Kite's thoughts on pergens: Difference between revisions
Wikispaces>TallKite **Imported revision 630628699 - Original comment: ** |
Wikispaces>TallKite **Imported revision 630628743 - Original comment: ** |
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<h2>IMPORTED REVISION FROM WIKISPACES</h2> | <h2>IMPORTED REVISION FROM WIKISPACES</h2> | ||
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: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-06-19 02: | : This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-06-19 02:33:59 UTC</tt>.<br> | ||
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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 color depth of 1 (i.e. exponent of ±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, the diminished temperament. 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>, the quadruple red temperament. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. | 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 color depth of 1 (i.e. exponent of ±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, the diminished temperament. 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>, the quadruple red temperament. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. | ||
Another method: if the generator's cents are known, look on the genchain for an interval that approximates a ratio of color depth ±1. Let the interval be I, and the genspan of this interval be | Another method: if the generator's cents are known, look on the genchain for an interval that approximates a ratio of color depth ±1. Let the interval be I, and the genspan of this interval be x. Then n<span class="nowrap">⋅x</span> gens = n<span class="nowrap">⋅</span>I = x<span class="nowrap">⋅</span>M, where M is the multigen and M/n is the generator. The comma can be found from this equation, if n and x are coprime. For example, suppose (P8, P5/5) has G = 140¢. The genchain is all multiples of 140¢. Looking at the cents, 280¢ is about 7/6. Thus 2G = 7/6, and 10G = 5<span class="nowrap">⋅</span>(7/6) = 2<span class="nowrap">⋅P5. Thus </span>2<span class="nowrap">⋅P</span>5 - 5<span class="nowrap">⋅</span>(7/6) = 0G = 0¢, and the comma is (3, 7, 0, -5). If the period is split and the generator isn't, use the perchain instead of the genchain. For example, (P8/7, P5) has a period of 141¢. 2 gens = 343¢, about 11/9. Thus 7<span class="nowrap">⋅</span>(11/9) = 2<span class="nowrap">⋅</span>P8, and the comma is (-2, -14, 0, 0, 7). | ||
If the pergen's notation is known, an even easier method is to simply assume that the up symbol equals a comma that maps to P1, such as 81/80 or 64/63 (see mapping commas in the next section). Thus for (P8/4, P5), if P = vm3, and ^1 = 64/63, P is 32/27 ÷ 64/63 = 7/6. This method is notation-dependent: (P8/2, P5) with P = vA4 and ^1 = 81/80 gives P = 45/32, but if P = ^4, then P = 27/20. | If the pergen's notation is known, an even easier method is to simply assume that the up symbol equals a comma that maps to P1, such as 81/80 or 64/63 (see mapping commas in the next section). Thus for (P8/4, P5), if P = vm3, and ^1 = 64/63, P is 32/27 ÷ 64/63 = 7/6. This method is notation-dependent: (P8/2, P5) with P = vA4 and ^1 = 81/80 gives P = 45/32, but if P = ^4, then P = 27/20. | ||
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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 color depth of 1 (i.e. exponent of ±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, the diminished temperament. 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>, the quadruple red temperament. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively.<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 color depth of 1 (i.e. exponent of ±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, the diminished temperament. 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>, the quadruple red temperament. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively.<br /> | ||
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Another method: if the generator's cents are known, look on the genchain for an interval that approximates a ratio of color depth ±1. Let the interval be I, and the genspan of this interval be | Another method: if the generator's cents are known, look on the genchain for an interval that approximates a ratio of color depth ±1. Let the interval be I, and the genspan of this interval be x. Then n<span class="nowrap">⋅x</span> gens = n<span class="nowrap">⋅</span>I = x<span class="nowrap">⋅</span>M, where M is the multigen and M/n is the generator. The comma can be found from this equation, if n and x are coprime. For example, suppose (P8, P5/5) has G = 140¢. The genchain is all multiples of 140¢. Looking at the cents, 280¢ is about 7/6. Thus 2G = 7/6, and 10G = 5<span class="nowrap">⋅</span>(7/6) = 2<span class="nowrap">⋅P5. Thus </span>2<span class="nowrap">⋅P</span>5 - 5<span class="nowrap">⋅</span>(7/6) = 0G = 0¢, and the comma is (3, 7, 0, -5). If the period is split and the generator isn't, use the perchain instead of the genchain. For example, (P8/7, P5) has a period of 141¢. 2 gens = 343¢, about 11/9. Thus 7<span class="nowrap">⋅</span>(11/9) = 2<span class="nowrap">⋅</span>P8, and the comma is (-2, -14, 0, 0, 7).<br /> | ||
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If the pergen's notation is known, an even easier method is to simply assume that the up symbol equals a comma that maps to P1, such as 81/80 or 64/63 (see mapping commas in the next section). Thus for (P8/4, P5), if P = vm3, and ^1 = 64/63, P is 32/27 ÷ 64/63 = 7/6. This method is notation-dependent: (P8/2, P5) with P = vA4 and ^1 = 81/80 gives P = 45/32, but if P = ^4, then P = 27/20.<br /> | If the pergen's notation is known, an even easier method is to simply assume that the up symbol equals a comma that maps to P1, such as 81/80 or 64/63 (see mapping commas in the next section). Thus for (P8/4, P5), if P = vm3, and ^1 = 64/63, P is 32/27 ÷ 64/63 = 7/6. This method is notation-dependent: (P8/2, P5) with P = vA4 and ^1 = 81/80 gives P = 45/32, but if P = ^4, then P = 27/20.<br /> | ||