Question about vals

If the generator for a rank-2 tuning is a 5/4, you can tell because
the generator mapping for 5 will be 1.

Is there some interesting property that all vals that have a 6/5
generator will share? I thought it would be that the generator mapping
for 5 would be one less than the generator mapping for 3, and the
period mapping would be one more, since 3/2 / 6/5 = 5/4. This is
apparently not the case in amity, which has a val of [<1, 3, 6], <0,
-5, -13]>. Is there something I'm missing...?

Assuming that's solvable, which is must be - is there some complicated
algebraic expression that you could set up which, when solved, would
yield all vals with a generator of 5/4 and those with 6/5?

Thanks,
Mike

On Fri, Feb 4, 2011 at 9:28 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
>
> Is there some interesting property that all vals that have a 6/5
> generator will share? I thought it would be that the generator mapping
> for 5 would be one less than the generator mapping for 3, and the
> period mapping would be one more, since 3/2 / 6/5 = 5/4. This is
> apparently not the case in amity, which has a val of [<1, 3, 6], <0,
> -5, -13]>. Is there something I'm missing...?

Sorry, I now realize that this is because amity's generator is not
actually 6/5. So there you go, numbers work again.

So then my only question is this:

> Assuming that's solvable, which is must be - is there some complicated
> algebraic expression that you could set up which, when solved, would
> yield all vals with a generator of 5/4 and those with 6/5?

Specifically I'd like to plot all of the 5-limit temperaments that
have generators that lie between 300 and 400 cents, but it would be
good to know how to work the math out first.

-Mike

Mike Battaglia <battaglia01@gmail.com> wrote:

> So then my only question is this:
>
> > Assuming that's solvable, which is must be - is there
> > some complicated algebraic expression that you could
> > set up which, when solved, would yield all vals with a
> > generator of 5/4 and those with 6/5?

If the generator's 5/4, you know that [-2, 0, 1> maps to
one generator and zero periods. If the generator's 6/5,
you know that [1, 1, -1> maps the same way. These give
Diophantine equations you have to solve.

> Specifically I'd like to plot all of the 5-limit
> temperaments that have generators that lie between 300
> and 400 cents, but it would be good to know how to work
> the math out first.

You can state the TE (TOP-RMS) generator size as an
algebraic expression, and set up an inequality that has to
be fulfilled. You can also take the relevant section of
the scale tree and find appropriate mappings.

Graham

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> > Assuming that's solvable, which is must be - is there some complicated
> > algebraic expression that you could set up which, when solved, would
> > yield all vals with a generator of 5/4 and those with 6/5?

If you are looking at the 5-limit, temperaments are defined by a single comma, |a b c>. If |b|=1 then the temperament will be linear (octave period) with generator a major third. If |b+c|=1 then the temperament will be linear with generator a minor third.

Some commas with the first condition are 393216/390625, 3125/3072, 25/24 and 16/15. Some commas with the second condition satisfied are |-36 -52 51>, |8 14 -13>, 15625/15552, 25/24, 27/25.

Nice! -C.

Gene wrote:

>> > Assuming that's solvable, which is must be - is there some complicated
>> > algebraic expression that you could set up which, when solved, would
>> > yield all vals with a generator of 5/4 and those with 6/5?
>
>If you are looking at the 5-limit, temperaments are defined by a
>single comma, |a b c>. If |b|=1 then the temperament will be linear
>(octave period) with generator a major third. If |b+c|=1 then the
>temperament will be linear with generator a minor third.
>
>Some commas with the first condition are 393216/390625, 3125/3072,
>25/24 and 16/15. Some commas with the second condition satisfied are
>|-36 -52 51>, |8 14 -13>, 15625/15552, 25/24, 27/25.
>

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:

> Some commas with the first condition are 393216/390625, 3125/3072, 25/24 and 16/15. Some commas with the second condition satisfied are |-36 -52 51>, |8 14 -13>, 15625/15552, 25/24, 27/25.

I wasn't aware of |-36 -52 51> until now other than the fact that it was on my list and Tonalsoft has it cataloged as "egads". It is, of course, too complex to be of any obvious use, but I note that tossing in 4375/4374 as well gives a pretty natural extension to the 7-limit. That one has wedgie <<51 52 149 -36 93 200||. Taking 51 6/5 generators and then down 13 octaves gives 4/3, another 6/5 gives us 8/5. Since 2500/2187 = (4/3)^3/(6/5)^4 and (2500/2187)/(8/7) = 4375/4374, there's our 8/7.

On Fri, Feb 4, 2011 at 9:54 AM, Graham Breed <gbreed@gmail.com> wrote:
>
> You can state the TE (TOP-RMS) generator size as an
> algebraic expression, and set up an inequality that has to
> be fulfilled.

I see. And this doesn't play into the actual mapping, right? Rather
it's a separate expression denoting the tuning of the generator.

-Mike

On Fri, Feb 4, 2011 at 2:41 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> If you are looking at the 5-limit, temperaments are defined by a single comma, |a b c>. If |b|=1 then the temperament will be linear (octave period) with generator a major third. If |b+c|=1 then the temperament will be linear with generator a minor third.

I'm not even going to bother asking how it is that a single comma
defining a linear temperament turns into two vals defining a linear
temperament, or how bra-ket notation works with two bra's and only one
ket. But if you have some bival

<a b c]
<d e f]

in reduced form, I guess the translation is that if the generator is a
major third, f will be 1, and if it's a minor third, f-e will be 1. So
if I want to generate an algebraic expression that will yield all vals
of period 2/1 that have generators of 5/4 or 6/5, the following will
do the trick, right?

<1 b c |
<0 e f |

f * (f-e) = 1
e = f-1/f

So here we go:

<1 b c |
<0 f-1/f f |

And it actually works like that, it's that simple? And hence there you
go, a huge chunk of the 3L1s scale tree spectrum is represented by the
above? Am I doing it right here?

-Mike

Mike wrote:
>But if you have some bival
>
><a b c]
><d e f]

That's looks like two vals, not a bival, which is the wedge
product of two vals. -C.

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> On Fri, Feb 4, 2011 at 2:41 PM, genewardsmith
> <genewardsmith@...> wrote:
> >
> > If you are looking at the 5-limit, temperaments are defined by a single comma, |a b c>. If |b|=1 then the temperament will be linear (octave period) with generator a major third. If |b+c|=1 then the temperament will be linear with generator a minor third.
>
> I'm not even going to bother asking how it is that a single comma
> defining a linear temperament turns into two vals defining a linear
> temperament, or how bra-ket notation works with two bra's and only one
> ket. But if you have some bival
>
> <a b c]
> <d e f]

That's not a bival. If |a b c> is the comma, the corresponding bival is +- <<c -b a||. From that you can deduce the claims I made above about what the comma tells you. Since the comma is not a power, you have GCD(a,b,c)=1, and if GCD(b,c)=N, the period is 1/N octaves. Also, if |b|=1, the period is an octave and the generator is a fifth/fourth, which goes with the rest of what I said.

On 6 February 2011 06:16, Mike Battaglia <battaglia01@gmail.com> wrote:
> On Fri, Feb 4, 2011 at 9:54 AM, Graham Breed <gbreed@gmail.com> wrote:
>>
>> You can state the TE (TOP-RMS) generator size as an
>> algebraic expression, and set up an inequality that has to
>> be fulfilled.
>
> I see. And this doesn't play into the actual mapping, right? Rather
> it's a separate expression denoting the tuning of the generator.

The optimal generator size is a function of the mapping. The TE error
is a function of both the mapping and the generator size. I forget
which is relevant here.

Graham