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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????
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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,
>
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,
>>
> 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.
>
> -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 - 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.
>
>
> Graham
I 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.
>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 - 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.
>> 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: 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
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