This is an
**
Opt In Archive
.
**
We would like to hear from you if you want your posts included. For the contact address see
About this archive. All posts are copyright (c).

10000
10050
10100
10150
10200
10250
10300
10350
10400
**10450**
10500
10550
10600
10650
10700
10750
10800
10850
10900
10950

**10450 -**
10475 -

Message: 10450 - Contents - Hide Contents Date: Mon, 01 Mar 2004 23:28:26 Subject: Re: 9-limit stepwise From: Gene Ward Smith --- 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: >> >> On the other end of the size scale we have these. Paul, have you > ever>> considered Pajara[6] as a possible melody scale? >> I'm confused -- I thought the largest step was supposed to be less > than 200 cents?I moved it up to 8/7. Carl thinks I could go even higher.

Message: 10451 - Contents - Hide Contents 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 * [with cont.] (Wayb.) 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 - Type Ok * [with cont.] (Wayb.) and calls it "superaugmented subminor 7th 9th augmented 11th", which is even worse.

Message: 10452 - Contents - Hide Contents 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 - Contents - Hide Contents 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/ * [with cont.] <*> 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 * [with cont.] (Wayb.)

Message: 10455 - Contents - Hide Contents Date: Tue, 02 Mar 2004 19:03:58 Subject: Re: Canonical generators for 7-limit planar temperaments From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul G Hjelmstad" <paul.hjelmstad@u...> wrote:>> Taking the square root defines the symmetric lattice distance to the >> origin. We get a bilinear form, and hence a dot product, from B (a,b) > =>> (Q(ab)-Q(a)-Q(b))/2 (writing it multiplicitively, additively, in > terms>> of row vectors, (Q(a+b)-Q(a)-Q(b))/2.) >> >> So, which one are you using?It depends on whether you feed the program a monzo or a ratio. You should get the same result using either <-4 4 -1| or 81/80, since they refer to the same thing.>> I have Q(81/80)=13, so I get a unit vector in the 81/80 direction by >> dividing by sqrt(13): [4/sqrt(13), -1/sqrt(13), 0]. Taking the dot >> product of this with [a,b,c] and substracting, I get the orthogonal >> projection of [a,b,c]: >> > > Subtracting what?The unit vector times its dot product with the other vector. In other words, if u is a unit vector and v is any vector, we can take v - (u.v)u and then the dot product of this with u will be zero, since u.(v - (u.v)u) = u.v - (u.v)(u.u) = 0 since u.u = 1. Hence, it will be in the subspace orthogonal to u, and be the orthogonal projection to that subspace.>> 0 {1} >> [1] >> [0] >> >> 3 {3/2, 4/3} >> [1, 4/3, 3/2] >> [-1, 0, 1] >>> So, would this one be Q81((3/2)^a (4/3)^b)? Sorry if this is an > obvious question...These are just scales of 81/80 planar, along with their reduction to septimal meantone. They are whatever lies inside of balls of various sizes around the unison, according to the Q81 metric.

Message: 10456 - Contents - Hide Contents Date: Tue, 02 Mar 2004 19:05:00 Subject: Re: Hanzos From: Gene Ward Smith --- 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.

Message: 10457 - Contents - Hide Contents Date: Tue, 02 Mar 2004 00:00:22 Subject: Re: Canonical generators for 7-limit planar temperaments From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul G Hjelmstad" <paul.hjelmstad@u...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > wrote:>> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul G Hjelmstad" >>> >>> Thanks. So for example - 81/80, has generators 4/3 and 9/7. How > does>>> one project onto the plane to obtain 1/7 and 5/7 respectively?I transformed coordinates first to face centered cubic when I did this before, but you don't have to. Define Q(3^a 5^b 7^c) = a^2+b^2+c^2+ab+ac+bc Then this is a positive definite quadratic form on note classes http://164.8.13.169/Enciklopedija/math/math/p/p520.htm * [with cont.] (Wayb.) Taking the square root defines the symmetric lattice distance to the origin. We get a bilinear form, and hence a dot product, from B(a,b)= (Q(ab)-Q(a)-Q(b))/2 (writing it multiplicitively, additively, in terms of row vectors, (Q(a+b)-Q(a)-Q(b))/2.) I have Q(81/80)=13, so I get a unit vector in the 81/80 direction by dividing by sqrt(13): [4/sqrt(13), -1/sqrt(13), 0]. Taking the dot product of this with [a,b,c] and substracting, I get the orthogonal projection of [a,b,c]: [-1/13*a-4/13*b-6/13*c, 14/13*b+3/26*c+7/26*a, c] I now get a positive semidefinite quadratic form by substituting this in Q; if I normalize by clearing denominators, I get Q81(3^a 5^b 7^c) = 3a^2+48b^2+43c^2+24ab+10ac+40bc Positive semidefinite: http://164.8.13.169/Enciklopedija/math/math/p/p524.htm * [with cont.] (Wayb.) Q81 is semidefinite, since Q81(81/80)=0. We can get a positive definite form, and hence a lattice, by going to two dimensions with basis 3/2 and 9/7: Q81((3/2)^a (9/7)^b) = 3a^2 + 2ab + 35b^2 If we take the TM reduction of intervals in the octave we can sort them by size according to the above distance measure. Below I give the intervals in order of size, the scales you get by combining them (which should be tempered by "didymus", or 81/80 planar) and the meantone reduction of the scales, where 7 is mapped to ten generators of a fifth. 0 {1} [1] [0] 3 {3/2, 4/3} [1, 4/3, 3/2] [-1, 0, 1] 12 {9/5, 9/8} [1, 9/8, 4/3, 3/2, 9/5] [-2, -1, 0, 1, 2] 27 {5/3, 6/5} [1, 9/8, 6/5, 4/3, 3/2, 5/3, 9/5] [-3, -2, -1, 0, 1, 2, 3] 35 {14/9, 9/7} [1, 9/8, 6/5, 9/7, 4/3, 3/2, 14/9, 5/3, 9/5] [-8, -3, -2, -1, 0, 1, 2, 3, 8] 36 {7/6, 12/7} [1, 9/8, 7/6, 6/5, 9/7, 4/3, 3/2, 14/9, 5/3, 12/7, 9/5] [-9, -8, -3, -2, -1, 0, 1, 2, 3, 8, 9] 40 {21/20, 27/14} [1, 21/20, 9/8, 7/6, 6/5, 9/7, 4/3, 3/2, 14/9, 5/3, 12/7, 9/5, 27/14] [-9, -8, -7, -3, -2, -1, 0, 1, 2, 3, 7, 8, 9] 43 {8/7, 7/4} [1, 21/20, 9/8, 8/7, 7/6, 6/5, 9/7, 4/3, 3/2, 14/9, 5/3, 12/7, 7/4, 9/5, 27/14] [-10, -9, -8, -7, -3, -2, -1, 0, 1, 2, 3, 7, 8, 9, 10] 48 {5/4, 8/5} [1, 21/20, 9/8, 8/7, 7/6, 6/5, 5/4, 9/7, 4/3, 3/2, 14/9, 8/5, 5/3, 12/7, 7/4, 9/5, 27/14] [-10, -9, -8, -7, -4, -3, -2, -1, 0, 1, 2, 3, 4, 7, 8, 9, 10] 51 {7/5, 10/7} [1, 21/20, 9/8, 8/7, 7/6, 6/5, 5/4, 9/7, 4/3, 7/5, 10/7, 3/2, 14/9, 8/5, 5/3, 12/7, 7/4, 9/5, 27/14] [-10, -9, -8, -7, -6, -4, -3, -2, -1, 0, 1, 2, 3, 4, 6, 7, 8, 9, 10] 56 {21/16, 32/21} [1, 21/20, 9/8, 8/7, 7/6, 6/5, 5/4, 9/7, 21/16, 4/3, 7/5, 10/7, 3/2, 32/21, 14/9, 8/5, 5/3, 12/7, 7/4, 9/5, 27/14] [-11, -10, -9, -8, -7, -6, -4, -3, -2, -1, 0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 11] 68 {15/14, 28/15} [1, 21/20, 15/14, 9/8, 8/7, 7/6, 6/5, 5/4, 9/7, 21/16, 4/3, 7/5, 10/7, 3/2, 32/21, 14/9, 8/5, 5/3, 12/7, 7/4, 9/5, 28/15, 27/14] [-11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] 75 {15/8, 35/18, 16/15, 36/35} [1, 36/35, 21/20, 16/15, 15/14, 9/8, 8/7, 7/6, 6/5, 5/4, 9/7, 21/16, 4/3, 7/5, 10/7, 3/2, 32/21, 14/9, 8/5, 5/3, 12/7, 7/4, 9/5, 28/15, 15/8, 27/14, 35/18] [-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] 91 {56/45, 45/28} [1, 36/35, 21/20, 16/15, 15/14, 9/8, 8/7, 7/6, 6/5, 56/45, 5/4, 9/7, 21/16, 4/3, 7/5, 10/7, 3/2, 32/21, 14/9, 8/5, 45/28, 5/3, 12/7, 7/4, 9/5, 28/15, 15/8, 27/14, 35/18] [-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] 100 {48/35, 35/24} [1, 36/35, 21/20, 16/15, 15/14, 9/8, 8/7, 7/6, 6/5, 56/45, 5/4, 9/7, 21/16, 4/3, 48/35, 7/5, 10/7, 35/24, 3/2, 32/21, 14/9, 8/5, 45/28, 5/3, 12/7, 7/4, 9/5, 28/15, 15/8, 27/14, 35/18] [-13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13] 108 {36/25, 25/18} [1, 36/35, 21/20, 16/15, 15/14, 9/8, 8/7, 7/6, 6/5, 56/45, 5/4, 9/7, 21/16, 4/3, 48/35, 25/18, 7/5, 10/7, 36/25, 35/24, 3/2, 32/21, 14/9, 8/5, 45/28, 5/3, 12/7, 7/4, 9/5, 28/15, 15/8, 27/14, 35/18] [-13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13] 120 {25/21, 42/25} [1, 36/35, 21/20, 16/15, 15/14, 9/8, 8/7, 7/6, 25/21, 6/5, 56/45, 5/4, 9/7, 21/16, 4/3, 48/35, 25/18, 7/5, 10/7, 36/25, 35/24, 3/2, 32/21, 14/9, 8/5, 45/28, 5/3, 42/25, 12/7, 7/4, 9/5, 28/15, 15/8, 27/14, 35/18] [-13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13] 131 {35/32, 64/35} [1, 36/35, 21/20, 16/15, 15/14, 35/32, 9/8, 8/7, 7/6, 25/21, 6/5, 56/45, 5/4, 9/7, 21/16, 4/3, 48/35, 25/18, 7/5, 10/7, 36/25, 35/24, 3/2, 32/21, 14/9, 8/5, 45/28, 5/3, 42/25, 12/7, 7/4, 9/5, 64/35, 28/15, 15/8, 27/14, 35/18] [-14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14] 139 {49/27, 54/49} [1, 36/35, 21/20, 16/15, 15/14, 35/32, 54/49, 9/8, 8/7, 7/6, 25/21, 6/5, 56/45, 5/4, 9/7, 21/16, 4/3, 48/35, 25/18, 7/5, 10/7, 36/25, 35/24, 3/2, 32/21, 14/9, 8/5, 45/28, 5/3, 42/25, 12/7, 7/4, 9/5, 49/27, 64/35, 28/15, 15/8, 27/14, 35/18] [-17, -14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 17] 140 {49/40, 80/49} [1, 36/35, 21/20, 16/15, 15/14, 35/32, 54/49, 9/8, 8/7, 7/6, 25/21, 6/5, 49/40, 56/45, 5/4, 9/7, 21/16, 4/3, 48/35, 25/18, 7/5, 10/7, 36/25, 35/24, 3/2, 32/21, 14/9, 8/5, 45/28, 80/49, 5/3, 42/25, 12/7, 7/4, 9/5, 49/27, 64/35, 28/15, 15/8, 27/14, 35/18] [-17, -16, -14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17]

Message: 10459 - Contents - Hide Contents Date: Tue, 02 Mar 2004 00:04:27 Subject: Re: 9-limit stepwise From: Gene Ward Smith --- 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 * [with cont.] (Wayb.) > > 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 - Type Ok * [with cont.] (Wayb.) > > 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.

Message: 10461 - Contents - Hide Contents 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 * [with cont.] (Wayb.) >> >> 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 - Type Ok * [with cont.] (Wayb.) >> >> 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 - Contents - Hide Contents 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 - Contents - Hide Contents 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. > > -CarlYou 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 - Contents - Hide Contents 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. > > > GrahamI care, and I hope Gene does too. I'd like to see this revisited.

Message: 10465 - Contents - Hide Contents Date: Tue, 02 Mar 2004 14:01:20 Subject: Re: Hanzos From: Carl Lumma>> > 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 - Contents - Hide Contents Date: Tue, 02 Mar 2004 14:04:05 Subject: Re: Hanzos From: Carl Lumma>>>>> >o 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.

Message: 10467 - Contents - Hide Contents 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.

Message: 10468 - Contents - Hide Contents Date: Tue, 02 Mar 2004 22:30:55 Subject: Re: Hanzos From: Gene Ward Smith --- 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; it doesn't require remarking on. 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.

Message: 10469 - Contents - Hide Contents Date: Tue, 02 Mar 2004 14:37:36 Subject: Re: Hanzos From: Carl Lumma>>>> > 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: 10470 - Contents - Hide Contents Date: Tue, 02 Mar 2004 22:39:43 Subject: Re: Hanzos From: Gene Ward Smith --- 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. 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.

Message: 10471 - Contents - Hide Contents 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 - Contents - Hide Contents 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 - Contents - Hide Contents 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 - Contents - Hide Contents 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

10000
10050
10100
10150
10200
10250
10300
10350
10400
**10450**
10500
10550
10600
10650
10700
10750
10800
10850
10900
10950

**10450 -**
10475 -