HI there,

for anyone following this, I've found the scale tree

generating formula for the general case of r integer

- probably. Also found interesting data for r/s

rational but can't yet see the pattern there.

In all this, it is entirely experimental, but

to avoid continually saying "probably", I'll

just present the (pro-tem) conclusions, without comment.

Suppose one wants to find location for (k+m)/k

in the under / over series scales with repeat

at r/s.

The general format seems to be:

For x = (a+n)/a, (k+m)/k occurs at c(a+n)/(da+en)

where c, d and e depend on k and m, and need to be

determined.

4/3 occurs at

x 1.1 1.2 1.3

r 2 11/13 12/16 .. c = 1, d = 1, e = 3

r 3 11/16 12/22 .. c = 1, d = 1, e = 6

r 4 11/19 12/28 .. c = 1, d = 1, e = 9

5/4 at

r 2 11/14 12/18 .. c = 1, d = 1, e = 4

r 3 11/18 12/26 .. c = 1, d = 1, e = 8

r 4 11/22 12/34 .. c = 1, d = 1, e = 12

So we see:

If scale repeat is r (integer), and the ratio

is superparticular, e = (r-1)*k

...................................

Now try a couple of others:

5/3 (k = 3, m = 2) at

x 1.1 1.2 1.3

r 2 22/23 24/26 26/29 .. c = 2, d = 2, e = 3

r 3 22/26 24/32 24/38 .. c = 2, d = 2, e = 6

5/2 (k = 2, m = 3) at

r 2

r 3 33/34 36/38 .. c = 3, d = 3, e = 4

r 4 33/36 .. c = 3, d = 3, e = 6

So we see:

If scale repeat is r (integer),

c = m, d = m, and e = (r-1)*k

This is okay.

...................................

What about general case of repeat at r/s rational:

Try superparticular first

r/s = 7/2,

x = 1.1 1.2

3/2 at 22/30 24/40 .. c = 2, d = 2, e = 10

4/3 at 22/35 24/50 .. c = 2, d = 2, e = 15

r/s = 7/3,

x = 1.1

3/2 at 33/38 .. c = 3, d = 3, e = 8

4/3 at 33/42 .. c = 3, d = 3, e = 12

5/4 at 33/46 .. c = 3, d = 3, e = 16

6/5 at 33/50 .. c = 3, d = 3, e = 20

r/s = 7/4,

x = 1.1

3/2 at 44/46 .. c = 4, d = 4, e = 6

4/3 at 44/49 .. c = 4, d = 4, e = 9

So we see:

c = s, d = s, e = k(r-s)

Now the general rational case:

r/s 7/3

x = 1.1 1.2 1.3

5/3 at 33/36 36/42 39/48 .. c = 3, d = 3, e = 6

7/5 at 33/40 36/50 39/60 .. c = 3, d = 3, e = 10

11/7 at 33/37 36/44 39/51 .. c = 3, d = 3, e = 7

11/5 at 99/100 108/110 117/120 .. c = 9, d = 9, e = 10

11/6 at 55/58 60/66 65/74 .. c = 5, d = 5, e = 8

9/5 at 33/35 9/10 13/15 21/25

r/s = 7/4

x = 1.1 1.2 1.3

8/5 at 44/45 48/50 52/55 .. c = 4, d = 4, e = 5

I can't see the pattern here.

...................................

Constructing the scale tree lookup

For r = 2,

(k+m)/k occurs at m(a+n)/(a+kn)

So 3/2 occurs at (a+n)/(a+2n)

Then if 1/1 is at 0/v and 2/1 is at w/w

3/2 is at v/(v+w) so we can find v and w here.

...................................

General case of r integer,

To make the lookup tree, want to find (r+1)/2

Putting this in the form (k+m)/k

k = 2, m = r-1 then putting that into

m(a+n)/(ma+k(r-1)n)

we get (r-1)(a+n)/((r-1)a+2(r-1)n)

=

(r-1)(a+n)/((r-1)(a+2n))

Then if 1/1 is at 0/v and r/1 is at w/w

(r+1)/2 is at t/(s+t) so we can find v and w here.

as before.

...................................

General case:

To construct the scale tree lookup,

then you need to find a value for

(r+1)/(s+1), from which you can get

r/s and 1/1 by looking for them

in the form 1/1 = 0/v and r/s = w/w

as before, but here we need the formula

for (r+1)/(s+1) which we don't know yet.

However, to make the scale tree - the et one

to use to look up the ratios one, we don't need

the general case, we only need to

know the formula for (r+1)/(s+1) so

let's try that out.

x = 1.1:

r/s = 3/2, 4/3 at 22/23 c = d = 2, e = 3

r/s = 4/3, 5/4 at 33/34 c = d = 3, e = 4

...

r/s = 10/9 11/10 at 99/100 c = d = 9, e = 10

r/s = 7/4, 8/5 at 44/45 c = 4, e = 5

r/s = 7/2, 8/3 at 22/23 c = 2, e = 3

I don't get the pattern yet...

Anyone see it?

The page is now making Dan's scale tree for scales

with integer repeats.

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

Robert