Partch scores ?

Does anyone know where to get copies of

scores for Harry Partch's productions ?

He _did_ use scores, didn't he ?

Thanks.

Will






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On Thu, 21 Nov 1996, Kami Rousseau wrote:
> We use 31TET, 53TET and 72TET to approximate just intervals, like
> 5/4
> and 7/4. But what are we _really_ doing? Stacking up an infinity of
> 3/2's! In
> a certain sense, using a ET scale means playing in an extended 3-limit
> tuning.

I can't quite agree--for example, if you try to derive 72TET from a
spiral of fifths it closes after twelve steps. You need to throw in at
least one higher-limit interval to make a two-dimensional web.

--pH (manynote@library.wustl.edu or http://library.wustl.edu/~manynote)
O
/\ "How about that? The guy can't run six balls,
-\-\-- o and they make him president."

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My thanks to Kami Rousseau for an interesting post; since I have
recently been working along the same lines, using int[n log2(3)], I'd
like to make a few comments.


> M={ (3^x / 2^y) / (81/80)^(x/4) | a=log2(3), x E Z, y=[ax]}.
>
> The expression can be simplifed to
>
> M={ 5^x / 2^(y-x) | a=log2(3), x E Z, y=[ax]}.

This simplification is not a correct (a typo?); the correct expression
is:

M = {5^(x/4) / 2^(y-x) | a = log2(3), x E Z, y = [ax]}

Or if I may re-write this in my own preferred notational form:

M = {2^(n-m).5^(n/4) : n E Z, m = [x log2(3)]}

where n will be assigned values of 1 to 12 for a 12-notes per octave
keyboard. Without an upper limit for n, the cardinality of M will of
course be aleph-null.

There is no such handy simplification for the other regular syntonic
comma-based temperaments (i.e. what we loosely call meantone
temperaments in addition to the 1/4-comma temperament). The members of
the 1/6-comma set are all of the form:

(2^(2-m).3.5^(1/2))^(n/3)

For 1/5-comma:

(2^(4-m).3.5)^(n/5)

For 2/9-comma:

(2^(8-m).3.5^2)^(n/9)

The notation including the root of the syntonic comma is therefore
better suited to the task of generalising the meantone-related
temperaments.

> T={ (3^x / 2^y) / (3^12 / 2^19)^(x/12) | a=log2(3), x E Z, y = [ax]}

This can be simplified thus (I'll retain my notation):

12TET = {2^(19-m).3^(n-12) : n E Z, m = [n log2(3)]}

or still further, but rather inelegantly:

12TET = {2^(19 - [n log2(3)]).3(n-12) : n E Z}

2^(n log2(3)) = 3^n of course, but the whole exponent doesn't simplify
thus.

The cardinality is 12 because, as Kami implied, for n = 12, the
exponents are both 0, and the cycle begins again.

> By studying this procedure, we discover that each ET is generated by a
> special enharmonic relation.
>
> 5TET B=C
> 7TET C#=C
> 12TET B#=C
> 19TET BX=C
>
> As the number of degrees increases, the comma gets smaller and the
> fifths get purer.

Sorry, but the comma does not become smaller as the number of notes to
the octave increases. However the amount by which each 3/2 is tempered
does not also increase, because if t is the amount by which each 3/2 is
tempered and k is the comma , then t = k^(1/n) for a given nTET, so
increasing values of n and k will not yield increasing values of t. For
instance, for 31TET (the aural equivalent of meantone):

k = 2^-[31log2(3)].3^31 = 2^-49 . 3^31 = 160.6 cents to one place

but t = k^(1/31) = 5.2 cents to one place

Or for 74TET (the aural equivalent of 2/9-comma temperament):

k = 2^-[74log2(3)].3^74 = 2^-117 . 3^74 = 344.7 cents to one place

but t = k^(1/74) = 4.7 cents to one place

You asked for comments, Kami, so I hope the above is of some use.
--
Jonathan Walker
Queen's University Belfast
mailto:kollos@cavehill.dnet.co.uk
http://www.music.qub.ac.uk/~walker/

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