Tuning-Math Digests messages 10451 - 10475

This is an Opt In Archive . We would like to hear from you if you want your posts included. For the contact address see About this archive. All posts are copyright (c).

Contents Hide Contents S 11

Previous Next

10000 10050 10100 10150 10200 10250 10300 10350 10400 10450 10500 10550 10600 10650 10700 10750 10800 10850 10900 10950

10450 - 10475 -



top of page bottom of page down


Message: 10451

Date: Mon, 01 Mar 2004 23:31:44

Subject: Re: 9-limit stepwise

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...>
> wrote:
> 
> > > 116 [49/45, 10/9, 9/8, 8/7] [1, 1, 2, 2] 1.568046 6
> > Pajara[6]
> 
> It's six notes of pajara, but not Pajara[6].

Oh!

> The 49/45 and 10/9 are
> seven fifths up and four octaves down--the apotome, 49/45 and 10/9 
all
> being the same in pajara. The 9/8 and 8/7 of course are the same, 
both
> two fifths up and an octave down. So in terms of generators, it 
looks
> like [2,2,7,2,2,7], which adds up to 22 fifths. In 22-equal, it 
would
> be 443443,

Ah. Well you can see on Manuel's list:

List of musical modes *

that I already gave its second mode a bland, technical name. Dave 
Keenan shows it as a chord on his page:

http://www.uq.net.au/~zzdkeena/Music/ErlichDecChords.gif - Ok *

and calls it "superaugmented subminor 7th 9th augmented 11th", which 
is even worse.


top of page bottom of page up down


Message: 10452

Date: Mon, 01 Mar 2004 23:33:24

Subject: Re: Hanzos

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
> >> One of us is still misunderstanding Paul Hahn's 9-limit approach.
> >> In the unweighted version 3, 5, 7 and 9 are all the same length.
> >
> >In this system you don't exactly, have 7-limit notes and intervals.
> >You do have "hanzos", with basis 2,3,5,7,9.
> 
> The usual point of odd-limit is to get octave equivalence, and
> therefore I'd say the 2s should be dropped from the basis.

Bad move -- for one thing, you can't detect torsion.

> >The hahnzo |0 -2 0 0 1> is a comma, 9/3^2, which obviously would
> >play a special role. Hahnzos map onto 7-limit intervals, but not
> >1-1. Are you happy with the idea that two scales could be
> >different, since they have steps and notes which are distinct as
> >hahnzos, even though they have exactly the same steps and notes
> >in the 7-limit?
> 
> I think the answer here is yes, though I'm at a loss for why
> you're mapping hanzos to the 7-limit.

7-prime-limit.

> >We've got three hahnzos corresponding to 81/80;
> 
> My recollection is that Paul H.'s algorithm assigns a unique
> lattice route (and therefore hanzo) to each 9-limit interval.

Unique "hanzo" but not unique lattice route -- even monzos don't 
assign a unique lattice route.


top of page bottom of page up down


Message: 10453

Date: Mon, 01 Mar 2004 23:36:18

Subject: Re: Hanzos

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
> >> My recollection is that Paul H.'s algorithm assigns a unique
> >> lattice route (and therefore hanzo) to each 9-limit interval.
> >
> >So what? You still get an infinite number representing each 
interval,
> >since you can multiply by arbitary powers of the dummy comma 9/3^2.
> 
> An infinite number from where?  If you look at the algorithm, that
> dummy comma has zero length.

How do you get that????



________________________________________________________________________
________________________________________________________________________



------------------------------------------------------------------------
Yahoo! Groups Links

<*> To visit your group on the web, go to:
     Yahoo groups: /tuning-math/ *

<*> To unsubscribe from this group, send an email to:
     tuning-math-unsubscribe@xxxxxxxxxxx.xxx

<*> Your use of Yahoo! Groups is subject to:
     Yahoo! Terms of Service *


top of page bottom of page up down


Message: 10461

Date: Tue, 02 Mar 2004 00:32:22

Subject: Re: 9-limit stepwise

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> > --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" 
<gwsmith@s...> 
> > wrote:
> > > --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" 
<gwsmith@s...>
> > > wrote:
> > > 
> > > > > 116 [49/45, 10/9, 9/8, 8/7] [1, 1, 2, 2] 1.568046 6
> > > > Pajara[6]
> > > 
> > > It's six notes of pajara, but not Pajara[6].
> > 
> > Oh!
> > 
> > > The 49/45 and 10/9 are
> > > seven fifths up and four octaves down--the apotome, 49/45 and 
10/9 
> > all
> > > being the same in pajara. The 9/8 and 8/7 of course are the 
same, 
> > both
> > > two fifths up and an octave down. So in terms of generators, it 
> > looks
> > > like [2,2,7,2,2,7], which adds up to 22 fifths. In 22-equal, it 
> > would
> > > be 443443,
> > 
> > Ah. Well you can see on Manuel's list:
> > 
> > List of musical modes *
> > 
> > that I already gave its second mode a bland, technical name. Dave 
> > Keenan shows it as a chord on his page:
> > 
> > http://www.uq.net.au/~zzdkeena/Music/ErlichDecChords.gif - Ok *
> > 
> > and calls it "superaugmented subminor 7th 9th augmented 11th", 
which 
> > is even worse.
> 
> Yow. I'll take 9-limit consonant whole-tone, thank you. Of course 
the
> other whole tone scale counts also.

It doesn't, because 4+4+4=12 is not a 9-limit consonance in 22-equal.


top of page bottom of page up down


Message: 10462

Date: Tue, 02 Mar 2004 20:38:24

Subject: Re: Hanzos

From: Graham Breed

Carl:
>>The usual point of odd-limit is to get octave equivalence, and
>>therefore I'd say the 2s should be dropped from the basis.

Paul E:
> Bad move -- for one thing, you can't detect torsion.

Can't you?  I'm not sure it's even relevant here, but I thought I went 
through this in excruciating detail some time ago and showed that you 
can detect torsion with octave-equivalent vectors.  In fact, I seem to 
remember running the calculation in parallel using both methods, and 
showing that I always got the same results.

No, the outstanding problem is that we can't go from the 
octave-equivalent mapping to an optimized generator size.  But I'm sure 
it can be done if anybody cared.  The situation as I left it was that 
nobody did.


                  Graham


top of page bottom of page up down


Message: 10463

Date: Tue, 02 Mar 2004 20:58:25

Subject: Re: Hanzos

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
> >> >> My recollection is that Paul H.'s algorithm assigns a unique
> >> >> lattice route (and therefore hanzo) to each 9-limit interval.
> >> >
> >> >So what? You still get an infinite number representing each
> >> >interval, since you can multiply by arbitary powers of the dummy
> >> >comma 9/3^2.
> >> 
> >> An infinite number from where?  If you look at the algorithm, 
that
> >> dummy comma has zero length.
> >
> >How do you get that????
> 
> >Given a Fokker-style interval vector (I1, I2, . . . In):
> 
> [-2 0 0 1]
> 
> >1.  Go to the rightmost nonzero exponent; add its absolute value
> >to the total.
> 
> T=1
> 
> >2.  Use that exponent to cancel out as many exponents of the 
opposite
> >sign as possible, starting to its immediate left and working right;
> >discard anything remaining of that exponent.
> 
> [-1 0 0 0]
> T=1
> 
> >3.  If any nonzero exponents remain, go back to step one, otherwise
> >stop.
> 
> T=... whoops, I forgot an absolute value here.  The correct
> value is 2.
> 
> -Carl

You mean 1, right? It's certainly not zero, though. So you have an 
infinite chain of duplicates for each pitch, spaced at increments of 
1 unit . . .


top of page bottom of page up down


Message: 10464

Date: Tue, 02 Mar 2004 21:03:24

Subject: Re: Hanzos

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:
> Carl:
> >>The usual point of odd-limit is to get octave equivalence, and
> >>therefore I'd say the 2s should be dropped from the basis.
> 
> Paul E:
> > Bad move -- for one thing, you can't detect torsion.
> 
> Can't you?  I'm not sure it's even relevant here, but I thought I 
went 
> through this in excruciating detail some time ago and showed that 
you 
> can detect torsion with octave-equivalent vectors.  In fact, I seem 
to 
> remember running the calculation in parallel using both methods, 
and 
> showing that I always got the same results.

Right, but that was a specific set of calculations, not a general 
proof. You were just looking at 'linear' temperaments, if I recall 
correctly, but torsion can afflict all types of temperaments. Plus it 
seemed your method was far less elegant.

> No, the outstanding problem is that we can't go from the 
> octave-equivalent mapping to an optimized generator size.  But I'm 
sure 
> it can be done if anybody cared.  The situation as I left it was 
that 
> nobody did.
> 
> 
>                   Graham

I care, and I hope Gene does too. I'd like to see this revisited.


top of page bottom of page up down


Message: 10465

Date: Tue, 02 Mar 2004 14:01:20

Subject: Re: Hanzos

From: Carl Lumma

>> I don't know where sqrt would be coming from.  I thought everything
>> would have to have whole number lengths.
>
>I'm using Euclidean distances.

I wish you had said that.

-C.


top of page bottom of page up down


Message: 10466

Date: Tue, 02 Mar 2004 14:04:05

Subject: Re: Hanzos

From: Carl Lumma

>> >> >So what? You still get an infinite number representing each
>> >> >interval, since you can multiply by arbitary powers of the dummy
>> >> >comma 9/3^2.
>> >> 
>> >> An infinite number from where?  If you look at the algorithm, 
>> >> that dummy comma has zero length.
>> >
>> >How do you get that????
>> 
>> >Given a Fokker-style interval vector (I1, I2, . . . In):
>> 
>> [-2 0 0 1]
>> 
>> >1.  Go to the rightmost nonzero exponent; add its absolute value
>> >to the total.
>> 
>> T=1
>> 
>> >2.  Use that exponent to cancel out as many exponents of the 
>> >opposite sign as possible, starting to its immediate left and
>> >working right; discard anything remaining of that exponent.
>> 
>> [-1 0 0 0]
>> T=1
>> 
>> >3.  If any nonzero exponents remain, go back to step one, otherwise
>> >stop.
>> 
>> T=... whoops, I forgot an absolute value here.  The correct
>> value is 2.
>
>You mean 1, right?

No, T=1 and you've got to add what's left of the 3 exponent, abs(-1),
to it.

>It's certainly not zero, though. So you have an 
>infinite chain of duplicates for each pitch, spaced at increments of 
>1 unit . . .

...

-C.


top of page bottom of page up down


Message: 10467

Date: Tue, 02 Mar 2004 22:08:13

Subject: Re: Hanzos

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
> >> >> >So what? You still get an infinite number representing each
> >> >> >interval, since you can multiply by arbitary powers of the 
dummy
> >> >> >comma 9/3^2.
> >> >> 
> >> >> An infinite number from where?  If you look at the algorithm, 
> >> >> that dummy comma has zero length.
> >> >
> >> >How do you get that????
> >> 
> >> >Given a Fokker-style interval vector (I1, I2, . . . In):
> >> 
> >> [-2 0 0 1]
> >> 
> >> >1.  Go to the rightmost nonzero exponent; add its absolute value
> >> >to the total.
> >> 
> >> T=1
> >> 
> >> >2.  Use that exponent to cancel out as many exponents of the 
> >> >opposite sign as possible, starting to its immediate left and
> >> >working right; discard anything remaining of that exponent.
> >> 
> >> [-1 0 0 0]
> >> T=1
> >> 
> >> >3.  If any nonzero exponents remain, go back to step one, 
otherwise
> >> >stop.
> >> 
> >> T=... whoops, I forgot an absolute value here.  The correct
> >> value is 2.
> >
> >You mean 1, right?
> 
> No, T=1 and you've got to add what's left of the 3 exponent, abs(-
1),
> to it.

Oh yeah, you're right.

> >It's certainly not zero, though. So you have an 
> >infinite chain of duplicates for each pitch, spaced at increments 
of 
> >1 unit . . .
> 
> ...

2 units.


top of page bottom of page up down


Message: 10469

Date: Tue, 02 Mar 2004 14:37:36

Subject: Re: Hanzos

From: Carl Lumma

>> >> I don't know where sqrt would be coming from.  I thought everything
>> >> would have to have whole number lengths.
>> >
>> >I'm using Euclidean distances.
>> 
>> I wish you had said that.
>
>If I'm talking about symmetric lattices, that would be assumed. All
>these diagrams of hexagons and tetrahedrons and octahedrons that
>people draw are Euclidean;

I also like to think of them as graphs.

>it doesn't require remarking on.

Obviously it did.

>However,
>another thing to bear in mind is this--positive definite quadradic
>forms entails Euclidean, and vice-versa. You can identify Euclidean
>lattices with quadratic forms.

That I know.

-Carl


top of page bottom of page up down


Message: 10471

Date: Tue, 02 Mar 2004 22:41:48

Subject: Re: Hanzos

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
> > >> I don't know where sqrt would be coming from.  I thought 
everything
> > >> would have to have whole number lengths.
> > >
> > >I'm using Euclidean distances.
> > 
> > I wish you had said that.
> 
> If I'm talking about symmetric lattices, that would be assumed.

You didn't assume it the other week, when it seemed (even upon 
clarification) that you referred to *two* possible metrics in the 
symmetric lattice -- the euclidean one, and the 'taxicab' one. The 
latter is what Paul Hahn's algorithm gives you.

But now you say this. So now I wonder what you were *really* talking 
about.

> All
> these diagrams of hexagons and tetrahedrons and octahedrons that
> people draw are Euclidean;

That's a bogus assumption. We live in a nearly euclidean universe so 
people have no choice in the matter when it comes to diagrams. If I 
draw a Tenney lattice, which assumes a taxicab metric, how am I 
supposed to avoid drawing it in a way that you'd interpret as 
euclidean??


top of page bottom of page up down


Message: 10472

Date: Tue, 02 Mar 2004 22:42:17

Subject: Re: Hanzos

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
> >> >> I don't know where sqrt would be coming from.  I thought 
everything
> >> >> would have to have whole number lengths.
> >> >
> >> >I'm using Euclidean distances.
> >> 
> >> I wish you had said that.
> >
> >If I'm talking about symmetric lattices, that would be assumed. All
> >these diagrams of hexagons and tetrahedrons and octahedrons that
> >people draw are Euclidean;
> 
> I also like to think of them as graphs.

Exactly; Paul Hahn apparently did too.


top of page bottom of page up down


Message: 10473

Date: Tue, 02 Mar 2004 22:44:55

Subject: Re: Hanzos

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> 
> > I care, and I hope Gene does too. I'd like to see this revisited.
> 
> You can do anything with octave-equivalent vectors that you can do
> with monzos if you can recover the monzos. This means you can't do
> some things, but since commas are small, you can recover the commas,
> and hence can eg take two octave-equivalent 7-limit vectors and get
> the corresponding temperament.

Sure. That's how I generated most of those graphs of commas -- a 
large search in n-1 dimensional space, filling in the 2 component 
later by assuming each interval was between -600 and 600 cents. But I 
don't think this is what Graham was doing.

> What I don't understand is why anyone would want to. You can also
> throw away the 7, and keep the 2,3 and 5. Would anyone propose doing
> that? It strikes me as an absurd proceedure. Where octave-equivalent
> vectors are useful is in octave-equivalent contexts.

Which would seem to be most musical contexts of interest on these 
lists.


top of page bottom of page up down


Message: 10474

Date: Tue, 02 Mar 2004 15:00:50

Subject: Re: Hanzos

From: Carl Lumma

>What I don't understand is why anyone would want to. You can also
>throw away the 7, and keep the 2,3 and 5. Would anyone propose doing
>that? It strikes me as an absurd proceedure. Where octave-equivalent
>vectors are useful is in octave-equivalent contexts.

We weren't talking about temperaments, we were talking about finding
chord sequences in JI; clearly an occasion for octave equivalence.

-Carl


top of page bottom of page up

Previous Next

10000 10050 10100 10150 10200 10250 10300 10350 10400 10450 10500 10550 10600 10650 10700 10750 10800 10850 10900 10950

10450 - 10475 -

top of page