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MOS Theory

🔗John Chalmers <non12@...>

12/30/1997 11:11:31 AM
MOS: Finding the generators of M-toned MOS is not a simple task, but
I'll try to explain my procedure. MOS are generated by intervals which
do not divide the octave or other Interval of Equivalence evenly. They
are also characterized by a quasi-periodic pattern of two intervals
(A and B) whose sizes may have any proportion, but the most familiar
ones have 1/2 <= A/B <= 2 as these are Rothenberg-proper. The two
intervals A and B are arranged "maximally-evenly" (see papers by
John Clough and colleagues, Clampitt, Carey, Douthett et al.) and
any given M, there are several possible patterns such that the numbers
P of A and Q of B sum to M and have no common factors. Thus for
7 tone MOS (M=7), the possible patterns are 6A+1B (and 1A +6B),
5A+2B (2A +5B) and 4A+3B (3A+4B). In an N-tone ET, PA +QB = N,
and P +Q =M. Both G and the complement of G with respect to the
IE generate the same MOS, though modally rotated. Hence one can
obtain the diatonic scale in 12-tet by a cycle of either 4ths or
5ths.

To find the Generator G of each class of MOS of M tones, I do the
following. For the simplest case where P= M-1 and Q =1, the limits
are 0 to N/M in cents or logs and from 0 to [N/M] where [] is the
integer <= N/M in the case of equal temperaments. The logic is that
if A/B were infinite and A>B, then B-> 0 and there would be only P=M-1
tones per octave (or IE). In the case of M=7, the largest possible
generator
would be 200 cents as that would yield 6 whole tones and a zero-width
interval for B. In the A+6B case, if G were infinitesimal, then there
would be 6 very small intervals of G and a large remainder. Hence for
the M-1, 1 type of MOS, G lies between 0 and Octave/(M-1).

For the 5A+2B=7 MOS case. the logic is similar. IF A/B is infinite, then
there are only 5 tones per octave (IE), if infinitesimal, then only 2.
Hence intervals between 600 and 720 cents generate 7-tone MOS of the
5A+2B and 2A+5B form. In 9-tet, G= 5 degrees or 666.667 cents and a
cycle of 5 degrees produces a MOS 1 1 1 2 1 1 2. The corresponding
scale in 12-tet is 2 2 2 1 2 2 (as ascending successive degrees).
The 4+3 scales are produced by cycles of neutral thirds (or 6ths) such
as 7 degrees of 24-tet.

To find the generator(s) of a particular class of MOS in a given
N-tone ET, one simply chooses the interval(s) which lie in the ranges
calculated above. Needless to say, not every N-tone ET contains every
M-tone MOS even when M < N. Remember that PA + QB = N and P+Q = M.

A few examples:

If one wishes a 9-tone MOS of the 5+4 class and whose intervals are
in the ratio of 2/1, then 5*2 + 4*1= 14 and the MOS exists in 14-tet.
One may find the MOS by distributing the intervals as evenly as possible
by inspection: 2 1 2 1 2 1 2 1 2 and immediately see that G must be
3 degrees as the repeating block in the MOS spans 3 degrees of 14 (2+1).
Alternatively, one can use the fact that G must lie between 1200/5
and 1200/4 cents (from 5+4=9) or 240-300 cents. The only interval in
14-tet in this range is 3 degrees (257 cents) (one could also use 11
degrees).

--John


SMTPOriginator: tuning@eartha.mills.edu
From: John Chalmers
Subject: Stellate Hexanies (14-anies)
PostedDate: 30-12-97 20:12:10
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