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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.

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

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

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!

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

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

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?

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

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.

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

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

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.

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

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.

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.

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.]

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.

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

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.

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

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

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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