Kite's thoughts on pergens: Difference between revisions

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==Tipping points==

Removing the ups and downs from an enharmonic interval makes ~~a "bare"~~ enharmonic, a conventional 3-limit interval which vanishes in certain edos. For example, (P8/2, P5)'s enharmonic interval is ^^d2, the ~~bare~~ enharmonic is d2, and d2 vanishes in 12-edo. Every rank-2 temperament has a "sweet spot" for tuning the 5th, usually a narrow range of about 5-10¢. 12-edo's fifth is the "tipping point": if the temperament's 5th is flatter than 12-edo's, d2 is ascending, and if it's sharper, it's descending. The ups and downs are meant to indicate that the enharmonic interval vanishes. Thus if d2 is ascending, it should be downed, and if it's descending, upped. Therefore '''up may need to be swapped with down, depending on the size of the 5th''' in the particular rank-2 tuning you are using. In the above table, this is shown explicitly for (P8/2, P5), and implied for all the other pergens. In the table, the other pergens' enharmonic intervals are upped or downed as if the 5th were just.

Removing the ups and downs from an enharmonic interval makes an '''uninflected''' enharmonic, a conventional 3-limit interval which vanishes in certain edos. For example, (P8/2, P5)'s enharmonic interval is ^^d2, the uninflected enharmonic is d2, and d2 vanishes in 12-edo. Every rank-2 temperament has a "sweet spot" for tuning the 5th, usually a narrow range of about 5-10¢. 12-edo's fifth is the "tipping point": if the temperament's 5th is flatter than 12-edo's, d2 is ascending, and if it's sharper, it's descending. The ups and downs are meant to indicate that the enharmonic interval vanishes. Thus if d2 is ascending, it should be downed, and if it's descending, upped. Therefore '''up may need to be swapped with down, depending on the size of the 5th''' in the particular rank-2 tuning you are using. In the above table, this is shown explicitly for (P8/2, P5), and implied for all the other pergens. In the table, the other pergens' enharmonic intervals are upped or downed as if the 5th were just.

Heptatonic 5th-based notation is only possible if the 5th ranges from 600¢ to 720¢. For every ~~bare~~ enharmonic, the following table shows in what parts of this range this interval should be upped or downed. The tipping point edo is simply the 3-exponent of the ~~bare~~ enharmonic.

Heptatonic 5th-based notation is only possible if the 5th ranges from 600¢ to 720¢. For every uninflected enharmonic, the following table shows in what parts of this range this interval should be upped or downed. The tipping point edo is simply the 3-exponent of the uninflected enharmonic.

{| class="wikitable" style="text-align:center;"

|-

! colspan="2" | ~~bare~~ enharmonic

! colspan="2" | uninflected enharmonic

interval

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vM7/2: C - ^F - vB (vM7 = 15/8, probably more harmonious than M7 = 243/128)

More remote intervals include A1, d4, d7 and d10. These unfortunately require a very long genchain. The most interesting melodically is A1: C - ^C - vC# - C#. From C to C# is seven 5ths, which equals 21 generators, so the genchain would have to contain 22 notes if it had no gaps. Note that A1 is the ~~bare~~ enharmonic of third-4th. The ~~bare~~ enharmonic is always a secondary split.

More remote intervals include A1, d4, d7 and d10. These unfortunately require a very long genchain. The most interesting melodically is A1: C - ^C - vC# - C#. From C to C# is seven 5ths, which equals 21 generators, so the genchain would have to contain 22 notes if it had no gaps. Note that A1 is the uninflected enharmonic of third-4th. The uninflected enharmonic is always a secondary split.

For a pergen (P8, (a,b)/n), any interval generated by n octaves and the multigen splits into at least n parts. For a pergen (P8/m, P5), any interval generated by the octave and m 5ths splits into at least m parts. Thus any naturally occurring split of m parts occurs in all voicings of that interval. For example, M9 naturally splits into two 5ths, therefore (P8/2, P5) splits all voicings of M9, including M2.

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For example, consider the half-5th pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = m3. The enharmonic can also be found using gedras: xEI = M - n⋅G = P5 - 2⋅m3 = [7,4] - 2⋅[3,2] = [7,4] - [6,4] = [1,0] = A1.

Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xEI = P8 - mP = P8 - 5⋅M2 = [12,7] - 5⋅[2,1] = [2,2] = 2⋅[1,1] = 2⋅m2. Because x = 2, EI will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2⋅m2 = d3). The enharmonic's '''count''' is 2. The ~~bare~~ enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus EI = v5m2. Since P8 = 5⋅P + 2⋅EI, the period must be ^^M2, to make the ups and downs come out even. The number of the period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:

Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xEI = P8 - mP = P8 - 5⋅M2 = [12,7] - 5⋅[2,1] = [2,2] = 2⋅[1,1] = 2⋅m2. Because x = 2, EI will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2⋅m2 = d3). The enharmonic's '''count''' is 2. The uninflected enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus EI = v5m2. Since P8 = 5⋅P + 2⋅EI, the period must be ^^M2, to make the ups and downs come out even. The number of the period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:

P1 -- ^^M2=v3m3 -- v4 -- ^5 -- ^3M6=vvm7 -- P8

C -- ^^D=v3Eb -- vF -- ^G -- ^3A=vvBb -- C

Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has ~~a bare~~ generator [5,3]/5 = [1,1] = m2. The ~~bare~~ enharmonic is P4 - 5⋅m2 = [5,3] - 5⋅[1,1] = [5,3] - [5,5] = [0,-2] = -2⋅[0,1] = two descending d2's. The d2 must be upped, and EI = ^5d2. Since P4 = 5⋅G - 2⋅EI, G must be ^^m2. The genchain is:

Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has an uninflected generator [5,3]/5 = [1,1] = m2. The uninflected enharmonic is P4 - 5⋅m2 = [5,3] - 5⋅[1,1] = [5,3] - [5,5] = [0,-2] = -2⋅[0,1] = two descending d2's. The d2 must be upped, and EI = ^5d2. Since P4 = 5⋅G - 2⋅EI, G must be ^^m2. The genchain is:

P1 -- ^^m2=v3A1 -- vM2 -- ^m3 -- ^3d4=vvM3 -- P4C -- ^^Db -- vD -- ^Eb -- vvE -- F

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To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multigen as before. Then deduce the period from the enharmonic. If the multigen was changed by unreducing, find the original generator from the period and the alternate generator.

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The ~~bare~~ alternate generator G' is [1,1]/10 = [0,0] = P1. The ~~bare~~ enharmonic is m2 - 10⋅P1 = m2. It must be downed, thus EI = v10m2. Since m2 = 10⋅G' + EI, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x⋅m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5⋅P + 2⋅EI, and P = ^4M2. Next, find the original half-4th generator. P = P8/5 = ~240¢, and G = P4/2 = ~250¢. Because P < G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^4M2 + ^1 = ^5M2. While the alternate multigen is more complex than the original multigen, the alternate generator is usually simpler than the original generator.

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The uninflected alternate generator G' is [1,1]/10 = [0,0] = P1. The uninflected enharmonic is m2 - 10⋅P1 = m2. It must be downed, thus EI = v10m2. Since m2 = 10⋅G' + EI, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x⋅m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5⋅P + 2⋅EI, and P = ^4M2. Next, find the original half-4th generator. P = P8/5 = ~240¢, and G = P4/2 = ~250¢. Because P < G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^4M2 + ^1 = ^5M2. While the alternate multigen is more complex than the original multigen, the alternate generator is usually simpler than the original generator.

P1- - ^4M2=v6m3 -- vv4 -- ^^5 -- ^6M6=v4m7 -- P8

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A false-double pergen can optionally use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is not a unison or a 2nd, as with (P8/2, P4/3).

Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅M3 = [1,2] = dd3. So EI = v4dd3, and G = ^M3. But by using double-pair notation, we can avoid that. We find P11/2, which equals two generators: P11/2 = 2⋅G = [17,10]/2 = [8,5] = m6. The ~~bare~~ enharmonic is P11 - 2⋅m6 = [1,0] = A1. For this second enharmonic, we use the second pair of accidentals: EI' = \\A1 and 2⋅G = /m6 or \M6. The sum or difference of two enharmonic intervals is also an enharmonic: EI + EI' = v4\\d3 = 2·vv\m2, and EI - EI' = v4//ddd3 = 2·vv/d2. Thus vv\m2 and vv/d2 are equivalent enharmonics, and v\4 and v/d4 are equivalent generators. Here is the genchain:

Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅M3 = [1,2] = dd3. So EI = v4dd3, and G = ^M3. But by using double-pair notation, we can avoid that. We find P11/2, which equals two generators: P11/2 = 2⋅G = [17,10]/2 = [8,5] = m6. The uninflected enharmonic is P11 - 2⋅m6 = [1,0] = A1. For this second enharmonic, we use the second pair of accidentals: EI' = \\A1 and 2⋅G = /m6 or \M6. The sum or difference of two enharmonic intervals is also an enharmonic: EI + EI' = v4\\d3 = 2·vv\m2, and EI - EI' = v4//ddd3 = 2·vv/d2. Thus vv\m2 and vv/d2 are equivalent enharmonics, and v\4 and v/d4 are equivalent generators. Here is the genchain:

P1 -- ^M3=v\4 -- /m6=\M6 -- ^/8=vm9 -- P11

C -- ^E=v\F -- /Ab=\A -- ^/C=vDb -- 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 EI = 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 EI' = /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 EI + EI' and EI - 2·EI'. 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.

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 EI = v12m3. Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the uninflected enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus EI' = /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 EI + EI' and EI - 2·EI'. 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.

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

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==Alternate enharmonics==

Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, ccM6/12), a false double. The ~~bare~~ alternate generator is ccM6/12 = [33,19]/12 = [3,2] = m3. The ~~bare~~ enharmonic is [33,19] - 12·[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12·[4,2] = [-15,-5] = -5·[3,1] = -5·v12A2, which is an improvement but still awkward. The period is ^4m3 and the generator is v3M2. ^1 = 25¢ + 0.75·c, about an eighth-tone.

Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, ccM6/12), a false double. The uninflected alternate generator is ccM6/12 = [33,19]/12 = [3,2] = m3. The uninflected enharmonic is [33,19] - 12·[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12·[4,2] = [-15,-5] = -5·[3,1] = -5·v12A2, which is an improvement but still awkward. The period is ^4m3 and the generator is v3M2. ^1 = 25¢ + 0.75·c, about an eighth-tone.

P1 -- ^4m3 -- v4M6 -- P8

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To search for alternate enharmonics, convert EI to a gedra, then multiply it by the count to get the multi-EI (multi-enharmonic). The count is always the number of ups or downs in the generator (or period). The count is positive only if G (or P) is upped and EI is downed, or vice versa. Add or subtract the splitting fraction n (or m) to/from either half of the gedra as desired to get a new multi-EI. If the stepspan becomes negative, or if it's zero and the keyspan becomes negative, invert the gedra. If the two halves of the new gedra have a common factor, simplify the gedra by this factor, which becomes the new count. Convert the simplified gedra to a 3-limit interval. Add n (or m) ups or downs, this is the new EI. Choose between ups and downs according to whether the 5th falls in the enharmonic's upping or downing range, see [[pergen#Applications-Tipping points|tipping points]] above. Add n'''·'''count ups or downs to the new multi-EI. Add or subtract the new multi-EI from the multigen (or the octave) to get an interval which splits cleanly into n (or m) parts. Each part is the new generator (or period).

For example, (P8, P5/3) has n = 3, G = ^M2, and EI = v3m2 = [1,1]. G is upped only once, so the count is 1, and the multi-EI is also [1,1]. Subtract n from the gedra's keyspan to make a new multi-EI [-2,1]. This can't be simplified, so the new EI is also [-2,1] = d32. Assuming a reasonably just 5th, EI needs to be upped, so EI = ^3d32. Add the multi-EI ^3[-2,1] to the multigen P5 = [7,4] to get ^3[5,3]. This isn't divisible by n, so we must subtract instead: [7,4] - ^3[-2,1] = v3[9,3] = 3·v[3,1], and G = vA2. Use the equation M = n·G + ~~y·E~~ to check: 3·vA2 + 1·^3d32 = 3·M2 + 1·m2 = P5. (Diminish three A2's once and augment one d32 three times.) Here are the genchains: P1 -- vA2=^^dd3 -- ^d4 -- P5 and C -- vD# -- ^Fb -- G. Because d32 = -200¢ - 26·c, ^ = (-d32) / 3 = 67¢ + 8.67·c, about a third-tone.

For example, (P8, P5/3) has n = 3, G = ^M2, and EI = v3m2 = [1,1]. G is upped only once, so the count is 1, and the multi-EI is also [1,1]. Subtract n from the gedra's keyspan to make a new multi-EI [-2,1]. This can't be simplified, so the new EI is also [-2,1] = d32. Assuming a reasonably just 5th, EI needs to be upped, so EI = ^3d32. Add the multi-EI ^3[-2,1] to the multigen P5 = [7,4] to get ^3[5,3]. This isn't divisible by n, so we must subtract instead: [7,4] - ^3[-2,1] = v3[9,3] = 3·v[3,1], and G = vA2. Use the equation M = n·G + y·EI to check: 3·vA2 + 1·^3d32 = 3·M2 + 1·m2 = P5. (Diminish three A2's once and augment one d32 three times.) Here are the genchains: P1 -- vA2=^^dd3 -- ^d4 -- P5 and C -- vD# -- ^Fb -- G. Because d32 = -200¢ - 26·c, ^ = (-d32) / 3 = 67¢ + 8.67·c, about a third-tone.

Approaching pergens in a higher-primes-agnostic way, independently of specific commas or temperaments, one enharmonic (and one notation) will usually be obviously superior. But sometimes the temperament being notated implies a certain enharmonic. Specifically, the comma tempered out should map to EI, or the multi-EI if the count is > 1. For example, consider Semaphore aka Zozo (2.3.7 and 49/48), which is half-4th. Assuming 7/4 is a m7, the comma is a m2, and a vvm2 enharmonic makes sense. G = ^M2 and the genchain is C -- ^D=vEb -- F. But consider Lala-yoyo, which tempers out (-22, 11, 2). This temperament is also half-4th. The comma is a descending dd2, thus EI = ^^dd2, G = vA2, and the genchain is C -- vD#=^Ebb -- F. This is the best notation because the (-10, 5, 1) generator is an augmented 2nd.

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The tipping point depends on the choice of enharmonic. It's not the temperament that tips, it's the notation. Half-8ve could be notated with an EI of vvM2. The tipping point becomes 600¢, a very unlikely 5th, and tipping is impossible. For single-comma temperaments, EI usually equals the 3-limit mapping of the comma. Thus for 5-limit and 7-imit temperaments, the choice of EI is a given. However, the mapping of primes 11 and 13 is not agreed on.

The notation's tipping point is determined by the ~~bare~~ enharmonic, which is implied by the vanishing comma. For example, Porcupine aka Triyo's 250/243 comma is an A1 = (-11,7), which implies ~~a bare~~ EI of A1, which implies 7-edo, and a 685.7¢ tipping point. Dicot aka Yoyo's 25/24 comma is also an A1, and has the same tipping point. Semaphore aka Zozo's 49/48 comma is a minor 2nd = (8,-5), implying 5-edo and a 720¢ tipping point. See [[pergen#Applications-Tipping points|tipping points]] above for a more complete list.

The notation's tipping point is determined by the uninflected enharmonic, which is implied by the vanishing comma. For example, Porcupine aka Triyo's 250/243 comma is an A1 = (-11,7), which implies an uninflected EI of A1, which implies 7-edo, and a 685.7¢ tipping point. Dicot aka Yoyo's 25/24 comma is also an A1, and has the same tipping point. Semaphore aka Zozo's 49/48 comma is a minor 2nd = (8,-5), implying 5-edo and a 720¢ tipping point. See [[pergen#Applications-Tipping points|tipping points]] above for a more complete list.

Double-pair notation has two enharmonics, and two tipping points to be avoided. (P8/2, P4/2) has three possible notations. The two enharmonics can be either A1, m2 or d2. One can choose to use whichever two enharmonics best avoid tipping.

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== Addenda ==

Ups and downs are collectively called [[Arrow|'''arrows''']]. Lifts and drops are collectively called '''slants'''. Arrows and slants are collectively called '''inflections'''. Thus a better term for "bare" enharmonic is uninflected enharmonic.

[[Category:Regular temperament theory]]

[[Category:Notation]]

@@ Line 864: / Line 864: @@
 ==Tipping points==
-Removing the ups and downs from an enharmonic interval makes a "bare" enharmonic, a conventional 3-limit interval which vanishes in certain edos. For example, (P8/2, P5)'s enharmonic interval is ^^d2, the bare enharmonic is d2, and d2 vanishes in 12-edo. Every rank-2 temperament has a "sweet spot" for tuning the 5th, usually a narrow range of about 5-10¢. 12-edo's fifth is the "tipping point": if the temperament's 5th is flatter than 12-edo's, d2 is ascending, and if it's sharper, it's descending. The ups and downs are meant to indicate that the enharmonic interval vanishes. Thus if d2 is ascending, it should be downed, and if it's descending, upped. Therefore <u>'''up may need to be swapped with down, depending on the size of the 5th'''</u> in the particular rank-2 tuning you are using. In the above table, this is shown explicitly for (P8/2, P5), and implied for all the other pergens. In the table, the other pergens' enharmonic intervals are upped or downed as if the 5th were just.
+Removing the ups and downs from an enharmonic interval makes an '''uninflected''' enharmonic, a conventional 3-limit interval which vanishes in certain edos. For example, (P8/2, P5)'s enharmonic interval is ^^d2, the uninflected enharmonic is d2, and d2 vanishes in 12-edo. Every rank-2 temperament has a "sweet spot" for tuning the 5th, usually a narrow range of about 5-10¢. 12-edo's fifth is the "tipping point": if the temperament's 5th is flatter than 12-edo's, d2 is ascending, and if it's sharper, it's descending. The ups and downs are meant to indicate that the enharmonic interval vanishes. Thus if d2 is ascending, it should be downed, and if it's descending, upped. Therefore <u>'''up may need to be swapped with down, depending on the size of the 5th'''</u> in the particular rank-2 tuning you are using. In the above table, this is shown explicitly for (P8/2, P5), and implied for all the other pergens. In the table, the other pergens' enharmonic intervals are upped or downed as if the 5th were just.
-Heptatonic 5th-based notation is only possible if the 5th ranges from 600¢ to 720¢. For every bare enharmonic, the following table shows in what parts of this range this interval should be upped or downed. The tipping point edo is simply the 3-exponent of the bare enharmonic.
+Heptatonic 5th-based notation is only possible if the 5th ranges from 600¢ to 720¢. For every uninflected enharmonic, the following table shows in what parts of this range this interval should be upped or downed. The tipping point edo is simply the 3-exponent of the uninflected enharmonic.
 {| class="wikitable" style="text-align:center;"
 |-
-! colspan="2" | bare enharmonic
+! colspan="2" | uninflected enharmonic
 interval
@@ Line 1,023: / Line 1,023: @@
 vM7/2: C - ^F - vB (vM7 = 15/8, probably more harmonious than M7 = 243/128)
-More remote intervals include A1, d4, d7 and d10. These unfortunately require a very long genchain. The most interesting melodically is A1: C - ^C - vC# - C#. From C to C# is seven 5ths, which equals 21 generators, so the genchain would have to contain 22 notes if it had no gaps. Note that A1 is the bare enharmonic of third-4th. The bare enharmonic is always a secondary split.
+More remote intervals include A1, d4, d7 and d10. These unfortunately require a very long genchain. The most interesting melodically is A1: C - ^C - vC# - C#. From C to C# is seven 5ths, which equals 21 generators, so the genchain would have to contain 22 notes if it had no gaps. Note that A1 is the uninflected enharmonic of third-4th. The uninflected enharmonic is always a secondary split.
 For a pergen (P8, (a,b)/n), any interval generated by n octaves and the multigen splits into at least n parts. For a pergen (P8/m, P5), any interval generated by the octave and m 5ths splits into at least m parts. Thus any naturally occurring split of m parts occurs in all voicings of that interval. For example, M9 naturally splits into two 5ths, therefore (P8/2, P5) splits all voicings of M9, including M2.
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 For example, consider the half-5th pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = m3. The enharmonic can also be found using gedras: xEI = M - n<span style="">⋅</span>G = P5 - 2<span style="">⋅</span>m3 = [7,4] - 2<span style="">⋅</span>[3,2] = [7,4] - [6,4] = [1,0] = A1.
-Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xEI = P8 - mP = P8 - 5<span style="">⋅</span>M2 = [12,7] - 5<span style="">⋅</span>[2,1] = [2,2] = 2<span style="">⋅</span>[1,1] = 2<span style="">⋅</span>m2. Because x = 2, EI will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2<span style="">⋅</span>m2 = d3). The enharmonic's '''count''' is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus EI = v<span style="vertical-align: super;">5</span>m2. Since P8 = 5<span style="">⋅</span>P + 2<span style="">⋅</span>EI, the period must be ^^M2, to make the ups and downs come out even. The number of the period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:
+Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xEI = P8 - mP = P8 - 5<span style="">⋅</span>M2 = [12,7] - 5<span style="">⋅</span>[2,1] = [2,2] = 2<span style="">⋅</span>[1,1] = 2<span style="">⋅</span>m2. Because x = 2, EI will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2<span style="">⋅</span>m2 = d3). The enharmonic's '''count''' is 2. The uninflected enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus EI = v<span style="vertical-align: super;">5</span>m2. Since P8 = 5<span style="">⋅</span>P + 2<span style="">⋅</span>EI, the period must be ^^M2, to make the ups and downs come out even. The number of the period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:
 <span style="display: block; text-align: center;">P1 -- ^^M2=v<span style="vertical-align: super;">3</span>m3 -- v4 -- ^5 -- ^<span style="vertical-align: super;">3</span>M6=vvm7 -- P8</span>
 <span style="display: block; text-align: center;">C -- ^^D=v<span style="vertical-align: super;">3</span>Eb -- vF -- ^G -- ^<span style="vertical-align: super;">3</span>A=vvBb -- C</span>
-Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [1,1] = m2. The bare enharmonic is P4 - 5<span style="">⋅</span>m2 = [5,3] - 5<span style="">⋅</span>[1,1] = [5,3] - [5,5] = [0,-2] = -2<span style="">⋅</span>[0,1] = two descending d2's. The d2 must be upped, and EI = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5<span style="">⋅</span>G - 2<span style="">⋅</span>EI, G must be ^^m2. The genchain is:
+Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has an uninflected generator [5,3]/5 = [1,1] = m2. The uninflected enharmonic is P4 - 5<span style="">⋅</span>m2 = [5,3] - 5<span style="">⋅</span>[1,1] = [5,3] - [5,5] = [0,-2] = -2<span style="">⋅</span>[0,1] = two descending d2's. The d2 must be upped, and EI = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5<span style="">⋅</span>G - 2<span style="">⋅</span>EI, G must be ^^m2. The genchain is:
 <span style="display: block; text-align: center;">P1 -- ^^m2=v<span style="vertical-align: super;">3</span>A1 -- vM2 -- ^m3 -- ^<span style="vertical-align: super;">3</span>d4=vvM3 -- P4</span><span style="display: block; text-align: center;">C -- ^^Db -- vD -- ^Eb -- vvE -- F</span>
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 To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multigen as before. Then deduce the period from the enharmonic. If the multigen was changed by unreducing, find the original generator from the period and the alternate generator.
-For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10<span style="">⋅</span>P1 = m2. It must be downed, thus EI = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10<span style="">⋅</span>G' + EI, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x<span style="">⋅</span>m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5<span style="">⋅</span>P + 2<span style="">⋅</span>EI, and P = ^<span style="vertical-align: super;">4</span>M2. Next, find the original half-4th generator. P = P8/5 = ~240¢, and G = P4/2 = ~250¢. Because P &lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^<span style="vertical-align: super;">4</span>M2 + ^1 = ^<span style="vertical-align: super;">5</span>M2. While the alternate multigen is more complex than the original multigen, the alternate generator is usually simpler than the original generator.
+For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The uninflected alternate generator G' is [1,1]/10 = [0,0] = P1. The uninflected enharmonic is m2 - 10<span style="">⋅</span>P1 = m2. It must be downed, thus EI = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10<span style="">⋅</span>G' + EI, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x<span style="">⋅</span>m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5<span style="">⋅</span>P + 2<span style="">⋅</span>EI, and P = ^<span style="vertical-align: super;">4</span>M2. Next, find the original half-4th generator. P = P8/5 = ~240¢, and G = P4/2 = ~250¢. Because P &lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^<span style="vertical-align: super;">4</span>M2 + ^1 = ^<span style="vertical-align: super;">5</span>M2. While the alternate multigen is more complex than the original multigen, the alternate generator is usually simpler than the original generator.
 <span style="display: block; text-align: center;">P1- - ^<span style="vertical-align: super;">4</span>M2=v<span style="vertical-align: super;">6</span>m3 -- vv4 -- ^^5 -- ^<span style="vertical-align: super;">6</span>M6=v<span style="vertical-align: super;">4</span>m7 -- P8</span>
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 A false-double pergen can optionally use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is not a unison or a 2nd, as with (P8/2, P4/3).
-Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅M3 = [1,2] = dd3. So EI = v<span style="vertical-align: super;">4</span>dd3, and G = ^M3. But by using double-pair notation, we can avoid that. We find P11/2, which equals two generators: P11/2 = 2⋅G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2⋅m6 = [1,0] = A1. For this second enharmonic, we use the second pair of accidentals: EI' = \\A1 and 2⋅G = /m6 or \M6. The sum or difference of two enharmonic intervals is also an enharmonic: EI + EI' = v<span style="vertical-align: super;">4</span>\\d3 = 2·vv\m2, and EI - EI' = v<span style="vertical-align: super;">4</span>//ddd3 = 2·vv/d2. Thus vv\m2 and vv/d2 are equivalent enharmonics, and v\4 and v/d4 are equivalent generators. Here is the genchain:
+Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅M3 = [1,2] = dd3. So EI = v<span style="vertical-align: super;">4</span>dd3, and G = ^M3. But by using double-pair notation, we can avoid that. We find P11/2, which equals two generators: P11/2 = 2⋅G = [17,10]/2 = [8,5] = m6. The uninflected enharmonic is P11 - 2⋅m6 = [1,0] = A1. For this second enharmonic, we use the second pair of accidentals: EI' = \\A1 and 2⋅G = /m6 or \M6. The sum or difference of two enharmonic intervals is also an enharmonic: EI + EI' = v<span style="vertical-align: super;">4</span>\\d3 = 2·vv\m2, and EI - EI' = v<span style="vertical-align: super;">4</span>//ddd3 = 2·vv/d2. Thus vv\m2 and vv/d2 are equivalent enharmonics, and v\4 and v/d4 are equivalent generators. Here is the genchain:
 <span style="display: block; text-align: center;">P1 -- ^M3=v\4 -- /m6=\M6 -- ^/8=vm9 -- P11</span>
 <span style="display: block; text-align: center;">C -- ^E=v\F -- /Ab=\A -- ^/C=vDb -- 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 EI = 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 EI' = /<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 EI + EI' and EI - 2·EI'. 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.
+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 EI = v<span style="vertical-align: super;">12</span>m3. Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the uninflected enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus EI' = /<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 EI + EI' and EI - 2·EI'. 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.
 <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
@@ Line 1,228: / Line 1,228: @@
 ==Alternate enharmonics==
-Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, ccM6/12), a false double. The bare alternate generator is ccM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12·[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12·[4,2] = [-15,-5] = -5·[3,1] = -5·v<span style="vertical-align: super;">12</span>A2, which is an improvement but still awkward. The period is ^<span style="vertical-align: super;">4</span>m3 and the generator is v<span style="vertical-align: super;">3</span>M2. ^1 = 25¢ + 0.75·c, about an eighth-tone.
+Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, ccM6/12), a false double. The uninflected alternate generator is ccM6/12 = [33,19]/12 = [3,2] = m3. The uninflected enharmonic is [33,19] - 12·[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12·[4,2] = [-15,-5] = -5·[3,1] = -5·v<span style="vertical-align: super;">12</span>A2, which is an improvement but still awkward. The period is ^<span style="vertical-align: super;">4</span>m3 and the generator is v<span style="vertical-align: super;">3</span>M2. ^1 = 25¢ + 0.75·c, about an eighth-tone.
 <span style="display: block; text-align: center;">P1 -- ^<span style="vertical-align: super;">4</span>m3 -- v<span style="vertical-align: super;">4</span>M6 -- P8
@@ Line 1,246: / Line 1,246: @@
 To search for alternate enharmonics, convert EI to a gedra, then multiply it by the count to get the multi-EI (multi-enharmonic). The count is always the number of ups or downs in the generator (or period). The count is positive only if G (or P) is upped and EI is downed, or vice versa. Add or subtract the splitting fraction n (or m) to/from either half of the gedra as desired to get a new multi-EI. If the stepspan becomes negative, or if it's zero and the keyspan becomes negative, invert the gedra. If the two halves of the new gedra have a common factor, simplify the gedra by this factor, which becomes the new count. Convert the simplified gedra to a 3-limit interval. Add n (or m) ups or downs, this is the new EI. Choose between ups and downs according to whether the 5th falls in the enharmonic's upping or downing range, see [[pergen#Applications-Tipping points|tipping points]] above. Add n'''·'''count ups or downs to the new multi-EI. Add or subtract the new multi-EI from the multigen (or the octave) to get an interval which splits cleanly into n (or m) parts. Each part is the new generator (or period).
-For example, (P8, P5/3) has n = 3, G = ^M2, and EI = v<span style="vertical-align: super;">3</span>m2 = [1,1]. G is upped only once, so the count is 1, and the multi-EI is also [1,1]. Subtract n from the gedra's keyspan to make a new multi-EI [-2,1]. This can't be simplified, so the new EI is also [-2,1] = d<span style="vertical-align: super;">3</span>2. Assuming a reasonably just 5th, EI needs to be upped, so EI = ^<span style="vertical-align: super;">3</span>d<span style="vertical-align: super;">3</span>2. Add the multi-EI ^<span style="vertical-align: super;">3</span>[-2,1] to the multigen P5 = [7,4] to get ^<span style="vertical-align: super;">3</span>[5,3]. This isn't divisible by n, so we must subtract instead: [7,4] - ^<span style="vertical-align: super;">3</span>[-2,1] = v<span style="vertical-align: super;">3</span>[9,3] = 3·v[3,1], and G = vA2. Use the equation M = n·G + y·E to check: 3·vA2 + 1·^3d<span style="vertical-align: super;">3</span>2 = 3·M2 + 1·m2 = P5. (Diminish three A2's once and augment one d<span style="vertical-align: super;">3</span>2 three times.) Here are the genchains: P1 -- vA2=^^dd3 -- ^d4 -- P5 and C -- vD# -- ^Fb -- G. Because d<span style="vertical-align: super;">3</span>2 = -200¢ - 26·c, ^ = (-d<span style="vertical-align: super;">3</span>2) / 3 = 67¢ + 8.67·c, about a third-tone.
+For example, (P8, P5/3) has n = 3, G = ^M2, and EI = v<span style="vertical-align: super;">3</span>m2 = [1,1]. G is upped only once, so the count is 1, and the multi-EI is also [1,1]. Subtract n from the gedra's keyspan to make a new multi-EI [-2,1]. This can't be simplified, so the new EI is also [-2,1] = d<span style="vertical-align: super;">3</span>2. Assuming a reasonably just 5th, EI needs to be upped, so EI = ^<span style="vertical-align: super;">3</span>d<span style="vertical-align: super;">3</span>2. Add the multi-EI ^<span style="vertical-align: super;">3</span>[-2,1] to the multigen P5 = [7,4] to get ^<span style="vertical-align: super;">3</span>[5,3]. This isn't divisible by n, so we must subtract instead: [7,4] - ^<span style="vertical-align: super;">3</span>[-2,1] = v<span style="vertical-align: super;">3</span>[9,3] = 3·v[3,1], and G = vA2. Use the equation M = n·G + y·EI to check: 3·vA2 + 1·^3d<span style="vertical-align: super;">3</span>2 = 3·M2 + 1·m2 = P5. (Diminish three A2's once and augment one d<span style="vertical-align: super;">3</span>2 three times.) Here are the genchains: P1 -- vA2=^^dd3 -- ^d4 -- P5 and C -- vD# -- ^Fb -- G. Because d<span style="vertical-align: super;">3</span>2 = -200¢ - 26·c, ^ = (-d<span style="vertical-align: super;">3</span>2) / 3 = 67¢ + 8.67·c, about a third-tone.
 Approaching pergens in a higher-primes-agnostic way, independently of specific commas or temperaments, one enharmonic (and one notation) will usually be obviously superior. But sometimes the temperament being notated implies a certain enharmonic. Specifically, the comma tempered out should map to EI, or the multi-EI if the count is &gt; 1. For example, consider Semaphore aka Zozo (2.3.7 and 49/48), which is half-4th. Assuming 7/4 is a m7, the comma is a m2, and a vvm2 enharmonic makes sense. G = ^M2 and the genchain is C -- ^D=vEb -- F. But consider Lala-yoyo, which tempers out (-22, 11, 2). This temperament is also half-4th. The comma is a descending dd2, thus EI = ^^dd2, G = vA2, and the genchain is C -- vD#=^Ebb -- F. This is the best notation because the (-10, 5, 1) generator is an augmented 2nd.
@@ Line 1,284: / Line 1,284: @@
 The tipping point depends on the choice of enharmonic. It's not the temperament that tips, it's the notation. Half-8ve could be notated with an EI of vvM2. The tipping point becomes 600¢, a <u>very</u> unlikely 5th, and tipping is impossible. For single-comma temperaments, EI usually equals the 3-limit mapping of the comma. Thus for 5-limit and 7-imit temperaments, the choice of EI is a given. However, the mapping of primes 11 and 13 is not agreed on.
-The notation's tipping point is determined by the bare enharmonic, which is implied by the vanishing comma. For example, Porcupine aka Triyo's 250/243 comma is an A1 = (-11,7), which implies a bare EI of A1, which implies 7-edo, and a 685.7¢ tipping point. Dicot aka Yoyo's 25/24 comma is also an A1, and has the same tipping point. Semaphore aka Zozo's 49/48 comma is a minor 2nd = (8,-5), implying 5-edo and a 720¢ tipping point. See [[pergen#Applications-Tipping points|tipping points]] above for a more complete list.
+The notation's tipping point is determined by the uninflected enharmonic, which is implied by the vanishing comma. For example, Porcupine aka Triyo's 250/243 comma is an A1 = (-11,7), which implies an uninflected EI of A1, which implies 7-edo, and a 685.7¢ tipping point. Dicot aka Yoyo's 25/24 comma is also an A1, and has the same tipping point. Semaphore aka Zozo's 49/48 comma is a minor 2nd = (8,-5), implying 5-edo and a 720¢ tipping point. See [[pergen#Applications-Tipping points|tipping points]] above for a more complete list.
 Double-pair notation has two enharmonics, and two tipping points to be avoided. (P8/2, P4/2) has three possible notations. The two enharmonics can be either A1, m2 or d2. One can choose to use whichever two enharmonics best avoid tipping.
@@ Line 4,417: / Line 4,417: @@
 == Addenda ==
-Ups and downs are collectively called [[Arrow|'''arrows''']]. Lifts and drops are collectively called '''slants'''. Arrows and slants are collectively called '''inflections'''. Thus a better term for "bare" enharmonic is uninflected enharmonic.
 [[Category:Regular temperament theory]]
 [[Category:Notation]]