Tuning-Math Digests messages 10975 - 10999

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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. :)



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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






 



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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.



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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 . . .



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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 *


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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 . . .


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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 . . .



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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.


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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


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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


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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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