Yahoo Tuning Groups Ultimate Backup tuning-math From Commas to Generators

"Kalle Aho" <kalleaho@mappi.helsinki.fi> writes:

> Hi!
>
> Let's say I have a set of commas that define a 2D temperament. How
> can I get the period-generator-mappings of the primes from it?
>
> I believe they can be found by manipulating a set of linear
> equations but I guess you folks have a more straightforward method.
>
> Maybe a simple 7-limit example would be nice, thank you.

What a coincidence, I just worked this out myself the other night.
And all my notes are at home... so let's see if I can reconstruct.

I'll give you your 7-limit, but I'll take away your 5. That is, I'll
use the primes 2, 3, and 7. Using four primes (2, 3, 5, and 7) puts
you in a higher dimension and complicates things, but I'm guessing the
principles are similar. If not, we'll hear about it, I'm sure.

In cents, these are 1200, 1901.96, and 3368.83.

Here's our comma: 64/63 = 2^6 * 3^-2 * 7^-1 = | 6 -2 -1 >. In cents,
it's 27.26.

To temper this comma out, we make the 2 smaller and the 3 and 7 larger
(since the signs of the exponents are +, -, and -). The amount we
temper by is

2 smaller by 27.26 * ln(2) / ln (64*63) = 2.28 cents
3 larger by 27.26 * ln(3) / ln (64*63) = 3.61 cents
7 larger by 27.26 * ln(7) / ln (64*63) = 6.39 cents

So the tempered values are

2 : 1200 - 2.28 = 1197.72
3 : 1901.96 + 3.61 = 1905.56
7 : 3368.83 + 6.39 = 3375.22

To verify: 6 * 1197.72 - 2 * 1905.56 - 1 * 3375.22 = 0 .

All the above is following Paul Erlich's new paper.

Now to get the period and generator. The period is easy. Note the
exponents of 3 and 7 in our comma: -2 and -1. The greatest common
divisor of (the absolute value of) these is 1, so the period is the
tempered octave, 1197.72, divided by 1. The tempered octave is the
period in this case.

For the generator, we have two equations of the form

(tempered interval) = (integer) * period + (integer) * generator

namely

1905.56 = n3 * 1197.72 + m3 * g
3375.22 = n7 * 1197.72 + m7 * g

For m3 and m7 use the exponents of 7 and 3, respectively, in the
comma, with the sign of one changed: -1 and 2. (I think it's
legitimate to change the sign of either.)

1905.56 = n3 * 1197.72 + -1 * g
3375.22 = n7 * 1197.72 + 2 * g

Eliminate g to get

(2*1905.56+3375.22)/1197.72 = 6 = 2 * n3 + n7

Different choices of integers for n3 and n7 give different generators,
all related by period equivalence (note you don't need n7 to compute
g, directly, but you do need the relationship between n3 and n7 to
discover which integers are valid values for n3, in general -- though
in this case all integers are):

n3 n7 = 6 - 2 * n3 g = n3 * 1197.72 - 1905.56
1 4 -707.84
2 2 489.88
3 0 1687.61

So we can use e.g. period = 1197.72, generator = 489.88.

I think I got that right.

- Rich Holmes

🔗monz <monz@tonalsoft.com>

--- In tuning-math@yahoogroups.com, Rich Holmes<rsholmes@m...> wrote:

> I'll give you your 7-limit, but I'll take away your 5.
> That is, I'll use the primes 2, 3, and 7. Using four
> primes (2, 3, 5, and 7) puts you in a higher dimension
> and complicates things, but I'm guessing the principles
> are similar. If not, we'll hear about it, I'm sure.
>
> In cents, these are 1200, 1901.96, and 3368.83.
>
> Here's our comma: 64/63 = 2^6 * 3^-2 * 7^-1 = | 6 -2 -1 >.

i think it would be really good to standardize the monzo
notation. any objections to writing that as [6 -2, 0 -1> ?

i suppose the straight pipe symbol | is preferred to
the square bracket [ on the left, so as to make the
bra-ket implication clear ... but i really like to see
the comma-mark after the 3-exponent, as well as the zero
holding the 5-exponent place.

-monz

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, Rich Holmes<rsholmes@m...> wrote:

> "Kalle Aho" <kalleaho@m...> writes:
>
> > Hi!
> >
> > Let's say I have a set of commas that define a 2D temperament.
How
> > can I get the period-generator-mappings of the primes from it?
> >
> > I believe they can be found by manipulating a set of linear
> > equations but I guess you folks have a more straightforward
method.
> >
> > Maybe a simple 7-limit example would be nice, thank you.
>
> What a coincidence, I just worked this out myself the other night.
> And all my notes are at home... so let's see if I can reconstruct.
>
> I'll give you your 7-limit, but I'll take away your 5. That is,
I'll
> use the primes 2, 3, and 7. Using four primes (2, 3, 5, and 7) puts
> you in a higher dimension and complicates things, but I'm guessing
the
> principles are similar.

You're right.

> If not, we'll hear about it, I'm sure.
>
> In cents, these are 1200, 1901.96, and 3368.83.
>
> Here's our comma: 64/63 = 2^6 * 3^-2 * 7^-1 = | 6 -2 -1 >. In
cents,
> it's 27.26.
>
> To temper this comma out, we make the 2 smaller and the 3 and 7
larger
> (since the signs of the exponents are +, -, and -). The amount we
> temper by is
>
> 2 smaller by 27.26 * ln(2) / ln (64*63) = 2.28 cents
> 3 larger by 27.26 * ln(3) / ln (64*63) = 3.61 cents
> 7 larger by 27.26 * ln(7) / ln (64*63) = 6.39 cents
>
> So the tempered values are
>
> 2 : 1200 - 2.28 = 1197.72
> 3 : 1901.96 + 3.61 = 1905.56
> 7 : 3368.83 + 6.39 = 3375.22
>
> To verify: 6 * 1197.72 - 2 * 1905.56 - 1 * 3375.22 = 0 .
>
> All the above is following Paul Erlich's new paper.

Glad someone's following it :) :) :)

> Now to get the period and generator. The period is easy. Note the
> exponents of 3 and 7 in our comma: -2 and -1. The greatest common
> divisor of (the absolute value of) these is 1, so the period is the
> tempered octave, 1197.72, divided by 1. The tempered octave is the
> period in this case.

Yup!

> For the generator, we have two equations of the form
>
> (tempered interval) = (integer) * period + (integer) * generator
>
> namely
>
> 1905.56 = n3 * 1197.72 + m3 * g
> 3375.22 = n7 * 1197.72 + m7 * g
>
> For m3 and m7 use the exponents of 7 and 3, respectively, in the
> comma, with the sign of one changed: -1 and 2.

Yes, this is what I mentioned to Kalle on the tuning list.

> (I think it's
> legitimate to change the sign of either.)
>
> 1905.56 = n3 * 1197.72 + -1 * g
> 3375.22 = n7 * 1197.72 + 2 * g
>
> Eliminate g to get
>
> (2*1905.56+3375.22)/1197.72 = 6 = 2 * n3 + n7
>
> Different choices of integers for n3 and n7 give different
generators,
> all related by period equivalence (note you don't need n7 to compute
> g, directly, but you do need the relationship between n3 and n7 to
> discover which integers are valid values for n3, in general --
though
> in this case all integers are):
>
> n3 n7 = 6 - 2 * n3 g = n3 * 1197.72 - 1905.56
> 1 4 -707.84
> 2 2 489.88
> 3 0 1687.61
>
> So we can use e.g. period = 1197.72, generator = 489.88.
>
> I think I got that right.

Probably. What do Gene and Graham get?

Hey, Rich, BTW, great to have you on board, we're a pretty rarefied
group so every new member makes us all much richer.

🔗Rich Holmes <rsholmes@mailbox.syr.edu>

"monz" <monz@tonalsoft.com> writes:

> --- In tuning-math@yahoogroups.com, Rich Holmes<rsholmes@m...> wrote:
>
>
> > Here's our comma: 64/63 = 2^6 * 3^-2 * 7^-1 = | 6 -2 -1 >.
>
>
> i think it would be really good to standardize the monzo
> notation. any objections to writing that as [6 -2, 0 -1> ?
>
> i suppose the straight pipe symbol | is preferred to
> the square bracket [ on the left, so as to make the
> bra-ket implication clear ... but i really like to see
> the comma-mark after the 3-exponent, as well as the zero
> holding the 5-exponent place.

Well, dropping the unused exponents is something I've seen done.
Granted it has potential for confusion, if you don't make it clear
what you're doing; but I think it can also simplify discussion and
notation <-> computer code conversion in cases like this where
e.g. we're considering a 5-less system. Certainly the inclusion of
the 5-exponent in this case would obscure the fact that we're talking
about a 3-dimensional lattice. Really, to me it's neither clearly
good nor clearly bad to drop the unused exponent(s).

I'd never noticed use of the comma after the 3-exponent before. I'm
curious as to its function.

I'm used to bra-ket notation that uses a vertical line, but I suppose
actually one could argue in favor of bracket instead of pipe, since
we're often reading these things off computer screens and the contents
of the vector often are integers, and [ is arguably less likely than |
to be confused with 1, at least in some fonts.

- Rich Holmes

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...> wrote:

> Probably. What do Gene and Graham get?

I get confused; you are trying to find period and generator for a planar
(3D) temperament? 64/63 bridges the 5-limit to the 7-limit, so
generators can be taken as {2,3,5}; in other words, we are using a
retuned 64/9 for 7, so we want a sharp fifth to keep the 7 from being
too sharp. What 5 does is irrelevant, so we may as well leave it pure.

On the other hand, I suppose you could be asking for period and
generator for the corresponding no-fives system, which has a period of
an octave and a generator a fourth or fifth. You want two fourths to
be an approximate 7/4, so the fourths are flat and the fifths sharp,
but you know all this. Also a Pythagorean third of 81/64 is equated to
9/7. and if we take two of these, we get close to a 5/3, depending on
the tuning, and adding in (5/3)/(9/7)^2 = 245/243 to the commas makes
it into superpyth, which seems like a reasonable way to bring 5 into
the picture.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
>
> > Probably. What do Gene and Graham get?
>
> I get confused; you are trying to find period and generator for a
planar
> (3D) temperament?

No -- you must have missed all the discussion about prime 5 being
omitted and this being a temperament of {2,3,7}-JI, thus a rank-two
(2D) temperament is the desired result.

> On the other hand, I suppose you could be asking for period and
> generator for the corresponding no-fives system,

Yes, Rich couldn't have been more explicit that this is what he had
in mind, or at least so it seemed to me.

> which has a period of
> an octave and a generator a fourth or fifth. You want two fourths to
> be an approximate 7/4, so the fourths are flat and the fifths sharp,
> but you know all this.

I asked you because I thought you'd be a good candidate to both
verify the exact results Rich got, and to relate his method to any
more-abstract formulations you might have put forth at some point for
finding generators.

🔗Paul Erlich <perlich@aya.yale.edu>

--- In tuning-math@yahoogroups.com, "monz" <monz@t...> wrote:
>
> --- In tuning-math@yahoogroups.com, Rich Holmes<rsholmes@m...>
wrote:
>
>
> > I'll give you your 7-limit, but I'll take away your 5.
> > That is, I'll use the primes 2, 3, and 7. Using four
> > primes (2, 3, 5, and 7) puts you in a higher dimension
> > and complicates things, but I'm guessing the principles
> > are similar. If not, we'll hear about it, I'm sure.
> >
> > In cents, these are 1200, 1901.96, and 3368.83.
> >
> > Here's our comma: 64/63 = 2^6 * 3^-2 * 7^-1 = | 6 -2 -1 >.
>
>
> i think it would be really good to standardize the monzo
> notation. any objections to writing that as [6 -2, 0 -1> ?

The problem with this is that it makes remembering the how to do the
complement operation harder, if you know you're dealing with a system
without prime 5. Didn't you suggest the notation [6 -2, * -1> at some
point for this purpose?

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Kalle Aho <kalleaho@mappi.helsinki.fi>

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Kalle Aho" <kalleaho@m...> wrote:

> > First can you explain what you mean by 2D?
>
> Formerly known as "linear" temperament. :)

There are different ways to do this, and probably you would like the
ways other people do it better than my system. I first find the wedgie
for the temperament, and then find what I call the "subgroup vals"
belonging to it, which are vals defined in terms of the coordinates of
the wedgie which send one of the primes to zero. For instance, for
7-limit wedgies, this would take

<<l[1] l[2] l[3] l[4] l[5] l[6]||

and from it derive

<[0, -l[1], -l[2], -l[3]|, <l[1], 0, -l[4], -l[5]|,
<l[2], l[4], 0, -l[6]|, <l[3], l[5], l[6], 0]|

We want two independent reduced vals, one of which sends 2 to the
number of periods in an octave, and the other of which sends 2 to
zero. We can get this by taking the Hermite normal form of the above.
We then can find optimal generators using, for instance, rms
optimization, and then can standarize our result by making the
generator larger than one but less than sqrt(period).

This describes what my Maple program does, but it isn't the only way
to solve the problem, and for people not using a computer algebra
package, not likely to prove a good system. However, since I have such
a system it is very fast for me.

Simply by inspecting the wedgie, you can find a lot of relevant
information; if we have n primes in the prime limit, then the first n
entries in the wedgie define the generator val, and their gcd defines
the period. Thus from the wedgie

<<2 -4 -4 -11 -12 2||

we can see that <0 1 -2 -2| is the generator val, up to sign, and the
period is defined by the gcd of (2,-4,-4), or 2; in other words, a
half-octave. Hence the mapping will look like

[<2 * * *|, +-<0 1 -2 -2|]

Filling in the blanks involves finding a reduced size for the
generator, which can be done by hand, or coded.

🔗Ozan Yarman <ozanyarman@superonline.com>

Finally, I now understand how the exponent notation is written. But in order to distinguish primes, can I suggest something like this:

Septimal Comma = [ A6 B-2 D-1 >

Where the primes are represented by the letters of the latin alphabet.

Cordially,
Ozan

----- Original Message -----
From: Paul Erlich
To: tuning-math@yahoogroups.com
Sent: 08 Mart 2005 Salı 19:47
Subject: [tuning-math] Re: From Commas to Generators

🔗monz <monz@tonalsoft.com>

hi Rich,

--- In tuning-math@yahoogroups.com, Rich Holmes<rsholmes@m...> wrote:

> Well, dropping the unused exponents is something I've seen
> done. Granted it has potential for confusion, if you don't
> make it clear what you're doing; but I think it can also
> simplify discussion and notation <-> computer code conversion
> in cases like this where e.g. we're considering a 5-less system.
> Certainly the inclusion of the 5-exponent in this case would
> obscure the fact that we're talking about a 3-dimensional
> lattice. Really, to me it's neither clearly good nor clearly
> bad to drop the unused exponent(s).

yes, i can easily agree to everything you say here.

i've often used the form which drops unused exponents myself.
in talking about Boethius's theory, for instance, you
certainly don't want to include the unused exponents, because
his theory used prime-factors 2, 3, 19, and 499.

when i do this i make it a standard practice to label the
monzo clearly with the prime-factors involved, i.e.,
a monzo used to describe a ratio in Boethius would be
labeled a 2,3,19,499-monzo, and written [a b c d> where
those variables are the exponents of those four primes
in that sequence. it would be ridiculous to inlcude all
the unused prime-factors in this case. etc.

> I'd never noticed use of the comma after the 3-exponent
> before. I'm curious as to its function.

this was a proposal which i enthusiastically embraced
and put into the Encyclopedia definition.

http://tonalsoft.com/enc/monzo.htm

putting the comma-mark after 3 is a useful idea because
the exponent of 2 is so often ignored in many theoretical
concerns, and the comma-mark helps the reader to see
whether this is the case or not.

(notice that i'm always careful to write "comma-mark"
instead of just plain "comma" ... since "comma" is a word
loaded with musical tuning meaning, and monzos are often
used to describe "commas", it could become really confusing
in these discussions about monzos.)

putting subsequent comma-marks after every third exponent
simply helps the eye to pinpoint which prime-factor an
exponent belongs to, and it just so happens by coincidence
(or is it a coincidence?) that for a while in the prime series,
every third prime-factor after 3 is somewhat important in
the history / theory / practice of rational tunings. viz,
11, 19, 31, 43.

> I'm used to bra-ket notation that uses a vertical line,
> but I suppose actually one could argue in favor of bracket
> instead of pipe, since we're often reading these things off
> computer screens and the contents of the vector often are
> integers, and [ is arguably less likely than | to be confused
> with 1, at least in some fonts.

i prefer to use the bracket for exactly that reason.
when a complet bra-ket is notated, then fine, the pipe
symbol is OK. by for a bra or ket by itself, i think
the bracket is better.

-monz

🔗monz <monz@tonalsoft.com>

hi Paul,

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:

> --- In tuning-math@yahoogroups.com, "monz" <monz@t...> wrote:
> >
> >
> > i think it would be really good to standardize the monzo
> > notation. any objections to writing that as [6 -2, 0 -1> ?
>
> The problem with this is that it makes remembering the how
> to do the complement operation harder, if you know you're
> dealing with a system without prime 5. Didn't you suggest
> the notation [6 -2, * -1> at some point for this purpose?

hmm, this is funny ... yes, actually, that was exactly
how i wrote that post in the first place ... but the asterisk
is supposed to be a "wildcard" which may represent any integer
and it wouldn't affect the results, and i reasoned that since
5 was absent then that exponent should just be zero.

but i certainly preferred the notation with both the
comma-mark and the asterisk. then i guess this is valid
for the usage in this thread?

-monz

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗monz <monz@tonalsoft.com>

hi Ozan,

--- In tuning-math@yahoogroups.com, "Ozan Yarman" <ozanyarman@s...>
wrote:

> Finally, I now understand how the exponent notation is
> written. But in order to distinguish primes, can I suggest
> something like this:
>
> Septimal Comma = [ A6 B-2 D-1 >
>
> Where the primes are represented by the letters of the
> latin alphabet.

hmm, that's an interesting idea.

here's the mapping to the whole alphabet:

A = 2
B = 3
C = 5
D = 7
E = 11
F = 13
G = 17
H = 19
I = 23
J = 29
K = 31
L = 37
M = 41
N = 43
O = 47
P = 53
Q = 59
R = 61
S = 67
T = 71
U = 73
V = 79
W = 83
X = 89
Y = 97
Z = 101

of course, we run into a slight problem with composers
and theorists who have used prime-factors higher than 101,
and as i just posted, they do exist. a simple solution is
to use 2-digit letters after Z:

AA = 103
AB = 107
AC = 109
AD = 113
AE = 127
AF = 131
AG = 137
AH = 139
.
.
.
AZ = 239
BA = 241
.
.
.
BZ = 397
CA = 401

but it's not likely that anyone's going to remember the
mapping of more than just the first few ... so in the
end it's better to just use the prime-factors themselves,
i.e.:

2^a 3^b 5^c 7^d ... p^n

instead of [a b c d ...n>.

for me, the two best methods are the ones i always use:
the comma-marks and asterisk-wildcard if the prime-series
covers a spread that is not to far (i.e., most cases),
and the explicit prime-factor label for those cases in which
the spread is far (as with Boethius's 3,19,499 system).

-monz

🔗Ozan Yarman <ozanyarman@superonline.com>

Hi Monz,

If the intention is to skip the primes in between primes, then I suggest the following usage:

Septimal Comma = [ 6 -2, -1 >

Where the comma-mark (,) indicates one skip in the prime sequence and a semi colon (;) indicates indefinite amount of skips up to the next prime.

So, Boethius's prime factors can be represented like this:

2^a; 19^b; 499^c

----- Original Message -----
From: monz
To: tuning-math@yahoogroups.com
Sent: 09 Mart 2005 Çarşamba 11:37
Subject: [tuning-math] Re: From Commas to Generators

hi Ozan,

--- In tuning-math@yahoogroups.com, "Ozan Yarman" <ozanyarman@s...>
wrote:

hmm, that's an interesting idea.

here's the mapping to the whole alphabet:

A = 2
B = 3
C = 5
D = 7
E = 11
F = 13
G = 17
H = 19
I = 23
J = 29
K = 31
L = 37
M = 41
N = 43
O = 47
P = 53
Q = 59
R = 61
S = 67
T = 71
U = 73
V = 79
W = 83
X = 89
Y = 97
Z = 101

AA = 103
AB = 107
AC = 109
AD = 113
AE = 127
AF = 131
AG = 137
AH = 139
.
.
.
AZ = 239
BA = 241
.
.
.
BZ = 397
CA = 401

but it's not likely that anyone's going to remember the
mapping of more than just the first few ... so in the
end it's better to just use the prime-factors themselves,
i.e.:

2^a 3^b 5^c 7^d ... p^n

instead of [a b c d ...n>.

-monz

🔗Ozan Yarman <ozanyarman@superonline.com>

I do not wish to approximate temperaments by any EDOs at this stage. That is the easy part once you have a temperament. I rather liked the methodology used by Rich (at least things don’t get messed up once in order).

Now that I have understood how to prime factorize 4000:3993, let me review step by step how to temper it out.

1. 4000:3993 = (2*2*2*2*2*5*5*5) / (3*11*11*11) = (2^5*5^3) * (3^-1*11^-3)

2. In other words, 4000:3993 = [5 -1 3 0 -3> where zero represents the absent prime number 7.

3. If the primes 2,3,5,11 are represented by generators g1, g2, g3 and g5 respectfully, then these would have the just ratios 2:1, 3:2, 5:4 and 11:10. The cent values for g1 is 1200.000, g2 is 701.9550, g3 is 386.3137 and g5 is 165.0042. I admit that cent calculations appeal to me more.

4. What I fail to comprehend as yet is by what amount the tempering ought to be done. Can someone please give me a few examples from this point forward on how to do it?

And Paul, you ask:

Hi Ozan,

I'm impressed that you're sensitive to a 3 cent change in the
intonation of one note. What means are you using to render and listen
to these intervals?

I’m using a SBLive audio card with Scala. The piano sample sound is ok for theoretical verification of intervals.

(1) How do you know it's 9801/8000 and not 49/40 that your ear is
guiding you towards?

I find 49/40 to be too low when the hammer hits the string in the first micro-second and 27/22 to be too high likewise. It must be a matter of timbral artifacts as Monz pointed out earlier.

And I think I understand why log(a^b) = b*log(a). The logarithm of, say, 100 (10^2) is 2, which is equal to this exponent times log 10 (1):

(log 10^2 = 2*log 10)

You say:

OK . . . let's take a step back. Do you normally consider interval
sizes in cents? Or in some other system of units? Would you agree
than an octave is a 2:1 ratio, and that a triple-octave is an 8:1
ratio? Or does this mystify you?

Not at all.

You say:

I wasn't thinking of a piano, because the scale in question will have
too many notes per octave to be expressible on a piano. A piano has
stretched overtones, which is why stretched octaves sound great on
it. However, many other instruments, such as bowed strings,
brass/wind instruments, and the human voice have harmonic overtones.
For such instruments, a stretched octave won't sound much less
discordant than a similarly compressed octave, based on my experience
and knowledge. But perhaps you have found otherwise? Perhaps you're
only speaking of *melodic* octaves, not harmonic (simultaneously-
sounding) ones?

To me, they are more or less the same.

>Tempering out the syntonic comma will result in the upper-most
>dimension (prime) to collapse,

What do you mean uppermost? The largest? How do you know this? What
is it about the syntonic comma that allows you to reach this
conclusion?

Oops, I did something wrong again, didn’t I? Oh, forget it…

>creating a temperament of 2 dimensions. So, the first dimension is
>the octave with 2:1 which will remain untouched for my purposes, and
>the second dimension is the tempered perfect fifth with the ratio
>3:2 and the third dimension is the tempered pure third with the
>ratio 5:4.

I thought you said 2 dimensions?

No longer! I will say `rank` from now on…

Cordially,

Ozan

🔗Kalle Aho <kalleaho@mappi.helsinki.fi>

--- In tuning-math@yahoogroups.com, Rich Holmes<rsholmes@m...> wrote:
> "Kalle Aho" <kalleaho@m...> writes:
>
> > Hi!
> >
> > Let's say I have a set of commas that define a 2D temperament.
How
> > can I get the period-generator-mappings of the primes from it?
> >
> > I believe they can be found by manipulating a set of linear
> > equations but I guess you folks have a more straightforward
method.
> >
> > Maybe a simple 7-limit example would be nice, thank you.
>
> What a coincidence, I just worked this out myself the other night.
> And all my notes are at home... so let's see if I can reconstruct.

<snip>

> So we can use e.g. period = 1197.72, generator = 489.88.
>
> I think I got that right.
>
> - Rich Holmes

That's very nice, Rich. Thank you for this demonstration! Now, I
would very much like to understand the general case. Gene already
demonstrated his method (thank you, Gene!) but that is just pretty
hard to understand. I hope others will demonstrate their methods
too.

Kalle

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Ozan Yarman" <ozanyarman@s...> wrote:

> 4. What I fail to comprehend as yet is by what amount the tempering
ought to be done. Can someone please give me a few examples from this
point forward on how to do it?

When there is just one comma, as in the case of the 4000/3993
temperament, TOP tuning is particularly easy. However, there are other
ways to optimize, and a method which is useful and easy to caculate is
least-squares, so I'll explain that. We have five primes and one
comma, and therefore four generators. You gave these as 2,3/2,5/4, and
11/10, but this isn't quite right, as 3/2 is equivalent to
(3/2)(4000/3993) = 2000/1331, and we also need a generator for the 7
stuff. If we set g1=2, g3=5/4, g5=11/10 as you do, we can add g4=7/4,
and now we have a complete set of generators.

In terms of these, we represent the primes by

2 ~ g1
3 ~ g1^2*g5^(-3)
5 ~ g1^2*g3
7 ~ g1^2*g4
11 ~ g1^3*g3*g5

The exponents of this allow us to write down a matrix mapping from
generators to primes:

[[1 2 2 2 3], [0 0 1 0 1], [0 0 0 1 0], [0 -3 0 0 1]]

The vectors in the above being columns of the matrix, and representing
the exponents of g1, g3, g4 and g5 respectively. Now we take a finite
list of the consonances we want to approximate, and if we like, a
weighting on each consonance. If the consonances are c_i and the
weights are w_i, we define a function of g1, g3, g4, and g5 by

S(g1,g3,g4,g5) = sum w_i (rep(c_i)-cents(c_i))^2

Here rep(c_i) is the representation of an interval in terms of these
generators which we can define from the above matrix map by
multiplying it by its monzo; for instance to find rep(7/6), we
multiply |-1 -1 0 1 0>M = [1 0 1 3], where M is the mapping matrix.
This tells us rep(7/6) = g1+g4+3g5.

If now we take the partial derivatives of S with respect to g1, g3, g4
and g5, we get *linear* equations, which we may solve to obtain a
unique optimal solution by this definition of "optimal".

For instance, suppose we constrain g1 to be exactly 1200 cents, and
then take {3, 5, 7, 9, 11, 11/5, 5/3, 7/5, 11/3, 9/7, 9/5, 7/3, 11/9,
11/7} to be our set of consonances, weighted equally. Then

g3 = 385.912, g4 = 968.915, g5 = 165.985

are the optimal tunings of the other generators.

🔗Ozan Yarman <ozanyarman@superonline.com>

Gene, widening the fifth by 4000/3993 is enough to temper all instances of this kleisma?

I cannot make the proper correlations between temperament and matrix yet. Since you are math-savvy, I'm sure you can explain. I'm still in the dark as to the cent values that should be added to or subtracted from consonant ratios. Please remember that I am not knowledged in equations and algebra as you are! Gentler introductions are in order.

Cordially,
Ozan
----- Original Message -----
From: Gene Ward Smith
To: tuning-math@yahoogroups.com
Sent: 10 Mart 2005 Perşembe 1:01
Subject: [tuning-math] Re: From Commas to Generators

--- In tuning-math@yahoogroups.com, "Ozan Yarman" <ozanyarman@s...> wrote:

> 4. What I fail to comprehend as yet is by what amount the tempering
ought to be done. Can someone please give me a few examples from this
point forward on how to do it?

In terms of these, we represent the primes by

2 ~ g1
3 ~ g1^2*g5^(-3)
5 ~ g1^2*g3
7 ~ g1^2*g4
11 ~ g1^3*g3*g5

The exponents of this allow us to write down a matrix mapping from
generators to primes:

[[1 2 2 2 3], [0 0 1 0 1], [0 0 0 1 0], [0 -3 0 0 1]]

S(g1,g3,g4,g5) = sum w_i (rep(c_i)-cents(c_i))^2

If now we take the partial derivatives of S with respect to g1, g3, g4
and g5, we get *linear* equations, which we may solve to obtain a
unique optimal solution by this definition of "optimal".

For instance, suppose we constrain g1 to be exactly 1200 cents, and
then take {3, 5, 7, 9, 11, 11/5, 5/3, 7/5, 11/3, 9/7, 9/5, 7/3, 11/9,
11/7} to be our set of consonances, weighted equally. Then

g3 = 385.912, g4 = 968.915, g5 = 165.985

are the optimal tunings of the other generators.

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Ozan Yarman <ozanyarman@superonline.com>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Ozan Yarman <ozanyarman@superonline.com>

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Ozan Yarman <ozanyarman@superonline.com>

I surmise what you just said stands valid for `temperament`, not `specifically tempering an interval`. In this case, the question was: Is the 12-tone scale I produced a correct example where 4000:3993 is tempered regardless of a `just intonation reference`? Or does one need a simple integer ratio to compare a comma against?

----- Original Message -----
From: Gene Ward Smith
To: tuning-math@yahoogroups.com
Sent: 11 Mart 2005 Cuma 22:41
Subject: [tuning-math] Re: From Commas to Generators

--- In tuning-math@yahoogroups.com, "Ozan Yarman" <ozanyarman@s...> wrote:

> You lost me again. I thought you said widening the fifth by
4000:3993 was enough to temper all instances of this kleisma.

Yes, but tempering is generally conceived of as an adjustment of just
intonation. You can take anything in just intonation, widen the fifth by
4000/3993, leave everything else the same, and 4000/3993 will be
tempered out. Hwever, you didn't start with something in just intonation.

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Ozan Yarman" <ozanyarman@s...>
wrote:

> I surmise what you just said stands valid for `temperament`, not
`specifically tempering an interval`. In this case, the question was:
Is the 12-tone scale I produced a correct example where 4000:3993 is
tempered regardless of a `just intonation reference`? Or does one
need a simple integer ratio to compare a comma against?

Your scale is quite close to a scale on 80-equal, namely

! ozan80.scl
80-et version of Ozan Yarman scale
12
!
135.000000
210.000000
345.000000
420.000000
495.000000
630.000000
705.000000
840.000000
915.000000
1050.000000
1125.000000
1200.000000

Since 80-et is a 4000/3993 system if regarded as the regular
temperament tempering out 176/175, 540/539, 896/891 and 1331/1323
your scale can reasonably be considered as a 4000/3993 scale.

It is a MOS for the 63&80 temperament, which has a generator of a
sharp fifth. I don't think this has been discussed before, but it
does have some no-fives potential, as 5 is complex (of course, 3 is
just the opposite.)

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> It is a MOS for the 63&80 temperament, which has a generator of a
> sharp fifth. I don't think this has been discussed before, but it
> does have some no-fives potential, as 5 is complex (of course, 3 is
> just the opposite.)

Incidentally, this "yarman" temperament, to suggest a possible name
for it, really makes more sense if you go up-limit, at least as far
as 13, if not higher. As a 13-limit temperament it has the mapping

[<1 1 -20 -6 -3 -1|, <0 1 38 15 11 8|]

if we take the generator to be a fifth. It has a TM comma basis of
{169/168, 352/351, 364/363, 540/539}. If I take a chain of fifths and
reduce by these commas, I get

13/12, 9/8, 13/11, 14/11, 4/3, 13/9, 3/2, 11/7, 22/13, 16/9, 21/11, 2

🔗Gene Ward Smith <gwsmith@svpal.org>

🔗Ozan Yarman <ozanyarman@superonline.com>

Ah, but you still have not given me an example where 4000/3993 alone is tempered out. Please, I implore you, explain to me once more, in rudimentary math and in layman terms.... How can I discern which simple integer just ratios I tempered by sharpening the fifths and forming a chain?

----- Original Message -----
From: Gene Ward Smith
To: tuning-math@yahoogroups.com
Sent: 14 Mart 2005 Pazartesi 3:57
Subject: [tuning-math] Re: From Commas to Generators

You need to interpret what you did to decide; you gave a chain of
sharp fifths, and the most reasonable interpretation of that, it seems
to me, is by way of a temperament in which, among other things,
4000/3993 vanishes. However, this is a rank 2 temperament, and I was
thinking of a temperament which tempered out 4000/3993 and nothing else.