Yahoo Tuning Groups Ultimate Backup tuning "Flipping a monzo"

I've looked for answers to the following questions, but I don't see
them. I was wondering if anyone can help me understand wedgies and
monzos.

1. What are the rules for "flipping a monzo" to create a wedgie?

2. What, practically speaking, can I learn from a wedgie?

I'll write up what I learn and add it to the wedgie page on the wiki,
with examples.

Let me unpack those two questions a little bit.

==========

I ask #1 because it looks like "flipping" may be easier than
multiplying-out-the-(n,m)-shuffles stuff going on on the Monzo page --
but the flipping process isn't clear, because I see
similar-but-different flipping going on.

For instance, on the Wiki page for the Schismatic family (
http://xenharmonic.wikispaces.com/Schismatic+family ), I read the
following: "Its [the schisma, 32805/32768] monzo is |-15 8 1>, and
flipping that yields <<1 -8 -15|| for the wedgie." Reverse the
exponent order, flip the outer signs, keep the middle sign. But for,
say, the Wuerschmidt comma, I read: "Its monzo is |17 1 -8>, and
flipping that yields <<8 1 -17|| for the wedgie." Reverse the exponent
order, but keep all the signs.

So, is there a shortcut? Does it also work for 7-limit and higher commas?

Also, is this a reversible process? In other words, if I have a
wedgie, can I flip it to find the monzo of the comma(s) that's being
tempered?

==========

I ask #2 because I can make some sense of some wedgies, but not
others. Here are a few examples to show you the level of understanding
I have.

-----
WEDGIES WITH A 1 IN THE (2,n) PLACE.
-----

These are finally straightforward to me, at least with respect to the
(2,n) values.

Here's what the wiki page says about the schisma's wedgie, <<1 -8 -15||.

"This tells us the generator is a fifth and that we will need eight
fourths in succession to reach the pitch class of a major third. In
fact, 10 = (4/3)^8 * 32805/32768."

Following the wedgie page, and using "Schismatic" where it uses
"meantone", Schismatic(2,3) = 1, Schismatic(2,5) = -8, and
Schismatic(3,5) = -15. In other words, the first two numbers have a
period of 2, and the third has a period of 3. (I'm not sure why the
third one is useful, as we'll see.)

A. This means that 1 generator (on a period of 2) gets you to a
multiple of 3, so the generator must be (or, more precisely, I think,
must be reducible to) a 3/2.

B. -8 generators (on a period of 2) gets you to a multiple of 5 in the
numerator. The wiki uses 8 fourths up instead of 5 fourths down, but
it's the same thing under octave equivalence: 4/3^8 = 2^16 / 3^8, and
3/2^-8 = 2^8 / 3^8; these are both, when I multiply them by the
schisma, 5 * (2 to some power).

C. I don't understand the third number. It looks like -15 generators
on a period of 3 gets you to a multiple of 5 in the numerator. This
means ((3/2)^-15) * schisma = (2^15 / 3^15) * schisma = 5/2187. This
is 5 * 3^-6, and since 3 is the period, maybe I can ignore it the same
way I ignored powers of two above. The thing is, I don't understand
what to do with it. In some sense, we have a tritave-equivalent 5/3, I
guess. Since -8 fifths = -5 octaves + a 5/4, and -15 fifths = 2*(-8
fifths) + 1 fifth, it seems that we could also say this is -10 octaves
+ 2 5/4 + 1 fifth. I tried this in 12-tET, which tempers the schisma,
and it works: from C, it's an Eb. But Eb in 12-tET is a tempered 8/5,
so I'm not sure what that has to do with prime 3.

Anyway, none of this has anything to do with how the tempering is
done, of course, but if I temper a scale using Scala or an EDO
mapping, I know the relationships of my five-limit intervals. 12EDO
tempers the schisma, so I can test this on familiar terrain, and it
works. (I have all the details written down, and will use them on the
wiki if I do a write-up, but I won't bother you with them now.)

The schisma is a 5-limit comma, so I don't expect there to be any
information about 7-limit mappings in the wedgie. I'd expect the
schisma to leave the 7/4 completely untempered. I'm not sure how it
would affect something like 7/6, where the 7 is untempered and the 3
is tempered, or if such an operation is even defined with a 5-limit
wedgie.

I've proved to myself that this basic procedure works for, e.g., the
Wuerschmidt comma, which has a 1 in the (2,5) place. I assume that if
I had a 7-limit comma, it would have a (2,7) place, and a 1 in that
place would also give me the generator. I did a little checking using
the Giribaldi wedgie, and the checking I did looked good. (Of course,
I don't know whether my check was complete.)

-----
WEDGIES WITHOUT A 1 IN THEM ANYWHERE
-----

These seem more complicated, but not always completely opaque. I think
I understand the kleisma, but I get lost on handling the diaschisma.

KLEISMA: <<6 5 -6|| -- "This tells us the generator is a minor
third, and that to get to the interval class of major thirds will
require five of these, and so to get to fifths will require six. In
fact, (6/5)^5 = 5/2 * 15625/15552."

The generator on this isn't immediately obvious to me, but I think I
understand one way to get it. The 6 and 5 in the wedgie are
tantalizing when the wiki tells me that the generator's a minor third,
but I think that what matters isn't that they're 6 and 5 -- what
matters is that they differ by 1.

5 generators gets you to 5*(2^m), and 6 generators gets you to
3*(2^n), so 5*2^m * g = 3*2^n. I don't care what the factors of 2 are,
so I can see that g = (3 / 5) * (2^x). The octave-equivalent generator
of 3/5 is, of course, 6/5. Other intervals follow; e.g., 9/8 = 3*3
(ignoring the twos because of octave equivalence) = 2*6 generator
steps up.

15 tempers out the kleisma, so I can check my work, and it checks out:
using < 15 24 35 |, 6/5 = 320 cents and 9/8 = 240 cents. 12 generators
= 12 * 320 = 3840 cents, which, modulo 1200, is 240 cents.

Are there other ways to get the generator?

------

Diaschisma: <<2 -4 -11|| -- "This tells us the period is half an
octave, the GCD of 2 and -4, and that the generator is a fifth. Three
periods gives 1800 cents, and decreasing this by two fifths gives the
major third."

This one completely messes me up.

It's not intuitive to me why the period should be divided by the GCD
of (2,3) and (2,5) -- and my understanding goes downhill from there.

Diaschisma(2,3) = 2. That means two generators should get us to
3*(2^n). But this seems to contradict the statement, "the generator is
a fifth." If the generator were a fifth, it seems like Diaschisma(2,3)
should equal 1.

Also, two fifths on a period of an octave gets you to 9/8, or 3^2, not
to 3 as the wedgie suggests. So it seems like the period should
actually be half of a fifth.

Diaschisma(2,5) = -4. That means going down four generators should get
us to 5*(2^m).

So 6/5 should be 2 steps up (for the 3 in 3*2 -- we're ignoring the 2)
and -4 steps (for the 5) down (because the 5 is in the denominator),
or 2 - -4 = 6 steps up. But with the fifth as a generator in 12-tET
(which tempers out the diaschisma), 6 steps up gets me the tritone.

So let's say I think the wiki's wrong (which I don't), and I want to
determine what the generator is myself. Using the same logic as I did
with the kleisma, 2 generators gets you to 3*(2^n), and -4 gets you to
5*(2^m), so 5*(2^m) * g^6 = 3*(2^n). That means g^6 = (3/5)*(2^x), or
g = (3/5)^-6. Somehow I doubt it -- we've gone from having primes to
being in irrational number territory.

So what do I do? Take the GCD for Diaschisma(2,3) and Diaschisma(2,5)
and, in all my calculations, divide it by that number?

That means one generator gets to 3 (which at least tells me the
generator is a fifth), and two generators down gets to 5.

9/8 = two generators up. That works: 12-tET tempers the diaschisma,
and two fifths = C to G to D.

6/5 = 1 generator - -2 generators = 3 generators. That doesn't work:
In 12-tET, three fifths = C to G to D to A, but it should be Eb.

I'm clearly all kinds of screwed up. Why?

-----
WEDGIES WITH A 1, BUT NOT IN A (2,n) PLACE
-----

Porcupine: <<3 5 1|| -- "This tells us the generator is a minor whole
tone, the 10/9 interval, and that three of these add up to a fourth,
with two more giving the minor sixth. In fact, (10/9)^3 = 4/3 *
250/243, and (10/9)^5 = 8/5 * (250/243)^2."

This is weird: In previous calculations, the first number in the
wedgie gave me the number of steps to reach 3*(2^n). This time, I'm
being told that it's the number of steps to reach (2^n)/3. And the
second number was the steps needed to reach 5*(2^m), but this time
it's the number of steps to reach (2^m)/5. Why?

I also don't see why the generator is 10/9. The 1 in Porcupine(3,5)
makes it reasonable that the generator would be related to both 3 and
5, but why to 3 *squared* and 5?

If I didn't have the wiki's statement in front of me, here's what I
would guess this wedgie means.

By definition, Porcupine(2,3) = 3, Porcupine(2,5) = 5, and Porcupine(3,5) = 1.

The 1 in Porcupine(3,5) makes me think that it's the generator:
analogous to the case where schismatic(2,3) = 1 and one step gets you
to 3*(2^n), Porcupine(3,5) = 1 makes it look like 1 generator step
gets you to 5*(3^n). Now, I wonder whether the generator is supposed
to be based on a period of 2, i.e., octave-equivalent, or based on a
period of 3, i.e., tritave-equivalent. It doesn't matter, though,
since either way it has to be 5/3.

Now, 3 generator steps should get me to 3*(2^n), and 5 generator steps
should get you to 5*(2^n). 15 EDO tempers Porcupine, so using val <15
24 35| I get the generator 5/3 = 880 cents, 3 steps gets me to 240
cents and 5 steps gets me to 800 cents. Clearly, I'm wrong again.

So, what *is* right?

===================

And on top of all that, what obvious stuff am I leaving out? :)

My degree is in engineering, by the way, not mathematics. Hand-waving
is fine, as long as it's deterministic. :)

Thanks very much,
Jake

Jake Freivald <jdfreivald@...> wrote:
> I've looked for answers to the following questions, but I
> don't see them. I was wondering if anyone can help me
> understand wedgies and monzos.
>
> 1. What are the rules for "flipping a monzo" to create a
> wedgie?

It's the complement operation from exterior algebra.

This is our old reference that still comes out top in a
search:

http://gaupdate.wordpress.com/2010/05/12/jbrowne-book-and-mathematica-appl-for-grassmann-alg/

> 2. What, practically speaking, can I learn from a wedgie?

It's one way of representing the mapping of a regular
temperament, without choosing a specific tuning, or even
period-generator pair. One thing it tells you that you
couldn't get any easier way is the degree of torsion but
you don't need the complement for that.

> I ask #1 because it looks like "flipping" may be easier
> than multiplying-out-the-(n,m)-shuffles stuff going on on
> the Monzo page -- but the flipping process isn't clear,
> because I see similar-but-different flipping going on.
>
> For instance, on the Wiki page for the Schismatic family (
> http://xenharmonic.wikispaces.com/Schismatic+family ), I
> read the following: "Its [the schisma, 32805/32768] monzo
> is |-15 8 1>, and flipping that yields <<1 -8 -15|| for
> the wedgie." Reverse the exponent order, flip the outer
> signs, keep the middle sign. But for, say, the
> Wuerschmidt comma, I read: "Its monzo is |17 1 -8>, and
> flipping that yields <<8 1 -17|| for the wedgie." Reverse
> the exponent order, but keep all the signs.

I'm guessing that's a reliable recipe for the rank 2
5-limit case.

> So, is there a shortcut? Does it also work for 7-limit
> and higher commas?

There's no shortcut, but there are ways of abstracting it.
One comma in the 7-limit defines a rank 3 temperament
class. If you want to stay in rank 2 beyond the 5-limit,
you have to deal with multivectors either way -- or ignore
wedgies altogether, which is an option.

> Also, is this a reversible process? In other words, if I
> have a wedgie, can I flip it to find the monzo of the
> comma(s) that's being tempered?

Yes. Strictly speaking, it isn't perfectly reversible
because the sign may change. Applying the complement 4
times always gets you back to where you started.

> -----
> WEDGIES WITH A 1 IN THE (2,n) PLACE.
> -----

The (2,n) place tells you the number of generators to a
fifth, then? This can be useful because it helps you guess
what multiple of 12-equal you can use, if you want to use
such a thing.

> C. I don't understand the third number. It looks like -15
> generators on a period of 3 gets you to a multiple of 5
> in the numerator. This means ((3/2)^-15) * schisma =
> (2^15 / 3^15) * schisma = 5/2187. This is 5 * 3^-6, and
> since 3 is the period, maybe I can ignore it the same way
> I ignored powers of two above. The thing is, I don't
> understand what to do with it. In some sense, we have a
> tritave-equivalent 5/3, I guess. Since -8 fifths = -5
> octaves + a 5/4, and -15 fifths = 2*(-8 fifths) + 1
> fifth, it seems that we could also say this is -10 octaves
> + 2 5/4 + 1 fifth. I tried this in 12-tET, which tempers
> the schisma, and it works: from C, it's an Eb. But Eb in
> 12-tET is a tempered 8/5, so I'm not sure what that has
> to do with prime 3.

That number involves periods rather than generators.

> The schisma is a 5-limit comma, so I don't expect there
> to be any information about 7-limit mappings in the
> wedgie. I'd expect the schisma to leave the 7/4
> completely untempered. I'm not sure how it would affect
> something like 7/6, where the 7 is untempered and the 3
> is tempered, or if such an operation is even defined with
> a 5-limit wedgie.

What you get is a rank 3 temperament where the 7:1 is
orthogonal to the schismatic/helmholtz plane. It's
numerically like this:

http://x31eq.com/cgi-bin/rt.cgi?ets=53_171_130&limit=7

It looks like the 7:1 is left pure, but you could have
fixed it with 7:6 pure instead. (Maybe it doesn't matter
because schismatic is already so close to pure.)

> -----
> WEDGIES WITHOUT A 1 IN THEM ANYWHERE
> -----

Finding the generator isn't obvious. The best thing is to
follow an algorithm to find its optimal tuning. If you
want a rational representative, you can look for ratios
that approximate the optimal tuning.

> Diaschisma: <<2 -4 -11|| -- "This tells us the period
> is half an octave, the GCD of 2 and -4, and that the
> generator is a fifth. Three periods gives 1800 cents, and
> decreasing this by two fifths gives the major third."
>
> This one completely messes me up.
>
> It's not intuitive to me why the period should be divided
> by the GCD of (2,3) and (2,5) -- and my understanding
> goes downhill from there.

You always divide the period by the GCD of the octave
equivalent mapping. For a 5-limit wedgie,
those first two entries are the octave equivalent wedgie.

> Diaschisma(2,3) = 2. That means two generators should get
> us to 3*(2^n). But this seems to contradict the
> statement, "the generator is a fifth." If the generator
> were a fifth, it seems like Diaschisma(2,3) should equal
> 1.

That's because the octave equivalent wedgie isn't the same
as the generator mapping. You need to divide the former
through by its GCD.

> Also, two fifths on a period of an octave gets you to
> 9/8, or 3^2, not to 3 as the wedgie suggests. So it seems
> like the period should actually be half of a fifth.

No, the period's half of an octave because the period has
to divide an octave. That's how it's defined.

>
> -----
> WEDGIES WITH A 1, BUT NOT IN A (2,n) PLACE
> -----
>
> Porcupine: <<3 5 1|| -- "This tells us the generator is
> a minor whole tone, the 10/9 interval, and that three of
> these add up to a fourth, with two more giving the minor
> sixth. In fact, (10/9)^3 = 4/3 * 250/243, and (10/9)^5 =
> 8/5 * (250/243)^2."
>
> This is weird: In previous calculations, the first number
> in the wedgie gave me the number of steps to reach
> 3*(2^n). This time, I'm being told that it's the number
> of steps to reach (2^n)/3. And the second number was the
> steps needed to reach 5*(2^m), but this time it's the
> number of steps to reach (2^m)/5. Why?

The first number gives you the number of steps to a fifth
in octave-equivalent terms. Fourths and fifths are also
equivalent because the generator could be either ascending
or descending -- the wedgie doesn't specify it.

> I also don't see why the generator is 10/9. The 1 in
> Porcupine(3,5) makes it reasonable that the generator
> would be related to both 3 and 5, but why to 3 *squared*
> and 5?

You can extract the generator mapping as <0 3 5]. Then you
can write 10/9 as a ket vector: [1 -2 1>. The number of
generators to a 10/9 is the bracket of these two: <0 3 5|1
-2 1> = -3*2 + 5*1 = 5-6 = -1. (For a 10/9 generator, the
mapping should have been <0 -3 -5]. The wedgie doesn't
directly tell you that.)

> My degree is in engineering, by the way, not mathematics.
> Hand-waving is fine, as long as it's deterministic. :)

If you happen to know linear algebra, you don't need
wedgies.

Graham

"Flipping a monzo"

🔗Jake Freivald <jdfreivald@...>

🔗Graham Breed <gbreed@...>

🔗genewardsmith <genewardsmith@...>

🔗genewardsmith <genewardsmith@...>