## 5-limit ripple (12p&23p) notation

herman.miller
Posts: 33
Joined: Sun Sep 06, 2015 8:27 am

### 5-limit ripple (12p&23p) notation

I've been trying to work out Sagittal notations for each of the temperaments in Paul's Middle Path paper. 5-limit ripple, with the generator mapping [<1 2 3] <0 -5 -8]>, presents an interesting quirk. It's possible to notate it using the 5-limit Sagittals, but watch out! The comma symbols appear to be pointing the wrong direction.

As a general rule I don't like sharp symbols that sound like flats, and vice versa. So in this case I think the best approach might be to borrow symbols from a 7-limit ripple temperament, [<1 2 3 3|, <0 -5 -8 -2|],

You still get things like being sharper than , which might be a bit confusing, but it's not as bad as the pure 5-limit notation.
Attachments
ripple-12p-23p-5limit-a.png
ripple-12p-23p-5limit-b.png

Dave Keenan
Posts: 2092
Joined: Tue Sep 01, 2015 2:59 pm
Location: Brisbane, Queensland, Australia
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### Re: 5-limit ripple (12p&23p) notation

Ooh yeah. That's a tricky one.

Another approach would be to use 12 pseudo-nominals, so a 12-note chain of the approximately 101 cent generators would be notated the same as 12edo.:
A Bb B C C# D Eb E F F# G G#

Then we only need a sagittal for 12 generators, to represent an approximate 12 cent sharpening -- a Ripple chroma. As you point out, the 5-comma up is negative cents here (it is minus 12 generators) and so it seems we can't use it.

I'll just use . and ' to represent the 12 gen accidental for now. So a chain of 36 is

A. Bb. B. C. C. D. Eb. E. F. F#. G. G#. A Bb B C C# D Eb E F F# G G# A' Bb' B' C' C#' D' Eb' E' F' F#' G' G#'

It may be that a 12-pseudo-nominal scheme like this can only work in Evo flavour, not Revo flavour.

Most chroma-accidental-free perfect fifths (-5 gens) are spelled sensibly. Only the last is weird.
D:A
Eb:Bb
E:B
F:C
F#:C#
G:D
G#:Eb (spelled as a sixth)

One can respell G#:Eb as Ab:Eb but it then becomes an exception to the following rule.

Any fifth whose nominals are in alphabetical order is a wide wolf. +7 generators (approx 707c) instead of -5 gens (approx 695c):
A:E
Bb:F
B:F#
C:G
C#:G#
unless it is narrowed by a (12 gen 12 cent) ripple chroma. e.g. A:E. or A':E

The chroma-accidental-free majorish thirds (-8 generators) are spelled:
F:A
F#:Bb (spelled as a fourth)
G:B
G#:C (or Ab:C)

Other majorish thirds are spelled:
E:G#.
Eb:G.
D:F#.
C#:F. (spelled as a fourth)
C:E.

But that tells us that we can use the 5-comma symbols in a positive manner after all. The downward dot "." can be replaced by , and the upward tick ' can be replaced by . So the chain of 36 can be notated:

A B B C C D E E F F G G A B B C C D E E F F G G A B B C C D E E F F G G

or

A B B C C D E E F F G A A B B C C D E E F F G A A B B C C D E E F F G A

and I don't see why you couldn't substitute for etc, and use Revo.

herman.miller
Posts: 33
Joined: Sun Sep 06, 2015 8:27 am

### Re: 5-limit ripple (12p&23p) notation

You can respell G#:Eb as G#:D# without any trouble. It makes more sense to notate G#. as Ab -- I almost ended the sentence there, but that would be confusing since Ab would have looked like Ab. -- using . and ' as accidentals makes it a bit awkward to mention them in a sentence. You can see how that works out on this new chart. In practice I wouldn't use both A and G but either one is fine. This is a nice symmetrical arrangement of nominals and it works because the or works out to be (+0, +1) consistently.

Attachments
ripple-12p-23p-5limit-c.png

Dave Keenan
Posts: 2092
Joined: Tue Sep 01, 2015 2:59 pm
Location: Brisbane, Queensland, Australia
Contact:

### Re: 5-limit ripple (12p&23p) notation

Oh yes. That's beautiful. So G = A and G:E can be spelled A:E and everything obeys the rule that it's a properly -tempered 2:3 if it's spelled anti-alphabetically, or if it's spelled alphabetically and is narrowed by a 5-comma symbol. Same for tempered 4:5s.

This seems to violate all kinds of Sagittal rules about chains of fifths and mapping of commas, but it works so well!

I was misled into thinking G# = Ab by my thinking of it as using 12edo notation as names of 12 MOS nominals, or pseudo-nominals.

But the right way to look at it would seem to be:
CD=DE=FG=GA=AB = 2 gens
BC=EF = 1 gen
# = 1 gen
= 1 gen

In combination with the mapping, where prime 3 is -5 gens, this results in the conventional chain of fifths FCGDAEB no longer being a chain of fifths. Whenever two letters are in alphabetical order, it is broken. Ab×EbAb×FC×GDA×EB×F#C#×G#.

An actual chain of fifths looks like E B F C G D A E B F C But you'd need Ripple[51] to get a chain of fifths that long.

Is there some way we can resurrect some version of the maths where we calculate the number of gens that a symbol like corresponds to, after we do funny things to the nominals like this? How do we work out the new mapping?