Pergen names: Difference between revisions
Wikispaces>TallKite **Imported revision 624351415 - Original comment: ** |
Wikispaces>TallKite **Imported revision 624363937 - 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>2018-01-02 | : This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-02 13:03:05 UTC</tt>.<br> | ||
: The original revision id was <tt> | : The original revision id was <tt>624363937</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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Examples: Marvel (2.3.5.7 and 225/224) is (P8, P5, ^1) = unsplit with ups (^1 = 81/80). Deep reddish (2.3.5.7 and 50/49) is (P8/2, P5, /1) = half-octave with highs (/1 = 81/80). Triple bluish (2.3.5.7 and 1029/1000) is (P8, P11/3, ^1) = third-eleventh with ups. | Examples: Marvel (2.3.5.7 and 225/224) is (P8, P5, ^1) = unsplit with ups (^1 = 81/80). Deep reddish (2.3.5.7 and 50/49) is (P8/2, P5, /1) = half-octave with highs (/1 = 81/80). Triple bluish (2.3.5.7 and 1029/1000) is (P8, P11/3, ^1) = third-eleventh with ups. | ||
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. | 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. 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. | 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. 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. | ||
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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. | 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. | |||
There are also alternate enharmonics, see below. | There are also alternate enharmonics, see below. | ||
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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 basic concepts: | 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 basic terminology and concepts: | ||
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' | |||
The unreduced pergen is (P8/m, M'/n'), with M' = n'G' + zE" | |||
If M' = [a,b], then G' = [round(a/n'), round(b/n')] makes the smallest zE", but not always the smallest E" | |||
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. Again, this concept can be expanded to include pentatonicism, octotonicism, etc., but we'll assume heptatonicism for now. | 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. Again, this concept can be expanded to include pentatonicism, octotonicism, etc., but we'll assume heptatonicism for now. | ||
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Gedras can be manipulated exactly like monzos. Just as (a,b) + (a',b') = (a+a',b+b'), likewise [k,s] + [k',s'] = [k+k',s+s']. If (a,b) is a stack of n intervals, with (a,b) = (na', nb') = n(a',b'), and if (a,b) = [k,s], then [k,s] = [nk',ns'] = n[k',s']. | Gedras can be manipulated exactly like monzos. Just as (a,b) + (a',b') = (a+a',b+b'), likewise [k,s] + [k',s'] = [k+k',s+s']. If (a,b) is a stack of n intervals, with (a,b) = (na', nb') = n(a',b'), and if (a,b) = [k,s], then [k,s] = [nk',ns'] = n[k',s']. | ||
Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. For example | 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. 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, P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. Here, the bare enharmonic is P5 - 2*m3 = [7,4] - 2*[3,2] = [7,4] - [6,4] = [1,0] = A1. | ||
Next, consider (P8/5, P5). One fifth of an octave = [12,7]/5 = [2,1] = M2. The bare enharmonic is P8 - 5*M2 = [12,7] - 5*[2,1] = [2,2] = 2*[1,1] = 2*m2. Because the enharmonic E is multiplied by 2, it will occur twice in the perchain, even thought the comma only occurs once (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: | |||
<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 | 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: | ||
<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^^ | <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^^ -- Dv -- Eb^ -- Evv -- F</span> | ||
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. | ||
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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. | ||
A false-double pergen can 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 a | A false-double pergen can 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. 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=" | |||
</span> | <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 -- | |||
</span> | </span> | ||
<span style="display: block; text-align: center;">P1 -- | 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><span style="display: block; text-align: center;">C -- Ev -- Ab^ -- C | |||
</span><span style=" | <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;"> | </span><span style="display: block; text-align: center;">C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- 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 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> | </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== | ||
For (P8/ | 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, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [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 off to [4,2] = M3. The enharmonic becomes [33,19] - 12*[4,2] = [-15,-5] = -5*[3,1] = 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. | ||
<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 -- C | |||
C -- Eb^<span style="vertical-align: super;">4</span> -- Av<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>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 | |||
</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 (P8/3, P4/4). P = vM3, E = ^3d2, G = /m2, E' = /4dd2. | |||
<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;">P1 -- /m2 -- ``//``d3=\\A2 -- \M3 -- P4 | |||
</span><span style="display: block; text-align: center;">C -- Db/ -- Ebb``//``=D#\\ -- E\ -- F</span><span style="display: block; text-align: center;"> | |||
</span> | |||
The comma equals xE and/or yE'. | The comma equals xE and/or yE'. | ||
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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. | 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. | ||
For single-comma pergens, the enharmonic should equal the comma's mapping. For example, (P8, P5/2) might arise from 243/242, which splits the 5th into two 11/9 halves. | For single-comma pergens, the enharmonic should equal the comma's mapping. For example, (P8, P5/2) might arise from 243/242, which splits the 5th into two 11/9 halves. | ||
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[//Question: what if there are highs and lows?//]</pre></div> | [//Question: what if there are highs and lows?//]</pre></div> | ||
<h4>Original HTML content:</h4> | <h4>Original HTML content:</h4> | ||
<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html"><html><head><title>pergen names</title></head><body><!-- ws:start:WikiTextHeadingRule: | <div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html"><html><head><title>pergen names</title></head><body><!-- ws:start:WikiTextHeadingRule:31:&lt;h1&gt; --><h1 id="toc0"><!-- ws:end:WikiTextHeadingRule:31 --> </h1> | ||
<!-- ws:start:WikiTextTocRule: | <!-- ws:start:WikiTextTocRule:57:&lt;img id=&quot;wikitext@@toc@@normal&quot; class=&quot;WikiMedia WikiMediaToc&quot; title=&quot;Table of Contents&quot; src=&quot;/site/embedthumbnail/toc/normal?w=225&amp;h=100&quot;/&gt; --><div id="toc"><h1 class="nopad">Table of Contents</h1><!-- ws:end:WikiTextTocRule:57 --><!-- ws:start:WikiTextTocRule:58: --><div style="margin-left: 1em;"><a href="#toc0"> </a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:58 --><!-- ws:start:WikiTextTocRule:59: --><div style="margin-left: 1em;"><a href="#Definition">Definition</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:59 --><!-- ws:start:WikiTextTocRule:60: --><div style="margin-left: 1em;"><a href="#Derivation">Derivation</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:60 --><!-- ws:start:WikiTextTocRule:61: --><div style="margin-left: 1em;"><a href="#Applications">Applications</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:61 --><!-- ws:start:WikiTextTocRule:62: --><div style="margin-left: 1em;"><a href="#Further Discussion">Further Discussion</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:62 --><!-- ws:start:WikiTextTocRule:63: --><div style="margin-left: 2em;"><a href="#Further Discussion-Extremely large multi-gens">Extremely large multi-gens</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:63 --><!-- ws:start:WikiTextTocRule:64: --><div style="margin-left: 2em;"><a href="#Further Discussion-Singles and doubles">Singles and doubles</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:64 --><!-- ws:start:WikiTextTocRule:65: --><div style="margin-left: 2em;"><a href="#Further Discussion-Finding an example temperament">Finding an example temperament</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:65 --><!-- ws:start:WikiTextTocRule:66: --><div style="margin-left: 2em;"><a href="#Further Discussion-Finding a notation for a pergen">Finding a notation for a pergen</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:66 --><!-- ws:start:WikiTextTocRule:67: --><div style="margin-left: 2em;"><a href="#Further Discussion-Alternate enharmonics">Alternate enharmonics</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:67 --><!-- ws:start:WikiTextTocRule:68: --><div style="margin-left: 2em;"><a href="#Further Discussion-Alternate keyspans and stepspans">Alternate keyspans and stepspans</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:68 --><!-- ws:start:WikiTextTocRule:69: --><div style="margin-left: 2em;"><a href="#toc11"> </a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:69 --><!-- ws:start:WikiTextTocRule:70: --><div style="margin-left: 2em;"><a href="#Further Discussion-Combining pergens">Combining pergens</a></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:70 --><!-- ws:start:WikiTextTocRule:71: --></div> | ||
<!-- ws:end:WikiTextTocRule: | <!-- ws:end:WikiTextTocRule:71 --><!-- ws:start:WikiTextHeadingRule:33:&lt;h1&gt; --><h1 id="toc1"><a name="Definition"></a><!-- ws:end:WikiTextHeadingRule:33 --><u><strong>Definition</strong></u></h1> | ||
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Examples: Marvel (2.3.5.7 and 225/224) is (P8, P5, ^1) = unsplit with ups (^1 = 81/80). Deep reddish (2.3.5.7 and 50/49) is (P8/2, P5, /1) = half-octave with highs (/1 = 81/80). Triple bluish (2.3.5.7 and 1029/1000) is (P8, P11/3, ^1) = third-eleventh with ups.<br /> | Examples: Marvel (2.3.5.7 and 225/224) is (P8, P5, ^1) = unsplit with ups (^1 = 81/80). Deep reddish (2.3.5.7 and 50/49) is (P8/2, P5, /1) = half-octave with highs (/1 = 81/80). Triple bluish (2.3.5.7 and 1029/1000) is (P8, P11/3, ^1) = third-eleventh with ups.<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.<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 /> | ||
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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. 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 /> | 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. 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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<!-- ws:start:WikiTextHeadingRule: | <!-- ws:start:WikiTextHeadingRule:35:&lt;h1&gt; --><h1 id="toc2"><a name="Derivation"></a><!-- ws:end:WikiTextHeadingRule:35 --><u>Derivation</u></h1> | ||
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For any comma containing primes 2 and 3, let m = the GCD of all the monzo's exponents other than the 2-exponent, and let n = the GCD of all its higher-prime exponents. The comma will split the octave into m parts, and if n &gt; m, it will split some other 3-limit interval into n parts. Therefore a comma with only two primes, one of which is 2, always splits the octave (unless the other prime's exponent is ±1, e.g. 32/31). And a comma with only one higher prime will always split something, unless that prime's exponent is ±1.<br /> | For any comma containing primes 2 and 3, let m = the GCD of all the monzo's exponents other than the 2-exponent, and let n = the GCD of all its higher-prime exponents. The comma will split the octave into m parts, and if n &gt; m, it will split some other 3-limit interval into n parts. Therefore a comma with only two primes, one of which is 2, always splits the octave (unless the other prime's exponent is ±1, e.g. 32/31). And a comma with only one higher prime will always split something, unless that prime's exponent is ±1.<br /> | ||
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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. Assuming 7's notational comma is 64/63, 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 /> | 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. Assuming 7's notational comma is 64/63, 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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<!-- ws:start:WikiTextHeadingRule: | <!-- ws:start:WikiTextHeadingRule:37:&lt;h1&gt; --><h1 id="toc3"><a name="Applications"></a><!-- ws:end:WikiTextHeadingRule:37 --><u>Applications</u></h1> | ||
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One obvious application is to name regular temperaments in a logical, consistent manner, avoiding the need to memorize many arbitrary names. Many temperaments have pergen-like names: Hemififths is (P8, P5/2), semihemi is (P8/2, P4/2), triforce is (P8/3, P4/2), both tetracot and semihemififths are (P8, P5/4), fourfives is (P8/4, P5/5), pental is (P8/5, P5), and fifive is (P8/2, P5/5). Pergen names are an improvement over these because they specify more exactly what is split. Some temperament names are what might be called pseudo-pergens, because the multigen isn't 3-limit. Meantone (mean = average, tone = major 2nd) implies y3/2 = (5/4)/2. Semisixth (aka sensei) implies y6/2 = (5/3)/2. Sometimes the multi-gen isn't even a multi-gen: y3/2 isn't actually a generator, although y6/2 is.<br /> | One obvious application is to name regular temperaments in a logical, consistent manner, avoiding the need to memorize many arbitrary names. Many temperaments have pergen-like names: Hemififths is (P8, P5/2), semihemi is (P8/2, P4/2), triforce is (P8/3, P4/2), both tetracot and semihemififths are (P8, P5/4), fourfives is (P8/4, P5/5), pental is (P8/5, P5), and fifive is (P8/2, P5/5). Pergen names are an improvement over these because they specify more exactly what is split. Some temperament names are what might be called pseudo-pergens, because the multigen isn't 3-limit. Meantone (mean = average, tone = major 2nd) implies y3/2 = (5/4)/2. Semisixth (aka sensei) implies y6/2 = (5/3)/2. Sometimes the multi-gen isn't even a multi-gen: y3/2 isn't actually a generator, although y6/2 is.<br /> | ||
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<!-- ws:start:WikiTextHeadingRule: | <!-- ws:start:WikiTextHeadingRule:39:&lt;h1&gt; --><h1 id="toc4"><a name="Further Discussion"></a><!-- ws:end:WikiTextHeadingRule:39 --><u>Further Discussion</u></h1> | ||
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<!-- ws:start:WikiTextHeadingRule: | <!-- ws:start:WikiTextHeadingRule:41:&lt;h2&gt; --><h2 id="toc5"><a name="Further Discussion-Extremely large multi-gens"></a><!-- ws:end:WikiTextHeadingRule:41 -->Extremely large multi-gens</h2> | ||
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So far, the largest multi-gen has been a 12th. As the multi-gen fractions get larger, the multi-gen gets quite wide. To avoid cumbersome degree names like 16th or 22nd, for degrees above 12, the widening is indicated with one &quot;W&quot; per octave. Thus 32/9 = Wm7, 9/2 = WWM2 or WM9, etc. For (P8, M/n), valid multi-gens are any voicing of the fifth that is less than n/2 octaves. For (P8, M/6), the multi-gen can be P4, P5, P11, P12, WWP4 or WWP5.<br /> | So far, the largest multi-gen has been a 12th. As the multi-gen fractions get larger, the multi-gen gets quite wide. To avoid cumbersome degree names like 16th or 22nd, for degrees above 12, the widening is indicated with one &quot;W&quot; per octave. Thus 32/9 = Wm7, 9/2 = WWM2 or WM9, etc. For (P8, M/n), valid multi-gens are any voicing of the fifth that is less than n/2 octaves. For (P8, M/6), the multi-gen can be P4, P5, P11, P12, WWP4 or WWP5.<br /> | ||
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<!-- ws:start:WikiTextHeadingRule: | <!-- ws:start:WikiTextHeadingRule:43:&lt;h2&gt; --><h2 id="toc6"><a name="Further Discussion-Singles and doubles"></a><!-- ws:end:WikiTextHeadingRule:43 -->Singles and doubles</h2> | ||
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If a pergen has only one fraction, like (P8/2, P5) or (P8, P4/3), the pergen is a <strong>single-split</strong> pergen. If it has two fractions, it's a <strong>double-split</strong> pergen. A single-split pergens can result from tempering out only a single comma, although it can be created by multiple commas. A single-split pergen can be notated with only ups and downs, called <strong>single-pair</strong> notation because it adds only a single pair of accidentals to conventional notation. <strong>Double-pair</strong> notation uses both ups/downs and highs/lows. In general, single-pair notation is preferred, because it's simpler. However, double-pair notation may be preferred, especially if the enharmonic for single-pair notation is a 3rd or larger.<br /> | If a pergen has only one fraction, like (P8/2, P5) or (P8, P4/3), the pergen is a <strong>single-split</strong> pergen. If it has two fractions, it's a <strong>double-split</strong> pergen. A single-split pergens can result from tempering out only a single comma, although it can be created by multiple commas. A single-split pergen can be notated with only ups and downs, called <strong>single-pair</strong> notation because it adds only a single pair of accidentals to conventional notation. <strong>Double-pair</strong> notation uses both ups/downs and highs/lows. In general, single-pair notation is preferred, because it's simpler. However, double-pair notation may be preferred, especially if the enharmonic for single-pair notation is a 3rd or larger.<br /> | ||
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A pergen (P8/m, (a,b)/n) is a false double if and only if GCD (m,n) = |b|. The next section discusses an alternate test.<br /> | A pergen (P8/m, (a,b)/n) is a false double if and only if GCD (m,n) = |b|. The next section discusses an alternate test.<br /> | ||
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<!-- ws:start:WikiTextHeadingRule: | <!-- ws:start:WikiTextHeadingRule:45:&lt;h2&gt; --><h2 id="toc7"><a name="Further Discussion-Finding an example temperament"></a><!-- ws:end:WikiTextHeadingRule:45 -->Finding an example temperament</h2> | ||
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To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4*P and P8. If P is 6/5, the comma is 4*P - P8 = (6/5)^4 / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4*P = (2/1) * (7/6)^-4. Neither 13/11 nor 32/27 would work for P, too many and too few primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3*G - P4 = (10/9)^3 / (4/3) = 250/243.<br /> | To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4*P and P8. If P is 6/5, the comma is 4*P - P8 = (6/5)^4 / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4*P = (2/1) * (7/6)^-4. Neither 13/11 nor 32/27 would work for P, too many and too few primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3*G - P4 = (10/9)^3 / (4/3) = 250/243.<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 /> | ||
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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.<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 /> | |||
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There are also alternate enharmonics, see below.<br /> | There are also alternate enharmonics, see below.<br /> | ||
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<!-- ws:start:WikiTextHeadingRule: | <!-- 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 basic concepts:<br /> | 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 basic terminology and 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 false doubles using single-pair notation, E = E'<br /> | |||
The unreduced pergen is (P8/m, M'/n'), with M' = n'G' + zE&quot;<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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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. Again, this concept can be expanded to include pentatonicism, octotonicism, etc., but we'll assume heptatonicism for now.<br /> | 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. Again, this concept can be expanded to include pentatonicism, octotonicism, etc., but we'll assume heptatonicism for now.<br /> | ||
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Gedras can be manipulated exactly like monzos. Just as (a,b) + (a',b') = (a+a',b+b'), likewise [k,s] + [k',s'] = [k+k',s+s']. If (a,b) is a stack of n intervals, with (a,b) = (na', nb') = n(a',b'), and if (a,b) = [k,s], then [k,s] = [nk',ns'] = n[k',s'].<br /> | Gedras can be manipulated exactly like monzos. Just as (a,b) + (a',b') = (a+a',b+b'), likewise [k,s] + [k',s'] = [k+k',s+s']. If (a,b) is a stack of n intervals, with (a,b) = (na', nb') = n(a',b'), and if (a,b) = [k,s], then [k,s] = [nk',ns'] = n[k',s'].<br /> | ||
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Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. For example | 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. 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, P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. Here, the bare enharmonic is 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). One fifth of an octave = [12,7]/5 = [2,1] = M2. The bare enharmonic is P8 - 5*M2 = [12,7] - 5*[2,1] = [2,2] = 2*[1,1] = 2*m2. Because the enharmonic E is multiplied by 2, it will occur twice in the perchain, even thought the comma only occurs once (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 /> | |||
<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 | 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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<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^^ | <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^^ -- Dv -- Eb^ -- Evv -- F</span><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 /> | 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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| Line 2,142: | Line 2,171: | ||
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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A false-double pergen can 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 a | A false-double pergen can 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).<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 /> | |||
<span style=" | <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 -- | |||
</span><br /> | </span><br /> | ||
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<span style="display: block; text-align: center;">P1 -- | 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><span style="display: block; text-align: center;">C -- Ev -- Ab^ -- C<br /> | <br /> | ||
</span><span style=" | <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><br /> | |||
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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 /> | |||
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<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><br /> | </span><br /> | ||
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<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><br /> | |||
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<!-- ws:start:WikiTextHeadingRule: | <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><br /> | ||
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<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 /> | |||
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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> | |||
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For (P8/ | 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, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [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 off to [4,2] = M3. The enharmonic becomes [33,19] - 12*[4,2] = [-15,-5] = -5*[3,1] = 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.<br /> | ||
<span style="display: block; text-align: center;"><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 /> | |||
</span><span style="display: block; text-align: center;"><br /> | |||
<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 /> | |||
</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 (P8/3, P4/4). P = vM3, E = ^3d2, G = /m2, E' = /4dd2.<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;">P1 -- /m2 -- <!-- ws:start:WikiTextRawRule:029:``//`` -->//<!-- ws:end:WikiTextRawRule:029 -->d3=\\A2 -- \M3 -- P4<br /> | |||
</span><span style="display: block; text-align: center;">C -- Db/ -- Ebb<!-- ws:start:WikiTextRawRule:030:``//`` -->//<!-- ws:end:WikiTextRawRule:030 -->=D#\\ -- E\ -- F</span><span style="display: block; text-align: center;"><br /> | |||
</span><br /> | |||
The comma equals xE and/or yE'.<br /> | The comma equals xE and/or yE'.<br /> | ||
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| Line 2,173: | Line 2,219: | ||
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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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For single-comma pergens, the enharmonic should equal the comma's mapping. For example, (P8, P5/2) might arise from 243/242, which splits the 5th into two 11/9 halves.<br /> | For single-comma pergens, the enharmonic should equal the comma's mapping. For example, (P8, P5/2) might arise from 243/242, which splits the 5th into two 11/9 halves.<br /> | ||
| Line 2,180: | Line 2,224: | ||
(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4*G = [1,2] = dd3. It must be downed, thus E = v<span style="vertical-align: super;">4</span>dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.<br /> | (P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4*G = [1,2] = dd3. It must be downed, thus E = v<span style="vertical-align: super;">4</span>dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.<br /> | ||
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<!-- ws:start:WikiTextHeadingRule: | <!-- ws:start:WikiTextHeadingRule:51:&lt;h2&gt; --><h2 id="toc10"><a name="Further Discussion-Alternate keyspans and stepspans"></a><!-- ws:end:WikiTextHeadingRule:51 -->Alternate keyspans and stepspans</h2> | ||
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One might wonder, why 12-edo keyspans? Why heptatonic stepspans? Heptatonic is best because conventional notation is heptatonic, and we want to minimize the heptatonic stepspan of the enharmonic. In order for the matrix to be invertible, the edo must be connected to 7-edo on the scale tree. The only choices are 5-edo, 12-edo, 19-edo, 26-edo, 33-edo, 40-edo and 47-edo, and on the other side, 2-edo, 9-edo, 16-edo and 23-edo. These edos would also work, 12-edo is merely the most convenient choice.<br /> | One might wonder, why 12-edo keyspans? Why heptatonic stepspans? Heptatonic is best because conventional notation is heptatonic, and we want to minimize the heptatonic stepspan of the enharmonic. In order for the matrix to be invertible, the edo must be connected to 7-edo on the scale tree. The only choices are 5-edo, 12-edo, 19-edo, 26-edo, 33-edo, 40-edo and 47-edo, and on the other side, 2-edo, 9-edo, 16-edo and 23-edo. These edos would also work, 12-edo is merely the most convenient choice.<br /> | ||
<!-- ws:start:WikiTextHeadingRule: | <!-- ws:start:WikiTextHeadingRule:53:&lt;h2&gt; --><h2 id="toc11"><!-- ws:end:WikiTextHeadingRule:53 --> </h2> | ||
<!-- ws:start:WikiTextHeadingRule: | <!-- 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 makes third-fourth, and 49/48 makes half-fourth, and tempering out both commas makes sixth-fourth. Therefore (P8, P4/3) + (P8, P4/2) = (P8, P4/6). General rules for combining pergens:<br /> | 250/243 makes third-fourth, and 49/48 makes half-fourth, and tempering out both commas makes sixth-fourth. Therefore (P8, P4/3) + (P8, P4/2) = (P8, P4/6). General rules for combining pergens:<br /> | ||