Repost: A gentle introduction to Fokker periodicity blocks, part 1

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*A gentle introduction to Fokker periodicity blocks, part 1*
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Just intonation provides us with a potentially infinite set of pitches.
Whether we are considering a one-dimensional string of perfect fifths
(Pythagorean tuning or 3-limit JI), a two-dimensional grid of perfect fifths
and major thirds (5-limit JI), a three-dimensional lattice of perfect
fifths, major thirds, and harmonic sevenths (7-limit JI), or any other set
of mutually prime intervals, the set of pitches will be infinite, with
number of dimensions equal to the number of primary intervals.

Sometimes, for instance if we are building fixed-pitch JI instruments, we
may decide it is necessary to settle for a finite number of pitches. There
are of course many ways of doing this, but let's say we choose this
approach: We keep adding more and more pitches, using simple JI intervals,
until we find a pitch very close to one we already have. We have found a
"comma" or "unison vector". Then we decide that the new pitch is not worth
adding. In the 3-limit case, we are done. Because once we find a unison
vector, any further additions along the chain of fifths will just introduce
that same unison vector between the new notes and notes we already have.
Let's give some examples:

Here is a chain of fifths:

1024/729-256/243-128/81-32/27-16/9-4/3-1/1-3/2-9/8-27/16-81/64-243/128-729/5
12

Let's say we start in the middle (1/1) and start adding notes alternately to
the right and left on the chain of fifths (it doesn't matter; we could go in
just one direction if we wanted to). Soon we have both 9/8 (204.9 cents) and
32/27 (294.1 cents) in our scale, and the interval between them, about 90
cents, is the smallest interval in the scale. Let's say that we consider
this a "unison vector", and decide that we only want to keep one of those
pitches, say 9/8, in our scale. So we have a pentatonic (5-note) scale. If
we tried to add a fifth above 9/8, which is 27/16, to the scale, we would
find that that is 90 cents from a pitch we already have, 16/9. This could
have been anticipated by noting the regularity of the chain of fifths: all
fifths are the same size, so all intervals constructed from a given number
of fifths are the same size no matter where they occur in the chain. In this
case, an interval of 90 cents is always produced by a chain of 5 fifths. In
vector notation, we say

(5) = 90 cents.

This unison vector defines the pentatonic scale. As we have seen,
considering 90 cents to be a unison vector causes us to stop after we have 5
notes in the chain of fifths. The pentatonic scale is called a "periodicity
block" because the entire, infinite chain of fifths can be divided up into a
string of pentatonic scales, each transposed 90 cents relative to its
neighbor. Every note in the infinite chain can be seen as equivalent to one
and only one of the five pitches in our pentatonic scale, through
transpositions of 90 cents. So considering (5) to be a unison vector in a
sense collapses the infinite resources of Pythagorean tuning into a single
pentatonic scale.

Now let's say you consider 90 cents to be a large interval, a distinction
worth making on our instrument (most of us would!). Then you keep expanding
your chain of fifths. When you get all the pitches shown above, you find
that the interval between 1024/729 (588.3 cents) and 729/512 (612.7 cents)
is only about 23 cents. You might decide that this is too small a
distinction to keep on your instrument. So you choose only one of those two
pitches, and you have a 12-tone scale. The unison vector in question is thus

(12) = 23 cents.

Again, this unison vector defines a periodicity block of 12 notes: every
note in the infinite chain is equivalent to one and only one note in the
12-tone scale through transpositions of 23 cents.

Any questions? If not, I will continue with the 2-D (5-limit) case next
time.