Some omnitetrachordal scales

Call a scale "omnitetrachordal" if, within any octave span, the pattern of steps within one (tempered) 4/3 is replicated exactly in another 4/3, and the octave is filled in by a 9/8 which may be divided into smaller steps arbitrarily. (The 9/8 may be between the two 4/3s, or above or below.) This concept has a straightforward generalization to intervals of quasi-equivalence other than 4/3, but these examples are all for the original version with 4/3.

Clearly any MOS of a temperament with octave period and whose generator is a tempered 4/3, including for example meantone, mavila, and father, is omnitetrachordal. However, there are also omnitetrachordal scales associated with linear temperaments whose generator does not represent 4/3, or whose period is not an octave.

The following MODMOSes are omnitetrachordal. The temperament names given are merely examples, in that any other temperament with the same structure and the same mapping of 4/3 will also work. For example, Injera[6] LsLsss is omnitetrachordal because it has the same structure as pajara.

Pajara[6] LsLsss
Pajara[8] LssLssss
Liese[9] LssLsssss
Pajara[10] LsssLsssss (pentachordal decatonic)
Mohajira[10] LLLsLLLsLs
Liese[11] LsssLssssss
Pajara[12] LLLLsLLLLsLL
Shrutar[12] LsssLsssssss
Liese[13] LssssLsssssss
Tetracot[13] LLLssLLLssLss
Shrutar[14] LssssLssssssss
Tritonic[15] LssssLsssssssss

Note that these are not the most general omnitetrachordal scales, which may not even be rank 2. For example, the pentachordal decatonic pattern, LsssLsssss, could be generalized to LabaLababa where a and b may be different. I believe that omnitetrachordality implies rank <= 3 but I have no proof of this.

I derived these by hand, because I have no way to mathematically generate all possible omnitetrachordal scales yet. It's good to have nontrivial examples though.

Many of these scales are well known, but some are novel. Of particular note is the Liese[9] MODMOS, whose tetrachords strongly resemble those of the ancient Greek enharmonic genus. Thus this scale could be viewed as a filled-in or regularized version of the enharmonic scale.

Keenan

--- In tuning@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
>
> Call a scale "omnitetrachordal" if, within any octave span, the pattern of steps within one (tempered) 4/3 is replicated exactly in another 4/3, and the octave is filled in by a 9/8 which may be divided into smaller steps arbitrarily.

Has anyone ever tried concocting omni-thirds scales, three identical divisions of 5/4 plus the diesis?

On Mon, Sep 5, 2011 at 10:09 AM, Keenan Pepper <keenanpepper@...> wrote:
>
> Note that these are not the most general omnitetrachordal scales, which may not even be rank 2. For example, the pentachordal decatonic pattern, LsssLsssss, could be generalized to LabaLababa where a and b may be different. I believe that omnitetrachordality implies rank <= 3 but I have no proof of this.
>
> I derived these by hand, because I have no way to mathematically generate all possible omnitetrachordal scales yet. It's good to have nontrivial examples though.

Nice work! I did some work on this too a while ago, from most recent
(and hence most developed) to least recent

Formula:
/tuning-math/message/19130

Pretty pictures of what's going on:
/tuning-math/message/19129
/tuning-math/message/19085

The basic hallmark of omnitetrachordal scales are that they have a
"subperiod" at a certain interval. This subperiod isn't infinite in
extent like an actual period, but is repeated only as many times as it
can fit within the period.

I've since gotten the formula even simpler, so here it is:

1) Pick a sub-period (let's say 4/3).
2) Pick an MOS generated by that sub-period (let's say 2L3s).
3) Pick a generator, and make sure that this generator is smaller than
the smallest interval in the MOS (let's say a semaphore-tempered
28/27, which is also 64/63).
4) Take your MOS and copy and paste it shifted over by the generator.
5) Do #4 as many times as you want, but only as long as the total
number of generators you're shifting by doesn't end up being larger
than the side of the smallest interval in #2

If you screw up #5 and shift too much, to keep it omnitetrachordal
you'll have to wrap all of the "overstretched" notes around into the
other sub-periodic cells, and if you keep doing this until you get to
an omnitetrachordal scale you'll end up doing the whole thing again
with just the next-largest choice of MOS in #2.

This will generate, I think, rank-3 parallelodahsdfh periodicity
blocks, but I'm not sure if there might be an exception to that. I
conjecture that it will and that a correspondence exists between
omnitetrachordal scales and rank-3 periodicity blocks, just like one
exists between MOS's and rank-2 periodicity blocks. The question of
how rank-2 scales might be viewed as tempered versions of the above
blocks, kind of how blackwood[10] can be viewed as a tempered version
of pythagorean[5] with a generator of 81/80 and then with 256/243
eliminated, is something I'm leaving open as I'm not quite sure if my
conjecture is too overreaching.

It's noteworthy that a lot of temperaments generate omnitetrachordal
scales that aren't related to the MOS series in the temperament at
all. For example, semaphore in 19-equal has an omnitetrachordal scale
at 1 2 1 4 1 2 1 4 1 2, which is 10 notes. It can be viewed as a
chromatically altered version of "semaphore[10]", despite that
semaphore[10] isn't an MOS. It's what you get if you take the 1 2 1 3
1 3 1 3 1 3 near-MOS of semaphore and sharpen the 5th, 6th, 9th, and
10th degrees of the scale.

These scales might be described as forming "moments of
omnitetrachordality," but something like "moment of sub-symmetry"
might be better in that it doesn't form the unfortunate acronym MOO.

-Mike

"genewardsmith" <genewardsmith@...> wrote:

> Has anyone ever tried concocting omni-thirds scales,
> three identical divisions of 5/4 plus the diesis?

The tripod scale (or 9 note marvel hobbit) is one such.
Also, various MOS scales in magic temperament (7, 10, 13,
16, 19, or 22 notes).

Graham

--- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> I've since gotten the formula even simpler, so here it is:
>
> 1) Pick a sub-period (let's say 4/3).
> 2) Pick an MOS generated by that sub-period (let's say 2L3s).
> 3) Pick a generator, and make sure that this generator is smaller than
> the smallest interval in the MOS (let's say a semaphore-tempered
> 28/27, which is also 64/63).
> 4) Take your MOS and copy and paste it shifted over by the generator.
> 5) Do #4 as many times as you want, but only as long as the total
> number of generators you're shifting by doesn't end up being larger
> than the side of the smallest interval in #2
>
> If you screw up #5 and shift too much, to keep it omnitetrachordal
> you'll have to wrap all of the "overstretched" notes around into the
> other sub-periodic cells, and if you keep doing this until you get to
> an omnitetrachordal scale you'll end up doing the whole thing again
> with just the next-largest choice of MOS in #2.

Once I realized that it was allowed for some notes of the shifted copies to be coincident, I could see how this generated most of the scales I came up with. However, I can't see how to generate two of my scales using these rules:

Liese[13] LssssLsssssss
As 19edo steps, 4 5 6 7 8 12 13 14 15 16 17 18 19

Tetracot[13] LLLssLLLssLss
As 27edo steps, 3 6 9 10 11 14 17 20 21 22 25 26 27

Could you explain in detail how your formula generates these scales?

Keenan

On Tue, Sep 6, 2011 at 4:22 AM, Keenan Pepper <keenanpepper@...> wrote:
>
> Once I realized that it was allowed for some notes of the shifted copies to be coincident, I could see how this generated most of the scales I came up with. However, I can't see how to generate two of my scales using these rules:
>
> Liese[13] LssssLsssssss
> As 19edo steps, 4 5 6 7 8 12 13 14 15 16 17 18 19

Start with the mode ssssLssssLsss.

You get this by starting with the C-F-Bb-C MOS, in 19-tet meantone
notation, and then shifting it over by 1\19 4 times. You'll note that
after 3 shifts, the Bb on top will become C, which is a note already
in the scale, and then after 4 shifts, it'll become C#, which is also
a note already in the scale. This is because the liese scalar
structure subdivides the 9/8 into three equal steps. Maybe this will
break my conjecture about rank-3 periodicity blocks, or maybe it can
be viewed as a tempered version of a rank-3 PB in some way I'm not
seeing.

> Tetracot[13] LLLssLLLssLss
> As 27edo steps, 3 6 9 10 11 14 17 20 21 22 25 26 27

This one's pretty trippy. If we're considering the C-F-Bb-C MOS to be
the subperiod, it can be thought of using two generators - 3\27 and
1\27. Perhaps it'll be clearer if I explain where the formula came
from.

If you think about it, if you start with C-F-Bb-C and just start
shifting things over (not necessarily using the same shift size each
time), and make sure that the top note never goes over C, you'll end
up with something omnitetrachordal by definition. If you shift things
in a clever enough way that although you do go over C, the things that
you hit are exact duplicate notes of notes you already have, you'll
still end up with something omnitetrachordal. Any omnitetrachordal
scale can be laid out this way, so at least part of the mystique of
omnitetrachordal scales is laid to rest.

If you screw up and you do go over the C, thus creating a new note in
the lower tetrachord - and then you try to duplicate that note in the
other tetrachords, and keep repeating until you hit a stable
omnitetrachordal structure - you'll probably find that you've ended up
with the next-highest MOS, something like C-D-F-G-Bb-C shifted a few
times over. Thinking way back on it now, I don't remember if I ever
proved it mathematically, but if you try it you'll see what I mean.
Perhaps there's an exception I missed, but I vaguely remember working
out at least intuitively that this is how this situation would always
resolve itself.

There's really no reason you should have to shift by the same
generator every time, so long as your shifted copy of the top note in
the MOS doesn't exceed the octave and create a new note in the bottom
tetrachord. So I should have been a bit more specific in my
description of the formula.

There's a bit more going on here in that the generators are probably
related in size to one another somehow, much like the Liese generator
was exactly 1/3 of the whole tone, so perhaps there's an even simpler
pattern I'm missing.

-Mike

--- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> > Liese[13] LssssLsssssss
> > As 19edo steps, 4 5 6 7 8 12 13 14 15 16 17 18 19
>
> Start with the mode ssssLssssLsss.
>
> You get this by starting with the C-F-Bb-C MOS, in 19-tet meantone
> notation, and then shifting it over by 1\19 4 times. You'll note that
> after 3 shifts, the Bb on top will become C, which is a note already
> in the scale, and then after 4 shifts, it'll become C#, which is also
> a note already in the scale. This is because the liese scalar
> structure subdivides the 9/8 into three equal steps. Maybe this will
> break my conjecture about rank-3 periodicity blocks, or maybe it can
> be viewed as a tempered version of a rank-3 PB in some way I'm not
> seeing.

But with the 4th shift, you've shifted a total of more than 9/8, which your formula said not to do. There would have to be some kind of exception for the case where the new note (in this case C#) is coincident with an existing one.

Also, the formula says if this happens you should use the next larger MOS of pythagorean, C-D-F-G-Bb-C, but I don't see how to do it using this MOS at all.

> > Tetracot[13] LLLssLLLssLss
> > As 27edo steps, 3 6 9 10 11 14 17 20 21 22 25 26 27
>
> This one's pretty trippy. If we're considering the C-F-Bb-C MOS to be
> the subperiod, it can be thought of using two generators - 3\27 and
> 1\27. Perhaps it'll be clearer if I explain where the formula came
> from.
>
> If you think about it, if you start with C-F-Bb-C and just start
> shifting things over (not necessarily using the same shift size each
> time), and make sure that the top note never goes over C, you'll end
> up with something omnitetrachordal by definition. If you shift things
> in a clever enough way that although you do go over C, the things that
> you hit are exact duplicate notes of notes you already have, you'll
> still end up with something omnitetrachordal. Any omnitetrachordal
> scale can be laid out this way, so at least part of the mystique of
> omnitetrachordal scales is laid to rest.

This is good. I agree with all the above stuff, and it seems like a useful characterization of all omintetrachordal scales. The only thing that would make it simpler or more useful is a better understanding of exactly how it's possible to be "clever enough" to hit duplicate notes.

Keenan

On Tue, Sep 6, 2011 at 5:30 AM, Keenan Pepper <keenanpepper@...> wrote:
>
> --- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> But with the 4th shift, you've shifted a total of more than 9/8, which your formula said not to do. There would have to be some kind of exception for the case where the new note (in this case C#) is coincident with an existing one.

I thought I had described that in one of the threads I linked you to,
but at any rate my full rationale is at the bottom of the message
you're replying to. I did this a long time ago so I have to remember
what all of the nuances and exceptions are now.

> > > Tetracot[13] LLLssLLLssLss
> > > As 27edo steps, 3 6 9 10 11 14 17 20 21 22 25 26 27
> >
> > This one's pretty trippy. If we're considering the C-F-Bb-C MOS to be
> > the subperiod, it can be thought of using two generators - 3\27 and
> > 1\27. Perhaps it'll be clearer if I explain where the formula came
> > from.
> >
> > If you think about it, if you start with C-F-Bb-C and just start
> > shifting things over (not necessarily using the same shift size each
> > time), and make sure that the top note never goes over C, you'll end
> > up with something omnitetrachordal by definition. If you shift things
> > in a clever enough way that although you do go over C, the things that
> > you hit are exact duplicate notes of notes you already have, you'll
> > still end up with something omnitetrachordal. Any omnitetrachordal
> > scale can be laid out this way, so at least part of the mystique of
> > omnitetrachordal scales is laid to rest.
>
> This is good. I agree with all the above stuff, and it seems like a useful characterization of all omintetrachordal scales. The only thing that would make it simpler or more useful is a better understanding of exactly how it's possible to be "clever enough" to hit duplicate notes.

One way is that if you're using only one generator, then make sure
that it's some equal division of the smallest note in the sub-period
MOS. There might be more to it though.

What I really want to do is start understanding all of this in terms
of higher-dimensional periodicity blocks though, because we're talking
about 3+ generators here. If you use the C-F-Bb-C MOS, that's two
units out along the 4/3 axis, and then if you shift it 3 times by a
tempered 25/24, that's 3 units out on the 25/24 axis, thus giving you
a parallelogram with a total area of 6. Reconfiguring the basis in
terms of 3/1 and 5/1 should yield some kind of more intuitive
periodicity block with obvious unison vectors.

The case when you hit "duplicate" notes can then, in that paradigm, be
viewed as two of the generators being related to one another, i.e.
putting you at rank 2. Figuring all of the nuances and implications of
that out is what I'm having trouble understanding.

-Mike