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Message: 10451 Date: Mon, 01 Mar 2004 23:31:44 Subject: Re: 9-limit stepwise From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote: > --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > wrote: > > > > 116 [49/45, 10/9, 9/8, 8/7] [1, 1, 2, 2] 1.568046 6 > > Pajara[6] > > It's six notes of pajara, but not Pajara[6]. Oh! > The 49/45 and 10/9 are > seven fifths up and four octaves down--the apotome, 49/45 and 10/9 all > being the same in pajara. The 9/8 and 8/7 of course are the same, both > two fifths up and an octave down. So in terms of generators, it looks > like [2,2,7,2,2,7], which adds up to 22 fifths. In 22-equal, it would > be 443443, Ah. Well you can see on Manuel's list: List of musical modes * that I already gave its second mode a bland, technical name. Dave Keenan shows it as a chord on his page: http://www.uq.net.au/~zzdkeena/Music/ErlichDecChords.gif - Ok * and calls it "superaugmented subminor 7th 9th augmented 11th", which is even worse.
Message: 10452 Date: Mon, 01 Mar 2004 23:33:24 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote: > >> One of us is still misunderstanding Paul Hahn's 9-limit approach. > >> In the unweighted version 3, 5, 7 and 9 are all the same length. > > > >In this system you don't exactly, have 7-limit notes and intervals. > >You do have "hanzos", with basis 2,3,5,7,9. > > The usual point of odd-limit is to get octave equivalence, and > therefore I'd say the 2s should be dropped from the basis. Bad move -- for one thing, you can't detect torsion. > >The hahnzo |0 -2 0 0 1> is a comma, 9/3^2, which obviously would > >play a special role. Hahnzos map onto 7-limit intervals, but not > >1-1. Are you happy with the idea that two scales could be > >different, since they have steps and notes which are distinct as > >hahnzos, even though they have exactly the same steps and notes > >in the 7-limit? > > I think the answer here is yes, though I'm at a loss for why > you're mapping hanzos to the 7-limit. 7-prime-limit. > >We've got three hahnzos corresponding to 81/80; > > My recollection is that Paul H.'s algorithm assigns a unique > lattice route (and therefore hanzo) to each 9-limit interval. Unique "hanzo" but not unique lattice route -- even monzos don't assign a unique lattice route.
Message: 10453 Date: Mon, 01 Mar 2004 23:36:18 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote: > >> My recollection is that Paul H.'s algorithm assigns a unique > >> lattice route (and therefore hanzo) to each 9-limit interval. > > > >So what? You still get an infinite number representing each interval, > >since you can multiply by arbitary powers of the dummy comma 9/3^2. > > An infinite number from where? If you look at the algorithm, that > dummy comma has zero length. How do you get that???? ________________________________________________________________________ ________________________________________________________________________ ------------------------------------------------------------------------ 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: 10461 Date: Tue, 02 Mar 2004 00:32:22 Subject: Re: 9-limit stepwise From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote: > --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: > > --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > > wrote: > > > --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > > > wrote: > > > > > > > > 116 [49/45, 10/9, 9/8, 8/7] [1, 1, 2, 2] 1.568046 6 > > > > Pajara[6] > > > > > > It's six notes of pajara, but not Pajara[6]. > > > > Oh! > > > > > The 49/45 and 10/9 are > > > seven fifths up and four octaves down--the apotome, 49/45 and 10/9 > > all > > > being the same in pajara. The 9/8 and 8/7 of course are the same, > > both > > > two fifths up and an octave down. So in terms of generators, it > > looks > > > like [2,2,7,2,2,7], which adds up to 22 fifths. In 22-equal, it > > would > > > be 443443, > > > > Ah. Well you can see on Manuel's list: > > > > List of musical modes * > > > > that I already gave its second mode a bland, technical name. Dave > > Keenan shows it as a chord on his page: > > > > http://www.uq.net.au/~zzdkeena/Music/ErlichDecChords.gif - Ok * > > > > and calls it "superaugmented subminor 7th 9th augmented 11th", which > > is even worse. > > Yow. I'll take 9-limit consonant whole-tone, thank you. Of course the > other whole tone scale counts also. It doesn't, because 4+4+4=12 is not a 9-limit consonance in 22-equal.
Message: 10462 Date: Tue, 02 Mar 2004 20:38:24 Subject: Re: Hanzos From: Graham Breed Carl: >>The usual point of odd-limit is to get octave equivalence, and >>therefore I'd say the 2s should be dropped from the basis. Paul E: > Bad move -- for one thing, you can't detect torsion. Can't you? I'm not sure it's even relevant here, but I thought I went through this in excruciating detail some time ago and showed that you can detect torsion with octave-equivalent vectors. In fact, I seem to remember running the calculation in parallel using both methods, and showing that I always got the same results. No, the outstanding problem is that we can't go from the octave-equivalent mapping to an optimized generator size. But I'm sure it can be done if anybody cared. The situation as I left it was that nobody did. Graham
Message: 10463 Date: Tue, 02 Mar 2004 20:58:25 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote: > >> >> My recollection is that Paul H.'s algorithm assigns a unique > >> >> lattice route (and therefore hanzo) to each 9-limit interval. > >> > > >> >So what? You still get an infinite number representing each > >> >interval, since you can multiply by arbitary powers of the dummy > >> >comma 9/3^2. > >> > >> An infinite number from where? If you look at the algorithm, that > >> dummy comma has zero length. > > > >How do you get that???? > > >Given a Fokker-style interval vector (I1, I2, . . . In): > > [-2 0 0 1] > > >1. Go to the rightmost nonzero exponent; add its absolute value > >to the total. > > T=1 > > >2. Use that exponent to cancel out as many exponents of the opposite > >sign as possible, starting to its immediate left and working right; > >discard anything remaining of that exponent. > > [-1 0 0 0] > T=1 > > >3. If any nonzero exponents remain, go back to step one, otherwise > >stop. > > T=... whoops, I forgot an absolute value here. The correct > value is 2. > > -Carl You mean 1, right? It's certainly not zero, though. So you have an infinite chain of duplicates for each pitch, spaced at increments of 1 unit . . .
Message: 10464 Date: Tue, 02 Mar 2004 21:03:24 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote: > Carl: > >>The usual point of odd-limit is to get octave equivalence, and > >>therefore I'd say the 2s should be dropped from the basis. > > Paul E: > > Bad move -- for one thing, you can't detect torsion. > > Can't you? I'm not sure it's even relevant here, but I thought I went > through this in excruciating detail some time ago and showed that you > can detect torsion with octave-equivalent vectors. In fact, I seem to > remember running the calculation in parallel using both methods, and > showing that I always got the same results. Right, but that was a specific set of calculations, not a general proof. You were just looking at 'linear' temperaments, if I recall correctly, but torsion can afflict all types of temperaments. Plus it seemed your method was far less elegant. > No, the outstanding problem is that we can't go from the > octave-equivalent mapping to an optimized generator size. But I'm sure > it can be done if anybody cared. The situation as I left it was that > nobody did. > > > Graham I care, and I hope Gene does too. I'd like to see this revisited.
Message: 10465 Date: Tue, 02 Mar 2004 14:01:20 Subject: Re: Hanzos From: Carl Lumma >> I don't know where sqrt would be coming from. I thought everything >> would have to have whole number lengths. > >I'm using Euclidean distances. I wish you had said that. -C.
Message: 10466 Date: Tue, 02 Mar 2004 14:04:05 Subject: Re: Hanzos From: Carl Lumma >> >> >So what? You still get an infinite number representing each >> >> >interval, since you can multiply by arbitary powers of the dummy >> >> >comma 9/3^2. >> >> >> >> An infinite number from where? If you look at the algorithm, >> >> that dummy comma has zero length. >> > >> >How do you get that???? >> >> >Given a Fokker-style interval vector (I1, I2, . . . In): >> >> [-2 0 0 1] >> >> >1. Go to the rightmost nonzero exponent; add its absolute value >> >to the total. >> >> T=1 >> >> >2. Use that exponent to cancel out as many exponents of the >> >opposite sign as possible, starting to its immediate left and >> >working right; discard anything remaining of that exponent. >> >> [-1 0 0 0] >> T=1 >> >> >3. If any nonzero exponents remain, go back to step one, otherwise >> >stop. >> >> T=... whoops, I forgot an absolute value here. The correct >> value is 2. > >You mean 1, right? No, T=1 and you've got to add what's left of the 3 exponent, abs(-1), to it. >It's certainly not zero, though. So you have an >infinite chain of duplicates for each pitch, spaced at increments of >1 unit . . . ... -C.
Message: 10467 Date: Tue, 02 Mar 2004 22:08:13 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote: > >> >> >So what? You still get an infinite number representing each > >> >> >interval, since you can multiply by arbitary powers of the dummy > >> >> >comma 9/3^2. > >> >> > >> >> An infinite number from where? If you look at the algorithm, > >> >> that dummy comma has zero length. > >> > > >> >How do you get that???? > >> > >> >Given a Fokker-style interval vector (I1, I2, . . . In): > >> > >> [-2 0 0 1] > >> > >> >1. Go to the rightmost nonzero exponent; add its absolute value > >> >to the total. > >> > >> T=1 > >> > >> >2. Use that exponent to cancel out as many exponents of the > >> >opposite sign as possible, starting to its immediate left and > >> >working right; discard anything remaining of that exponent. > >> > >> [-1 0 0 0] > >> T=1 > >> > >> >3. If any nonzero exponents remain, go back to step one, otherwise > >> >stop. > >> > >> T=... whoops, I forgot an absolute value here. The correct > >> value is 2. > > > >You mean 1, right? > > No, T=1 and you've got to add what's left of the 3 exponent, abs(- 1), > to it. Oh yeah, you're right. > >It's certainly not zero, though. So you have an > >infinite chain of duplicates for each pitch, spaced at increments of > >1 unit . . . > > ... 2 units.
Message: 10469 Date: Tue, 02 Mar 2004 14:37:36 Subject: Re: Hanzos From: Carl Lumma >> >> I don't know where sqrt would be coming from. I thought everything >> >> would have to have whole number lengths. >> > >> >I'm using Euclidean distances. >> >> I wish you had said that. > >If I'm talking about symmetric lattices, that would be assumed. All >these diagrams of hexagons and tetrahedrons and octahedrons that >people draw are Euclidean; I also like to think of them as graphs. >it doesn't require remarking on. Obviously it did. >However, >another thing to bear in mind is this--positive definite quadradic >forms entails Euclidean, and vice-versa. You can identify Euclidean >lattices with quadratic forms. That I know. -Carl
Message: 10471 Date: Tue, 02 Mar 2004 22:41:48 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote: > --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote: > > >> I don't know where sqrt would be coming from. I thought everything > > >> would have to have whole number lengths. > > > > > >I'm using Euclidean distances. > > > > I wish you had said that. > > If I'm talking about symmetric lattices, that would be assumed. You didn't assume it the other week, when it seemed (even upon clarification) that you referred to *two* possible metrics in the symmetric lattice -- the euclidean one, and the 'taxicab' one. The latter is what Paul Hahn's algorithm gives you. But now you say this. So now I wonder what you were *really* talking about. > All > these diagrams of hexagons and tetrahedrons and octahedrons that > people draw are Euclidean; That's a bogus assumption. We live in a nearly euclidean universe so people have no choice in the matter when it comes to diagrams. If I draw a Tenney lattice, which assumes a taxicab metric, how am I supposed to avoid drawing it in a way that you'd interpret as euclidean??
Message: 10472 Date: Tue, 02 Mar 2004 22:42:17 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote: > >> >> I don't know where sqrt would be coming from. I thought everything > >> >> would have to have whole number lengths. > >> > > >> >I'm using Euclidean distances. > >> > >> I wish you had said that. > > > >If I'm talking about symmetric lattices, that would be assumed. All > >these diagrams of hexagons and tetrahedrons and octahedrons that > >people draw are Euclidean; > > I also like to think of them as graphs. Exactly; Paul Hahn apparently did too.
Message: 10473 Date: Tue, 02 Mar 2004 22:44:55 Subject: Re: Hanzos From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote: > --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote: > > > I care, and I hope Gene does too. I'd like to see this revisited. > > You can do anything with octave-equivalent vectors that you can do > with monzos if you can recover the monzos. This means you can't do > some things, but since commas are small, you can recover the commas, > and hence can eg take two octave-equivalent 7-limit vectors and get > the corresponding temperament. Sure. That's how I generated most of those graphs of commas -- a large search in n-1 dimensional space, filling in the 2 component later by assuming each interval was between -600 and 600 cents. But I don't think this is what Graham was doing. > What I don't understand is why anyone would want to. You can also > throw away the 7, and keep the 2,3 and 5. Would anyone propose doing > that? It strikes me as an absurd proceedure. Where octave-equivalent > vectors are useful is in octave-equivalent contexts. Which would seem to be most musical contexts of interest on these lists.
Message: 10474 Date: Tue, 02 Mar 2004 15:00:50 Subject: Re: Hanzos From: Carl Lumma >What I don't understand is why anyone would want to. You can also >throw away the 7, and keep the 2,3 and 5. Would anyone propose doing >that? It strikes me as an absurd proceedure. Where octave-equivalent >vectors are useful is in octave-equivalent contexts. We weren't talking about temperaments, we were talking about finding chord sequences in JI; clearly an occasion for octave equivalence. -Carl
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