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Date: Sat, 28 Jul 2001.
Subject: Re: Hey Carl

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Here's why the hypothesis should work.

Take an n-dimensional lattice, and pick n independent unison vectors.
Use these to divide the lattice into parallelograms or parallelepipeds
or hyperparallelepipeds, Fokker style. Each one contains an identical
copy of a single scale (the PB) with N notes. Any vector in the lattice
now corresponds to a single generic interval in this scale no matter
where the vector is placed (if the PB is CS, which it normally should
be). Now suppose all but one of the unison vectors are tempered out. The
"wolves" now divide the lattice into parallel strips, or layers, or
hyper-layers. The "width" of each of these, along the direction of the
chromatic unison vector (the one that remains untempered), is equal to
the length of exactly one of this chromatic unison vector.

Now let's go back to "any vector in the lattice". This vector, added to
itself over and over, will land one back at a pitch in the same
equivalence class as the pitch one started with, after N iterations (and
more often if the vector represents a generic interval whose cardinality
is not relatively prime with N). In general, the vector will have a
length that is some fraction M/N of the width of one
strip/layer/hyperlayer, measured in the direction of this vector (NOT in
the direction of the chromatic unison vector). M must be an integer,
since after N iterations, you're guaranteed to be in a point in the same
equivalence class as where you started, hence you must be an exact
integer M strips/layers/hyperlayers away. As a special example, the
generator has length 1/N of the width of one strip/layer/hyperlayer,
measured in the direction of the generator. Anyhow, each occurence of
the vector will cross either floor(M/N) or ceiling(M/N) boundaries
between strips/layers/hyperlayers. Now, each time one crosses one of
these boundaries in a given direction, one shifts by a chromatic unison
vector. Hence each specific occurence of the generic interval in
question will be shifted by either floor(M/N) or ceiling(M/N) chromatic
unison vectors. Thus there will be only two specific sizes of the
interval in question, and their difference will be exactly 1 of the
chromatic unison vector. And since the vectors in the chain are equally
spaced and the boundaries are equally spaced, the pattern of these two
sizes will be an MOS pattern.

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