Pergen names: 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>

==Extremely large multi-gens==  

==Singles and doubles==  

==Finding an example temperament==  

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, simply find a ratio for P or G that contains only one higher prime, 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*P and P8. If P is 6/5, the comma is 4*P - P8 = (6/5)^4 / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4*P = (2/1) * (7/6)^-4. Neither 13/11 nor 32/27 would work for P, too many and too few primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3*G - P4 = (10/9)^3 / (4/3) = 250/243.

For example, half-octave (P8/2, P5) has generator P5, alternate generators P4 and vA1, period vA4, and equivalent period ^d5. (P8, P5/2) has generator ^m3 and equivalent generator vM3. Of the two equivalent generators, "the" generator is always the smaller one (smaller degree, or if the degrees are the same, more diminished). This is because the equivalent generator can be more easily found by adding the enharmonic, rather than subtracting it. For example, ^m3 + vvA1 = vM3 is an easier calculation than vM3 - vvA1 = ^m3. This is particularly true with complex enharmonics like ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;dd2.

==Finding a notation for a pergen==  

b = 7a - 12b

Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. For example, consider (P8/5, P5). One fifth of an octave [12,7] is approximately [round(12/5), round(7/5)] = [2,1] = (-3,2) = M2. This approximation will __always__ produce an enharmonic interval with the smallest possible keyspan and stepspan, which is usually the preferred enharmonic. In this example, the bare enharmonic is the difference between P8 and 5*M2, which is [12,7] - 5*[2,1] = [12,7] - [10,5] = [2,2] = 2*[1,1] = 2*m2. Because the enharmonic E is multiplied by 2, it will occur twice in the perchain, even thought the comma only occurs once. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. Equipped with the period and the enharmonic, the perchain is easily found:

==Alternate enharmonics==  

For single-comma pergens, the enharmonic should equal the comma's mapping. For example, (P8, P5/2) might arise from 243/242, which splits the 5th into two 11/9 halves.

(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4*G = [1,2] = dd3. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.

One might wonder, why 12-edo keyspans? Why heptatonic stepspans? Heptatonic is best because conventional notation is heptatonic, and we want to minimize the heptatonic stepspan of the enharmonic. In order for the matrix to be invertible, the edo must be connected to 7-edo on the scale tree. The only choices are 5-edo, 12-edo, 19-edo, 26-edo, 33-edo, 40-edo and 47-edo, and on the other side, 2-edo, 9-edo, 16-edo and 23-edo.

__**Extra paragraphs:**__

Not all enharmonics work with all pergens. There are two logical restrictions on the enharmonic, one based on its degree and the other on its 3-exponent (implied edo).

For {P8/M, multi-gen/N}, there are two conditions on the enharmonic. If T is the 3-exponent of the multi-gen, the conditions are edo = Mx and edo = Ny ± T. For {P8/2, P4/2}, the two conditions are mutually exclusive: the edo must be both even and odd. Therefore there must be two accidental pairs, each with its own enharmonic interval. In the main table, this pergen is notated with both ups/downs and highs/lows. Since the 8ve and 4th are split, the 5th is too. Each interval has its own genchain. One of these is notated with ups/downs, another with highs/lows, and the third with both. The 3 possible ways of allocating the two accidental pairs are all listed. Furthermore, ups/downs can be exchanged for highs/lows.

For **{P8/M, P5}**, the implied edo = Mx

&lt;!-- ws:end:WikiTextTocRule:51 --&gt;&lt;!-- ws:start:WikiTextTocRule:52: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Finding a notation for a pergen"&gt;Finding a notation for a pergen&lt;/a&gt;&lt;/div&gt;

&lt;!-- ws:end:WikiTextTocRule:52 --&gt;&lt;!-- ws:start:WikiTextTocRule:53: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Alternate enharmonics"&gt;Alternate enharmonics&lt;/a&gt;&lt;/div&gt;

&lt;!-- ws:end:WikiTextTocRule:54 --&gt;&lt;!-- ws:start:WikiTextTocRule:55: --&gt;&lt;/div&gt;

&lt;!-- ws:end:WikiTextTocRule:55 --&gt;&lt;!-- ws:start:WikiTextHeadingRule:23:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc1"&gt;&lt;a name="Definition"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:23 --&gt;&lt;u&gt;&lt;strong&gt;Definition&lt;/strong&gt;&lt;/u&gt;&lt;/h1&gt;

&lt;!-- ws:start:WikiTextHeadingRule:31:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc5"&gt;&lt;a name="Further Discussion-Extremely large multi-gens"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:31 --&gt;Extremely large multi-gens&lt;/h2&gt;

  &lt;br /&gt;

&lt;br /&gt;

&lt;!-- ws:start:WikiTextHeadingRule:33:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc6"&gt;&lt;a name="Further Discussion-Singles and doubles"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:33 --&gt;Singles and doubles&lt;/h2&gt;

&lt;!-- ws:start:WikiTextHeadingRule:35:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc7"&gt;&lt;a name="Further Discussion-Finding an example temperament"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:35 --&gt;Finding an example temperament&lt;/h2&gt;

  &lt;br /&gt;

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, simply find a ratio for P or G that contains only one higher prime, 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*P and P8. If P is 6/5, the comma is 4*P - P8 = (6/5)^4 / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4*P = (2/1) * (7/6)^-4. Neither 13/11 nor 32/27 would work for P, too many and too few primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3*G - P4 = (10/9)^3 / (4/3) = 250/243.&lt;br /&gt;

&lt;br /&gt;

For example, half-octave (P8/2, P5) has generator P5, alternate generators P4 and vA1, period vA4, and equivalent period ^d5. (P8, P5/2) has generator ^m3 and equivalent generator vM3. Of the two equivalent generators, &amp;quot;the&amp;quot; generator is always the smaller one (smaller degree, or if the degrees are the same, more diminished). This is because the equivalent generator can be more easily found by adding the enharmonic, rather than subtracting it. For example, ^m3 + vvA1 = vM3 is an easier calculation than vM3 - vvA1 = ^m3. This is particularly true with complex enharmonics like ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;dd2.&lt;br /&gt;

&lt;br /&gt;

&lt;br /&gt;

&lt;!-- ws:start:WikiTextHeadingRule:37:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc8"&gt;&lt;a name="Further Discussion-Finding a notation for a pergen"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:37 --&gt;Finding a notation for a pergen&lt;/h2&gt;

b = 7a - 12b&lt;br /&gt;

&lt;br /&gt;

&lt;br /&gt;

Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. For example, consider (P8/5, P5). One fifth of an octave [12,7] is approximately [round(12/5), round(7/5)] = [2,1] = (-3,2) = M2. This approximation will &lt;u&gt;always&lt;/u&gt; produce an enharmonic interval with the smallest possible keyspan and stepspan, which is usually the preferred enharmonic. In this example, the bare enharmonic is the difference between P8 and 5*M2, which is [12,7] - 5*[2,1] = [12,7] - [10,5] = [2,2] = 2*[1,1] = 2*m2. Because the enharmonic E is multiplied by 2, it will occur twice in the perchain, even thought the comma only occurs once. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. Equipped with the period and the enharmonic, the perchain is easily found:&lt;br /&gt;

&lt;!-- ws:start:WikiTextHeadingRule:39:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc9"&gt;&lt;a name="Further Discussion-Alternate enharmonics"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:39 --&gt;Alternate enharmonics&lt;/h2&gt;

  &lt;br /&gt;

For single-comma pergens, the enharmonic should equal the comma's mapping. For example, (P8, P5/2) might arise from 243/242, which splits the 5th into two 11/9 halves.&lt;br /&gt;

&lt;br /&gt;

(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4*G = [1,2] = dd3. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.&lt;br /&gt;

&lt;br /&gt;

  &lt;br /&gt;

One might wonder, why 12-edo keyspans? Why heptatonic stepspans? Heptatonic is best because conventional notation is heptatonic, and we want to minimize the heptatonic stepspan of the enharmonic. In order for the matrix to be invertible, the edo must be connected to 7-edo on the scale tree. The only choices are 5-edo, 12-edo, 19-edo, 26-edo, 33-edo, 40-edo and 47-edo, and on the other side, 2-edo, 9-edo, 16-edo and 23-edo.&lt;br /&gt;

&lt;br /&gt;

&lt;u&gt;&lt;strong&gt;Extra paragraphs:&lt;/strong&gt;&lt;/u&gt;&lt;br /&gt;

&lt;br /&gt;

Not all enharmonics work with all pergens. There are two logical restrictions on the enharmonic, one based on its degree and the other on its 3-exponent (implied edo).&lt;br /&gt;

For {P8/M, multi-gen/N}, there are two conditions on the enharmonic. If T is the 3-exponent of the multi-gen, the conditions are edo = Mx and edo = Ny ± T. For {P8/2, P4/2}, the two conditions are mutually exclusive: the edo must be both even and odd. Therefore there must be two accidental pairs, each with its own enharmonic interval. In the main table, this pergen is notated with both ups/downs and highs/lows. Since the 8ve and 4th are split, the 5th is too. Each interval has its own genchain. One of these is notated with ups/downs, another with highs/lows, and the third with both. The 3 possible ways of allocating the two accidental pairs are all listed. Furthermore, ups/downs can be exchanged for highs/lows.&lt;br /&gt;

&lt;br /&gt;

&lt;br /&gt;

&lt;br /&gt;

For &lt;strong&gt;{P8/M, P5}&lt;/strong&gt;, the implied edo = Mx&lt;br /&gt;