An RI version of Wendell Well

If we look at the brat ratios of Wendell Well, they are
approximately [r, 3/2, 3/2, r, 3/2, 3/2, 3/2, r, 3/2, q, r, r]
where both r and q are nearly 2, but r is a little less and q is
a little more. If we solve for these under the condition that
the product of the fifths comes to 128 (octave closure) then q is

320*(-27*r+18*r^2+40)/r/(2187*r^4-5184*r^3-7728*r+5760+5216*r^2)

We can relate r and q in various ways; for instance making q exactly
2, or having the average of r and q be 2. The simplest way however
is to make r exactly 2, which makes q to be 580/293. This gives us
the following scale:

[1, 1215/1144, 321/286, 68187/57200, 180/143, 3823/2860, 405/286,
214/143, 22729/14300, 963/572, 5107/2860, 270/143]

It is a rational intonation version of Robert's temperament.

Gene wrote:

>320*(-27*r+18*r^2+40)/r/(2187*r^4-5184*r^3-7728*r+5760+5216*r^2)

^ ^ two divisions?

I was wondering about the result of r = q. So it wasn't as
hairy as I thought!

>It is a rational intonation version of Robert's temperament.

Where 1/1 = A.

Manuel

--- In tuning-math@yahoogroups.com, manuel.op.de.coul@e... wrote:
> Gene wrote:
>
> >320*(-27*r+18*r^2+40)/r/(2187*r^4-5184*r^3-7728*r+5760+5216*r^2)
>
> ^ ^ two divisions?

Blame Maple.

> I was wondering about the result of r = q. So it wasn't as
> hairy as I thought!

What's really hairy, it turns out, is trying to sort out all the
possibilities. I have more than 41 but less than 510 (depending on how
many duplications) examples of RI scales *all* of whose brats are
either exactly 2 or exactly 3/2, and with either 6 or 7 pure fifths.
How do I evaluate these?

Gene wrote:
>Blame Maple.

Can it be written out as a polynomial? Then I could find
the result for r = q myself.

>How do I evaluate these?

Brute force? I find it a little easier to use the beat
rate quotient of 3/2 and 5/4 which is 0 and ~1/5 (say 1/m)
for 6 tempered fifths. What would be the m nearest to an
integer for 5 tempered fifths in any order, or 7?
Is Robert's order the best order for 6 tempered fifths?
(I suppose so).

Manuel

--- In tuning-math@yahoogroups.com, manuel.op.de.coul@e... wrote:
> Gene wrote:
> >Blame Maple.
>
> Can it be written out as a polynomial? Then I could find
> the result for r = q myself.

Maple thinks these associate left to right; that is,

A/B/C = (A/B)/C = A/(B*C)

> Is Robert's order the best order for 6 tempered fifths?
> (I suppose so).

It's hardly clear. You tell me how to decide.

>Maple thinks these associate left to right; that is,

>A/B/C = (A/B)/C = A/(B*C)

Ok, when I solve for q=r I get r=1.99780788361.

>> Is Robert's order the best order for 6 tempered fifths?
>> (I suppose so).

>It's hardly clear. You tell me how to decide.

Yeah, that was a vague question, never mind.

Manuel

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, manuel.op.de.coul@e... wrote:
> > Gene wrote:
> >
> > >320*(-27*r+18*r^2+40)/r/(2187*r^4-5184*r^3-7728*r+5760+5216*r^2)
> >
> > ^ ^ two divisions?
>
> Blame Maple.
>
> > I was wondering about the result of r = q. So it wasn't as
> > hairy as I thought!
>
> What's really hairy, it turns out, is trying to sort out all the
> possibilities. I have more than 41 but less than 510 (depending on
how
> many duplications) examples of RI scales *all* of whose brats are
> either exactly 2 or exactly 3/2, and with either 6 or 7 pure fifths.
> How do I evaluate these?

please ask bob wendell. some of these may end up being published
right alongside his!