Pergen names: Difference between revisions
Wikispaces>TallKite **Imported revision 624363937 - Original comment: ** |
Wikispaces>TallKite **Imported revision 624379539 - Original comment: ** |
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<h2>IMPORTED REVISION FROM WIKISPACES</h2> | <h2>IMPORTED REVISION FROM WIKISPACES</h2> | ||
This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br> | This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br> | ||
: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01- | : This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-03 00:59:00 UTC</tt>.<br> | ||
: The original revision id was <tt> | : The original revision id was <tt>624379539</tt>.<br> | ||
: The revision comment was: <tt></tt><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> | The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br> | ||
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A rank-4 temperament has a pergen of four intervals, rank-5 has five intervals, etc. A rank-1 temperament could have a pergen of one, such as (P8/12) for 12-edo or (P12/13) for 13-ed3, but there's no particular reason to do so. In fact, edos are simply rank-1 pergens, and what the concept of edos does for rank-1 temperaments, the concept of pergens does for temperaments of rank 2 or higher. | A rank-4 temperament has a pergen of four intervals, rank-5 has five intervals, etc. A rank-1 temperament could have a pergen of one, such as (P8/12) for 12-edo or (P12/13) for 13-ed3, but there's no particular reason to do so. In fact, edos are simply rank-1 pergens, and what the concept of edos does for rank-1 temperaments, the concept of pergens does for temperaments of rank 2 or higher. | ||
The preferred pergen for untempered just intonation is the octave, the fifth, and a list of commas, each containing only one higher prime: (P8, P5, 81/80, 64/63, ...). The higher prime's exponent in the monzo must be 1 or -1. | The preferred pergen for untempered just intonation is the octave, the fifth, and a list of commas, each containing only one higher prime: (P8, P5, 81/80, 64/63, ...). The higher prime's exponent in the monzo must be 1 or -1. More on these commas later. | ||
=__Derivation__= | =__Derivation__= | ||
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||~ 3/1 ||= 0 ||= 1 ||= -1 || || | ||~ 3/1 ||= 0 ||= 1 ||= -1 || || | ||
||~ 7/1 ||= 0 ||= 0 ||= 2 || /2 || | ||~ 7/1 ||= 0 ||= 0 ||= 2 || /2 || | ||
Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multi-gen, the multi-gen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multi-gen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. | Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multi-gen, the multi-gen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multi-gen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. We can let ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is > 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible. | ||
=__Applications__= | =__Applications__= | ||
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A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively. | A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively. | ||
Unreducing replaces the generator with an alternate generator. Any number of periods plus or minus a single generator makes an **alternate** generator. A generator or period plus or minus any number of enharmonics makes an **equivalent** generator or period. An equivalent generator is always the same size in cents, since the enharmonic is always 0¢. An equivalent generator is the same interval, merely notated differently. 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. | Unreducing replaces the generator with an alternate generator. Any number of periods plus or minus a single generator makes an **alternate** generator. A generator or period plus or minus any number of enharmonics makes an **equivalent** generator or period. An equivalent generator is always the same size in cents, since the enharmonic is always 0¢. An equivalent generator is the same interval, merely notated differently. 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 preferred 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 ^<span style="vertical-align: super;">6</span>dd2. | Of the two equivalent generators, the preferred 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 ^<span style="vertical-align: super;">6</span>dd2. | ||
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==Finding a notation for a pergen== | ==Finding a notation for a pergen== | ||
There are multiple notations for a given pergen, depending on the enharmonic interval(s). Preferably, the enharmonic's degree will be a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting. If it's a unison, it will always be an A1. (P1 would be pointless, d1 would be inverted to A1, and AA1 would be split into two A1's.) If it's a 2nd, preferably it will be a m2 or a d2 or a dd2, and not a M2 or an A2 or a ddd2. There is an easy method for finding such a pergen, if one exists. First, some | There are multiple notations for a given pergen, depending on the enharmonic interval(s). Preferably, the enharmonic's degree will be a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting. If it's a unison, it will always be an A1. (P1 would be pointless, d1 would be inverted to A1, and AA1 would be split into two A1's.) If it's a 2nd, preferably it will be a m2 or a d2 or a dd2, and not a M2 or an A2 or a ddd2. There is an easy method for finding such a pergen, if one exists. First, some terminology and basic concepts: | ||
For (P8/m, M/n), P8 = mP + xE and M = nG + yE', with 0 < |x| <= m/2 and 0 < |y| <= n/2 | For (P8/m, M/n), P8 = mP + xE and M = nG + yE', with 0 < |x| <= m/2 and 0 < |y| <= n/2 | ||
For false doubles using single-pair notation, E = E' | For false doubles using single-pair notation, E = E', but x and y are usually different | ||
The unreduced pergen is (P8/m, M'/n'), with M' = n'G' + zE" | The unreduced pergen is (P8/m, M'/n'), with M' = n'G' + zE", and P8 = mP + xE" | ||
The **keyspan** of an interval is the number of keys or frets or semitones that the interval spans in 12-edo. Most musicians know that a minor 2nd is one key or fret and a major 2nd is two keys or frets. The keyspans of larger intervals aren't as well known. The concept can easily be expanded to other edos, but we'll assume 12-edo for now. The **stepspan** of an interval is simply the degree minus one. M2, m2, A2 and d2 all have a stepspan of 1. P5, d5 and A5 all have stepspan 4. The stepspan can be thought of as the 7-edo keyspan. | The **keyspan** of an interval is the number of keys or frets or semitones that the interval spans in 12-edo. Most musicians know that a minor 2nd is one key or fret and a major 2nd is two keys or frets. The keyspans of larger intervals aren't as well known. The concept can easily be expanded to other edos, but we'll assume 12-edo for now. The **stepspan** of an interval is simply the degree minus one. M2, m2, A2 and d2 all have a stepspan of 1. P5, d5 and A5 all have stepspan 4. The stepspan can be thought of as the 7-edo keyspan. This concept can be expanded to include pentatonicism, octotonicism, etc., but we'll assume heptatonicism for now. | ||
Every 3-limit interval can be uniquely expressed as the combination of a keyspan and a stepspan. This combination is called a **gedra**, analogous to a monzo, but written in brackets not parentheses: 3/2 = (-1,1) is a 7-semitone 5th, thus (-1,1) = [7,4]. 9/8 = (-3,2) = [2,1] = a 2-semitone 1-step interval. The octave 2/1 = [12,7]. For any 3-limit interval with a monzo (a,b), there is a unique gedra [k,s], and vice versa: | Every 3-limit interval can be uniquely expressed as the combination of a keyspan and a stepspan. This combination is called a **gedra**, analogous to a monzo, but written in brackets not parentheses: 3/2 = (-1,1) is a 7-semitone 5th, thus (-1,1) = [7,4]. 9/8 = (-3,2) = [2,1] = a 2-semitone 1-step interval. The octave 2/1 = [12,7]. For any 3-limit interval with a monzo (a,b), there is a unique gedra [k,s], and vice versa: | ||
k = 12a + 19b | <span style="display: block; text-align: center;">k = 12a + 19b</span><span style="display: block; text-align: center;">s = 7a + 11b</span> | ||
s = 7a + 11b | |||
The matrix ((12,19) (7,11)) is unimodular, and can be inverted, and (a,b) can be derived from [k,s]: | The matrix ((12,19) (7,11)) is unimodular, and can be inverted, and (a,b) can be derived from [k,s]: | ||
a = -11k + 19b | <span style="display: block; text-align: center;">a = -11k + 19b</span><span style="display: block; text-align: center;">b = 7a - 12b</span> | ||
b = 7a - 12b | Gedras can be manipulated exactly like monzos. Just as adding two intervals (a,b) and (a',b') gives us (a+a',b+b'), likewise [k,s] added to [k',s'] equals [k+k',s+s']. If the GCD of a and b is n, then (a,b) is a stack of n identical intervals, with (a,b) = (na', nb') = n(a',b'), and if (a,b) is converted to [k,s], then the GCD of k and s is also n, and [k,s] = [nk',ns'] = n[k',s']. | ||
Gedras | 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. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. As noted above, the smaller of two equivalent periods or generators is preferred, so fractions of the form N/2 should be rounded down. For example, consider the half-fifth pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. Here xE = 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. xE = P8 - m*P = P8 - 5*M2 = [12,7] - 5*[2,1] = [2,2] = 2*[1,1] = 2*m2. Because x = 2, E 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 E = v<span style="vertical-align: super;">5</span>m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. 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). | |||
<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^^=Ebv<span style="vertical-align: super;">3</span> -- Fv -- G^ -- A^<span style="vertical-align: super;">3</span>=Bbvv -- C</span> | <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^^=Ebv<span style="vertical-align: super;">3</span> -- Fv -- G^ -- A^<span style="vertical-align: super;">3</span>=Bbvv -- 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*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 E = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is: | 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 E = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is: | ||
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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 multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, find the original generator from the period and the alternate generator. | 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 multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen 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 E = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10*G' + E, 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*E, and P = ^<span style="vertical-align: super;">4</span>M2. Next, find the original half-fourth 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 ^<span style="vertical-align: super;">4</span>M2 + ^1 = ^<span style="vertical-align: super;">5</span>M2. The alternate generator is | 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 E = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10*G' + E, 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*E, and P = ^<span style="vertical-align: super;">4</span>M2. Next, find the original half-fourth 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 ^<span style="vertical-align: super;">4</span>M2 + ^1 = ^<span style="vertical-align: super;">5</span>M2. The alternate generator is usually simpler than the original generator, and the alternate multigen is usually more complex than the original multigen. | ||
<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 -- vvP4 -- ^^P5 -- ^<span style="vertical-align: super;">6</span>M6=v<span style="vertical-align: super;">4</span>m7 -- P8</span><span style="display: block; text-align: center;">C -- D^<span style="vertical-align: super;">4</span>=Ebv<span style="vertical-align: super;">6</span> -- Fvv -- G^^ -- A^<span style="vertical-align: super;">6</span>=Bbv<span style="vertical-align: super;">4</span> -- C</span><span style="display: block; text-align: center;">P1 -- ^<span style="vertical-align: super;">5</span>M2=v<span style="vertical-align: super;">5</span>m3 -- P4</span><span style="display: block; text-align: center;">C -- D^<span style="vertical-align: super;">5</span>=Ebv<span style="vertical-align: super;">5</span> -- F</span> | <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 -- vvP4 -- ^^P5 -- ^<span style="vertical-align: super;">6</span>M6=v<span style="vertical-align: super;">4</span>m7 -- P8</span><span style="display: block; text-align: center;">C -- D^<span style="vertical-align: super;">4</span>=Ebv<span style="vertical-align: super;">6</span> -- Fvv -- G^^ -- A^<span style="vertical-align: super;">6</span>=Bbv<span style="vertical-align: super;">4</span> -- C</span><span style="display: block; text-align: center;">P1 -- ^<span style="vertical-align: super;">5</span>M2=v<span style="vertical-align: super;">5</span>m3 -- P4</span><span style="display: block; text-align: center;">C -- D^<span style="vertical-align: super;">5</span>=Ebv<span style="vertical-align: super;">5</span> -- F</span> | ||
To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2), the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2. | To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2), the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2. | ||
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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. But by using double-pair notation, we can get enharmonics that are 2nds. First 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, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6. But ^m6 has an up sign, and there's no such thing as half an up. The answer is to double all ups and downs: P11/2 = ^^m6 or vvM6, and E = v<span style="vertical-align: super;">4</span>A1. The bare generator is ^^m6/2 = ^^[8,5]/2 = ^[4,2] = ^M3, and the bare enharmonic is ^^m6 - 2*(^M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = //d2, and G = ^\M3 or ^/d4. 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. But by using double-pair notation, we can get enharmonics that are 2nds. First 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, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6. But ^m6 has an up sign, and there's no such thing as half an up. The answer is to double all ups and downs: P11/2 = ^^m6 or vvM6, and E = v<span style="vertical-align: super;">4</span>A1. The bare generator is ^^m6/2 = ^^[8,5]/2 = ^[4,2] = ^M3, and the bare enharmonic is ^^m6 - 2*(^M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = //d2, and G = ^\M3 or ^/d4. Here is the genchain: | ||
<span style="display: block; text-align: center;">P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11 | <span style="display: block; text-align: center;">P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11 | ||
</span><span style="display: block; text-align: center;">C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F | </span><span style="display: block; text-align: center;">C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F | ||
</span> | </span> | ||
Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4. | Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4. | ||
<span style="display: block; text-align: center;">P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11 | <span style="display: block; text-align: center;">P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11 | ||
</span><span style="display: block; text-align: center;">C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F | </span><span style="display: block; text-align: center;">C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F | ||
</span> | </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 proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. 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. Before we can divide by 4, we must quadruple all ups and downs: E = ^<span style="vertical-align: super;">12</span>d2 and 4*G' = ^<span style="vertical-align: super;">4</span>m2. The bare alt-generator is ^<span style="vertical-align: super;">4</span>[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^<span style="vertical-align: super;">4</span>m2 - 4*(^1) = m2. Thus E' = \<span style="vertical-align: super;">4</span>m2 and G' = ^/1.The period can be deduced from 4*G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4*G' = P4 - ^<span style="vertical-align: super;">4</span>m2 = v<span style="vertical-align: super;">4</span>M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = v4M3 + ^/1 = v3/M3. | 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 proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. 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. Before we can divide by 4, we must quadruple all ups and downs: E = ^<span style="vertical-align: super;">12</span>d2 and 4*G' = ^<span style="vertical-align: super;">4</span>m2. The bare alt-generator is ^<span style="vertical-align: super;">4</span>[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^<span style="vertical-align: super;">4</span>m2 - 4*(^1) = m2. Thus E' = \<span style="vertical-align: super;">4</span>m2 and G' = ^/1.The period can be deduced from 4*G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4*G' = P4 - ^<span style="vertical-align: super;">4</span>m2 = v<span style="vertical-align: super;">4</span>M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = v4M3 + ^/1 = v3/M3. | ||
<span style="display: block; text-align: center;">P1 -- v<span style="vertical-align: super;">4</span>M3 -- v<span style="vertical-align: super;">8</span>A5=^<span style="vertical-align: super;">4</span>m6-- P8 | <span style="display: block; text-align: center;">P1 -- v<span style="vertical-align: super;">4</span>M3 -- v<span style="vertical-align: super;">8</span>A5=^<span style="vertical-align: super;">4</span>m6-- P8 | ||
</span> | </span><span style="display: block; text-align: center;">C -- Ev<span style="vertical-align: super;">4</span> -- Ab^<span style="vertical-align: super;">4</span> -- C</span><span style="display: block; text-align: center;">P1 -- v<span style="vertical-align: super;">3</span>/M3 -- v<span style="vertical-align: super;">6</span>``//``A5=^<span style="vertical-align: super;">6</span>``//``m6=^<span style="vertical-align: super;">6</span>\\d7 -- ^<span style="vertical-align: super;">3</span>\m9 -- F</span><span style="display: block; text-align: center;">C -- Ev<span style="vertical-align: super;">3</span>/ -- G#v<span style="vertical-align: super;">6</span>``//``=Ab^<span style="vertical-align: super;">6</span>``//``=Bbb^<span style="vertical-align: super;">6</span>\\ -- Db^<span style="vertical-align: super;">3</span>\ -- F</span> | ||
<span style="display: block; text-align: center;">C -- Ev<span style="vertical-align: super;">4</span> -- Ab^<span style="vertical-align: super;">4</span> -- C</span> | |||
<span style="display: block; text-align: center;">P1 -- v<span style="vertical-align: super;">3</span>/M3 -- v<span style="vertical-align: super;">6</span>``//``A5=^<span style="vertical-align: super;">6</span>``//``m6=^<span style="vertical-align: super;">6</span>\\d7 -- ^<span style="vertical-align: super;">3</span>\m9 -- F</span> | |||
<span style="display: block; text-align: center;"> C -- Ev<span style="vertical-align: super;">3</span>/ -- G#v<span style="vertical-align: super;">6</span>``//``=Ab^<span style="vertical-align: super;">6</span>``//``=Bbb^<span style="vertical-align: super;">6</span>\\ -- Db^<span style="vertical-align: super;">3</span>\ -- F</span> | |||
==Alternate enharmonics== | ==Alternate enharmonics== | ||
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P1 -- ^<span style="vertical-align: super;">4</span>m3 -- v<span style="vertical-align: super;">4</span>M6 -- C | P1 -- ^<span style="vertical-align: super;">4</span>m3 -- v<span style="vertical-align: super;">4</span>M6 -- C | ||
C -- Eb^<span style="vertical-align: super;">4</span> -- Av<span style="vertical-align: super;">4</span> -- C | C -- Eb^<span style="vertical-align: super;">4</span> -- Av<span style="vertical-align: super;">4</span> -- C | ||
P1 -- v<span style="vertical-align: super;">3</span>M2 -- v<span style="vertical-align: super;">6</span>M3=^<span style="vertical-align: super;">6</span>m2 -- ^<span style="vertical-align: super;">3</span>m3 -- P4 | P1 -- v<span style="vertical-align: super;">3</span>M2 -- v<span style="vertical-align: super;">6</span>M3=^<span style="vertical-align: super;">6</span>m2 -- ^<span style="vertical-align: super;">3</span>m3 -- P4 | ||
C -- Dv<span style="vertical-align: super;">3</span> -- Ev<span style="vertical-align: super;">6</span>=Db^<span style="vertical-align: super;">6</span> -- Eb^<span style="vertical-align: super;">3</span> -- F | C -- Dv<span style="vertical-align: super;">3</span> -- Ev<span style="vertical-align: super;">6</span>=Db^<span style="vertical-align: super;">6</span> -- Eb^<span style="vertical-align: super;">3</span> -- F | ||
</span> | </span> | ||
Because G is a M2 and E is an A2, the equivalent generator G - E is a descending A1. Ascending intervals that look descending are best avoided, and double-pair notation is better for | Because G is a M2 and E is an A2, the equivalent generator G - E is a descending A1. Ascending intervals that look descending are best avoided, and double-pair notation is better for this pergen. We have P = vM3, E = ^3d2, G = /m2, and E' = /4dd2. | ||
<span style="display: block; text-align: center;">P1 -- vM3 -- ^m6 -- P8 | <span style="display: block; text-align: center;">P1 -- vM3 -- ^m6 -- P8 | ||
</span><span style="display: block; text-align: center;">C -- Ev -- Ab^ -- C | </span><span style="display: block; text-align: center;">C -- Ev -- Ab^ -- C | ||
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</span> | </span> | ||
The comma equals xE and/or yE'. | The comma equals xE and/or yE'. | ||
If M' = [a,b], then G' = [round(a/n'), round(b/n')] makes the smallest zE", but not always the smallest E" | |||
bbT = 49/48 = m2 = 36¢ = half-fourth. E = vvm2. | bbT = 49/48 = m2 = 36¢ = half-fourth. E = vvm2. | ||
(-22, 11, 2) = -dd2 = 94¢ is LLyyT = half-fourth. E = ^^dd2, and the genchain is C -- D#v=Eb^ -- F. | (-22, 11, 2) = -dd2 = 94¢ is LLyyT = half-fourth. E = ^^dd2, and the genchain is C -- D#v=Eb^ -- F. | ||
These commas are called **notational commas**. They are not necessarily tempered out, they merely determine how a higher prime is mapped to a 3-limit interval, and thus how ratios containing higher primes are notated on the staff. By definition, the only commas that map to P1 are notational ones, and those that are the sum or difference of notational ones. There is widespread agreement that 5's notational comma is 81/80. But the choice of notational commas for other primes, especially 11 and 13, is somewhat arbitrary. For example, if 11's notational comma is 33/32, 11/8 is notated as a perfect 4th. But if it's 729/704, 11/8 is an augmented 4th. | |||
An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - 2*G = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1. | An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - 2*G = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1. | ||
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==Combining pergens== | ==Combining pergens== | ||
250/243 | Tempering out 250/243 creates third-fourth, and 49/48 creates half-fourth, and tempering out both commas creates sixth-fourth. Therefore we can talk of adding pergens: (P8, P4/3) + (P8, P4/2) = (P8, P4/6). | ||
(P8/m, P5) + (P8/m', P5) = (P8/m", P5), where m" = LCM (m,m') | |||
(P8, M/n) + (P8, M/n') = (P8, M/n"), where n" = LCM (n,n') | General rules for combining pergens: | ||
(P8/m, P5) + (P8, M/n) = (P8/m, M/n) | * (P8/m, P5) + (P8/m', P5) = (P8/m", P5), where m" = LCM (m,m') | ||
* (P8, M/n) + (P8, M/n') = (P8, M/n"), where n" = LCM (n,n') | |||
* (P8/m, P5) + (P8, M/n) = (P8/m, M/n) | |||
However, (P8/2, M2/4) + (P8, P4/2) = (P8/4, P4/2), so the sum isn't always obvious. | |||
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A rank-4 temperament has a pergen of four intervals, rank-5 has five intervals, etc. A rank-1 temperament could have a pergen of one, such as (P8/12) for 12-edo or (P12/13) for 13-ed3, but there's no particular reason to do so. In fact, edos are simply rank-1 pergens, and what the concept of edos does for rank-1 temperaments, the concept of pergens does for temperaments of rank 2 or higher.<br /> | A rank-4 temperament has a pergen of four intervals, rank-5 has five intervals, etc. A rank-1 temperament could have a pergen of one, such as (P8/12) for 12-edo or (P12/13) for 13-ed3, but there's no particular reason to do so. In fact, edos are simply rank-1 pergens, and what the concept of edos does for rank-1 temperaments, the concept of pergens does for temperaments of rank 2 or higher.<br /> | ||
<br /> | <br /> | ||
The preferred pergen for untempered just intonation is the octave, the fifth, and a list of commas, each containing only one higher prime: (P8, P5, 81/80, 64/63, ...). The higher prime's exponent in the monzo must be 1 or -1. | The preferred pergen for untempered just intonation is the octave, the fifth, and a list of commas, each containing only one higher prime: (P8, P5, 81/80, 64/63, ...). The higher prime's exponent in the monzo must be 1 or -1. More on these commas later.<br /> | ||
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</table> | </table> | ||
Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multi-gen, the multi-gen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multi-gen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. | Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multi-gen, the multi-gen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multi-gen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. We can let ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is &gt; 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.<br /> | ||
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A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively.<br /> | A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively.<br /> | ||
<br /> | <br /> | ||
Unreducing replaces the generator with an alternate generator. Any number of periods plus or minus a single generator makes an <strong>alternate</strong> generator. A generator or period plus or minus any number of enharmonics makes an <strong>equivalent</strong> generator or period. An equivalent generator is always the same size in cents, since the enharmonic is always 0¢. An equivalent generator is the same interval, merely notated differently. 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. <br /> | Unreducing replaces the generator with an alternate generator. Any number of periods plus or minus a single generator makes an <strong>alternate</strong> generator. A generator or period plus or minus any number of enharmonics makes an <strong>equivalent</strong> generator or period. An equivalent generator is always the same size in cents, since the enharmonic is always 0¢. An equivalent generator is the same interval, merely notated differently. 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.<br /> | ||
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Of the two equivalent generators, the preferred 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 ^<span style="vertical-align: super;">6</span>dd2.<br /> | Of the two equivalent generators, the preferred 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 ^<span style="vertical-align: super;">6</span>dd2.<br /> | ||
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<!-- ws:start:WikiTextHeadingRule:47:&lt;h2&gt; --><h2 id="toc8"><a name="Further Discussion-Finding a notation for a pergen"></a><!-- ws:end:WikiTextHeadingRule:47 -->Finding a notation for a pergen</h2> | <!-- ws:start:WikiTextHeadingRule:47:&lt;h2&gt; --><h2 id="toc8"><a name="Further Discussion-Finding a notation for a pergen"></a><!-- ws:end:WikiTextHeadingRule:47 -->Finding a notation for a pergen</h2> | ||
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There are multiple notations for a given pergen, depending on the enharmonic interval(s). Preferably, the enharmonic's degree will be a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting. If it's a unison, it will always be an A1. (P1 would be pointless, d1 would be inverted to A1, and AA1 would be split into two A1's.) If it's a 2nd, preferably it will be a m2 or a d2 or a dd2, and not a M2 or an A2 or a ddd2. There is an easy method for finding such a pergen, if one exists. First, some | There are multiple notations for a given pergen, depending on the enharmonic interval(s). Preferably, the enharmonic's degree will be a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting. If it's a unison, it will always be an A1. (P1 would be pointless, d1 would be inverted to A1, and AA1 would be split into two A1's.) If it's a 2nd, preferably it will be a m2 or a d2 or a dd2, and not a M2 or an A2 or a ddd2. There is an easy method for finding such a pergen, if one exists. First, some terminology and basic concepts:<br /> | ||
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For (P8/m, M/n), P8 = mP + xE and M = nG + yE', with 0 &lt; |x| &lt;= m/2 and 0 &lt; |y| &lt;= n/2<br /> | For (P8/m, M/n), P8 = mP + xE and M = nG + yE', with 0 &lt; |x| &lt;= m/2 and 0 &lt; |y| &lt;= n/2<br /> | ||
For false doubles using single-pair notation, E = E'<br /> | For false doubles using single-pair notation, E = E', but x and y are usually different<br /> | ||
The unreduced pergen is (P8/m, M'/n'), with M' = n'G' + zE&quot; | The unreduced pergen is (P8/m, M'/n'), with M' = n'G' + zE&quot;, and P8 = mP + xE&quot;<br /> | ||
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The <strong>keyspan</strong> of an interval is the number of keys or frets or semitones that the interval spans in 12-edo. Most musicians know that a minor 2nd is one key or fret and a major 2nd is two keys or frets. The keyspans of larger intervals aren't as well known. The concept can easily be expanded to other edos, but we'll assume 12-edo for now. The <strong>stepspan</strong> of an interval is simply the degree minus one. M2, m2, A2 and d2 all have a stepspan of 1. P5, d5 and A5 all have stepspan 4. The stepspan can be thought of as the 7-edo keyspan. | The <strong>keyspan</strong> of an interval is the number of keys or frets or semitones that the interval spans in 12-edo. Most musicians know that a minor 2nd is one key or fret and a major 2nd is two keys or frets. The keyspans of larger intervals aren't as well known. The concept can easily be expanded to other edos, but we'll assume 12-edo for now. The <strong>stepspan</strong> of an interval is simply the degree minus one. M2, m2, A2 and d2 all have a stepspan of 1. P5, d5 and A5 all have stepspan 4. The stepspan can be thought of as the 7-edo keyspan. This concept can be expanded to include pentatonicism, octotonicism, etc., but we'll assume heptatonicism for now.<br /> | ||
<br /> | <br /> | ||
Every 3-limit interval can be uniquely expressed as the combination of a keyspan and a stepspan. This combination is called a <strong>gedra</strong>, analogous to a monzo, but written in brackets not parentheses: 3/2 = (-1,1) is a 7-semitone 5th, thus (-1,1) = [7,4]. 9/8 = (-3,2) = [2,1] = a 2-semitone 1-step interval. The octave 2/1 = [12,7]. For any 3-limit interval with a monzo (a,b), there is a unique gedra [k,s], and vice versa:<br /> | Every 3-limit interval can be uniquely expressed as the combination of a keyspan and a stepspan. This combination is called a <strong>gedra</strong>, analogous to a monzo, but written in brackets not parentheses: 3/2 = (-1,1) is a 7-semitone 5th, thus (-1,1) = [7,4]. 9/8 = (-3,2) = [2,1] = a 2-semitone 1-step interval. The octave 2/1 = [12,7]. For any 3-limit interval with a monzo (a,b), there is a unique gedra [k,s], and vice versa:<br /> | ||
k = 12a + 19b< | <span style="display: block; text-align: center;">k = 12a + 19b</span><span style="display: block; text-align: center;">s = 7a + 11b</span><br /> | ||
s = 7a + 11b< | |||
<br /> | |||
The matrix ((12,19) (7,11)) is unimodular, and can be inverted, and (a,b) can be derived from [k,s]:<br /> | The matrix ((12,19) (7,11)) is unimodular, and can be inverted, and (a,b) can be derived from [k,s]:<br /> | ||
a = -11k + 19b< | <span style="display: block; text-align: center;">a = -11k + 19b</span><span style="display: block; text-align: center;">b = 7a - 12b</span><br /> | ||
b = 7a - 12b<br /> | Gedras can be manipulated exactly like monzos. Just as adding two intervals (a,b) and (a',b') gives us (a+a',b+b'), likewise [k,s] added to [k',s'] equals [k+k',s+s']. If the GCD of a and b is n, then (a,b) is a stack of n identical intervals, with (a,b) = (na', nb') = n(a',b'), and if (a,b) is converted to [k,s], then the GCD of k and s is also n, and [k,s] = [nk',ns'] = n[k',s'].<br /> | ||
<br /> | <br /> | ||
Gedras | 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. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. As noted above, the smaller of two equivalent periods or generators is preferred, so fractions of the form N/2 should be rounded down. For example, consider the half-fifth pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. Here xE = M - n*G = P5 - 2*m3 = [7,4] - 2*[3,2] = [7,4] - [6,4] = [1,0] = A1.<br /> | ||
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Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xE = P8 - m*P = P8 - 5*M2 = [12,7] - 5*[2,1] = [2,2] = 2*[1,1] = 2*m2. Because x = 2, E will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2*m2 = d3). The enharmonic's <strong>count</strong> is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v<span style="vertical-align: super;">5</span>m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. 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:<br /> | |||
Next, consider (P8/5, P5). | |||
<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^^=Ebv<span style="vertical-align: super;">3</span> -- Fv -- G^ -- A^<span style="vertical-align: super;">3</span>=Bbvv -- C</span><br /> | <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^^=Ebv<span style="vertical-align: super;">3</span> -- Fv -- G^ -- A^<span style="vertical-align: super;">3</span>=Bbvv -- C</span><br /> | ||
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 E = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is:<br /> | 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 E = ^<span style="vertical-align: super;">5</span>d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is:<br /> | ||
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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 multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, find the original generator from the period and the alternate generator.<br /> | 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 multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, find the original generator from the period and the alternate generator.<br /> | ||
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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 E = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10*G' + E, 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*E, and P = ^<span style="vertical-align: super;">4</span>M2. Next, find the original half-fourth 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. The alternate generator is | 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 E = v<span style="vertical-align: super;">10</span>m2. Since m2 = 10*G' + E, 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*E, and P = ^<span style="vertical-align: super;">4</span>M2. Next, find the original half-fourth 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. The alternate generator is usually simpler than the original generator, and the alternate multigen is usually more complex than the original multigen.<br /> | ||
<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 -- vvP4 -- ^^P5 -- ^<span style="vertical-align: super;">6</span>M6=v<span style="vertical-align: super;">4</span>m7 -- P8</span><span style="display: block; text-align: center;">C -- D^<span style="vertical-align: super;">4</span>=Ebv<span style="vertical-align: super;">6</span> -- Fvv -- G^^ -- A^<span style="vertical-align: super;">6</span>=Bbv<span style="vertical-align: super;">4</span> -- C</span><span style="display: block; text-align: center;">P1 -- ^<span style="vertical-align: super;">5</span>M2=v<span style="vertical-align: super;">5</span>m3 -- P4</span><span style="display: block; text-align: center;">C -- D^<span style="vertical-align: super;">5</span>=Ebv<span style="vertical-align: super;">5</span> -- F</span><br /> | <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 -- vvP4 -- ^^P5 -- ^<span style="vertical-align: super;">6</span>M6=v<span style="vertical-align: super;">4</span>m7 -- P8</span><span style="display: block; text-align: center;">C -- D^<span style="vertical-align: super;">4</span>=Ebv<span style="vertical-align: super;">6</span> -- Fvv -- G^^ -- A^<span style="vertical-align: super;">6</span>=Bbv<span style="vertical-align: super;">4</span> -- C</span><span style="display: block; text-align: center;">P1 -- ^<span style="vertical-align: super;">5</span>M2=v<span style="vertical-align: super;">5</span>m3 -- P4</span><span style="display: block; text-align: center;">C -- D^<span style="vertical-align: super;">5</span>=Ebv<span style="vertical-align: super;">5</span> -- F</span><br /> | ||
To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2), the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.<br /> | To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2), the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.<br /> | ||
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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. But by using double-pair notation, we can get enharmonics that are 2nds. First 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, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6. But ^m6 has an up sign, and there's no such thing as half an up. The answer is to double all ups and downs: P11/2 = ^^m6 or vvM6, and E = v<span style="vertical-align: super;">4</span>A1. The bare generator is ^^m6/2 = ^^[8,5]/2 = ^[4,2] = ^M3, and the bare enharmonic is ^^m6 - 2*(^M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = //d2, and G = ^\M3 or ^/d4. Here is the genchain:<br /> | 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. But by using double-pair notation, we can get enharmonics that are 2nds. First 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, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6. But ^m6 has an up sign, and there's no such thing as half an up. The answer is to double all ups and downs: P11/2 = ^^m6 or vvM6, and E = v<span style="vertical-align: super;">4</span>A1. The bare generator is ^^m6/2 = ^^[8,5]/2 = ^[4,2] = ^M3, and the bare enharmonic is ^^m6 - 2*(^M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = //d2, and G = ^\M3 or ^/d4. Here is the genchain:<br /> | ||
<span style="display: block; text-align: center;">P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11<br /> | <span style="display: block; text-align: center;">P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11<br /> | ||
</span><span style="display: block; text-align: center;">C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F<br /> | </span><span style="display: block; text-align: center;">C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F<br /> | ||
</span><br /> | </span><br /> | ||
Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.<br /> | Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.<br /> | ||
<span style="display: block; text-align: center;">P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11<br /> | <span style="display: block; text-align: center;">P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11<br /> | ||
</span><span style="display: block; text-align: center;">C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F<br /> | </span><span style="display: block; text-align: center;">C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F<br /> | ||
</span><br /> | </span><br /> | ||
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 proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. 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. Before we can divide by 4, we must quadruple all ups and downs: E = ^<span style="vertical-align: super;">12</span>d2 and 4*G' = ^<span style="vertical-align: super;">4</span>m2. The bare alt-generator is ^<span style="vertical-align: super;">4</span>[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^<span style="vertical-align: super;">4</span>m2 - 4*(^1) = m2. Thus E' = \<span style="vertical-align: super;">4</span>m2 and G' = ^/1.The period can be deduced from 4*G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4*G' = P4 - ^<span style="vertical-align: super;">4</span>m2 = v<span style="vertical-align: super;">4</span>M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = v4M3 + ^/1 = v3/M3.<br /> | 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 proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. 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. Before we can divide by 4, we must quadruple all ups and downs: E = ^<span style="vertical-align: super;">12</span>d2 and 4*G' = ^<span style="vertical-align: super;">4</span>m2. The bare alt-generator is ^<span style="vertical-align: super;">4</span>[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^<span style="vertical-align: super;">4</span>m2 - 4*(^1) = m2. Thus E' = \<span style="vertical-align: super;">4</span>m2 and G' = ^/1.The period can be deduced from 4*G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4*G' = P4 - ^<span style="vertical-align: super;">4</span>m2 = v<span style="vertical-align: super;">4</span>M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = v4M3 + ^/1 = v3/M3.<br /> | ||
<span style="display: block; text-align: center;">P1 -- v<span style="vertical-align: super;">4</span>M3 -- v<span style="vertical-align: super;">8</span>A5=^<span style="vertical-align: super;">4</span>m6-- P8<br /> | <span style="display: block; text-align: center;">P1 -- v<span style="vertical-align: super;">4</span>M3 -- v<span style="vertical-align: super;">8</span>A5=^<span style="vertical-align: super;">4</span>m6-- P8<br /> | ||
</span> | </span><span style="display: block; text-align: center;">C -- Ev<span style="vertical-align: super;">4</span> -- Ab^<span style="vertical-align: super;">4</span> -- C</span><span style="display: block; text-align: center;">P1 -- v<span style="vertical-align: super;">3</span>/M3 -- v<span style="vertical-align: super;">6</span><!-- ws:start:WikiTextRawRule:025:``//`` -->//<!-- ws:end:WikiTextRawRule:025 -->A5=^<span style="vertical-align: super;">6</span><!-- ws:start:WikiTextRawRule:026:``//`` -->//<!-- ws:end:WikiTextRawRule:026 -->m6=^<span style="vertical-align: super;">6</span>\\d7 -- ^<span style="vertical-align: super;">3</span>\m9 -- F</span><span style="display: block; text-align: center;">C -- Ev<span style="vertical-align: super;">3</span>/ -- G#v<span style="vertical-align: super;">6</span><!-- ws:start:WikiTextRawRule:027:``//`` -->//<!-- ws:end:WikiTextRawRule:027 -->=Ab^<span style="vertical-align: super;">6</span><!-- ws:start:WikiTextRawRule:028:``//`` -->//<!-- ws:end:WikiTextRawRule:028 -->=Bbb^<span style="vertical-align: super;">6</span>\\ -- Db^<span style="vertical-align: super;">3</span>\ -- F</span><br /> | ||
<span style="display: block; text-align: center;">C -- Ev<span style="vertical-align: super;">4</span> -- Ab^<span style="vertical-align: super;">4</span> -- C</span> | |||
<span style="display: block; text-align: center;">P1 -- v<span style="vertical-align: super;">3</span>/M3 -- v<span style="vertical-align: super;">6</span><!-- ws:start:WikiTextRawRule:025:``//`` -->//<!-- ws:end:WikiTextRawRule:025 -->A5=^<span style="vertical-align: super;">6</span><!-- ws:start:WikiTextRawRule:026:``//`` -->//<!-- ws:end:WikiTextRawRule:026 -->m6=^<span style="vertical-align: super;">6</span>\\d7 -- ^<span style="vertical-align: super;">3</span>\m9 -- F</span> | |||
<span style="display: block; text-align: center;"> C -- Ev<span style="vertical-align: super;">3</span>/ -- G#v<span style="vertical-align: super;">6</span><!-- ws:start:WikiTextRawRule:027:``//`` -->//<!-- ws:end:WikiTextRawRule:027 -->=Ab^<span style="vertical-align: super;">6</span><!-- ws:start:WikiTextRawRule:028:``//`` -->//<!-- ws:end:WikiTextRawRule:028 -->=Bbb^<span style="vertical-align: super;">6</span>\\ -- Db^<span style="vertical-align: super;">3</span>\ -- F</span> | |||
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<!-- ws:start:WikiTextHeadingRule:49:&lt;h2&gt; --><h2 id="toc9"><a name="Further Discussion-Alternate enharmonics"></a><!-- ws:end:WikiTextHeadingRule:49 -->Alternate enharmonics</h2> | <!-- ws:start:WikiTextHeadingRule:49:&lt;h2&gt; --><h2 id="toc9"><a name="Further Discussion-Alternate enharmonics"></a><!-- ws:end:WikiTextHeadingRule:49 -->Alternate enharmonics</h2> | ||
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P1 -- ^<span style="vertical-align: super;">4</span>m3 -- v<span style="vertical-align: super;">4</span>M6 -- C<br /> | P1 -- ^<span style="vertical-align: super;">4</span>m3 -- v<span style="vertical-align: super;">4</span>M6 -- C<br /> | ||
C -- Eb^<span style="vertical-align: super;">4</span> -- Av<span style="vertical-align: super;">4</span> -- C<br /> | C -- Eb^<span style="vertical-align: super;">4</span> -- Av<span style="vertical-align: super;">4</span> -- C<br /> | ||
P1 -- v<span style="vertical-align: super;">3</span>M2 -- v<span style="vertical-align: super;">6</span>M3=^<span style="vertical-align: super;">6</span>m2 -- ^<span style="vertical-align: super;">3</span>m3 -- P4<br /> | P1 -- v<span style="vertical-align: super;">3</span>M2 -- v<span style="vertical-align: super;">6</span>M3=^<span style="vertical-align: super;">6</span>m2 -- ^<span style="vertical-align: super;">3</span>m3 -- P4<br /> | ||
C -- Dv<span style="vertical-align: super;">3</span> -- Ev<span style="vertical-align: super;">6</span>=Db^<span style="vertical-align: super;">6</span> -- Eb^<span style="vertical-align: super;">3</span> -- F<br /> | C -- Dv<span style="vertical-align: super;">3</span> -- Ev<span style="vertical-align: super;">6</span>=Db^<span style="vertical-align: super;">6</span> -- Eb^<span style="vertical-align: super;">3</span> -- F<br /> | ||
</span><br /> | </span><br /> | ||
Because G is a M2 and E is an A2, the equivalent generator G - E is a descending A1. Ascending intervals that look descending are best avoided, and double-pair notation is better for | Because G is a M2 and E is an A2, the equivalent generator G - E is a descending A1. Ascending intervals that look descending are best avoided, and double-pair notation is better for this pergen. We have P = vM3, E = ^3d2, G = /m2, and E' = /4dd2.<br /> | ||
<span style="display: block; text-align: center;">P1 -- vM3 -- ^m6 -- P8<br /> | <span style="display: block; text-align: center;">P1 -- vM3 -- ^m6 -- P8<br /> | ||
</span><span style="display: block; text-align: center;">C -- Ev -- Ab^ -- C<br /> | </span><span style="display: block; text-align: center;">C -- Ev -- Ab^ -- C<br /> | ||
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The comma equals xE and/or yE'.<br /> | The comma equals xE and/or yE'.<br /> | ||
If M' = [a,b], then G' = [round(a/n'), round(b/n')] makes the smallest zE&quot;, but not always the smallest E&quot;<br /> | |||
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bbT = 49/48 = m2 = 36¢ = half-fourth. E = vvm2.<br /> | bbT = 49/48 = m2 = 36¢ = half-fourth. E = vvm2.<br /> | ||
(-22, 11, 2) = -dd2 = 94¢ is LLyyT = half-fourth. E = ^^dd2, and the genchain is C -- D#v=Eb^ -- F.<br /> | (-22, 11, 2) = -dd2 = 94¢ is LLyyT = half-fourth. E = ^^dd2, and the genchain is C -- D#v=Eb^ -- F.<br /> | ||
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These commas are called <strong>notational commas</strong>. They are not necessarily tempered out, they merely determine how a higher prime is mapped to a 3-limit interval, and thus how ratios containing higher primes are notated on the staff. By definition, the only commas that map to P1 are notational ones, and those that are the sum or difference of notational ones. There is widespread agreement that 5's notational comma is 81/80. But the choice of notational commas for other primes, especially 11 and 13, is somewhat arbitrary. For example, if 11's notational comma is 33/32, 11/8 is notated as a perfect 4th. But if it's 729/704, 11/8 is an augmented 4th.<br /> | |||
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An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - 2*G = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1.<br /> | An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - 2*G = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1.<br /> | ||
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<!-- ws:start:WikiTextHeadingRule:55:&lt;h2&gt; --><h2 id="toc12"><a name="Further Discussion-Combining pergens"></a><!-- ws:end:WikiTextHeadingRule:55 -->Combining pergens</h2> | <!-- ws:start:WikiTextHeadingRule:55:&lt;h2&gt; --><h2 id="toc12"><a name="Further Discussion-Combining pergens"></a><!-- ws:end:WikiTextHeadingRule:55 -->Combining pergens</h2> | ||
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250/243 | Tempering out 250/243 creates third-fourth, and 49/48 creates half-fourth, and tempering out both commas creates sixth-fourth. Therefore we can talk of adding pergens: (P8, P4/3) + (P8, P4/2) = (P8, P4/6). <br /> | ||
(P8/m, P5) + (P8/m', P5) = (P8/m&quot;, P5), where m&quot; = LCM (m,m')< | <br /> | ||
(P8, M/n) + (P8, M/n') = (P8, M/n&quot;), where n&quot; = LCM (n,n')< | General rules for combining pergens:<br /> | ||
(P8/m, P5) + (P8, M/n) = (P8/m, M/n)<br /> | <ul><li>(P8/m, P5) + (P8/m', P5) = (P8/m&quot;, P5), where m&quot; = LCM (m,m')</li><li>(P8, M/n) + (P8, M/n') = (P8, M/n&quot;), where n&quot; = LCM (n,n')</li><li>(P8/m, P5) + (P8, M/n) = (P8/m, M/n)</li></ul><br /> | ||
However, (P8/2, M2/4) + (P8, P4/2) = (P8/4, P4/2), so the sum isn't always obvious.<br /> | |||
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