Designing a Canasta guitar (was: Designing a Blackjack guitar)

--- In tuning@y..., "Paul Erlich" <paul@s...> wrote:
> Hey Dave, I think the fretting should be Canasta, since that doesn't
> introduce any smaller intervals . . .

Yeah. I guess so. And those smallest intervals are no smaller than they
would be on a 36-EDO guitar. Eek.

> and there should be at least
> one Blackjack scale playable in full on any string.

That's easy. It only means that the open string "chord" must not span more
than 10 generators in a chain. However, I think we should try to keep it to
a span of 8 generators. This can still accomodate 1:5:7 and 1:3:7 (but not
1:3:5:7 which would require 13 gens). If these triads are available on open
strings it should make more chords playable, whether by using open strings
or barres or partial barres.

We have the following possible intervals between open strings.

apprx
ratio cents abbr gens
---------------------
7:8 233c SM2 +2
9:11 350c n3 +3
diss 467c s4 +4
5:7 583c sd5 +5
2:3 700c P5 +6

7:10 633c d5 -5
3:4 500c P4 -6
4:5 383c M3 -7
6:7 267c sm3 -8

The above otonal triads can be expressed on adjacent open strings, in the
following inversions.
4:5:7, 5:7:8, 7:8:10
4:6:7, 6:7:8, 7:8:12

Here are the possible ways of joining these triads with an additional
interval between them while staying within a span of 8 generators.

6:7:8 |9:11|7:8:10
6:7:8 |467c|7:8:10
4:6:7 |5:7 |7:8:10 = 4:6:7|5:7:8:10
4:6:7 |2:3 |7:8:10
6:7:8 |5:7 |4:5:7
6:7:8 |2:3 |4:5:7 = 6:7:8:12:15:21
4:5:7 |7:10|7:8:12 = 4:5:7:10|7:8:12
4:5:7 |3:4 |7:8:12
5:7:8 |7:10|4:6:7
5:7:8 |3:4 |4:6:7 = 5:7:8|3:4:6:7
7:8:10|7:8 |4:6:7 = 7:8:10|7:8:12:14
4:5:7 |7:8 |6:7:8 = 4:5:7:8|6:7:8
4:5:7 |9:11|6:7:8
5:7:8 |4:5 |7:8:12 = 5:7:8:10|7:8:12
5:7:8 |6:7 |7:8:12 = 5:7:8|6:7:8:12
7:8:12|6:7 |5:7:8 = 7:8:12:14|5:7:8
7:8:12|4:5 |5:7:8 = 7:8:12:15:21:24

Of course you can also join the triads directly and put the extra interval
before or after them. Keeping within 8 gens, they can only be joined as

7:8:10|7:8:12
4:6:7 |5:7:8
5:7:8 |6:7:8
7:8:12|4:5:7

The possible ways of completing these fore and aft are

7:8:10|7:8:12|7:10
7:8:10|7:8:12|3:4
7:8:10|7:8:12|4:5
7:8:10|7:8:12|6:7
7:8 |7:8:10|7:8:12
9:11|7:8:10|7:8:12
467c|7:8:10|7:8:12
5:7 |7:8:10|7:8:12

4:6:7 |5:7:8|7:10
4:6:7 |5:7:8|3:4
4:6:7 |5:7:8|4:5
7:8 |4:6:7 |5:7:8
7:10|4:6:7 |5:7:8
3:4 |4:6:7 |5:7:8 = 3:4:6:7 |5:7:8
5:7:8 |6:7:8 |7:8
5:7:8 |6:7:8 |9:11
5:7:8 |6:7:8 |467c
5:7:8 |6:7:8 |5:7
5:7:8 |6:7:8 |2:3 = 5:7:8 |6:7:8:12
7:10|5:7:8 |6:7:8
3:4 |5:7:8 |6:7:8
4:5 |5:7:8 |6:7:8 = 4:5:7:8 |6:7:8
6:7 |5:7:8 |6:7:8

7:8:12|4:5:7 |7:8 = 7:8:12:15:21:24
7:8:12|4:5:7 |7:10 = 7:8:12:15:21:30
7:8:12|4:5:7 |3:4 = 7:8:12:15:21|3:4
7:10|7:8:12|4:5:7 = 7:10|7:8:12:15:21
3:4 |7:8:12|4:5:7 = 3:4 |7:8:12:15:21
4:5 |7:8:12|4:5:7 = 4:5 |7:8:12:15:21
6:7 |7:8:12|4:5:7 = 6:7:8:12:15:21

There are also those cases where there are only 1:3:7 triads or only 1:5:7
triads on the open strings, rather than one of each.

4:6:7:8:12
6:7:8:12:14
7:8:12:14:16
4:5:7:8:10
5:7:8:10:14
7:8:10:14:16

Which can be extended fore or aft in various ways.

Notice that everything so far contains a 7:8. It would ne nice to avoid
such a small step between strings.

There are probably some other ways to do it, such as where a triad does not
occupy 3 successive strings. Such as taking the 4:6 in 4:6:7 as two stages
of 9:11. i.e. 4[9:11|9:11]6:7.
Also 4[6:7|2:3]7:10.

Still too many to choose from. Can anyone suggest other criteria to prune
the search?

I guess we can insist that there is no more than one 7:8?

Maybe the lowest string should be a 1 identity (i.e a 4 in the above) in a
large otonality, to facilitate dronality?

Maybe the highest pair of strings should be a 9:11 apart to facilitate
playing hexads?

It looks like it might be impossible to achieve the last two simultaneously
within a span of 8 generators.

-- Dave Keenan
Brisbane, Australia
http://dkeenan.com