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Message: 7150

Date: Wed, 30 Jul 2003 23:47:36

Subject: Re: Creating a Temperment /Comma

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:
> Carl Lumma wrote:
> 
> > Only one?  Doesn't the number missing depend on the space?
> 
> A linear temperament always needs one unison vector less than a 
> periodicity block.  An equal temperament needs the same number, 
hence 
> Fokker's 31 note periodicity blocks. 

hence? they have nothing to do with equal temperament, since he 
didn't temper any of the unison vectors out.


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Message: 7151

Date: Wed, 30 Jul 2003 18:47:04

Subject: Re: Creating a Temperment /Comma

From: Carl Lumma

>> Whoa.  Is there a friendly guide to temperaments somewhere?
>
>I should get busy and put more stuff up on xenharmony, but I don't 
>guarantee friendly.

The thing is, everybody is approaching this from different angles.
There's the PB paradigm, the generators/MOS paradigm, the wedgie/val
paradigm.  However well I understand them individual, I understand
how they fit together far less well.

-Carl


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Message: 7152

Date: Wed, 30 Jul 2003 18:53:13

Subject: Re: Creating a Temperment /Comma

From: Carl Lumma

>>>A linear temperament always needs one unison vector less than a 
>>>periodicity block.  An equal temperament needs the same number,
>>>a planar temperament needs two less, and so on.
>> 
>> So this formulation does depend on the space, since the number
>> needed for a block depends on the space!
>
>but the *difference* is always 1, regardless of the space. it's 
>called the "codimension".

Ok, but what's happening to this comma?  It's the one that doesn't
vanish?  Why is it any less "defining" than the others?

I thought it's what *does* vanish that defines things.  Wait --
is it that in a PB, everything 'vanishes', even though it doesn't,
and Graham's ambiguous grammar (above) means to say that ets and
pbs have the same number?  Then it makes sense.

-Carl


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Message: 7154

Date: Thu, 31 Jul 2003 19:25:24

Subject: Re: Creating a Temperment /Comma

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
> >> >>>A linear temperament always needs one unison vector less than 
a 
> >> >>>periodicity block.  An equal temperament needs the same 
number,
> >> >>>a planar temperament needs two less, and so on.
> >> >> 
> >> >> So this formulation does depend on the space, since the number
> >> >> needed for a block depends on the space!
> >> >
> >> >but the *difference* is always 1, regardless of the space. it's 
> >> >called the "codimension".
> >> 
> >> Ok, but what's happening to this comma?
> >
> >what's happening to *which* comma?
> 
> The one that the pb needs but the lt doesn't.

well, that comma's presence in defining the system makes the set of 
notes finite. without it, the set of notes is infinite, since linear 
temperaments contain an infinite number of notes per octave. does 
that answer your question?

> >> It's the one that doesn't vanish?
> >> Why is it any less "defining" than the others?
> >
> >let's get straight which comma we're talking about. how about an 
> >example.
> 
> Howabout a 7-limit block defined by 81:80, 25:24, and 36:35?

ok, so if you temper out two of these, and ignore the third, you get 
a linear temperament with an infinite number of notes. is there one 
of the three you wanted to use for this purpose?

> >> I thought it's what *does* vanish that defines things.
> >
> >no, for example a 3-d ji pb is defined by 3 independent unison 
> >vectors, none of which vanish.
> 
> But meantone is defined by 81:80, which does vanish.

yes, but we could just as easily define a "slice" of ji using 81:80 
in non-vanishing form -- leading to constructs like monz's and 
terpstra's "just interpretations" of meantone (which are of course 
deficient since they have 27:20 "wolves" and the like). "defining" 
means "delimiting" when it comes to unison vectors -- they section 
off a region of the space, but if they vanish, one is effectively 
free to move from this region to any another, since one will simply 
be using the same set of pitches anyway!


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Message: 7155

Date: Thu, 31 Jul 2003 20:42:43

Subject: Re: Creating a Temperment /Comma

From: Graham Breed

Carl Lumma wrote:

> I was confused by Graham calling it a library too.  It's *his*
> contribution, available from his web site.

Yes, it's a third party library, where I'm the third party.  What if I 
called it a "module" instead?  I still call it a script on the website, 
which is wrong because it isn't runnable, although it used to be.

> BTW, Python 2.3 final was released 2 days ago.  But don't try to
> put a whitespace in the install path on Windows!

Whitespace in Windows paths is always trouble!!


                            Graham


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Message: 7156

Date: Thu, 31 Jul 2003 13:31:20

Subject: Re: Creating a Temperment /Comma

From: Carl Lumma

Hey Paul E., you back already?  How was/is Washington?

I'm hoping you'll get a chance to look at the harmonic
entropy thread.

As far as this temperament stuff goes, I'm not sure what
I didn't understand, if anything.  I'll just voice again
a lack of introductory materials that explain the basic
tools needed to do scale building with temperaments.

If a linear temperament has an infinite number of notes,
what happens when we cast it into a scale?  If a 5-limit
lt has chromatic uv 25:24 and commatic uv 81:80, do we
have a name for the "wolf" comma formed between the ends
of the chain?

-Carl


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Message: 7157

Date: Thu, 31 Jul 2003 20:40:46

Subject: Re: Creating a Temperment /Comma

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
> Hey Paul E., you back already?

yup!

> How was/is Washington?

hot. nice.

> I'm hoping you'll get a chance to look at the harmonic
> entropy thread.

i've already replied to everything, so i'm not sure what you're 
referring to.

> As far as this temperament stuff goes, I'm not sure what
> I didn't understand, if anything.  I'll just voice again
> a lack of introductory materials that explain the basic
> tools needed to do scale building with temperaments.

_the forms of tonality_?

> If a linear temperament has an infinite number of notes,
> what happens when we cast it into a scale?  If a 5-limit
> lt has chromatic uv 25:24 and commatic uv 81:80, do we
> have a name for the "wolf" comma formed between the ends
> of the chain?
> 
> -Carl

the chromatic uv. that is, if you put the last note on one end of the 
chain, it will differ by the chromatic uv compared to where it would 
be if you put it on the other end instead. is this what you had in 
mind?


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Message: 7158

Date: Thu, 31 Jul 2003 14:15:25

Subject: Re: Creating a Temperment /Comma

From: Carl Lumma

>> How was/is Washington?
>
>hot. nice.

'dyou get a recording of the gig?

>> I'm hoping you'll get a chance to look at the harmonic
>> entropy thread.
>
>i've already replied to everything, so i'm not sure what
>you're referring to.

Mm, maybe nothing then.

>> As far as this temperament stuff goes, I'm not sure what
>> I didn't understand, if anything.  I'll just voice again
>> a lack of introductory materials that explain the basic
>> tools needed to do scale building with temperaments.
>
>_the forms of tonality_?

Is fantastic.  But I don't think it covers many of the
topics on this list.  It isn't online, and doesn't have any
automation.

>>A linear temperament always needs one unison vector less
>>than periodicity block.  An equal temperament needs the same
>>number, a planar temperament needs two less, and so on.
//
>does that answer your question?

First off, how many does an et need:

(a) Same as pb.
(b) Same as lt.

??????????????

I think (a), which makes my blurb from a few posts back
correct.

>> thought it's what *does* vanish that defines things.
>
>no, for example a 3-d ji pb is defined by 3 independent unison
>vectors, none of which vanish.

But a pb isn't a temperament.  Temperaments are defined by
the commas that vanish, no?

>> If a linear temperament has an infinite number of notes,
>> what happens when we cast it into a scale?  If a 5-limit
>> lt has chromatic uv 25:24 and commatic uv 81:80, do we
>> have a name for the "wolf" comma formed between the ends
>> of the chain?
>
>the chromatic uv. that is, if you put the last note on one
>end of the chain, it will differ by the chromatic uv compared
>to where it would be if you put it on the other end instead.
>is this what you had in mind?

I was speaking of the (generator + chromatic uv) interval,
which I will call the wolf.

1. Does the wolf have any significance apart from the
chromatic uv?  Gene's T[n] stuff seems to show it does.

2. How is the difference between the pitches of a new note
at either end of the chain the same regardless of the length
of the chain?  If I chain three 700-cent intervals, this
"chromatic uv" as you define it above is 100 cents.  If
chain two of them it is 300 cents!

-Carl


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Message: 7159

Date: Thu, 31 Jul 2003 21:32:08

Subject: Re: Creating a Temperment /Comma

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
> >> How was/is Washington?
> >
> >hot. nice.
> 
> 'dyou get a recording of the gig?

yup, dave recorded it on minidisc . . . we did the tunes on our demo 
plus 4 more, which we should be laying down in the studio soon . . .

> 
> >>A linear temperament always needs one unison vector less
> >>than periodicity block.  An equal temperament needs the same
> >>number, a planar temperament needs two less, and so on.
> //
> >does that answer your question?
> 
> First off, how many does an et need:
> 
> (a) Same as pb.

yes, that's what graham wrote in what you quoted above.

> (b) Same as lt.

no, a lt needs one less, as graham and i told you.

> >> thought it's what *does* vanish that defines things.
> >
> >no, for example a 3-d ji pb is defined by 3 independent unison
> >vectors, none of which vanish.
> 
> But a pb isn't a temperament.  Temperaments are defined by
> the commas that vanish, no?

yes.

> >> If a linear temperament has an infinite number of notes,
> >> what happens when we cast it into a scale?  If a 5-limit
> >> lt has chromatic uv 25:24 and commatic uv 81:80, do we
> >> have a name for the "wolf" comma formed between the ends
> >> of the chain?
> >
> >the chromatic uv. that is, if you put the last note on one
> >end of the chain, it will differ by the chromatic uv compared
> >to where it would be if you put it on the other end instead.
> >is this what you had in mind?
> 
> I was speaking of the (generator + chromatic uv) interval,
> which I will call the wolf.

ok!

> 1. Does the wolf have any significance apart from the
> chromatic uv?  Gene's T[n] stuff seems to show it does.

how so?

> 2. How is the difference between the pitches of a new note
> at either end of the chain the same regardless of the length
> of the chain?

of course it isn't! if you'd chosen a different chromatic unison 
vector, you'd most likely end up with a different number of notes per 
octave (the fokker determinant tells you that), and thus a different 
length for each chain.

> If I chain three 700-cent intervals, this
> "chromatic uv" as you define it above is 100 cents.

how do you get that? that seems way too small. and since the scale is 
not even an MOS (or what we used to call MOS), you're not logically 
justified to call it a chromatic uv at all. i wasn't defining it 
above, i was explaining what happens when you use one. and, now that 
i look, you made an incorrect statement in your question, which i 
should have corrected:

"If a 5-limit lt has chromatic uv 25:24 and commatic uv 81:80,"

that's not an lt at all -- it's an MOS scale (in fact, it's the 
meantone diatonic scale).

> If
> chain two of them it is 300 cents!

yes, that looks right, you have C-G-D vs. G-D-A, and A-C is 300 
cents. 32:27 is what you might call this unison vector. this 3-tone 
scale is nice and useful, but not entirely common.


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Message: 7160

Date: Thu, 31 Jul 2003 22:37:44

Subject: Re: Creating a Temperment /Comma

From: Graham Breed

Gene Ward Smith wrote:

> Not true any more, and I think I posted this. Here is the Maple code 
> for 7-limit geometric complexity:

Sorry, yes, you have shown the derivation of real numbers.  But still no 
general algorithm or full explanation of the indexing rules.

> It's unfortunately true that these results are not immediate from the 
> definition of geometric complexity, but must themselves be computed, 
> so only for 5-limit is this straightforward.

Yes, I chased this down, and found a definition here:

Yahoo groups: /tuning-math/message/5546 * [with cont.] 

Given that it is so straightforward, why do you exclude it from your 
other listings?

So, I've got as far as converting that into Python:

import math, temper

def complexity5(u, primes=None):
   primes = primes or temper.primes[:u.maxBasis()]
   return math.log(2) * math.sqrt(
       primes[0]**2 * u[1,]**2 +
       primes[0]**2 * u[1,] * u[2,] +
       primes[1]**2 * u[2,]**2)


>>If I had worked out your numbering rule, I've forgotten it now.
>>
>>Every time you try to explain something, you bring in more jargon 
> 
> terms 
> 
>>that I don't understand (I can't speak for anybody else).
>>
>>The word "metric" in particular is something that's important but 
> 
> you 
> 
>>haven't defined.
> 
> 
> I've posted this before; it's standard math:
> 
> Metric -- from MathWorld * [with cont.] 

*That's* standard math, yes, but it doesn't say anything about applying 
a metric to an exterior algebra.  And in the geometric complexity definiton:

Yahoo groups: /tuning-math/message/4533 * [with cont.] 

you give what is, as far as I can work out, a function of a single 
rational number as the "metric".  Yet the definition you pointed to says 
a metric is a function of two variables!  It looks more like a norm than 
a metric to me, but I still don't know how to apply a norm to an 
exterior algebra.


>>You keep missing out important steps in explanations, like the need 
> 
> to 
> 
>>take the complement at certain points when using wedge products.
> 
> 
> As I've explained before, the way I wrote my Maple code I don't need 
> to. All that is built into the functions themselves.

Do we have to go through this again?  I thought I finally pinned you 
down before.  Here:

Yahoo groups: /tuning-math/message/5673 * [with cont.] 

you say "...I store the wedgies as lists, and reverse the ordering when 
I compute from commas, etc...".  Doesn't that mean that reversing the 
ordering is your implementation of the complement?

Besides which, whatever your Maple code does, your explanations only 
defined a wedge product.  Not a complement operation or a dual space or 
however you claim to get around using a complement.  So a step is missing.

Except for here:

Yahoo groups: /tuning-math/message/5583 * [with cont.] 

where you give different formulae for the wedge products of intervals 
and vals.  And say "These two types of wedge product can be
indentified with each other, by `Poincare duality'. It does not matter
whether the wedgie comes from commas or vals, therefore."  Not only do 
you use an undefined jargon term that you can't expect us to know, but 
you use it to dismiss something that clearly does matter, because you 
(or I, at least) don't get the right results without it.

Or maybe you meant that the geometric complexity of a wedgie is equal to 
that of its complement?

And yes, Poincaré Duality is in Mathworld, but it's completely impenetrable.


                           Graham


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Message: 7161

Date: Thu, 31 Jul 2003 14:46:32

Subject: Re: Creating a Temperment /Comma

From: Carl Lumma

>yup, dave recorded it on minidisc . . . we did the tunes on
>our demo plus 4 more, which we should be laying down in the
>studio soon . . .

Suh-weet.

>> >>A linear temperament always needs one unison vector less
>> >>than periodicity block.  An equal temperament needs the same
>> >>number, a planar temperament needs two less, and so on.
>> //
>> >does that answer your question?
>> 
>> First off, how many does an et need:
>> 
>> (a) Same as pb.
>
>yes, that's what graham wrote in what you quoted above.

His gramar was ambiguous, as I said before!

>> (b) Same as lt.
>
>no, a lt needs one less, as graham and i told you.

You never mentioned ets that I saw.

>> >> thought it's what *does* vanish that defines things.
>> >
>> >no, for example a 3-d ji pb is defined by 3 independent unison
>> >vectors, none of which vanish.
>> 
>> But a pb isn't a temperament.  Temperaments are defined by
>> the commas that vanish, no?
>
>yes.

Thank god.

>> >> If a linear temperament has an infinite number of notes,
>> >> what happens when we cast it into a scale?  If a 5-limit
>> >> lt has chromatic uv 25:24 and commatic uv 81:80, do we
>> >> have a name for the "wolf" comma formed between the ends
>> >> of the chain?
>> >
>> >the chromatic uv. that is, if you put the last note on one
>> >end of the chain, it will differ by the chromatic uv compared
>> >to where it would be if you put it on the other end instead.
>> >is this what you had in mind?
>> 
>> I was speaking of the (generator + chromatic uv) interval,
>> which I will call the wolf.
>
>ok!
>
>> 1. Does the wolf have any significance apart from the
>> chromatic uv?  Gene's T[n] stuff seems to show it does.
>
>how so?

Did you catch that thread.  Gene's probably the better one to
explain it.  Or maybe search for "T[n]".

He basically looked at not-necessarily-MOS n for popular
temperaments T, and found that the wolf was sometimes itself
consonant, which added to the utility of the tuning.  Or
something.  I've been struggling to understand how this fits
into the temperament terminology.  For 5-limit lts, does it
represent a 3rd comma?  Does the use of it to form consonant
intervals break regularity/consistency?  Etc.

>> 2. How is the difference between the pitches of a new note
>> at either end of the chain the same regardless of the length
>> of the chain?
>
>of course it isn't! if you'd chosen a different chromatic unison 
>vector, you'd most likely end up with a different number of
>notes per octave (the fokker determinant tells you that), and
>thus a different length for each chain.

So the chromatic uv tells you the length!  So maybe the T[n]
stuff was just about the chromatic uv!

>> If I chain three 700-cent intervals, this
>> "chromatic uv" as you define it above is 100 cents.
>
>how do you get that? that seems way too small.

(F)-C-G-D-A-(E); F-E=100cents

>since the scale is not even an MOS (or what we used to call MOS),
>you're not logically justified to call it a chromatic uv at all.

Ok, this is a bombshell.  How can we flesh this out a bit?

>i wasn't defining it above, i was explaining what happens when
>you use one. and, now that i look, you made an incorrect statement
>in your question, which i should have corrected:
>
>"If a 5-limit lt has chromatic uv 25:24 and commatic uv 81:80,"
>
>that's not an lt at all -- it's an MOS scale (in fact, it's the 
>meantone diatonic scale).

Is this because lts don't have chromatic uvs?

>> If chain two of them it is 300 cents!
>
>yes, that looks right, you have C-G-D vs. G-D-A, and A-C is 300 
>cents. 32:27 is what you might call this unison vector. this
>3-tone scale is nice and useful, but not entirely common.

It is MOS, IIRC.

-Carl


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Message: 7162

Date: Thu, 31 Jul 2003 22:43:05

Subject: Re: Creating a Temperment /Comma

From: Paul Erlich

> >> 1. Does the wolf have any significance apart from the
> >> chromatic uv?  Gene's T[n] stuff seems to show it does.
> >
> >how so?
> 
> Did you catch that thread.

yes.

> found that the wolf was sometimes itself
> consonant, which added to the utility of the tuning.

ok.

> Or
> something.  I've been struggling to understand how this fits
> into the temperament terminology.  For 5-limit lts, does it
> represent a 3rd comma?

3rd? 5-limit lts are defined by 1 comma.

> Does the use of it to form consonant
> intervals break regularity/consistency?  Etc.

consistency is only defined for ets. it doesn't break regularity, 
though it may break the universality of the mapping, if the wolf is 
another form of a consonant interval already mapped, and you want to 
use it as a consonance.

> >> 2. How is the difference between the pitches of a new note
> >> at either end of the chain the same regardless of the length
> >> of the chain?
> >
> >of course it isn't! if you'd chosen a different chromatic unison 
> >vector, you'd most likely end up with a different number of
> >notes per octave (the fokker determinant tells you that), and
> >thus a different length for each chain.
> 
> So the chromatic uv tells you the length!  So maybe the T[n]
> stuff was just about the chromatic uv!

yes (in many cases it is), apparently you're the one who missed a lot 
of that thread. in fact, gene didn't even understand what a chromatic 
unison vector was until he posted his T[n] idea and i explained the 
relationship to him.

> 
> >> If I chain three 700-cent intervals, this
> >> "chromatic uv" as you define it above is 100 cents.
> >
> >how do you get that? that seems way too small.
> 
> (F)-C-G-D-A-(E); F-E=100cents

that's not three 700-cent intervals; it's a pentatonic scale, so four 
700-cent intervals. this is just the gentle introduction to fokker 
periodicity blocks, part 1, revisited.

> 
> >since the scale is not even an MOS (or what we used to call MOS),
> >you're not logically justified to call it a chromatic uv at all.
> 
> Ok, this is a bombshell.  How can we flesh this out a bit?

well, your scale above *is* an MOS, so we're fine. the point is that 
the chromatic uv would have to be capable of producing a ji variant 
of the scale when it, along with any commatic uvs defining the 
temperament, are used to define a fokker periodicity block.

> >i wasn't defining it above, i was explaining what happens when
> >you use one. and, now that i look, you made an incorrect statement
> >in your question, which i should have corrected:
> >
> >"If a 5-limit lt has chromatic uv 25:24 and commatic uv 81:80,"
> >
> >that's not an lt at all -- it's an MOS scale (in fact, it's the 
> >meantone diatonic scale).
> 
> Is this because lts don't have chromatic uvs?

not until you start choosing finite scales from the lt. before that, 
you've got an infinite number of *potential* chromatic unison 
vectors, but none that are operative as yet.

> >> If chain two of them it is 300 cents!
> >
> >yes, that looks right, you have C-G-D vs. G-D-A, and A-C is 300 
> >cents. 32:27 is what you might call this unison vector. this
> >3-tone scale is nice and useful, but not entirely common.
> 
> It is MOS, IIRC.

yes, and as always, for any generic interval that comes in two 
different sizes, the difference between the sizes is the chromatic 
uv.


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Message: 7163

Date: Thu, 31 Jul 2003 15:55:01

Subject: Re: Creating a Temperment /Comma

From: Carl Lumma

>> Or something.  I've been struggling to understand how this fits
>> into the temperament terminology.  For 5-limit lts, does it
>> represent a 3rd comma?
>
>3rd? 5-limit lts are defined by 1 comma.

Eep!  2nd, that is!  

>> Does the use of it to form consonant
>> intervals break regularity/consistency?  Etc.
>
>consistency is only defined for ets.

What Gene said about regular temperaments sounded
equivalent to me.

>it may break the universality of the mapping, if the wolf is 
>another form of a consonant interval already mapped, and you
>want to use it as a consonance.

Yeah, that's what I'm thinkin'.  Let's take kleismic[8] as an
example.  It's non-MOS but a good scale.  It's proper, too.

>> So the chromatic uv tells you the length!  So maybe the T[n]
>> stuff was just about the chromatic uv!
>
>yes (in many cases it is), apparently you're the one who missed
>a lot of that thread. in fact, gene didn't even understand what
>a chromatic unison vector was until he posted his T[n] idea and
>i explained the relationship to him.

In fact I think this is coming back.

>>>since the scale is not even an MOS (or what we used to call MOS),
>>>you're not logically justified to call it a chromatic uv at all.
>> 
>> Ok, this is a bombshell.  How can we flesh this out a bit?
>
>well, your scale above *is* an MOS, so we're fine. the point is that 
>the chromatic uv would have to be capable of producing a ji variant 
>of the scale when it, along with any commatic uvs defining the 
>temperament, are used to define a fokker periodicity block.

And what happens in the case of kleismic[8]?

>> >i wasn't defining it above, i was explaining what happens when
>> >you use one. and, now that i look, you made an incorrect statement
>> >in your question, which i should have corrected:
>> >
>> >"If a 5-limit lt has chromatic uv 25:24 and commatic uv 81:80,"
>> >
>> >that's not an lt at all -- it's an MOS scale (in fact, it's the 
>> >meantone diatonic scale).
>> 
>> Is this because lts don't have chromatic uvs?
>
>not until you start choosing finite scales from the lt. before that, 
>you've got an infinite number of *potential* chromatic unison 
>vectors, but none that are operative as yet.

Got it.  Thanks!

-Carl


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Message: 7164

Date: Thu, 31 Jul 2003 23:17:22

Subject: Re: Creating a Temperment /Comma

From: Paul Erlich

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

> >consistency is only defined for ets.
> 
> What Gene said about regular temperaments sounded
> equivalent to me.

a definition of consistency for regular temperaments? doesn't seem 
possible, as they're infinite, so you can always find better and 
better approximations to anything.

> >it may break the universality of the mapping, if the wolf is 
> >another form of a consonant interval already mapped, and you
> >want to use it as a consonance.
> 
> Yeah, that's what I'm thinkin'.  Let's take kleismic[8] as an
> example.  It's non-MOS but a good scale.  It's proper, too.

ok, and what does the wolf do here? i'm in a rush so can't figure it 
out right now . . .

> And what happens in the case of kleismic[8]?

i'll have to answer this from the office tomorrow.


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Message: 7165

Date: Thu, 31 Jul 2003 01:27:07

Subject: Re: Creating a Temperment /Comma

From: Gene Ward Smith

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

> Whoa.  Is there a friendly guide to temperaments somewhere?

I should get busy and put more stuff up on xenharmony, but I don't 
guarantee friendly.


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Message: 7166

Date: Thu, 31 Jul 2003 01:31:05

Subject: Re: Creating a Temperment /Comma

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:

> but the *difference* is always 1, regardless of the space. it's 
> called the "codimension".

Codimension -- from MathWorld * [with cont.] 


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Message: 7167

Date: Thu, 31 Jul 2003 04:05:09

Subject: Re: Creating a Temperment /Comma

From: Paul Erlich

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

> >>>A linear temperament always needs one unison vector less than a 
> >>>periodicity block.  An equal temperament needs the same number,
> >>>a planar temperament needs two less, and so on.
> >> 
> >> So this formulation does depend on the space, since the number
> >> needed for a block depends on the space!
> >
> >but the *difference* is always 1, regardless of the space. it's 
> >called the "codimension".
> 
> Ok, but what's happening to this comma?

what's happening to *which* comma?

> It's the one that doesn't
> vanish?
> Why is it any less "defining" than the others?

let's get straight which comma we're talking about. how about an 
example.

> I thought it's what *does* vanish that defines things.

no, for example a 3-d ji pb is defined by 3 independent unison 
vectors, none of which vanish.


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Message: 7168

Date: Thu, 31 Jul 2003 01:11:24

Subject: Re: Creating a Temperment /Comma

From: Carl Lumma

>> >>>A linear temperament always needs one unison vector less than a 
>> >>>periodicity block.  An equal temperament needs the same number,
>> >>>a planar temperament needs two less, and so on.
>> >> 
>> >> So this formulation does depend on the space, since the number
>> >> needed for a block depends on the space!
>> >
>> >but the *difference* is always 1, regardless of the space. it's 
>> >called the "codimension".
>> 
>> Ok, but what's happening to this comma?
>
>what's happening to *which* comma?

The one that the pb needs but the lt doesn't.

>> It's the one that doesn't vanish?
>> Why is it any less "defining" than the others?
>
>let's get straight which comma we're talking about. how about an 
>example.

Howabout a 7-limit block defined by 81:80, 25:24, and 36:35?

>> I thought it's what *does* vanish that defines things.
>
>no, for example a 3-d ji pb is defined by 3 independent unison 
>vectors, none of which vanish.

But meantone is defined by 81:80, which does vanish.

-Carl


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Message: 7169

Date: Fri, 01 Aug 2003 16:45:18

Subject: Re: Calculating Commas from a Wedgie

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "paulhjelmstad" 
<paul.hjelmstad@u...> wrote:
writes}

> If the wedgie is [u1,u2,u3,u4,u5,u6] then we have commas given by
> 2^u6 3^(-u2) 5^u1
> 2^u5 3^u3 7^(-u1)
> 2^u4 5^(-u3) 7^u2
> 3^u4 5^u5 7^u6 
> 
> I see that the wedgie was calculated from the two commas or ets, 
but 
> here (at the end of Gene's message) we also have 4 commas derived 
> from the wedgie, how does this work? How are the particular commas 
> above derived from that particular wedgie. Thanks
> 
> Paul

Let's look at some of my Maple code

print(a7int);

This gives the product of two intervals (written as monzos) "down" to 
a 7-limit wedgie

proc(r, s)
    [r[3]*s[4] - s[3]*r[4], r[4]*s[2] - r[2]*s[4], -r[3]*s[2] + r[2]*s
[3],
    r[1]*s[4] - s[1]*r[4], r[3]*s[1] - r[1]*s[3], -r[2]*s[1] + r[1]*s
[2]]
end proc

print(a7val);

This gives the product of two vals up to a 7-limit wedgie

proc(r, s)
    [-r[2]*s[1] + r[1]*s[2], r[1]*s[3] - r[3]*s[1], r[1]*s[4] - s[1]*r
[4],
    -r[3]*s[2] + r[2]*s[3], r[2]*s[4] - r[4]*s[2], r[3]*s[4] - s[3]*r
[4]]
end proc

print(a7down);

This gives the product of a wedgie and an interval "down" to a val

proc(x, w)
    2^(x[4]*w[4] - x[5]*w[3] + x[6]*w[2])*
    3^(-x[2]*w[4] + x[3]*w[3] - x[6]*w[1])*
    5^(x[1]*w[4] - x[3]*w[2] + x[5]*w[1])*
    7^(-x[1]*w[3] + x[2]*w[2] - w[1]*x[4])
end proc

print(a7up);

This gives the product of a wedgie and a val "up" to an interval

proc(l, w)
    [l[3]*w[4] + l[2]*w[3] + l[1]*w[2], l[5]*w[4] + w[3]*l[4] - w[1]*l
[1],
    w[4]*l[6] - l[4]*w[2] - l[2]*w[1], -w[3]*l[6] - l[5]*w[2] - l[3]*w
[1]]
end proc

We want to find commas belonging to a wedgie whose factors contain 
only three out of the possible four primes. However, we can't use any 
6-vector of integers as a wedgie, since a wedgie is a "blade", or 
wedge product. As such it satisfies an algebraic condition 
("Pfaffian" is zero) which, rather than dealing with, I will get 
around by taking the product of two intervals, and then taking a 
further product and solving for the condition that the resulting val 
is zero.

Wedge product up of two monzos:
u := a7int([x1,x2,x3,x4], [y1,y2,y3,y4]);

u := [x3 y4 - y3 x4, x4 y2 - x2 y4, -x3 y2 + x2 y3, x1 y4 - y1 x4,

    x3 y1 - x1 y3, -x2 y1 + x1 y2]

Wedge product up of this with a monzo having its "7" coefficient zero:
v := a7down(u, [a,b,c,0]);

v := [(x4 y2 - x2 y4) c + (x3 y4 - y3 x4) b,

    c (x1 y4 - y1 x4) - a (x3 y4 - y3 x4),

    -(x1 y4 - y1 x4) b - (x4 y2 - x2 y4) a,

    -c (-x2 y1 + x1 y2) - (x3 y1 - x1 y3) b - (-x3 y2 + x2 y3) a]

Now solve for a, b, and c:
solve(convert(v,set), {a,b,c});


                 c (x1 y4 - y1 x4)      (-x4 y2 + x2 y4) c
            {a = -----------------, b = ------------------, c = c}
                   x3 y4 - y3 x4          x3 y4 - y3 x4

By comparison with "u" above, we see we can clear denominators and 
set a = x1 y4 - y1 x4 = u[4], b = x2 y4 - y2 x4 = -u[2], c = x3y4 - 
y3 x4 = u[1]. Hence 2^u[4] 3^(-u[2]) 5^u[1] is a 7-free comma 
belonging to the wedgie u. The same calculation works for 2, 3, and 5-
free commas, and for other types of wedgies.


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Message: 7170

Date: Fri, 01 Aug 2003 19:51:51

Subject: Re: Calculating Commas from a Wedgie

From: Graham Breed

paulhjelmstad wrote:

> Let r be the mapping to primes of an equal temperament given
> by r = [u1, u2, u3, u4], and s be given by [v1, v2, v3, v4]. This
> means r has u1 notes to the octave, u2 notes in the approximation of 
> 3, and so forth; hence [12, 19, 28, 24] would be the usual 12-equal, 
> and [31, 49, 72, 87] the usual 31-et. The wedge now is

That's the second time I've seen you post the wrong val for 12-equal. 
It should be [12, 19, 28, 34].  You'll get a strange pair of unison 
vectors from the vals you posted!


                           Graham


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Message: 7171

Date: Fri, 01 Aug 2003 12:21:37

Subject: master system

From: Carl Lumma

I know this topic comes up from time to time, but what's
everyone's favorite et for a master system encompassing
all of JI?

12, 31, 41, 53, 72, 94, 171, 311, 612?

-Carl


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Message: 7173

Date: Fri, 01 Aug 2003 13:00:08

Subject: Re: Creating a Temperment /Comma

From: Carl Lumma

>>>consistency is only defined for ets.
>> 
>>What Gene said about regular temperaments sounded
>>equivalent to me.
>
>a definition of consistency for regular temperaments?
>doesn't seem possible, as they're infinite, so you can
>always find better and better approximations to anything.

Yahoo groups: /tuning-math/message/3330 * [with cont.] 

Somewhere Gene gave the definition for regular
temperament, but it doesn't seem to have made it into
monz's dictionary.

-Carl


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