Re: Dan's over / under scales

Hi there,

Just seen it for the case s>1, was obvious really
but I was looking for something too compliciated

For repeat r/s, and s > 1,
(r+1)/(s+1) occurs at c = d = s, e = s+1.

So for x = (a+n)/a, (r+1)/(s+1) occurs at
(sa+sn)/(sa+(s+1)n)

Testing several values in the tree confirms this

It doesn't depend on r, and if you fix s, then
the ratio you are looking for is always in
the same place.

e.g. trying s = 4,
for x = 11/10
5/4: 6/5 occurs at 44/45,
3/2: 7/5 at 44/45
7/4: 8/5 at 44/45
2/1: 9/5 at 44/45
9/4: 2/1 at 44/45
...

For ratios of form r/1, the formula reduces to
(a+n)/(a+2n)

e.g. for x = 11/10, the required ratio is always
found at 11/12

e.g. r = 2

2/1: 3/2 at 11/12
3/1: 2/1 at 11/12
4/1: 5/2 at 11/12
5/1: 3/1 at 11/12
etc.

My previous observation was that for r integer,
(r+1)/2 occurs at (r-1)(a+n)/((r-1)(a+2n))
- which reduces to this formula if you
cancel the (r-1)s, as you can since the tree
has the same ratio at all integer multiples
of the degree, e.g. if it occurs in
11th position in 12th row, it will also
occur at 22nd position in 24th row etc.

Anyway I've written up the observations now on
the page, and will try to prove it after the
release of FTS.

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

Robert