Re: utonal and otonal

>Each has an utonal 1/9 : 1/7 : 1/6 : 1/5 : 1/4 followed by an otonal
>4 : 5 : 6 : 7 : 9, in sines and sawtooth waves, respectively. 16
>seconds per file at a 22050 Hz sampling rate, so be prepared for a
>1.34Mb download (sorry -- but mp3s of the sawtooth file sound fried).

Daniel, those files are a bit extreme. Try these...

http://lumma.org/488-01.mp3
http://lumma.org/488-02.mp3

-Carl

Hi Paul,

>Right -- with smaller numbers in the _bottom_ of the ratios than if you
>tried to express the triad as an otonal one. For example, the major triad is
>4:5:6 otonal and 1/15:1/12:1/10 utonal, so it's considered otonal, while the
>reverse is true for the minor triad.

Thanks, that's fitted in the last piece of the puzzle, and I understand how
the dekany triads are classified as utonal and otonal, and why changing
from 2)5 to 3)5 swaps one to the other.

- if you have a triad such as
5*9, 1*5, 1*9
= 45, 5, 9
(in the 1,3,5,7,9 decany, which I'm using for ref because of
Dave Keenan's clear graph of it)

then it is utonal because it is far simpler as
1*5*9/1, 1*5*9/9, 1*5*9/5
= 1/1 1/9 1/5

while
7*5, 1*5, 3*5 is otonal because of the common factor of 5,
so that 7, 1, 3, is simpler than
1*3*7*5/(1*3), 1*3*7*5/(3*7), 1*3*7*5/(1*7)
= 1/3 1/21 1/7

Also explains why there are 5 tetrahedra - one for
each of the factors. So for instance the tetrad for the 3 factor will be
1*3, 3*5, 3*7, 3*9.

You'll then get five hexanies as well, for instance,
one with all sets of 3 factors chosen from 1, 3, 5, 7
or 1,3, 5, 9, or 1,3,7,9, or 1,5,7,9, or 3,5,7,9.

Also each hexany is a kind of complement of one of the tetrads.

For instance the 1,5,7,9 hexany uses all the vertices that have no
3 factor in them, i.e. all the vertices of the dekany that aren't
part of the 1*3, 3*5, 3*7, 3*9 tetrad.

So it is all fitting together now.

In the 3)5 situation is reversed - the complement of 5*9, 1*5, 1*9 is
1*3*7, 3*7*9, 3*5*7, = 1, 9, 5, so otonal because of two common factors 3 and 7

- simpler than
1*3*5*7*9/(5*9),... = 1/45, 1/5, 1/9

while complement of 7*5, 1*5, 3*5 is
1*3*9, 3*7*9, 1*7*9 - only one common factor 9,
and obviously we are just getting 1/ the ratios for the 2)5 decany,
so 1/7 1/1 1/3 is simpler than 3, 21, 7.

I'm assuming you get the 3)5 dekany by this complementation process
by replacing 5*9 by 1*3*7, etc. in the case of the 1*3*5*7*9 decany
since it works so well. Do say if that's wrong. I'll look out the
Partch book you recommend,

thanks,

Robert

--- In tuning@egroups.com, "Robert Walker" <robert_walker@r...> wrote:

> I'm assuming you get the 3)5 dekany by this complementation process
> by replacing 5*9 by 1*3*7, etc. in the case of the 1*3*5*7*9 decany
> since it works so well. Do say if that's wrong.

That's a perfectly valid way of constructing it.

> I'll look out the
> Partch book you recommend,

Partch, followed by Wilson, invented this whole language that we're
speaking here.