Kite's thoughts on pergens: Difference between revisions

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C -- E^=Fv\ -- Ab/=A\ -- C^/=Dbv -- F

One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. We have [3,2]/12 = [0,0] = P1, and G' = ^1 and E = v12m3. Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus E' = /3d2, and 4·G' = /m2. The period can be deduced from 4·G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4·G' = P4 - /m2 = \M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = \M3 + ^1 = ^\M3. Equivalent enharmonics are found from E + E' and E - 2·E'. Equivalent periods and generators are found from the many enharmonics, which also allow much freedom in chord spelling.

One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. We have [3,2]/12 = [0,0] = P1, and G' = ^1 and E = v12m3. Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus E' = /3d2, and 4·G' = /m2. The period can be deduced from 4·G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4·G' = P4 - /m2 = \M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = \M3 + ^1 = ^\M3. Equivalent enharmonics are found from E + E' and E - 2·E'. Equivalent periods and generators are found from the many enharmonics, which also allow much freedom in chord spelling. Enharmonic = v12m3 = /3d2 = v4/m2 = v4\\A1. Period = \M3 = v44 = //d4. Generator = ^\M3 = v34 = ^//d4.

Enharmonic = v12m3 = /3d2 = v4/m2 = v4\\A1. Period = \M3 = v44 = //d4. Generator = ^\M3 = v34 = ^//d4.

P1 — \M3 — \\A5=/m6 — P8C — E\ — Ab/ — CP1 — ^\M3 — ^^\\A5=^^/m6=vv\M6 — ^38=v/m9 — P11

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It's not yet known if every pergen can avoid large enharmonics (those of a 3rd or more) with double-pair notation. One situation in which very large enharmonics occur is the "half-step glitch". This is when the stepspan of the multigen is half (or a third, a quarter, etc.) of the multigen's splitting fraction. For example, in sixth-4th, six generators must cover three scale steps, and each one must cover a half-step. Each generator is either a unison or a 2nd, which causes the enharmonic's stepspan to equal the multigen's stepspan.

Sixth-4th with single-pair notation has an awkward ^6d64 enharmonic. This pergen might result from combining third-4th and half-4th (e.g. tempering out both the Porcupine and Semaphore commas, aka Triyo & Zozo), and its double-pair notation can also combine both. Third-4th has E = v3A1 and G'= vM2 = ^^m2. Half-4th has E' = \\m2 and G' = /M2 = \m3. G' - G = P4/2 - P4/3 = P4/6. Thus the sixth-4th generator is G' - G = /M2 - vM2 = ^/1. Equivalent enharmonics are

Sixth-4th with single-pair notation has an awkward ^6d64 enharmonic. This pergen might result from combining third-4th and half-4th (e.g. tempering out both the Porcupine and Semaphore commas, aka Triyo & Zozo), and its double-pair notation can also combine both. Third-4th has E = v3A1 and G'= vM2 = ^^m2. Half-4th has E' = \\m2 and G' = /M2 = \m3. G' - G = P4/2 - P4/3 = P4/6. Thus the sixth-4th generator is G' - G = /M2 - vM2 = ^/1. Equivalent enharmonics are v3\\M2 and ^3\\d2.

P1 — ^/1=^\m2 — ^^m2=vM2 — /M2=\m3 — ^m3=vvM3 — v/M3=v\4 — P4

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 <span style="display: block; text-align: center;">C -- E^=Fv\ -- Ab/=A\ -- C^/=Dbv -- F</span>
-One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. We have [3,2]/12 = [0,0] = P1, and G' = ^1 and E = v<span style="vertical-align: super;">12</span>m3. Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus E' = /<span style="vertical-align: super;">3</span>d2, and 4·G' = /m2. The period can be deduced from 4·G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4·G' = P4 - /m2 = \M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = \M3 + ^1 = ^\M3. Equivalent enharmonics are found from E + E' and E - 2·E'. Equivalent periods and generators are found from the many enharmonics, which also allow much freedom in chord spelling.
+One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. We have [3,2]/12 = [0,0] = P1, and G' = ^1 and E = v<span style="vertical-align: super;">12</span>m3. Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus E' = /<span style="vertical-align: super;">3</span>d2, and 4·G' = /m2. The period can be deduced from 4·G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4·G' = P4 - /m2 = \M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = \M3 + ^1 = ^\M3. Equivalent enharmonics are found from E + E' and E - 2·E'. Equivalent periods and generators are found from the many enharmonics, which also allow much freedom in chord spelling. Enharmonic = v<span style="vertical-align: super;">12</span>m3 = /<span style="vertical-align: super;">3</span>d2 = v<span style="vertical-align: super;">4</span>/m2 = v<span style="vertical-align: super;">4</span>\\A1. Period = \M3 = v<span style="vertical-align: super;">4</span>4 = //d4. Generator = ^\M3 = v<span style="vertical-align: super;">3</span>4 = ^//d4.
-Enharmonic = v<span style="vertical-align: super;">12</span>m3 = /<span style="vertical-align: super;">3</span>d2 = v<span style="vertical-align: super;">4</span>/m2 = v<span style="vertical-align: super;">4</span>\\A1. Period = \M3 = v<span style="vertical-align: super;">4</span>4 = //d4. Generator = ^\M3 = v<span style="vertical-align: super;">3</span>4 = ^//d4.
 <span style="display: block; text-align: center;">P1 — \M3 — \\A5=/m6 — P8</span><span style="display: block; text-align: center;">C — E\ — Ab/ — C</span><span style="display: block; text-align: center;">P1 — ^\M3 — ^^\\A5=^^/m6=vv\M6 — ^<span style="vertical-align: super;">3</span>8=v/m9 — P11
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 It's not yet known if every pergen can avoid large enharmonics (those of a 3rd or more) with double-pair notation. One situation in which very large enharmonics occur is the "half-step glitch". This is when the stepspan of the multigen is half (or a third, a quarter, etc.) of the multigen's splitting fraction. For example, in sixth-4th, six generators must cover three scale steps, and each one must cover a half-step. Each generator is either a unison or a 2nd, which causes the enharmonic's stepspan to equal the multigen's stepspan.
-Sixth-4th with single-pair notation has an awkward ^<span style="vertical-align: super;">6</span>d<span style="vertical-align: super;">6</span>4 enharmonic. This pergen might result from combining third-4th and half-4th (e.g. tempering out both the Porcupine and Semaphore  commas, aka Triyo & Zozo), and its double-pair notation can also combine both. Third-4th has E = v<span style="vertical-align: super;">3</span>A1 and G'= vM2 = ^^m2. Half-4th has E' = \\m2 and G' = /M2 = \m3. G' - G = P4/2 - P4/3 = P4/6. Thus the sixth-4th generator is G' - G = /M2 - vM2 = ^/1. Equivalent enharmonics are
+Sixth-4th with single-pair notation has an awkward ^<span style="vertical-align: super;">6</span>d<span style="vertical-align: super;">6</span>4 enharmonic. This pergen might result from combining third-4th and half-4th (e.g. tempering out both the Porcupine and Semaphore  commas, aka Triyo & Zozo), and its double-pair notation can also combine both. Third-4th has E = v<span style="vertical-align: super;">3</span>A1 and G'= vM2 = ^^m2. Half-4th has E' = \\m2 and G' = /M2 = \m3. G' - G = P4/2 - P4/3 = P4/6. Thus the sixth-4th generator is G' - G = /M2 - vM2 = ^/1. Equivalent enharmonics are v<sup>3</sup>\\M2 and ^<sup>3</sup>\\d2.
 <span style="display: block; text-align: center;">P1 — ^/1=^\m2 — ^^m2=vM2 — /M2=\m3 — ^m3=vvM3 — v/M3=v\4 — P4