Kite's thoughts on pergens: Difference between revisions

<h2>IMPORTED REVISION FROM WIKISPACES</h2>

This is an imported revision from Wikispaces. The revision metadata is included below for reference:<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>

Just as a pair of edos and a prime subgroup can specify a rank-2 temperament, a pair of edos can specify a rank-2 pergen. For N-edo &amp; N'-edo, m = GCD (N,N'). The period P equals both (N/m)\N and (N'/m)\N'. For example, for 12edo and 16edo, m = 4, and the period is both 3\12 and 4\16, i.e. a quarter-8ve.

For example, 12edo's 3-limit edomapping is (12, 19), and 16edo's is (16, 25). The determinant of [(12 19) (16 25)] is -4, and the pergen for 12edo and 16edo is (P8/4, P5). To make the calculations easier, octave-reduced edomappings can be used, which indicate the number of edosteps that 3/2 maps to, not 3/1. For 12 and 16, we have [(12 7) (16 9)], and d is again -4.

If |d| ≠ m, we must find the multigen. Take the ratio of the two edos N/N' and reduce it by m. In the scale tree, find the two ancestors and the two descendants of this ratio. Choose one of the four (more on this below), and let this new ratio be g/g'. The generator G maps to both g\N and g'\N'. The two edos come closest to coinciding here.

Next we must find a comma that both edos temper out, and find the pergen from that comma. To do this, we extend the edomappings to three entries. Rather than finding the mapping of 5/4 or 7/4, we use the generator we found earlier, without concerning ourselves about which ratio it corresponds to, in keeping with the higher-prime-agnostic nature of pergens. Treating the two edomappings as 3-D vectors, take the cross product of them, whch makes a new vector which is perpendicular to both. Thus the dot product of this vector with either edomapping is zero. Treating this new vector as a monzo, both edos map this monzo to zero edosteps. In other words, they temper out this monzo, and this monzo is the comma we're looking for.

To verify the validity of this approach, one can find a specific ratio that maps to both 2\7 and 5\17 and use it to construct a comma. The ratio must contain only one higher prime, and must have color depth 1. If desired, a ratio of the form p/t can always be found, where p is a higher prime and t is a power of two. Any ratio between 3\14 and 5\14 maps to 2\7, and any ratio between 9\34 and 11\34 maps to 5\17. (The formula is (2n±1)/2d.) This gives us the ranges 257-429¢ and 318-388¢. Thus 5/4 barely works. The comma is (0, -1, 2) dot (2/1, 3/2, 5/4) = 25/24 = 71¢. A smaller comma can be found with a ratio more in the center of the range, such as 11/9, which yeilds 242/243. Because the direction of the cross product vector depends on the order of the two vectors multiplied, the sign of the comma is arbitrary, and the comma may be a descending interval. In that case, invert the comma to make 243/242. Using unreduced edomappings gives the same result. If the basis is (2/1, 3/1, 5/1), we have (7, 11, 16) x (17, 27, 39) = (-3, -1, 2) = 25/24.

Just as a pair of edos and a prime subgroup can specify a rank-2 temperament, a pair of edos can specify a rank-2 pergen. For N-edo &amp;amp; N'-edo, m = GCD (N,N'). The period P equals both (N/m)\N and (N'/m)\N'. For example, for 12edo and 16edo, m = 4, and the period is both 3\12 and 4\16, i.e. a quarter-8ve.&lt;br /&gt;

&lt;br /&gt;

&lt;br /&gt;

For example, 12edo's 3-limit edomapping is (12, 19), and 16edo's is (16, 25). The determinant of [(12 19) (16 25)] is -4, and the pergen for 12edo and 16edo is (P8/4, P5). To make the calculations easier, octave-reduced edomappings can be used, which indicate the number of edosteps that 3/2 maps to, not 3/1. For 12 and 16, we have [(12 7) (16 9)], and d is again -4.&lt;br /&gt;

If |d| ≠ m, we must find the multigen. Take the ratio of the two edos N/N' and reduce it by m. In the scale tree, find the two ancestors and the two descendants of this ratio. Choose one of the four (more on this below), and let this new ratio be g/g'. The generator G maps to both g\N and g'\N'. The two edos come closest to coinciding here.&lt;br /&gt;

&lt;br /&gt;

&lt;br /&gt;

Next we must find a comma that both edos temper out, and find the pergen from that comma. To do this, we extend the edomappings to three entries. Rather than finding the mapping of 5/4 or 7/4, we use the generator we found earlier, without concerning ourselves about which ratio it corresponds to, in keeping with the higher-prime-agnostic nature of pergens. Treating the two edomappings as 3-D vectors, take the cross product of them, whch makes a new vector which is perpendicular to both. Thus the dot product of this vector with either edomapping is zero. Treating this new vector as a monzo, both edos map this monzo to zero edosteps. In other words, they temper out this monzo, and this monzo is the comma we're looking for.&lt;br /&gt;

&lt;br /&gt;

&lt;br /&gt;

To verify the validity of this approach, one can find a specific ratio that maps to both 2\7 and 5\17 and use it to construct a comma. The ratio must contain only one higher prime, and must have color depth 1. If desired, a ratio of the form p/t can always be found, where p is a higher prime and t is a power of two. Any ratio between 3\14 and 5\14 maps to 2\7, and any ratio between 9\34 and 11\34 maps to 5\17. (The formula is (2n±1)/2d.) This gives us the ranges 257-429¢ and 318-388¢. Thus 5/4 barely works. The comma is (0, -1, 2) dot (2/1, 3/2, 5/4) = 25/24 = 71¢. A smaller comma can be found with a ratio more in the center of the range, such as 11/9, which yeilds 242/243. Because the direction of the cross product vector depends on the order of the two vectors multiplied, the sign of the comma is arbitrary, and the comma may be a descending interval. In that case, invert the comma to make 243/242. Using unreduced edomappings gives the same result. If the basis is (2/1, 3/1, 5/1), we have (7, 11, 16) x (17, 27, 39) = (-3, -1, 2) = 25/24.&lt;br /&gt;