source file: mills2.txt
Date: Wed, 28 Feb 1996 08:08:50 -0800

Subject: Re: 7:4 as a structural interval

From: Gary <71670.2576@compuserve.com>

Kerr Jamieson asks:

> >Re Mr. Tsuji's question "Can we meet the fifth somewhere?" If I understand
> >this to mean "will some number of 7:4's = some number of 3:2's," the answer
> >is "not in JI."
> 
> Surely seven of the one is half-a-dozen of the other in any context?

   I think that David Doty's point is that a stack of 7:4s atop one another
never produces any composite interval exactly the same as any interval produced
by a stack of 3:2s, no matter how high you pile up either stack.  

   You can think of it as a little like a ruler with inches on one side and
centimeters on the other.  It seems like no even centimeter mark ever exactly
aligns with any even inch mark, other than at the zero-point of ruler of course.
(Actually, you will find two that exactly align if you had a 2.54m-long ruler.
But in the case 7:4s and 3:2s, they will absolutely NEVER exactly align.)

   That's similar with the fact that a circle of just perfect fifths never
exactly closes.  No power of 3:2 is ever exactly equal to any power 2:1.  As it
happens, a stack of seven octaves is not far from a stack of twelve fifths, the
error being the Pythagorean comma (about 23.5 cents).  The basis for 12TET
tuning is that you can distribute that error evenly across all those fifths and
land exactly on seven octaves up.  But those would no longer be exact 3:2s
anymore.  

   And, as it turns out, a stack of eight 7:4s is about as tall as a stack of
eleven 3:2s, the error being about 29 cents out of about 7750 cents.  If you
approximate those ratios as does 88CET tuning (7:4 at 968 cents rather than
968.826 cents, and 32 at 704 cents rather than 701.955 cents), the two stacks
end up the same height.  But as with the circle of fifths, those are no longer
exact representations of 7:4 and 3:2 harmonies.  


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