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Message: 10975 - Contents - Hide Contents

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 * [with cont.]  (Wayb.)

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 - Contents - Hide Contents

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 * [with cont.] 
> 
> 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: 10977 - Contents - Hide Contents

Date: Fri, 21 May 2004 22:44:16

Subject: Re: Adding wedgies?

From: Gene Ward Smith

--- 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 * [with cont.]  (Wayb.)
> 
> 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.
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Message: 10978 - Contents - Hide Contents

Date: Fri, 21 May 2004 22:47:11

Subject: Re: tratios and yantras

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "monz" <monz@a...> wrote:



> 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.



I think my definition is preferable, as it gives a 1-1 mapping. In any
case, they are equivalent.
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Message: 10979 - Contents - Hide Contents

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 * [with cont.]  (Wayb.)
>> 
>> 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 - Contents - Hide Contents

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 * [with cont.] 
> 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 - Contents - Hide Contents

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 encycl... * [with cont.]  (Wayb.)
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Message: 10986 - Contents - Hide Contents

Date: Wed, 26 May 2004 05:54:18

Subject: Omnitetrachordal Scales

From: Kalle Aho

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'm speculating that there might be a way to investigate the possible 
harmonic properties of such scales after giving all kinds of 
different values to L and s. 

Kalle
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Message: 10987 - Contents - Hide Contents

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 - Contents - Hide Contents

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 encycl... * [with cont.]  (Wayb.)



fixed "just intonation" a bit too . . .



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Message: 10989 - Contents - Hide Contents

Date: Mon, 31 May 2004 09:10:40

Subject: 100 members

From: Gene Ward Smith

Count 'em!



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Message: 10990 - Contents - Hide Contents

Date: Tue, 01 Jun 2004 18:39:59

Subject: Nexials

From: Gene Ward Smith

Suppose p>5 is prime and q is the next largest prime. For any p-limit
linear temperament wedgie, we can project down to a q-limit wedgie by
for instance taking the wedge product of the truncated mapping. Its
elements will be a subset of certain specific p-limit elements, and in
the case which will interest us, we suppose these are all relatively
prime and we have a well-behaved temperament.

If w is a p-limit val belonging to this temperament, and 
u = <0 0 ... 1| is p-limit, then both truncated u and tuncated w
belong to the ttemperament, and therefore u^w belongs to the projected
temperament. Adding or subtracting multiples of it will lead to
wedgies which define different p-limit temperaments, but the same
q-limit temperament. We can say they are the same, mod the truncated w.

Given two p-limit wedgies a and b which have the same q-limit
projected temperament, the normalized wedgie for a-b will be of the
u^w form. We can therefore toss the zeros and get a q-limit val, which
I will call the nexus of a and b.

In the 7-limit, the nexus of two wedgies a and b with the same 5-limit
projection is the normalization of <c[3], c[5], c[6]| where
c = a-b. In the 11-limit, this becomes <c[4], c[7], c[9], c[10]|. We
are simply taking the nonzero elements here, which are the ones whose
basis involves a p.

I'm putting this forward in connection with the question of naming and
classifying p-limit temperaments in connection with already named
q-limit temperaments.
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Message: 10991 - Contents - Hide Contents

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: 10992 - Contents - Hide Contents

Date: Tue, 01 Jun 2004 19:49:43

Subject: Re: Nexials

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Carl Lumma" <ekin@l...> wrote:


>> 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.
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Message: 10993 - Contents - Hide Contents

Date: Tue, 01 Jun 2004 19:54:06

Subject: The meantone family

From: Gene Ward Smith

This gives part of the family tree of the meantone family; I don't go
into cousins such as the various 11-limit versions of flattone or
dominant sevenths. In each limit, I propose giving the name "meantone"
to the temperament with the same TOP generators as 5-limit meantone.
The numbers before the colon give the nexial val with the base
meantone, plus or minus depending on whether adding or subtracting is
required. Also, the TM basis and the TOP octave and fourth.

meantone 81/80 comma 
[1201.70, 504.13]

7-limit meantone family

0: <<1 4 10 4 13 12|| meantone {81/80, 126/125}
[1201.70, 504.13]

-5: <<1 4 5 4 5 0|| {15/14, 81/80}
[1185.31, 500.33]

-7: <<1 4 3 4 2 -4|| {21/20, 28/27}
[1214.25, 509.40]

-12: <<1 4 -2 4 -6 -16|| dominant seventh {36/35, 64/63}
[1195.23, 495.88]

-19: <<1 4 -9 4 -17 -32|| flattone {81/80, 525/512}
[1202.54, 507.14]

-31: <<1 4 -21 4 -36 -60|| {81/80, 65625/65536}
[1201.56, 503.96]


11-limit meantone family (no cousins)

0: <<1 4 10 -13 4 13 -24 12 -44 -71|| meantone {81/80, 126/125, 385/384}
[1201.70, 503.13]

19: <<1 4 10 6 4 13 6 12 0 -18|| {45/44, 56/55, 81/80}
[1198.56, 503.60]

31: <<1 4 10 18 4 13 25 12 28 16|| huygens {81/80, 99/98, 126/125}
[1201.61, 504.02]
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Message: 10994 - Contents - Hide Contents

Date: Tue, 01 Jun 2004 19:58:37

Subject: Re: Nexials

From: Carl Lumma


>>> >or 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: 10995 - Contents - Hide Contents

Date: Tue, 01 Jun 2004 21:28:16

Subject: The magic family

From: Gene Ward Smith

Here we have a 7-limit system with the same TOP generators as in the
5-limit, and an 11-limit system with nearly identical generators.
However, the -41-nexial system is also important in the 11-limit, and
the generators are pretty close.

magic 3125/3072 comma
[1201.28, 380.80]

7-limit magic family

0: <<5 1 12 -10 5 25|| magic {225/224, 245/243}
[1201.28, 380.80]

-3: <<5 1 9 -10 0 18|| {36/35, 1875/1792}
[1193.27, 377.33]

-16: <<5 1 -4 -10 -20 -12|| {21/20, 3125/3072}
[1211.19, 376.84]

-19: <<5 1 -7 -10 -25 -19|| muggles {126/125, 525/512}
[1203.15, 379.39]

-22: <<5 1 -10 -10 -30 -26|| {64/63, 3125/3024}
[1198.68, 381.42]

11-limit magic family

0: <<5 1 12 33 -10 5 35 25 73 51|| magic {225/224, 245/243, 441/440}
[1201.29, 380.75]

-19: <<5 1 12 14 -10 -5 -5 25 29 -2|| {55/54, 99/98, 5314/5108}
[1199.00, 381.39]

-22: <<5 1 12 11 -10 5 0 25 22 -11|| {45/44, 56/55, 245/243}
[1197.02, 379.17]

-41: <<5 1 12 -8 -10 5 -30 25 -22 -64|| {100/99, 225/224, 245/243}
[1200.75, 380.92]

-60: <<5 1 12 -27 -10 5 -60 25 -66 -117||  {225/224, 245/243, 1617/1600}
[1201.60, 380.50]
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Message: 10996 - Contents - Hide Contents

Date: Tue, 01 Jun 2004 23:03:25

Subject: The hanson family

From: Gene Ward Smith

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.

hanson 15625/15552 comma
[1200.29, 317.07]

7-limit hanson family

0: <<6 5 22 -6 18 37|| catakleismic? {225/224, 4375/4374}
[1200.54, 316.91]

-15: <<6 5 7 -6 -6 2|| {36/35, 375/343}
[1192.24, 314.59]

-19: <<6 5 3 -6 -12 -7|| kleismic? {49/48, 126/125}
[1203.19, 317.83]

-34: <<6 5 -12 -6 -36 -42|| {64/64, 15625/15309}

-53: <<6 5 -31 -6 -66 -86|| "Number 91" {5120/5103, 15625/15552}
[1199.98, 317.09]

11-limit catakleismic?

0: <<6 5 22 -21 -6 18 -54 37 -66 -135|| catakleismic? {225/224,
385/384, 4375/4374}
[1200.54, 316.91]

19: <<6 5 22 -2 -6 18 -24 37 -22 -82|| {100/99, 225/224, 864/847}
[1197.96, 316.75]

53: <<6 5 22 32 -6 18 30 37 57 14|| {99/98, 176/175, 2200/2187}
[1198.27, 316.76]

72: <<6 5 22 51 -6 18 60 37 101 67|| {225/224, 441/440, 4375/4374}
[1200.61, 316.85]

11-limit kleismic?

0: <<6 5 3 -2 -6 -12 -24 -7 -22 -16|| {49/48, 56/55, 100/99}
[1200.82, 318.18]

-4: <<6 5 3 -6 -6 -12 -30 -7 -31 -27|| {33/32, 49/48, 126/125}
[1205.30, 315.59]

15: <<6 5 3 13 -6 -12 0 -7 13 26|| {49/48, 55/54, 77/75}
[1200.98, 318.16]

19: <<6 5 3 17 -6 -12 -6 -7 22 37|| {45/44, 49/48, 126/125}
[1204.38, 315.93]

11-limit Number 91

0: <<6 5 -31 32 -6 -66 30 -86 57 197|| {385/384, 2200/2187, 3388/3375}
[1199.89, 317.15]
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Message: 10997 - Contents - Hide Contents

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: 10998 - Contents - Hide Contents

Date: Tue, 01 Jun 2004 23:18:36

Subject: Re: The hanson family

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Carl Lumma" <ekin@l...> wrote:



> 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.
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Message: 10999 - Contents - Hide Contents

Date: Tue, 01 Jun 2004 23:44:38

Subject: Re: The hanson family

From: Carl Lumma


>> >his 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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