RE: RE: A gentle introduction to Fokker periodicity bloc ks, part 2

Manuel wrote,

>Your idea about hexagonal periodicity blocks is brilliant.

Thanks.

>Still interested, I look forward to
>the third part.

Very busy with music and work this week, I'll do it next week.

In 3 dimensions (7-limit), the analogue to the hexagonal periodicity block
would be either a rhombic dodecahedron or the 14-sided cuboctahedron. I
can't decide which. Probably both are equally valid?

On Thu, 28 Oct 1999, Paul H. Erlich wrote:
> Manuel wrote,
>> Your idea about hexagonal periodicity blocks is brilliant.
>
> Thanks.

No offense, but--it's straightforward tiling/tessellation theory, isn't
it? I've been transposing pitches from one edge of a scale to the other
via unison vectors to make odd shapes for ages. In fact, long ago I
quit circumscribing Fokker-derived scales with parallelograms and
parallelepipeds, and just go for the best-connected, smallest-diameter
pitchsets I can find. For example, that was the basis for a message I
posted to the list in--good grief, 1994--about a alternate 5-limit
interpretation of the Indian 22-sruti scale.

> In 3 dimensions (7-limit), the analogue to the hexagonal periodicity block
> would be either a rhombic dodecahedron or the 14-sided cuboctahedron. I
> can't decide which. Probably both are equally valid?

?? Cuboctas don't fill space. I'd guess you want the rhombic dodec.

--pH <manynote@library.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "Well, so far, every time I break he runs out.
-\-\-- o But he's gotta slip up sometime . . . "

>?? Cuboctas don't fill space. I'd guess you want the rhombic dodec.

Yes they do. Look up "Kelvin."

On Thu, 28 Oct 1999, Paul H. Erlich wrote:
>> ?? Cuboctas don't fill space. I'd guess you want the rhombic dodec.
>
> Yes they do. Look up "Kelvin."

Those aren't cuboctas, they're--oh, hell, I can't remember the name.
But they're a different Archimedean solid; very similar in that they're
both formed by truncating an octahedron, but the cubocta has triangles
where Kelvin's solid has hexagons.

But I still think the rhombic dodec is what you want. I gotta go, but
I'll try to explain why in a later post.

--pH <manynote@library.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "Well, so far, every time I break he runs out.
-\-\-- o But he's gotta slip up sometime . . . "

>Those aren't cuboctas, they're--oh, hell, I can't remember the name.
>But they're a different Archimedean solid; very similar in that they're
>both formed by truncating an octahedron, but the cubocta has triangles
>where Kelvin's solid has hexagons.

I'm talking about cuboctahedra. You can see them fill space in the last
picture on this page: http://www.teleport.com/~pdx4d/sphpack.html. Kelvin's
solid did not have hexagons -- it was a slight deformation of the
couboctahedron.

>But I still think the rhombic dodec is what you want. I gotta go, but
>I'll try to explain why in a later post.

I'm looking forward to it.

On Thu, 28 Oct 1999, Paul H. Erlich wrote:
>> Those aren't cuboctas, they're--oh, hell, I can't remember the name.
>> But they're a different Archimedean solid; very similar in that they're
>> both formed by truncating an octahedron, but the cubocta has triangles
>> where Kelvin's solid has hexagons.
>
> I'm talking about cuboctahedra. You can see them fill space in the last
> picture on this page: http://www.teleport.com/~pdx4d/sphpack.html.

That is not cuboctahedra filling space. In fact, that's just the
tetrahedral lattice that we all know and love for 7-limit diagrams. The
edges form cuboctahedra, but you can't take the cuboctahedron and no
other shape and fill space with it.

If you take the Voronoi cells of the lattice points of that diagram, you
get rhombic dodecs--which, in a nutshell, is why I think that the
rhombic dodec is what you want.

> Kelvin's
> solid did not have hexagons -- it was a slight deformation of the
> couboctahedron.

I quote from <http://www.ics.uci.edu/~eppstein/junkyard/froth.html>:

: In 1887, Lord Kelvin pondered how to partition space into cells
: of equal volume with the least area of surface between them, i.e.
: the most efficient soap bubble froth. He came up with a 14-sided
: space-filling polyhedron with 6 square sides and 8 hexagonal sides.

--pH <manynote@library.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "Well, so far, every time I break he runs out.
-\-\-- o But he's gotta slip up sometime . . . "

Paul Hahn wrote,

>That is not cuboctahedra filling space. In fact, that's just the
>tetrahedral lattice that we all know and love for 7-limit diagrams.

No, you can clearly see that it's a bunch of cuboctahedra -- some of the
connections from the tetrahedral lattice are left out.

But you can see spaces between the cuboctahedra, leading me to believe that
you were actually right about this:

>you can't take the cuboctahedron and no
>other shape and fill space with it.

I must have confused Kepler's 14-sided shape with the cuboctahedron. Also, a
list member recently asked me if the 7-limit Tonality Diamond (a
cuboctahedron) fills space, why isn't it a periodicity block? I guess I
assumed that they were right about the premise but not the implication. But
I guess both are incorrect.

Manuel wrote,

>> Your idea about hexagonal periodicity blocks is brilliant.

I wrote,

> Thanks.

Paul Hahn wrote,

>No offense, but--it's straightforward tiling/tessellation theory, isn't
>it? I've been transposing pitches from one edge of a scale to the other
>via unison vectors to make odd shapes for ages. In fact, long ago I
>quit circumscribing Fokker-derived scales with parallelograms and
>parallelepipeds, and just go for the best-connected, smallest-diameter
>pitchsets I can find. For example, that was the basis for a message I
>posted to the list in--good grief, 1994--about a alternate 5-limit
>interpretation of the Indian 22-sruti scale.

Well, P.H., you're very brilliant yourself -- don't knock it. I'm surprised
you haven't had anything to contribute in the last few months. I'm certainly
interested in transforming Fokker periodicity blocks to "the best-connected,
smallest-diameter
pitchsets I can find." Do you have any suggestions as to searching for
examples of these where the smallest interval is smaller than the unison
vectors as well as those linear combinations of the unison vectors which
could result from adding one note to the pitchset? That would be a worthy
attempt at classifying all the natural closures of JI.

On Thu, 28 Oct 1999, Paul H. Erlich wrote:
> Paul Hahn wrote,
>> That is not cuboctahedra filling space. In fact, that's just the
>> tetrahedral lattice that we all know and love for 7-limit diagrams.
>
> No, you can clearly see that it's a bunch of cuboctahedra -- some of the
> connections from the tetrahedral lattice are left out.
>
> But you can see spaces between the cuboctahedra, leading me to believe that
> you were actually right about this:
>
>> you can't take the cuboctahedron and no
>> other shape and fill space with it.
>
> I must have confused Kepler's 14-sided shape with the cuboctahedron.

Yes, I think so. It's easily demonstrated that cuboctas don't fill
space: try it! If you take eight cuboctas and join them by the square
faces, you find them surrounding an octahedral hole. And there aren't
any other ways to do it--you can't very well join the square faces to
the triangular ones, and the highly symmetric shapes mean there aren't
any alternative orientations to try.

Kelvin's shape is just called a truncated octahedron, BTW. And if you
fit them together as above, the octahedral spaces that were left using
cuboctas expand just enough to butt up against each other at the
vertices and become truncated octas themselves.

Okay, now here's why I think you want the rhombic dodec instead of the
truncated octa. <deep breath> If you look at the way the truncated
octa fills space, you'll find that its symmetry group is that of a
body-centered cubic lattice. The rhombic dodec, OTOH, fills space such
that is _its_ symmetry group is that of a _face_-centered cubic
lattice.

Why is this important? Well, let's think about why we use FCC (= oc-tet
or triangulated lattice) for pitch diagrams in the first place. If you
ignore the edges and just look at the lattice points, it's equivalent to
the cubic lattice--it's just been subjected to a couple of affine
(shear) transformations. (Translation: we squished it a bit so that it
slants.) This makes sense because the lattice is actually a space whose
basis is the three vectors representing the 3/2, the 5/4, and the 7/4.
Right?

Now look at the way the shapes you want to use to tesselate space with.
They _also_ are related to each other by three basic vectors, it's just
that this time, it's the three unison vectors. Other than that, the
relationship is the same. But there's no way you can map the BCC to a
straightforward cubic lattice--you either have to leave some points out
of the cubic lattice, or interlock two cubic lattices together.

In sum: the hexagon is the Voronoi cell of the 2d triangulated lattice,
and the rhombic docec is the Voronoi cell of the 3d triangulated
lattice. Since the Voronoi cell is by definition the set of points
closer to a particular lattice point than to any other, it makes sense
that this is a good shape to start with when looking for "the
best-connected, smallest-diameter pitchsets".

On Thu, 28 Oct 1999, Paul H. Erlich wrote:
> Do you have any suggestions as to searching for
> examples of these where the smallest interval is smaller than the unison
> vectors as well as those linear combinations of the unison vectors which
> could result from adding one note to the pitchset? That would be a worthy
> attempt at classifying all the natural closures of JI.

I know I've been conspicuously silent on this question, but there are
several reasons for it:

(a) I've been busy. That, thankfully, is (I think) starting to ease up.

(b) It's a _hard_ problem, and I haven't really had any good ideas to
suggest besides brute-force searches.

(c) I'm not entirely sure that it's a _well-defined_ problem. It's not
just a question of taking your unison vectors and running. You might
find that with a given set of vectors, the scale circumscribed by the
parallelogram satisfied your conditions, but the one resulting from the
hexagon didn't--or vice versa. Even if you restricted yourself to the
best-connected, smallest-diameter pitchset for a given set of vectors,
that _still_ doesn't determine a unique scale--usually even after you've
gotten the connectivity maximized and the diameter minimized, there are
still several pitches you can transpose by one unison vector or another
to give you different scales, but not change the connectivity and
diameter. And each time that happens, you run the risk of one such
scale "working", and the other not.

In addition to this, I believe (although I have yet to prove it
rigorously) that _all_ sets of unison vectors which yield a given number
of pitches are, basically, equivalent; at least all those which don't
correspond to ludicrously large intervals. By "equivalent" I mean that
given two such sets, you can transform one to the other by (repeatedly
if necessary) adding or subtracting one vector within the set to one of
the others, which (from linear algebra) we know does not change the
value of the determinant. Assuming that my conjecture on this is true,
it seems to me that trying to pick the "best" set of vectors for a given
number of scale degrees is really sort of beside the point--the pitchset
is more important than the vectors. Which would tend to imply that no
method is going to be significantly better than brute force . . . [1]

[1] Other than my method of looking for "JI-like" ETs, i.e. ETs whose
diameter is less than or equal to their consistency level, which is no
doubt too severely restrictive for most people here.

--pH <manynote@library.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "Well, so far, every time I break he runs out.
-\-\-- o But he's gotta slip up sometime . . . "

Paul Hahn wrote,

>But the 7-limit diamond can be used to fill
>the lattice.

Wait a minute, Paul. The 7-limit diamond is a cuboctahedron (as Erv Wilson
fans will know). And you just finished convincing us that cuboctahedra don't
fill space. So aren't you contradicting yourself?

Paul H.,

Thanks for your extensive comments. Without going too deeply into them just
yet, let me bounce this off you: In 2-D, there will be some cases where the
parallelogram formed by two unison vectors is more compact in the triagular
lattice than the hexagons formed by treating either the sum or the
difference of the unison vectors equivalently to the UVs themselves. For
example, if the Fokker parallelogram looks like a square, "hexagonalizing"
it would only be detrimental to its compactness. So, in 3-D, is it not
conceivable that there would be cases where the parallelopiped (if its
nearly cubical) would be a more compact form for the periodicity block of a
given set of unison vectors than the rhombic dodecahedron?

-Paul E.

On Mon, 1 Nov 1999, Paul H. Erlich wrote:
> In 2-D, there will be some cases where the
> parallelogram formed by two unison vectors is more compact in the triagular
> lattice than the hexagons formed by treating either the sum or the
> difference of the unison vectors equivalently to the UVs themselves. For
> example, if the Fokker parallelogram looks like a square, "hexagonalizing"
> it would only be detrimental to its compactness. So, in 3-D, is it not
> conceivable that there would be cases where the parallelopiped (if its
> nearly cubical) would be a more compact form for the periodicity block of a
> given set of unison vectors than the rhombic dodecahedron?

Certainly, certainly. Depending on the vectors, that might certainly
happen. Which is why, as I said before, rather than circumscribing
scales with any particular geometric figure I have taken to just
aggregating the determinant-derived number of pitches in as compact a
shape as possible while disallowing pitches separated by unison vectors
(commatic intervals).

--pH <manynote@library.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

Paul Hahn wrote,

>Certainly, certainly. Depending on the vectors, that might certainly
>happen. Which is why, as I said before, rather than circumscribing
>scales with any particular geometric figure I have taken to just
>aggregating the determinant-derived number of pitches in as compact a
>shape as possible while disallowing pitches separated by unison vectors
>(commatic intervals).

And that geometric figure will automatically fill space. Can you give any
2-d examples where this process gives a shape other than a parallelogram or
hexagon?

On Mon, 1 Nov 1999, Paul H. Erlich wrote:
>> I have taken to just
>> aggregating the determinant-derived number of pitches in as compact a
>> shape as possible while disallowing pitches separated by unison vectors
>> (commatic intervals).
>
> And that geometric figure will automatically fill space. Can you give any
> 2-d examples where this process gives a shape other than a parallelogram or
> hexagon?

Actually, probably not--the most compact (best-connected) shape is
usually (I think always) circumscribable within a hexagon. However,
this is partly because geometrically speaking, there are arbitrarily
many hexagons which can be associated with a given unison vector pair.
My method allows me to not have to worry about _which_ hexagon or _which_
parallelogram; I just go directly to the pitches.

--pH <manynote@library.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

On Mon, 1 Nov 1999, Paul Hahn wrote:
>> Can you give any
>> 2-d examples where this process gives a shape other than a parallelogram or
>> hexagon?
>
> Actually, probably not--the most compact (best-connected) shape is
> usually (I think always) circumscribable within a hexagon.

If you drop the best-connectedness criterion, though, it's fairly simple
to construct an example. The harmonic minor, for example, while still
heptatonic and therefore based on the vectors (4 -1) and (-1 2), doesn't
quite fit within any convex shape:

5:3 15:8
4:3 1:1 3:2 9:8
6:5

--pH <manynote@library.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

On Mon, 1 Nov 1999, Paul Hahn wrote:
> The harmonic minor, for example, while still
> heptatonic and therefore based on the vectors (4 -1) and (-1 2), doesn't
> quite fit within any convex shape:
>
> 5:3 15:8
> 4:3 1:1 3:2 9:8
> 6:5

Sorry, that's the melodic minor ascending, isn't it? Oops.

--pH <manynote@library.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

Paul Hahn wrote,

>Actually, probably not--the most compact (best-connected) shape is
>usually (I think always) circumscribable within a hexagon.

What if the Fokker parallelogram is really big and very nearly a square? It
would seem that "hexagonalization" would be detrimental.

>My method allows me to not have to worry about _which_ hexagon or _which_
>parallelogram; I just go directly to the pitches.

Excellent!

>> 5:3 15:8
>> 4:3 1:1 3:2 9:8
>> 6:5

>Sorry, that's the melodic minor ascending, isn't it? Oops.

Right -- also known as jazz minor. Except that the II chord of jazz minor
should be a consonant minor triad, which would require replacing 9/8 with
10/9 (can we stick with using "/" when the ratios indicate pitches rather
than intervals?).

On Mon, 1 Nov 1999, Paul H. Erlich wrote:
>> Actually, probably not--the most compact (best-connected) shape is
>> usually (I think always) circumscribable within a hexagon.
>
> What if the Fokker parallelogram is really big and very nearly a square? It
> would seem that "hexagonalization" would be detrimental.

You can always construct a hexagon (see an earlier post about there
being arbitrarily many hexagons that can be used) with one pair of
parallel sides that are very short.

On Mon, 1 Nov 1999, Paul H. Erlich wrote:
>>> 5:3 15:8
>>> 4:3 1:1 3:2 9:8
>>> 6:5
>
>>Sorry, that's the melodic minor ascending, isn't it? Oops.
>
> Right -- also known as jazz minor. Except that the II chord of jazz minor
> should be a consonant minor triad, which would require replacing 9/8 with
> 10/9

In such cases I like to cheat a little by placing such a pitch exactly
on the border of the hexagon, so that both versions can be used.

> (can we stick with using "/" when the ratios indicate pitches rather
> than intervals?).

I thought the list had adopted a convention of : for pitches and / for
intervals; did I get it backwards? That way makes more sense to me,
because of all the /s we tend to use when drawing lattice diagrams;
using : for the pitches in the lattice makes for less confusion. But
whatever.

--pH <manynote@library.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

>I thought the list had adopted a convention of : for pitches and / for
>intervals; did I get it backwards?

I think so. "/" seems to be pretty engrained for pitches, probably due to
Harrry Partch's influence.

Paul Hahn wrote

>Yes--this gets into different categories of "unisonous" intervals.
>Since I am investigating how "generalized diatonic" scales embed within
>ETs (I'm not really into pure JI), I distinguish between "chromatic" and
>"commatic" unisons.

Some of my previous posts have made this distinction. See my post of Mon
10/4/99 10:00 PM. In the case of the decatonic scale, I basically stated
that 63:64 and 50:49 are the "commatic" unisons, and 49:48 is the
"chromatic" unison. A fine line in JI, to be sure, but the decatonic scale
was never really intended to operate in pure JI. In my post of Fri 10/15/99
6:34 PM, I essentially state that Bohlen's 9-tone-to-the-3:1 "diatonic"
scale has a "commatic" unison of 245:243, and a "chromatic" unison of
625:567.

I wrote,

>> Do you have any suggestions as to searching for
>> examples of these where the smallest interval is smaller than the unison
>> vectors as well as those linear combinations of the unison vectors which
>> could result from adding one note to the pitchset? That would be a worthy
>> attempt at classifying all the natural closures of JI.

Paul Hahn wrote,

>(c) I'm not entirely sure that it's a _well-defined_ problem. It's not
>just a question of taking your unison vectors and running. You might
>find that with a given set of vectors, the scale circumscribed by the
>parallelogram satisfied your conditions, but the one resulting from the
>hexagon didn't--or vice versa. Even if you restricted yourself to the
>best-connected, smallest-diameter pitchset for a given set of vectors,
>that _still_ doesn't determine a unique scale--usually even after you've
>gotten the connectivity maximized and the diameter minimized, there are
>still several pitches you can transpose by one unison vector or another
>to give you different scales, but not change the connectivity and
>diameter. And each time that happens, you run the risk of one such
>scale "working", and the other not.

Well, yes, I've mentioned that already, but doesn't that just make the
problem more difficult, rather than ill-defined?

Manuel Op de Coul wrote,

>The block you thought might be Ramis' scale in part 2 is in fact Ellis'
Duodene.

I wrote Ramos, not Ramis. Joe Monzo just referred to

>1482 (when
>Ramos first described a '5-limit' tuning system)

so I can't be completely off the wall. Anyone know more about Ramos, as
opposed to Ramis?

>The block below is Marpurg's Monochord nr.1 transposed by 3/2.

Thanks!

>Weaire-Phelan Foam? :-)

Almost coincidentally, that is mentioned in the page Paul Hahn referred to
in which Kepler's foam is mentioned. Unfortunately, neither type of foam is
suitable for constructing periodicity blocks.

On Tue, 2 Nov 1999, Paul H. Erlich wrote:
>> Weaire-Phelan Foam? :-)
>
> Almost coincidentally, that is mentioned in the page Paul Hahn referred to
> in which Kepler's foam is mentioned.

Shouldn't that be Kelvin?

--pH <manynote@library.wustl.edu> http://library.wustl.edu/~manynote
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

>>> Weaire-Phelan Foam? :-)
>
>> Almost coincidentally, that is mentioned in the page Paul Hahn referred
to
>> in which Kepler's foam is mentioned.

>Shouldn't that be Kelvin?

Ack! Of course, Kelvin. Whoops!

>I looked it up and both spellings are right.
>Ramos de Pareja Bartolomeo, or Ramis de Pareia Bartolome,
>(1440-c.1491), Spanish theorist.

And how did his 12-tone JI tuning differ from Ellis's duodene?

>>I looked it up and both spellings are right.
>>Ramos de Pareja Bartolomeo, or Ramis de Pareia Bartolome,
>>(1440-c.1491), Spanish theorist.

>And how did his 12-tone JI tuning differ from Ellis's duodene?

Five tones are different, it's:

135/128 10/9 32/27 5/4 4/3 45/32 3/2 128/81 5/3 16/9 15/8 2/1

Lattice:

* * * * * *
* * * * 0 *

Manuel Op de Coul coul@ezh.nl

I wrote,

>> In 2-D, there will be some cases where
the
>> parallelogram formed by two unison vectors is more compact in the
triagular
>> lattice than the hexagons formed by treating either the sum or the
>> difference of the unison vectors equivalently to the UVs themselves. For
>> example, if the Fokker parallelogram looks like a square,
"hexagonalizing"
>> it would only be detrimental to its compactness. So, in 3-D, is it not
>> conceivable that there would be cases where the parallelopiped (if its
>> nearly cubical) would be a more compact form for the periodicity block of
a
>> given set of unison vectors than the rhombic dodecahedron?

On Mon, 1 Nov 1999, Paul Hahn wrote

>Certainly, certainly. Depending on the vectors, that might certainly
>happen. Which is why, as I said before, rather than circumscribing
>scales with any particular geometric figure I have taken to just
>aggregating the determinant-derived number of pitches in as compact a
>shape as possible while disallowing pitches separated by unison vectors
>(commatic intervals).

There would seem to be a third possibility in three dimensions -- a
hexagonal prism. Right, Paul? In four dimensions, are there four
possibilities? What are they?