Paul's 'Gentle Introduction', part 1

I was rummaging around in some old TDs, to get the proper
citations into my webpages, when I found this:

> [Paul Erlich, TD 345.23]
> ... I was trying to point out that the entire infinitude
> of 3-limit JI (Pythagorean tuning) can be collapsed to a
> finite set by choosing just one small interval as your
> unison vector. Then, no matter how far out you go along
> the chain, any note you pick is equivalent to one and only
> one of the finite number of notes in your periodicity block.
> By way of a familiar geometrical model, the chain of fifths
> becomes a finite circle of fifths.

When I first read this it made perfect sense, but after
further reflection, I thought, is it really true?

If you carry the chain of '5ths' (and '4ths') out to the
24th note, it will be ~46.92 cents higher than the starting
note. That's darn near a 'quarter-tone', and I'd say
just about anyone would perceive a difference if the
comma did not vanish. If it is vanishing, can they
*really* be considered the same note? (of course, I know
a lot of one's perception depends on context).

And every dozen notes farther out that one goes, the
discrepancy between that dozen notes and their supposed
'equivalents' becomes greater by a Pythagorean comma
(~ 23.46 cents).

By the time you reach 3^48 [= ~93.84 cents] you have
a pitch that is much closer to the 'minor 2nd' of the
original scale [= 3^-5 = ~90.225 cents] than it is
to the original starting note.

I realize that this is mostly a mental exercise, as
there's probably not much practical use for a Pythagorean
system with 48 or more pitches to the 'octave' *as*
a Pythagorean system - that is, if it were to be used
at all, it would probably be for schismatic equivalents
of 5-limit pitches. But still, Paul's statement strikes
me as being less than totally correct.

Further comments appreciated.

-monz

Joseph L. Monzo Philadelphia monz@juno.com
http://www.ixpres.com/interval/monzo/homepage.html
|"...I had broken thru the lattice barrier..."|
| - Erv Wilson |
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I wrote,

>> ... I was trying to point out that the entire infinitude
>> of 3-limit JI (Pythagorean tuning) can be collapsed to a
>> finite set by choosing just one small interval as your
>> unison vector. Then, no matter how far out you go along
>> the chain, any note you pick is equivalent to one and only
>> one of the finite number of notes in your periodicity block.
>> By way of a familiar geometrical model, the chain of fifths
>> becomes a finite circle of fifths.

>When I first read this it made perfect sense, but after
>further reflection, I thought, is it really true?

>If you carry the chain of '5ths' (and '4ths') out to the
>24th note, it will be ~46.92 cents higher than the starting
>note. That's darn near a 'quarter-tone', and I'd say
>just about anyone would perceive a difference if the
>comma did not vanish. If it is vanishing, can they
>*really* be considered the same note? (of course, I know
>a lot of one's perception depends on context).

>And every dozen notes farther out that one goes, the
>discrepancy between that dozen notes and their supposed
>'equivalents' becomes greater by a Pythagorean comma
>(~ 23.46 cents).

>By the time you reach 3^48 [= ~93.84 cents] you have
>a pitch that is much closer to the 'minor 2nd' of the
>original scale [= 3^-5 = ~90.225 cents] than it is
>to the original starting note.

>I realize that this is mostly a mental exercise, as
>there's probably not much practical use for a Pythagorean
>system with 48 or more pitches to the 'octave' *as*
>a Pythagorean system - that is, if it were to be used
>at all, it would probably be for schismatic equivalents
>of 5-limit pitches. But still, Paul's statement strikes
>me as being less than totally correct.

Good thinking. Mathematically speaking, though, the statement is correct,
since once you choose a certain interval ('unison vector') as an equivalence
relation, stacking that interval on top of itself an indefinite number of
times results also in an equivalence relation. It would seem, though, that
you've pointed out an important practical shortcoming of this mathematical
model.

That's not really the case. In practice, once you choose your unison vector,
you're "cutting off" a little section of the chain of fifths and using that
as your entire pitch set (i.e., the periodicity block). No other pitches
come into the picture. So every time you go "around" the "circle" of 12
fifths, you actually keep skipping from the end of chain back to the
beginning of the chain, and you're simply ignoring the Pythagorean comma
itself, never any larger multiple of it.

As for what "true" JI relations you can actually perceive without reducing
them to a chain of simpler relations, 3-limit harmony implies one step along
the chain of fifths, 9-limit harmony implies two steps along the chain,
27-limit harmony implies three steps, etc. That's already beyond what I
think is normally possible. To perceive a double Pythagorean comma directly,
rather than as a construction from smaller units (e.g., 24 fifths), would
require a sensitivity to 3^24-limit, i.e., 282,429,536,481-limit harmony. We
don't really need to discuss the plausibility of that. So, since the chain
of 12 fifths allows one to construct a 3:2, 9:8, 27:16 if you like, etc.,
from any given note in the chain, with a maximum error of only _one_
Pythagorean comma, and because you've chosen the Pythagorean comma as your
unison vector (i.e., acceptable approximation of a unison), you have, for
all intents and purposes, collapsed the infinite chain of fifths into a
circle.

Clear enough?