Kite's thoughts on pergens: Difference between revisions
Wikispaces>TallKite **Imported revision 624585795 - Original comment: ** |
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To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4<span class="nowrap">⋅</span>P and P8. If P is 6/5, the comma is 4<span class="nowrap">⋅</span>P - P8 = (6/5)<span style="vertical-align: super;">4</span> / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4<span class="nowrap">⋅</span>P = (2/1) · (7/6)<span style="vertical-align: super;">-4</span>. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3<span class="nowrap">⋅</span>G - P4 = (10/9)^3 / (4/3) = 250/243. | To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4<span class="nowrap">⋅</span>P and P8. If P is 6/5, the comma is 4<span class="nowrap">⋅</span>P - P8 = (6/5)<span style="vertical-align: super;">4</span> / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4<span class="nowrap">⋅</span>P = (2/1) · (7/6)<span style="vertical-align: super;">-4</span>. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3<span class="nowrap">⋅</span>G - P4 = (10/9)^3 / (4/3) = 250/243. | ||
An even easier way is to simply assume that the up symbol equals a convenient comma such as 81/80 or 64/63. Thus for (P8/4, P5), since G = vm3, G is 32/27 ÷ 64/63 = 7/6. This method is notation-dependent: (P8/2, P5) with G = vA4 and ^1 = 81/80 gives G = 45/32, but if G = ^4, then G = 27/20. | |||
Finding the comma(s) for a double pergen is trickier. As previously noted, if a pergen's multigen is (a,b), the octave is split into at least |b| parts. Therefore if a pergen (P8/m, (a,b)/n) has m = |b|, it is **explicitly false**. If so, proceed as if the octave were unsplit: (P8/2, M2/4) requires G ~ 50¢, perhaps 33/32, and the comma is 4<span class="nowrap">⋅</span>G - M2 = (33/32)^4 / (9/8) = (-17, 2, 0, 0, 4). | Finding the comma(s) for a double pergen is trickier. As previously noted, if a pergen's multigen is (a,b), the octave is split into at least |b| parts. Therefore if a pergen (P8/m, (a,b)/n) has m = |b|, it is **explicitly false**. If so, proceed as if the octave were unsplit: (P8/2, M2/4) requires G ~ 50¢, perhaps 33/32, and the comma is 4<span class="nowrap">⋅</span>G - M2 = (33/32)^4 / (9/8) = (-17, 2, 0, 0, 4). | ||
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To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4<span class="nowrap">⋅</span>P and P8. If P is 6/5, the comma is 4<span class="nowrap">⋅</span>P - P8 = (6/5)<span style="vertical-align: super;">4</span> / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4<span class="nowrap">⋅</span>P = (2/1) · (7/6)<span style="vertical-align: super;">-4</span>. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3<span class="nowrap">⋅</span>G - P4 = (10/9)^3 / (4/3) = 250/243.<br /> | To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4<span class="nowrap">⋅</span>P and P8. If P is 6/5, the comma is 4<span class="nowrap">⋅</span>P - P8 = (6/5)<span style="vertical-align: super;">4</span> / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4<span class="nowrap">⋅</span>P = (2/1) · (7/6)<span style="vertical-align: super;">-4</span>. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3<span class="nowrap">⋅</span>G - P4 = (10/9)^3 / (4/3) = 250/243.<br /> | ||
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An even easier way is to simply assume that the up symbol equals a convenient comma such as 81/80 or 64/63. Thus for (P8/4, P5), since G = vm3, G is 32/27 ÷ 64/63 = 7/6. This method is notation-dependent: (P8/2, P5) with G = vA4 and ^1 = 81/80 gives G = 45/32, but if G = ^4, then G = 27/20.<br /> | |||
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Finding the comma(s) for a double pergen is trickier. As previously noted, if a pergen's multigen is (a,b), the octave is split into at least |b| parts. Therefore if a pergen (P8/m, (a,b)/n) has m = |b|, it is <strong>explicitly false</strong>. If so, proceed as if the octave were unsplit: (P8/2, M2/4) requires G ~ 50¢, perhaps 33/32, and the comma is 4<span class="nowrap">⋅</span>G - M2 = (33/32)^4 / (9/8) = (-17, 2, 0, 0, 4).<br /> | Finding the comma(s) for a double pergen is trickier. As previously noted, if a pergen's multigen is (a,b), the octave is split into at least |b| parts. Therefore if a pergen (P8/m, (a,b)/n) has m = |b|, it is <strong>explicitly false</strong>. If so, proceed as if the octave were unsplit: (P8/2, M2/4) requires G ~ 50¢, perhaps 33/32, and the comma is 4<span class="nowrap">⋅</span>G - M2 = (33/32)^4 / (9/8) = (-17, 2, 0, 0, 4).<br /> | ||
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<!-- ws:start:WikiTextHeadingRule:61:&lt;h2&gt; --><h2 id="toc15"><a name="Further Discussion-Misc notes"></a><!-- ws:end:WikiTextHeadingRule:61 -->Misc notes</h2> | <!-- ws:start:WikiTextHeadingRule:61:&lt;h2&gt; --><h2 id="toc15"><a name="Further Discussion-Misc notes"></a><!-- ws:end:WikiTextHeadingRule:61 -->Misc notes</h2> | ||
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Pergens were discovered by Kite Giedraitis in 2017, and developed with the help of Praveen Venkataramana. Earlier drafts of this article can be found at <!-- ws:start:WikiTextUrlRule: | Pergens were discovered by Kite Giedraitis in 2017, and developed with the help of Praveen Venkataramana. Earlier drafts of this article can be found at <!-- ws:start:WikiTextUrlRule:3137:http://xenharmonic.wikispaces.com/pergen+names --><a href="http://xenharmonic.wikispaces.com/pergen+names">http://xenharmonic.wikispaces.com/pergen+names</a><!-- ws:end:WikiTextUrlRule:3137 --><br /> | ||
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