)<br />
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| <br />
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| We can do the same calculations for 31 EDO.<br />
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| <br />
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| The patent val for 31 EDO is &lt;31 49 72 87 107 (etc) |. The 49 implies that it takes 49 steps to get to 3/1.<br />
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| 31 EDO steps are 38.70967742 cents each.<br />
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| 49 steps of 31 EDO = 49 steps * 38.70967742 cents/step = 1896.774194 cents.<br />
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| 1896.774194 cents =&gt; 2^(1896.774194/1200), or 2.991035765. This is what 31 EDO uses in place of prime 3.<br />
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| <br />
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| Again, as in 12 EDO, it's less than 3, so 9/8 and 3/2 will be somewhat flat, and 4/3 will be somewhat sharp. Note that 31 EDO's prime 3 is a little farther away from 3/1 than 12 EDO's 3/1 -- i.e., it has a greater error. That means 31 EDO's 3/2 will be even flatter, and its 4/3 will be even sharper, than in 12 EDO.<br />
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| <br />
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| That doesn't make 31 EDO better or worse than 12; it just means there's more error in the 3/1 ratio in 31 EDO than in 12 EDO. If you run these calculations for 5/1 using the patent vals for 12 EDO and 31 EDO, you'll find that 5/1 has more error in 12 EDO than in 31 EDO: 5.0396842 vs. 5.002262078, respectively. 31 EDO may therefore be preferred by people who like sweeter thirds (5/4 ratios) and are willing to have flatter fifths (3/2 ratios).<br />
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| <br />
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| <!-- ws:start:WikiTextHeadingRule:10:&lt;h1&gt; --><h1 id="toc5"><a name="How this relates to commas"></a><!-- ws:end:WikiTextHeadingRule:10 -->How this relates to commas</h1>
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| These deliberate errors ensure that certain commas get tempered out. The patent vals for both 12 EDO and 31 EDO temper out 81/80. Here are the calculations:<br />
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| <br />
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| 81 = 3*3*3*3. This can also be written as a power of a prime -- 3^4 -- or as a monzo -- | 0 4 &gt;.<br />
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| 80 = 2*2*2*2*5. This can also be written as a product of powers of primes -- (2^4)*(5^1) -- or as a monzo -- | 4 0 1 &gt;.<br />
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| <br />
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| Substitute in the values for 81/80 in 12 EDO and get this: (2.9966141538^4) / (2^4)*(5.0396842) = 80.6349472 / 80.6349472 = 1/1.<br />
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| Substitute in the values for 81/80 in 31 EDO and get this: (2.991035765^4) / (2^4)*(5.002262078) = 80.036193 / 80.036193 = 1/1.<br />
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| <br />
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| The lesson here is that even though the errors in the primes are different for each EDO + patent val in these cases, 81/80 is still tempered out. However, that's not true for all commas; for instance, 12 EDO tempers out 128/125, the diesis, while 31 EDO does not; and 31 EDO tempers out 393216/390625, the Wuerschmidt comma, while 12 EDO does not.<br />
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| <br />
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| By the way, there's a faster way to find out if a comma is being tempered out. Recall that multiplying by a number means adding logarithms: for instance, to go up an octave you can multiply a ratio by 2/1 or you can add 1200 to the cents value. Since the patent val shows you how many EDO steps it takes to get to a prime number, you can add that many steps instead of multiplying things out like we did above. Similarly, you can subtract if you need to divide.<br />
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| <br />
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| 81/80 = 3*3*3*3 / (2*2*2*2 * 5).<br />
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| <br />
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| To get 81, we need to multiply by three, four times. That means we add the number of steps to 3/1 in the patent val, four times. Using the 31 EDO patent val, that's 49+49+49+49, or 4*49, or 196.<br />
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| <br />
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| To get 80, we need to multiply by two, four times, and multiply again by five, once. So add together the number of steps to 2/1, four times, and then add the number of steps to 5/1. Using the 31 EDO patent val, that's 31+31+31+31+72, or 4*31+72, or 196.<br />
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| <br />
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| You're dividing 81 by 80, so (assuming we're starting at zero, though it works no matter where you start) you add the steps for 81 (+196) and subtract the steps for 80 (-196). 196-196 = 0. This means that it takes zero steps to reach 81/80 -- in other words, 81/80 &quot;vanishes&quot;.<br />
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| <br />
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| .</body></html></pre></div>
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