--- In tuning-math@y..., graham@m... wrote:

> Temper out the schisma from the periodicity block above. You end
up with
> a 24-note schismic scale. No way can that have two step sizes!
>
> That looks like a refutation with the definitions I have.

I think the problem is that, as you said before, the scale really has
12 pitch classes, not 24, due to the syntonic comma squared vanishing.

> How about a
> weaker hypothesis using propriety instead?

Blackjack it not proper.

> Schismic-24 is still proper,
> but not strictly proper. I'm sure some even hairier examples would
break
> this. Remember unison vectors don't even have to be small
intervals.

Exactly.
>
>
> > > Usually that comes out fine. The unison vectors define a
linear
> > > temperament, which forms an MOS with the right number of notes.
> >
> > Let's prove this.
>
> I'm sure you can always get the linear temperament. You can
describe it
> with fractions of the octave and chromatic unison vector if needs
be.
> Getting to the MOS is more difficult, if you have a formula for
that it
> would be useful anyway.
>
> Seeing as this is the mathematical list, I'll give the matrix
equation:
>
> (R1) (R2)
> (R2) (R2)
> (M1) (00)
> (. )H' = (. )H'
> (. ) (. )
> (. ) (. )
> (Mn) (00)
>
> Where R1 and R2 are the chromatic unison vectors (one of which will
> usually be the octave) as row vectors. M1 to Mn are the commatic
unison
> vectors. 00 is a row of zeros. So the things that look like
column
> matrices are actually square. H' is the tempered equivalent of the
list
> of prime axes, including 2.
>
> Multiply on the left by the inverse of the matrix with the unison
vectors
> in, and you have an equation defining H' in terms of itself. You
can
> then get your chromatic unison vectors in terms of H', and you have
a
> two-dimensional system.

OK, good so far.
>
> Usually the chromatic vectors are an octave and a twelfth.

Lost me there.

? For
> Miracle, the octave and twelfth both have to be divided by 6, so
you have
> to re-define it with a fifth as a unison vector.

??
> It probably would work for a sufficiently large MOS. But we could
frame
> the hypothesis so that the MOS is used as the default PB. I think
that
> would make sense.

All right, as long as we don't get circular.