Tuning-Math Digests messages 10501 - 10525

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Message: 10501

Date: Wed, 03 Mar 2004 21:57:31

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:
> 
> > > Hardly arbitrary if you accept the above 48 transformations as
> > > interesting, since they are orthogonal.
> > 
> > I don't get it. How does this list of transformations depend on 
your 
> > metric?
> 
> It defines a metric.

That seems impossible to me. Please convince me.

> > > I'm pointing out that Euclidean gives us more information.
> > 
> > But all kinds of other shapes, besides a sphere, would also imply 
> > these same distinct shells.
> 
> They imply distict shells, but not the *same* distinct shells.

But what's special about the shells you get from the Euclidean metric 
as opposed to the shells you get from using, instead of a sphere, 
some other rounded shape with all the symmetries of the rhombic 
dodecahedron?

> > > > > Is it in
> > > > > fact so obvious that 49/48<-->50/49 is a very different 
> > propostion
> > > > > than 15/14<-->21/20, and if so, why?
> > > > 
> > > > It's immediately evident from looking at the lattice.
> > > 
> > > It's immediately obvious from looking at the *Euclidean* 
lattice.
> > > Aren't you proving my point for me?
> > 
> > Absolutely not. For instance, for 50:49 or any other "para" dyad 
with 
> > Hahn distance 2, there's only one note consonant with both 
pitches in 
> > the dyad. So it's straightforward to take the "lattice" (or 
whatever 
> > you want to call it) and transform 50:49 to any dyad which has 
this 
> > property (such as 9:8, 25:18, 25:16, etc.) while it would take 
some 
> > severe gymnastics to transform it to one that doesn't have this 
> > property, 
> 
> Which is clear from the Euclidean lattice, which says that 9/8, 
25/18,
> 25/16, 50/49 are all located at a distance of 2 from the unison, and
> (this is harder; you can use the invariant stuff I mentioned as one
> method) that they are all in the same geometric relationship to the
> lattice. 

>It is hardly clear using Hahn distance,

I repeat: Hahn distance is not intended to give configuration 
information anyway. But configuration, not metrics, is what I was 
referring to above.

> which says all of the
> above are at a distance of 2, but also says 10/9, 16/15, 25/24, 
36/35
> etc for which the Euclidean distance is sqrt(3) are at a distance of
> 2, and even 15/14, 21/20 etc for which the Euclidean distance is
> sqrt(2) are at a distance of 2. The Hahn distance is making far less
> refined distictions, and is not providing the help in sorting these
> questions out that the Euclidean distance immediately gives.

We are going around in circles with this conversation.

> > > > > It's tough if I am going to be
> > > > > dumped on half the time for obscurity, and the other half 
for
> > > > > condescendingly explaining the blindingly obvious.
> > > > 
> > > > It's possible to explain the blindingly obvious without being 
> > > > condescending. Don't take it personally, but do try to 
improve.
> > > 
> > > First I need to learn what is blindingly obvious and what isn't.
> > 
> > No you don't -- just explain it without being condescending, and 
even 
> > if people already know it, they won't be offended.
> 
> Your claim is that I was being condescending. On what basis are you
> making it? What, exactly, was condescending about my remark?

I don't care to dwell on this any longer, but for the record, here it 
was:

"Yes, of course they are meaningful. If you want to
try to linearly transform 50/49 to 49/48, be my guest, but don't be 
surprised . . ."


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Message: 10507

Date: Thu, 04 Mar 2004 16:34:17

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:
> 
> > > Scales which have centers in holes--the octahedra and
> > > tetahedra--instead of lattice points, for starters.
> > 
> > OK. But why would a *sphere* be the most useful delimiting shape 
for 
> > constructing such scales? 
> 
> I didn't claim it would--in fact, you were the one expressing
> skepticism about the value of the Hahn metric for non-lattice 
points.

I retract that skepticism. But all your "shell" posts used the 
euclidean metric, and this question was aimed at that.


> I feel it wouldn't, since the resulting 
> > scales can actually have 'concavities' of a mild sort, indicating 
> > suboptimality in terms of consonances-per-note.
> 
> Concavities? You are claiming it is possible for the convex hull to
> contain lattice points not inside the sphere?

No -- that's why I put 'concavities' in quotes. But the smallest 
polyhedron large enough to contain all the edges in the scale will 
have concavities in general if you use a sphere (i.e., euclidean 
metric), which is indicative of suboptimality in terms of consonances-
per-note.

> In any case, with spheres the shells are less populous, so you get
> more scale possibilities--for good or ill, I don't know.

I think, beyond a certain shell, you begin to miss out on the most 
desirable scale possibilities entirely: those where the smallest 
polyhedron large enough to contain all the edges in the scale is 
convex.


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Message: 10508

Date: Thu, 04 Mar 2004 16:39:55

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:
> > --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" 
<gwsmith@s...> 
> > wrote:
> > > --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" 
<perlich@a...> 
> > wrote:
> > > 
> > > > > Hardly arbitrary if you accept the above 48 transformations 
as
> > > > > interesting, since they are orthogonal.
> > > > 
> > > > I don't get it. How does this list of transformations depend 
on 
> > your 
> > > > metric?
> > > 
> > > It defines a metric.
> > 
> > That seems impossible to me. Please convince me.
> 
> It's easiest to see in fcc coordinates, where the group consists of
> sign changes and permutations of [x, y, z]. What homogenous 
polynomial
> of degree two in x, y and z is invariant under the operations of the
> group? It is easy to check that x^2+y^2+z^2 fits the bill, and not
> difficult to show it is the *only* degree two polynomial, up to a
> constant factor, which does.

But not all metrics are degree two polynomials.

> That, however tells us that Euclidean
> distance from the unison is preserved by the operations of the 
group,
> or in other words, the group is a group of orthogonal 
transformations.
> The geometry comes right out of the group. One way to look at it is 
to
> put the origin at the center of a hexany instead, and what you are
> looking at are rotations and reflections--Euclidean
> self-congruences--of an octahedron.

Convince me that Hahn's metric doesn't yield the exact same list of 
48 orthogonal transformations.

> 
> > But what's special about the shells you get from the Euclidean 
metric 
> > as opposed to the shells you get from using, instead of a sphere, 
> > some other rounded shape with all the symmetries of the rhombic 
> > dodecahedron?
> 
> Obviously the sphere is far easier to manage, just for starters. You
> might try to get a polynomial of even degree in x, y, z which is 
not a
> power of x^2+y^2+z^2 but is invariant under the group operations and
> which has the same value on the twelve points [+-1, +1, 0] and its
> permutations, and use this to define a norm. There are independent
> invariants of degree four and six; the simplest possibility would be
> to use 
> 
> [(x^2+y^2+z^2)^2 + K (x+y+z)(-x+y+z)(x-y+z)(x+y-z)]^(1/4)
> 
> as the norm, where K is a constant. What values of K make this a 
norm?
> Hell if I know. :)

But you seem to be retracting your claim about that the list of 
orthogonal transformations defines a metric -- yes?


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Message: 10510

Date: Thu, 04 Mar 2004 18:17:32

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:
> 
> > No -- that's why I put 'concavities' in quotes. But the smallest 
> > polyhedron large enough to contain all the edges in the scale 
will 
> > have concavities in general if you use a sphere (i.e., euclidean 
> > metric), which is indicative of suboptimality in terms of 
consonances-
> > per-note.
> 
> Do you have an example of this?

I'm sure you've posted some already. Basically, if the convex hull is 
not a rhombic dodecahedron but has the same symmetry as one, you're 
suboptimal in terms of consonances-per-note -- you can improve the 
ratio by either adding or taking away notes until you do have a 
rhombic dodecahedron.


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Message: 10511

Date: Thu, 04 Mar 2004 18:21:39

Subject: Re: Hanzos

From: Paul Erlich

--- 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 "Paul Erlich" <perlich@a...> 
> wrote:
> > 
> > > No -- that's why I put 'concavities' in quotes. But the 
smallest 
> > > polyhedron large enough to contain all the edges in the scale 
> will 
> > > have concavities in general if you use a sphere (i.e., 
euclidean 
> > > metric), which is indicative of suboptimality in terms of 
> consonances-
> > > per-note.
> > 
> > Do you have an example of this?
> 
> I'm sure you've posted some already. Basically, if the convex hull 
is 
> not a rhombic dodecahedron but has the same symmetry as one, you're 
> suboptimal in terms of consonances-per-note -- you can improve the 
> ratio by either adding or taking away notes until you do have a 
> rhombic dodecahedron.

Or cuboctahedron (there may be other possibilities).

If you can create a depiction of the scales you obtained from 
spherical shells, you'll see the 'mild concavities' I'm talking about.

This would be a lot easier to depict in ASCII if we were talking 
about 5-limit instead of 7-limit . . .


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Message: 10513

Date: Thu, 04 Mar 2004 19:05:11

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:
> 
> > But not all metrics are degree two polynomials.
> 
> The point is that not all finite groups of matricies gives you such 
a
> thing, and that in this case we are dealing with a group which has 
the
> orthogonality property, which means the group reprsentation itself 
has  
> a Eucildean character.
> 
> > Convince me that Hahn's metric doesn't yield the exact same list 
of 
> > 48 orthogonal transformations.
> 
> When I talked about 49/48->50/49, you became upset with me, and now
> you ask me this! Let's see--a Hahn isometry would require that you
> send 3,5,7 to three independent 7-limit consonances, which gives you
> 12 choose 3 = 220 possibilities, clearly a lot more than 48.

Gene, you are wrong. A Hahn isometry would require that you send 
3,5,7 to three independent 7-limit consonances *which are consonant 
with one another*! Only 48 possibilities.

> but it certainly isn't clear how Hahn
> distinquishes the 48 orthogonal transformations among the rest.

As long as we're following the definition of orthogonal 
transformation that you gave, you're transforming the entire lattice, 
so this seems perfectly clear to me.

> > But you seem to be retracting your claim about that the list of 
> > orthogonal transformations defines a metric -- yes?
> 
> No, it clearly defines a metric--singling out the Euclidean metric
> because of the uniqueness of the degree two invariant; you seem to
> take that to mean a unique metric.

What else could "defines a metric" be taken to mean, if not "yields a 
unique choice of metric"?


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Message: 10515

Date: Thu, 04 Mar 2004 19:19:52

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:
> 
> > Gene, you are wrong. A Hahn isometry would require that you send 
> > 3,5,7 to three independent 7-limit consonances *which are 
consonant 
> > with one another*! Only 48 possibilities.
> 
> Yeah, that occurred to me after posting.
> 
> > What else could "defines a metric" be taken to mean, if 
not "yields a 
> > unique choice of metric"?
> 
> I don't think it's possible for a finite group to do that;

So you shouldn't say it "defines a metric", unless I'm missing 
something.


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Message: 10519

Date: Thu, 04 Mar 2004 01:34:56

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:
> > --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" 
<gwsmith@s...> 
> > wrote:
> > > --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" 
<perlich@a...> 
> > wrote:
> > > 
> > > > Why would anyone ever use it on non-lattice points?
> > > 
> > > Constructing scales, for starters.
> > 
> > Which scales involve non-lattice points? 
> 
> Scales which have centers in holes--the octahedra and
> tetahedra--instead of lattice points, for starters.

OK. But why would a *sphere* be the most useful delimiting shape for 
constructing such scales? I feel it wouldn't, since the resulting 
scales can actually have 'concavities' of a mild sort, indicating 
suboptimality in terms of consonances-per-note.


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Message: 10522

Date: Fri, 05 Mar 2004 19:29:26

Subject: Re: Hahn norm formula

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:
> 
> > For a note class represented by 3^a 5^b 7^c, I get that it should 
be
> > 
> > ||(a,b,c)||_Hahn = max(|a|,|b|,|c|,|b+c|,|a+c|,|a+b|,|a+b+c|)
> > 
> > This is a pretty simple formula, and applicable to non-lattice 
points.
> > I'll take a look at scales which arise from it.
> 
> The number of notes in Hahn shell n turns out to be 10n^2+2; about
> this the handbook of integer seqences says "Points on surface of
> cuboctahedron (or icosahedron): a(0) = 1, for n> 0, a(n) = 10n^2 + 2
> (coordination sequence for f.c.c. lattice)." So we seem to be doing
> things right here. If we take the union of the Hahn shells, we get
> lattice-point-centered Hahn scales. About the integer sequence for
> this the handbook says "Centered icosahedral (or cuboctahedral)
> numbers, also crystal ball sequence for f.c.c. lattice.
> Comments:  Called 'magic numbers' in some chemical contexts."
> 
> The latter stikes me as nifty. Crystal ball scales? Magic number
> scales? I like the first especially.
> 
> 
> Here are the first few Hahn shells:
> 
> Shell 0 
> [1]
> 
> Shell 1
> [8/7, 7/6, 6/5, 5/4, 4/3, 7/5, 10/7, 3/2, 8/5, 5/3, 12/7, 7/4]

Union of shell 0 and shell 1 is known as the 7-limit Tonality Diamond.

Thanks for your great work here, Gene.


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Message: 10524

Date: Fri, 05 Mar 2004 20:52:42

Subject: Re: Hahn norm formula

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> 
> > Union of shell 0 and shell 1 is known as the 7-limit Tonality 
Diamond.
> > 
> > Thanks for your great work here, Gene.
> 
> Thanks! Not only has Crystal Ball One been seen often before, 
Crystal
> Ball Two has already appeared--it is exactly the 55 note scale 
which I
> posted in message 9922, in connection with stepwise harmonizing 
scales.
> Despite its apparent theoretical interest, Crystal Ball Two should
> suffice as a name,

Paul Hahn might have called it the 7-limit radius 2 scale or perhaps 
more likely, the Level 2 7-limit Diamond.

Now what do you get if you center around a major tetrad (shallow 
hole)?


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