Commas and generators

If a comma can be written in the form 2^a U^b V^c, where U and V are
consonant ratios of odd integers, then we have a period-generator pair
for the group generated by {2,U,V} which is in some sense natural. For
instance, 225/224 = 2^(-5) 5^2 (9/7), and this leads to the
temperament with octave period, generator a 5/4, and for which 14/9 ~
(5/4)^2. If for some reason you are enamored of the 5/4-5/4-9/7
augmented triad, this could be a grand system for you. Other commas
cannot be deconstructed in this way, but even without consonances we
can find factorizations such as 2^a 3^b U^c.

>If a comma can be written in the form 2^a U^b V^c, where U and V are
>consonant ratios of odd integers, then we have a period-generator pair
>for the group generated by {2,U,V} which is in some sense natural.

You lost me. Is this something new?

>For instance, 225/224 = 2^(-5) 5^2 (9/7), and this leads to the
>temperament with octave period, generator a 5/4, and for which 14/9 ~
>(5/4)^2. If for some reason you are enamored of the 5/4-5/4-9/7
>augmented triad, this could be a grand system for you.

I don't understand how you got 14/9 ~ (5/4)^2 from this example.

>Other commas
>cannot be deconstructed in this way, but even without consonances we
>can find factorizations such as 2^a 3^b U^c.

Huh?

-Carl

Oh forget it, is there a book on temperaments you can recommend me?
----- Original Message -----
From: Carl Lumma
To: tuning-math@yahoogroups.com
Sent: 15 Mart 2005 Salı 18:41
Subject: Re: [tuning-math] Commas and generators

>If a comma can be written in the form 2^a U^b V^c, where U and V are
>consonant ratios of odd integers, then we have a period-generator pair
>for the group generated by {2,U,V} which is in some sense natural.

You lost me. Is this something new?

>For instance, 225/224 = 2^(-5) 5^2 (9/7), and this leads to the
>temperament with octave period, generator a 5/4, and for which 14/9 ~
>(5/4)^2. If for some reason you are enamored of the 5/4-5/4-9/7
>augmented triad, this could be a grand system for you.

I don't understand how you got 14/9 ~ (5/4)^2 from this example.

>Other commas
>cannot be deconstructed in this way, but even without consonances we
>can find factorizations such as 2^a 3^b U^c.

Huh?

-Carl

>Oh forget it, is there a book on temperaments you can recommend me?

Most of the stuff Gene and company are doing here is original.
There's Gene's website at xenharmony.org, which most readers
find impenetrable. For a gentler approach, try Paul's materials
that I pointed you to (I thought I did)...

http://lumma.org/tuning/erlich

...you can print these papers to read at your leisure. I recommend
you start with The Forms of Tonality.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >If a comma can be written in the form 2^a U^b V^c, where U and V are
> >consonant ratios of odd integers, then we have a period-generator pair
> >for the group generated by {2,U,V} which is in some sense natural.
>
> You lost me. Is this something new?

It won't be if it is a 5-limit comma, but in general, what you get is
a tempering of a subgroup of a p-limit, not the whole p-limit. The
subgroup is extracted from the comma, and used to temper.

> >For instance, 225/224 = 2^(-5) 5^2 (9/7), and this leads to the
> >temperament with octave period, generator a 5/4, and for which 14/9 ~
> >(5/4)^2. If for some reason you are enamored of the 5/4-5/4-9/7
> >augmented triad, this could be a grand system for you.
>
> I don't understand how you got 14/9 ~ (5/4)^2 from this example.

It's more or less the same as the way you'd get (3/2)^4 ~ 5 from
81/80 = 2^(-4) 3^4 5^(-1). We know that up to octave equivalence, we
have that 5^2 * (9/7) == 1, where "==" is octave equivalence. This is
the factorization of the comma I give. Hence, 5^2 == 7/9, so that
(since you can put in powers of two) (5/4)^2 == 14/9. It turns out
that you don't need any extra 2's in this, and (5/4)^2 ~ 14/9. The
generator is a major third, and two of them are a 14/9. In this
subgroup, you get only octaves and what you can cook up from thirds of
size 5/4 and 9/7, which you then temper from rank 3 down to rank 2.

> >Other commas
> >cannot be deconstructed in this way, but even without consonances we
> >can find factorizations such as 2^a 3^b U^c.
>
> Huh?

Divide out the 2 and 3, and see if what remains is a power. For
example, you can write 2401/2400 as 2^(-5) 3^(-1) (49/5)^2, so
3/2 == (49/40)^2, and in fact 3/2 ~ (49/40)^2. If 49/40 were more
interesting, like 11/9, this would be better.

Not every p-limit comma is deconstructable in terms of p-limit
consonances, even if we add odd limit up to the next prime. Those that
are could be called "deconstructable", I suppose. It doesn't depend on
complexity, though more complex commas are less likely to be
deconstructable. You can't deconstruct 2401/2400 or 4375/4374, but
250047/250000 = 2^(-4) (5/3)^(-6) 7^3; hence 7^3 == (5/3)^6, and
finally (8/7)^3 ~ (6/5)^6/2, so that a period of 1/3 octave and a
generator of ~ 8/7 works; it's a sort of fragment of ennealimmal, and
even more accurate.

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >Oh forget it, is there a book on temperaments you can recommend me?
>
> Most of the stuff Gene and company are doing here is original.
> There's Gene's website at xenharmony.org, which most readers
> find impenetrable. For a gentler approach, try Paul's materials
> that I pointed you to (I thought I did)...
>
> http://lumma.org/tuning/erlich

Paul's paper, so far only Part I, would be a good thing to have up on
the web.

I have read most of them Carl. Maybe it's my lack of competence with the technical terms. Can you recommend me a book on meantone and other temperaments?
----- Original Message -----
From: Carl Lumma
To: tuning-math@yahoogroups.com
Sent: 15 Mart 2005 Salı 20:55
Subject: Re: [tuning-math] Commas and generators

>Oh forget it, is there a book on temperaments you can recommend me?

Most of the stuff Gene and company are doing here is original.
There's Gene's website at xenharmony.org, which most readers
find impenetrable. For a gentler approach, try Paul's materials
that I pointed you to (I thought I did)...

http://lumma.org/tuning/erlich

...you can print these papers to read at your leisure. I recommend
you start with The Forms of Tonality.

-Carl

>> >Oh forget it, is there a book on temperaments you can recommend me?
>>
>> Most of the stuff Gene and company are doing here is original.
>> There's Gene's website at xenharmony.org, which most readers
>> find impenetrable. For a gentler approach, try Paul's materials
>> that I pointed you to (I thought I did)...
>>
>> http://lumma.org/tuning/erlich
>
>Paul's paper, so far only Part I, would be a good thing to have up on
>the web.

I'm working on it. I'm changing jobs and a whole bunch of stuff is
up in the air, but it'll get done. But the Forms of Tonality is
really much more accessible for beginners, I think.

-Carl

>>>Oh forget it, is there a book on temperaments you can recommend me?
>>
>>Most of the stuff Gene and company are doing here is original.
>>There's Gene's website at xenharmony.org, which most readers
>>find impenetrable. For a gentler approach, try Paul's materials
>>that I pointed you to (I thought I did)...
>>
>>http://lumma.org/tuning/erlich
>>
>>...you can print these papers to read at your leisure. I recommend
>>you start with The Forms of Tonality.
>
>I have read most of them Carl. Maybe it's my lack of competence with
>the technical terms. Can you recommend me a book on meantone and other
>temperaments?

Why not try asking specific questions, directly to Paul? I don't
want to speak for his time, but in the past he's been the most
helpful answerer of questions I've ever met on the internet.

-Carl

>> >If a comma can be written in the form 2^a U^b V^c, where U and V are
>> >consonant ratios of odd integers, then we have a period-generator pair
>> >for the group generated by {2,U,V} which is in some sense natural.
>>
>> You lost me. Is this something new?
>
>It won't be if it is a 5-limit comma, but in general, what you get is
>a tempering of a subgroup of a p-limit, not the whole p-limit. The
>subgroup is extracted from the comma, and used to temper.

Sounds like TOP...

-C.

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> >It won't be if it is a 5-limit comma, but in general, what you get is
> >a tempering of a subgroup of a p-limit, not the whole p-limit. The
> >subgroup is extracted from the comma, and used to temper.
>
> Sounds like TOP...

Not at all; TOP is a tuning. I'm talking about rank two temperaments
defined for subgroups of the p-limit. Of course, TOP could be used as
the tuning.

>> >It won't be if it is a 5-limit comma, but in general, what you get
>> >is a tempering of a subgroup of a p-limit, not the whole p-limit.
>> >The subgroup is extracted from the comma, and used to temper.
>>
>> Sounds like TOP...
>
>Not at all; TOP is a tuning. I'm talking about rank two temperaments
>defined for subgroups of the p-limit. Of course, TOP could be used as
>the tuning.

Er...

-Carl

>>>If a comma can be written in the form 2^a U^b V^c, where U and V are
>>>consonant ratios of odd integers, then we have a period-generator pair
>>>for the group generated by {2,U,V} which is in some sense natural.
>>>For instance, 225/224 = 2^(-5) 5^2 (9/7), and this leads to the
>>>temperament with octave period, generator a 5/4, and for which 14/9 ~
>>>(5/4)^2. If for some reason you are enamored of the 5/4-5/4-9/7
>>>augmented triad, this could be a grand system for you. Other commas
>>>cannot be deconstructed in this way, but even without consonances we
>>>can find factorizations such as 2^a 3^b U^c.
>>>
>>>For instance, 225/224 = 2^(-5) 5^2 (9/7), and this leads to the
>>>temperament with octave period, generator a 5/4, and for which
>>>14/9 ~ (5/4)^2. If for some reason you are enamored of the
>>>5/4-5/4-9/7 augmented triad, this could be a grand system for you.
>>
>> I don't understand how you got 14/9 ~ (5/4)^2 from this example.
>
>It's more or less the same as the way you'd get (3/2)^4 ~ 5 from
>81/80 = 2^(-4) 3^4 5^(-1). We know that up to octave equivalence, we
>have that 5^2 * (9/7) == 1, where "==" is octave equivalence. This is
>the factorization of the comma I give. Hence, 5^2 == 7/9, so that
>(since you can put in powers of two) (5/4)^2 == 14/9. It turns out
>that you don't need any extra 2's in this, and (5/4)^2 ~ 14/9. The
>generator is a major third, and two of them are a 14/9. In this
>subgroup, you get only octaves and what you can cook up from thirds
>of size 5/4 and 9/7, which you then temper from rank 3 down to
>rank 2.

Got it, thanks.

>>>Other commas cannot be deconstructed in this way, but even
>>>without consonances we can find factorizations such as
>>>2^a 3^b U^c.
>>
>>Huh?
>
>Divide out the 2 and 3, and see if what remains is a power. For
>example, you can write 2401/2400 as 2^(-5) 3^(-1) (49/5)^2, so
>3/2 == (49/40)^2, and in fact 3/2 ~ (49/40)^2. If 49/40 were more
>interesting, like 11/9, this would be better.

I see what you're doing.

>Not every p-limit comma is deconstructable in terms of p-limit
>consonances, even if we add odd limit up to the next prime. Those
>that are could be called "deconstructable", I suppose.

Kewl.

>It doesn't depend on complexity, though more complex commas are
>less likely to be deconstructable.

That's good.

-Carl