Re: Non octave over / under

Hi Dan,

> What happens is that *every* multiple of a given fraction of an octave
> will always be exactly the same ratio. So every value of x should
> theoretically give a complete Scale Tree that is unique that to that x
> (and, if generalized, to that period).

As you'll see from my last post, I just noticed that too!
We've got the pattern now as well.

> So what you have is the whole JI universe of ratios, uniquely
> articulated to ET scale degrees in giant, unbounded grids--each
> relative to the degree of bend and the complexity of x (the simpler
> the value of x, the simpler the fractions of the period).

Yes, it's amazing!

I wonder if you've had the thought I've just had:

scale repeat 2:

x = 20/10 21/10 22/10 ...
3/2 at 20/30 21/32 22/34 ...

Trying scale repeat 3:

x = 20/10 21/10 22/10 23/10 24/10 ...
3/2 at 20/50 21/54 22/58 23/62 24/66 ...

It's the same pattern, but you increment by 4
each time instead of 2.

Try scale repeat 4:

x = 20/10 21/10 22/10 23/10 ...
3/2 at 20/70 21/76 22/82 23/90 ...

increment by 6 each time.

Try scale repeat 5:

x = 20/10 21/10 22/10 ...
3/2 at 20/90 21/98 22/106 ...

Extrapolating all those series back to 10:

x = 10/10 11/10 12/10 ...
3/2 at 10/10 11/12 12/14 ... for scale repeat 2
3/2 at 10/10 11/14 12/18 ... for scale repeat 3
3/2 at 10/10 11/16 12/22 ... for scale repeat 4
etc.

Formula:

If scale repeat is r, and x = (a+n)/n
then the 3/2 occurs at (a+n)/(a+2(r-1)n)

I expect that will also work for r non integer.

Let's try r = 5/3, a = 6, n = 11, just for fun

x = 17/6, 3/2 should occur at 17/(6+2(2/3)11)
= 51/62

- no that doesn't work, the actual value is 13/4.

Can you see the pattern for r non integer I wonder?

The script wouldn't let you go back to those values
because it was handy to be able to enter 1 to
get the best value of r for the
undertone series (conversely the overtone),
but now I've added a button to do that, and
let it go down to just above 1.

http://tunesmithy.co.uk/uo_non_oct.htm

I know you originally thought it better to start
at the undertone series as the minimum value,
and that works well, but perhaps now this makes
sense so that we can explore beyond the edge
at the left side of the tree.

While you are off-line maybe I'll write this up
as a couple of web pages or something, as on
some of the other maths / micro sites.

Robert