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Message: 10975 Date: Wed, 19 May 2004 22:48:21 Subject: Re: Adding wedgies? From: Paul Erlich Though I'd still like to see this answered, it would appear to be true according to pages 7-8 of this paper: 403 Forbidden * where it seems that the Klein correspondence does indeed equate to the condition that the bivector be simple. But I could be reading it wrong. --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: > Aha -- does the klein condition equate with the bivector being a > simple bivector? > > --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > wrote: > > --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> > wrote: > > > What's going on here? > > > > > > [5, 13, -17, 9, -41, -76], TOP error 0.27611 > > > "plus" > > > [13, 14, 35, -8, 19, 42], TOP error 0.26193 > > > "equals" > > > [18, 27, 18, 1, -22, -34], TOP error 0.036378 > > > > Parakleismic + Amity = Ennealimmal. Both parakleismic and amity have > > 4375/4374 as a comma, and so does their sum (and difference, for > that > > matter.) > > > > I did talk about it before, though I can't recall what I said about > > it. It is related to the Klein stuff. For 7-limit wedgies, define > the > > Pfaffian as follows: let > > > > X = <<x1 x2 x3 x4 x5 x6|| > > Y = <<y1 y2 y3 y4 y5 y6|| > > > > Then > > > > Pf(X, Y) = y1x6 + x1y6 - y2x5 - x2y5 + y3x4 + x3y4 > > > > It is easily checked that we have the identity > > > > Pf(X+Y, X+Y) = Pf(X, X) + 2 Pf(X, Y) + Pf(Y, Y) > > > > The Klein condition for the wedgie X is Pf(X, X)=0. If X and Y both > > satisfy the Klein condition, and if Pf(X, Y)=0, then X+Y also > > satisfies the Klein condition, and hence is a wedgie. What > > Pf(X, Y)=0 means is that X and Y are related; they share a comma. > > > > Probably, you will not want to talk about this in the paper. :) ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: Yahoo groups: /tuning-math/ * <*> To unsubscribe from this group, send an email to: tuning-math-unsubscribe@xxxxxxxxxxx.xxx <*> Your use of Yahoo! Groups is subject to: Yahoo! Terms of Service *
Message: 10976 Date: Fri, 21 May 2004 09:12:35 Subject: Re: tratios and yantras From: monz hi Paul, --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: > --- In tuning-math@xxxxxxxxxxx.xxxx "monz" <monz@a...> wrote: > > hi Gene and Paul, > > > > > > definitions of tratio and yantra, please. > > thanks. > > > > > > > > -monz > > Hi Monz, > > Gene defined yantra here: > > Yahoo groups: /tuning-math/message/10095 * > > It's just the first N integers with no prime factors above P, for a > given N and P. > > Tratio was something you posted to tell me I should use the > terminology "proportion" for -- remember? If a temperament has > codimension 1, it can be described by a vanishing ratio. If a > temperament has codimension 2, it can be described by a vanishing > tratio (a three-term proportion, like 625:640:648). > > Apparently, many, but not all, of the important codimension-1 > temperaments can be described by vanishing tratios of three > consective terms in the yantra (for which P is either 5 or 7, and N > is infinite or simply "large enough"). thanks. Ernest McClain's definition of yantra (which i believe was Gene's source) always includes a terminating maximum value below which all the included integers must lie. References ---------- McClain, Ernest. _The Myth of Invariance_ McClain, Ernest. _The Pythagorean Plato_ -monz ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: Yahoo groups: /tuning-math/ * <*> To unsubscribe from this group, send an email to: tuning-math-unsubscribe@xxxxxxxxxxx.xxx <*> Your use of Yahoo! Groups is subject to: Yahoo! Terms of Service *
Message: 10979 Date: Fri, 21 May 2004 23:28:20 Subject: Re: Adding wedgies? From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote: > --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: > > Though I'd still like to see this answered, it would appear to be > > true according to pages 7-8 of this paper: > > > > 403 Forbidden * > > > > where it seems that the Klein correspondence does indeed equate to > > the condition that the bivector be simple. > > I thought I did answer that; in any case, it's true. You didn't answer that or any of the other approximately 10 questions I posted that day. ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: Yahoo groups: /tuning-math/ * <*> To unsubscribe from this group, send an email to: tuning-math-unsubscribe@xxxxxxxxxxx.xxx <*> Your use of Yahoo! Groups is subject to: Yahoo! Terms of Service *
Message: 10984 Date: Tue, 25 May 2004 18:18:28 Subject: Re: Cross-check for TOP 5-limit 12-equal From: Paul Erlich If someone could help explain this, and/or generalize it to higher dimensions, I'd be thrilled . . . pleeeeeeeeeeeease? Also would like to understand the mystery factors of exactly 2 and exactly 3 . . . --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: > Wedgie norm for 12-equal: > > Take the two unison vectors > > |7 0 -3> > |-4 4 -1> > > Now find the determinant, and the "area" it represents, in each of > the basis planes: > > |7 0| = 28*(e23) -> 28/lg2(5) = 12.059 > |-4 4| > > |7 -3| = -19*(e25) -> 19/lg2(3) = 11.988 > |-4 -1| > > |0 -3| = 12*(e35) -> 12 = 12 > |4 -1| > > sum = 36.047 > > If I just use the maximum (L_inf = 12.059) as a measure of notes per > acoustical octave, then I "predict" tempered octaves of 1194.1 cents. > If I use the sum (L_1), dividing by the "mystery constant" 3, > I "predict" tempered octaves of 1198.4 cents. Neither one is the TOP > value . . . :( . . . but what sorts of error criteria, if any, *do* > they optimize? > > So the cross-checking I found for the 3-limit case in "Attn: Gene 2" > Yahoo groups: /tuning-math/message/8799 * > doesn't seem to work in the 5-limit ET case for either the L_1 or > L_inf norms. > > However, if I just add the largest and smallest values above: > > 28/lg2(5)+19/lg2(3) > > I do predict the correct tempered octave (aside from a factor of 2), > > 1197.67406985219 cents. > > So what sort of norm, if any, did I use to calculate complexity this > time? It's related to how we temper for TOP . . . ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: Yahoo groups: /tuning-math/ * <*> To unsubscribe from this group, send an email to: tuning-math-unsubscribe@xxxxxxxxxxx.xxx <*> Your use of Yahoo! Groups is subject to: Yahoo! Terms of Service *
Message: 10985 Date: Wed, 26 May 2004 04:52:12 Subject: Equal Temperament wikipedia entry now improved . . . From: Paul Erlich . . . and linked to "Gene's" meantone entry: Equal temperament - Wikipedia, the free encyclopedia *
Message: 10987 Date: Wed, 26 May 2004 06:02:17 Subject: Re: Omnitetrachordal Scales From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Kalle Aho" <kalleaho@m...> wrote: > Hi, > > The following is pure "stream of consciousness" thinking so don't > take it too seriously. :) > > What if we started with melodic considerations instead of harmonic > when we search for interesting scales? > > Is there any kind of general algorithm which will generate all > omnitetrachordal scales, possibly in an abstract form of patterns of > L (larger step) and s (smaller step)? I've posted omnitetrachordal scales with *three* step sizes. But among those with two step sizes, there seem to be only two classes that I've found so far: 1. MOSs of temperaments where the period is an octave and the generator is the approximate fifth/fourth; 2. Non-MOSs of temperaments where the period is a half-octave and the generator can be expressed as the approximate fifth/fourth. In the first class we have the pentatonic and diatonic scales in meantone and mavila, 17- and 29-tone schismic scales, 19-tone meantone scales . . . In the second class, the 'pentachordal' and 'hexachordal' pajara scales, the 'hexachordal' and 'heptachordal' injera scales . . .
Message: 10988 Date: Wed, 26 May 2004 07:42:29 Subject: Re: Equal Temperament wikipedia entry now improved . . . From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: > . . . and linked to "Gene's" meantone entry: > > Equal temperament - Wikipedia, the free encyclopedia * fixed "just intonation" a bit too . . . ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ Yahoo! Groups Links <*> To visit your group on the web, go to: Yahoo groups: /tuning-math/ * <*> To unsubscribe from this group, send an email to: tuning-math-unsubscribe@xxxxxxxxxxx.xxx <*> Your use of Yahoo! Groups is subject to: Yahoo! Terms of Service *
Message: 10991 Date: Tue, 01 Jun 2004 18:52:16 Subject: Re: Nexials From: Carl Lumma >For any p-limit >linear temperament wedgie, we can project down to a q-limit wedgie Are p and q reversed here? -C.
Message: 10994 Date: Tue, 01 Jun 2004 19:58:37 Subject: Re: Nexials From: Carl Lumma >> >For any p-limit >> >linear temperament wedgie, we can project down to a q-limit >> >wedgie >> >> Are p and q reversed here? > >No; q is the next largest prime. If p is 7, q is 5, etc. Oh, by next largest I took you to mean bigger. -Carl
Message: 10997 Date: Tue, 01 Jun 2004 23:09:37 Subject: Re: The hanson family From: Carl Lumma > The 5 to 7 limit choice is not as clear as I was thinking, > because Number 91 intrudes itself, so I withdraw my objection > to keeping "hanson" only as a 5-limit name. It would be nice > to figure out a naming scheme which would help sort out these > family trees, however. This family stuff looks awesome. I wish I understood the half of it. I'm surprised you're using generator sizes. How do you standardize the generator representation? Forgive me if this is old stuff, I haven't kept up. -Carl
Message: 10999 Date: Tue, 01 Jun 2004 23:44:38 Subject: Re: The hanson family From: Carl Lumma >> This family stuff looks awesome. I wish I understood the >> half of it. I'm surprised you're using generator sizes. >> How do you standardize the generator representation? Forgive >> me if this is old stuff, I haven't kept up. > >It's just the TOP tuning for the generators. How do you get a unique set of generators out of the TOP tuning? My miserable notes offer... """ >>>it's easy to determine that [<1 x y z|, <0 6 -7 -2|] >>>is a possible mapping of miracle, as is [<x 1 y z|, >>><-6 0 -25 -20|], but I don't know how to get x, y, and z. >>>I've been trying to find something like this in the >>>archives, but I don't know where to look. >> >> I don't see that this was ever answered. Did I miss it? > >If you know the whole wedgie, finding x, y and z can be done by >solving a linear system. If you only know the period and >generator map, you first need to get the rest of the wedgie, >which will be the one which has a much lower badness than its >competitors. > >For instance, suppose I know the wedgie is <<1 4 10 4 13 12||. >Then I can set up the equations resulting from > ><1 x y z| ^ <0 1 4 10| = <<1 4 10 4 13 12|| > >We have <1 x y z| ^ <0 1 4 10| = <<1 4 10 4x-y 10x-z 10y-4z|| > >Solving this gives us y=4x-4, z=10x-13; we can pick any integer >for x so we choose one giving us generators in a range we like. >Since 3 is represented by [x 1] in terms of octave x and >generator, if we want 3/2 as a generator we pick x=1. """ ...which makes it sound as if 'tis predicated on mere fancy. -Carl
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