I'm a little insomniaic right now, so I thought I'd explain a neat

way to look at and calculate PB scales. The way it's been done might

be described as taking a basis for a notation in the p-limit, and

setting one basis element aside, finding the block from the rest. The

dual point of view says that we select out the corresponding val.

Suppose [v_1, ..., v_k] form a notation in the p-limit, where k=pi(p)

is the number of primes up to p, and suppose all the vals v_i are

positive, so that v_i(q) for q<=p a prime is a positive integer.

Suppose also that (b_1, ..., b_k) is a basis for the notation. We

create a block of n=v_1(2) scale steps in an octave by setting aside

b_1, and replacing the top row representing b_1 with [1 0 0 ... 0]

representing 2, which gives us a Graham matrix we can invert and use

to define the block. The inverted matrix can be written in terms of

our notation as

[(1/n)v_1, v_2 - (v_2(2)/n)v_1, ..., v_k - (v_k(2)/n)v_1],

which we may verify by applying it to 2 and the basis elements b_i

for i>1, obtaining the identity. This means that the condition for a

note q being in the block is 0 <= v_1(q) < n, and

-1/2 <= v_i(q) - (v_i(2)/n)v_i(q) < 1/2 for each i>1. Thus the m-th

scale step expressed in our notation can be calculated from

v_1(q) = m and v_i(q) = ceil(-1/2 + (v_i(2)/n)m). The notes are

strung out in an approximate line across a diagonal of the

rectangular prism defined by [v_1, ..., v_k], so it is a "scale along

a line" as I called it in the subject header.

I hope this was reasonably coherent.

--- In tuning-math@y..., genewardsmith@j... wrote:

>The notes are

> strung out in an approximate line across a diagonal of the

> rectangular prism defined by [v_1, ..., v_k], so it is a "scale

along

> a line" as I called it in the subject header.

I might add that given a notation [v_1, ..., v_k] we get C(k,1) = k

blocks, C(k,2) = k(k-1)/2 blocks with one comma tempered out, and so

forth down to C(k, k-2) = k(k-1)/2 blocks in various linear

temperaments, corresponding to faces, and C(k,k-1) = k equal

temperaments.