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.
5-limit ripple (12p&23p) notation
- herman.miller
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5-limit ripple (12p&23p) notation
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- Dave Keenan
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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.
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
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- 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.
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- Dave Keenan
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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?
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?