Re two-dimensional three-term BP scales

Hi Dan,

I've been away on holiday. I read some of your earlier posts while on holiday,
but haven't been posting much. Planned to do some reading around on this topic,
but lost all my photocopies in a carrier bag in a coach station. Had a really
refreshing break instead.

Let's see if I understand what you are saying:

> The Bohlen-Pierce Lambda scale is:
>
> 1/1 25/21 9/7 7/5 5/3 9/5 15/7 7/3 25/9 3/1
>

(follows: table of all the cyclic re-arrangements of the steps
of the scale to nearest cents)

>
> Generalized this as a [5,4] scale (where P = 1:3) allows for some
> interesting weighted Tribonacci interpretations.

I.e. the steps are approximately of two sizes, 5 small and 4 large:

steps: 25/21 27/25 49/45 25/21 27/25 25/21 49/45 25/21 27/25
where 49/45 ~= 27/25, (using ~= for approximately equal)
i.e.
L S S L S L S L S
= 5 * S and 4 * L.
L/S ~= log(25/21)/log(49/45) ~= 2

So one can approximate it as the steps
2 1 1 2 1 2 1 2 1
in 13 equal divisions of 3/1.

This is an ordinary Myhill scale - i.e. each interval class has two interval sizes.

(E.g. the 3 step intervals are either L S S or L S L)

Looking at the results in SCALA, the interval classes 2 and 7 are the ones
which have an interval with a single incidence, and the other intervals
are generators.

(For newbies to this topic:) that is by the standard result for Myhill scales:
2 is relatively prime with 9 (the number of notes in the scale), so the
two step intervals make a cycle of intervals that span the scale, and if
only one of them is left out, the cycle is broken, but the remaining intervals
still span the scale.

They are 3^(3/13) , which is 438.913 cents, and 3^(10/13). On repeating these
intervals, and reducing them into range 1/1 3/1, these do indeed generate
a scale with steps 2 1 1 2 1 2 1 2 1 in 13 equal divisions of 3/1.

As cents:
0 cents 292.608 cents 438.913 cents 585.217 cents 877.825 cents 1024.13 cents 1316.74
cents 1463.04 cents 1755.65 cents 1901.96 cents

Is this correct so far?

I lose you after that.

Here is what I was expecting you to do next:

- - - - - - idea - - - - - - - - - -

To simplify the numbers, let's use a modified cents notation c* where 100c* = one of these
new divisions, i.e. 1300c* = 3/1.

So the steps are L S S L S L S L S where L = 200c* and S = 100c*.

Then the generators are 300c* and 1000c*.

Idea is to temper alternate generators by a small increment.

E.g. just by way of example, one could use 299c* 301c* 299c* 301c* ... to generate the
scale,

This gives
3^(2.99/13) 3^(3.01/13) (alternating)
or 437.45 cents, 440.376 cents.

Yields scale (in step notation)
1/1 291.145 cents 146.304 cents 294.072 cents 146.304 cents 291.145 cents 146.304 cents
294.072 cents 146.304 cents 146.304 cents

Checking in SCALA, it is indeed trivalent.

Steps are: M S L S M S L S S
Or, rotating the steps cyclically to get the original mode
L S S M S L S M S
0 cents 294.072 cents 440.376 cents 586.68 cents 877.825 cents 1024.13 cents 1318.2 cents
1464.51 cents 1755.65 cents 1901.96 cents
What's happened is that all the S steps are the same size as before, and the
L steps are larger, and the M steps are smaller.

In the c* notation they are
M = 199c*, L = 201c*, S = 100c* as before.

It would be an interesting mathematical project to try to prove that this method will
always work for a Myhill scale with an odd number of notes in it, as seems to be a reasonable
conjecture at this point of time.

Conjecture would be that the method always converts the Myhill scale into a
trivalent scale if the increment is small enough.

One would also ask if it is pairwise well formed (see David Clampitt's post
http://www.egroups.com/message/tuning/16482)
L S S M S L S M S
Setting L = M -> original scale
Setting S = M -> L M M M M L M M M, Myhill
Setting S = L -> L L L M L L L M L - mode of previous scale - Myhill.
So this one is anyway.

- - - - - - end of idea - - - - - - - - - -

Your method also uses alternating generators, but I don't see how you obtain them.

Can you explain?

Also, have I followed you correctly as far as I got?

Robert Walker

Robert Walker wrote,

<< Let's see if I understand what you are saying: >>

Yes, we're in agreement with all this. One thing to perhaps note here
is that adjacent fractions of any GCD reduced [a,b] index will always
give the generator within a given periodicity (P). And when a chain of
these generators that is (a+b)-1 long results in a built-in ordering
rule. This is why I've felt that some three-term analogue to adjacent
fractions is what's really needed to completely generalize Trihill
scales. This seems to be a very difficult problem though, and as no
one I asked was able to help, I went about looking for some alternate
ways to go about tackling the problem.

What I did was to simply take the single generator model

1/1---3/2

and convert it into a generalized take on your initial two generator
hunch

5/4
/ \
/ \
1/1---3/2

by collapsing a single generator chain on itself:

8192/6561
/ \
/ \
1/1---3/2

And if limited to scales with an odd number of tones this works like a
charm. Different weightings of the adjacent fractions can be used to
optimize most anything you'd like. I tend to like clear delineation in
an [a,b,c] index -- in other words a real practical sense of
three-stepsize cardinality.

<< Your method also uses alternating generators, but I don't see how
you obtain them. Can you explain? >>

Looking at a two-term generator as a generalized fifth in a circle (or
spiral, or whatever) of fifths really helps, because if you multiply
it (I'll call "it", the generator, X) by the forth term of the
two-term/Fibonacci index, then the difference between this and P
(whatever your given periodicity is) will result in what I'd call a
generalized Pythagorean comma.

If P = 1:3 and your [a,b] index is [5,4], then if you multiply the
converted [a,b,c] generator by the weighted constant of 2.5 you get
the generalized Pythagorean comma of 1:3^(1/30).

Now if you look at the last example I gave in the "two-dimensional
three-term BP scales" post, the three-term index of [1,4,4] gives a
Tribonacci series of

1, 4, 4, 9, 17, 30, ...

So the single generator model is converted into the generalized take
on your initial two generator hunch by collapsing a single generator
chain on itself, and

0---23---16---9---2---25---18---11---4

becomes

26----19----12-----5
/ \ / \ / \ / \
/ \ / \ / \ / \
0----23----16-----9-----2

(BTW, this is also the three-term Fibonacci to Tribonacci Yasser
parallel that Paul Erlich initially posted about way back when.)

<< have I followed you correctly as far as I got? >>

Yes I think so, but please do ask questions if any of the rest of it
is still unclear.

thanks,

--Dan Stearns