Re: [tuning] Digest Number 3243

Hi Gene,

Thanks for the obeservation. Using Dave's picture,
I see that the pentads (quintads) form a square:

1/1

5/3 6/5
\ /
10/7 1/1 7/5
/ \
5/4 12/7 7/6 8/5
\ / \ /
10/9 3/2 7/7 4/3 9/5
/ \ / \
4/3 7/4 8/7 3/2
\ /
14/9 1/1 9/7
/ \
16/9 9/8

1/1

which makes a thirteen note scale with the
duplicated 1/1 in the middle and the duplicated
3/2s and 4/3s

> It is also a square scale in another sense--the qunitads, considered
> as in the cubic lattice of quintads, form a square. They are the
> quintads associated to
>
> [-1 0 0] [0 0 0]
> [-1 -1 -1][0 -1 -1]
>
> The horizontally related (major/minor pairings) and vertically related
> (a fifth apart) qunitads have two notes in common. [-1 0 0] and
> [0 -1 -1] have three notes in common, and [0 0 0] and [-1 -1 -1]
> (inverse to each other) one common note. All of the quintads, in fact,
> contain 1 as a note; this is a consequence of the fact that it is a
> subscale of the 9-limit tonality diamond, which can be described as
> the union of all quintads containing 1.
>

Ok so far except that I'm unfamiliar with the [-1 0 0] notation
- which I see gets followed up, so skipping forward:

> The lattice of tetrads works as follows: [a b c] refers to an otonal
> tetrad if a+b+c is even, then the root of the tetrad is
> r = 3^((-a+b+c)/2) 5^((a-b+c)/2) 7^((a+b-c)/2); the tetrad itself of
> course then is r*[1,3,5,7]. If a+b+c is odd, we have a utonal tetrad.
> The root (not guide tone!) of the tetrad is the root r of the
> corresponding otonal tetrad [a+1 b c], and the tetrad itself of course
> then is r*[1,3,3/5,3/7]. To get pentads, you merely take instead
> r*[1,3,5,7,9] or r*[1,3,3/5,3/7,1/3] instead. Of course the latter no
> longer has much of a claim to having 1 as a root, but that does not
> matter.

I'm lost here too. Well I could follow through the example
and check that you get unique pentads for each [a b c]
- but there may be many ways of devising such a notation
system - so what is the particular advantages of this
one?

What is the reason for the 3^((-a+b+c)/2) 5^((a-b+c)/2) 7^((a+b-c)/2
formulae? Perhaps it should be immediately obvious, but
I can't see it yet anyway.

Robert