On 26 Sep 2013, at 17:16, Phil Holmes <m...@philholmes.net> wrote: >> The section originates with me but I got diverted into trying to create a >> more elegant solution for how to rewrite accidentals in transposed music. It >> was all related to the need for an effective chromatic transposition >> solution that also worked well with arbitrary microtonal accidentals. >> >> I was also rather discouraged by the fact that the quarter-tone arrow >> notation issue didn't find a solution -- see: >> https://code.google.com/p/lilypond/issues/detail?id=1278 >> > > I think it's waiting for someone to propose how it could be represented in > LilyPond.
For one microtonal accidental, one needs, in addition to the minor/major seconds m and M, a neutral second n. For a pitch x = r*m + s*M + t*n, compute its degree deg(x) := r + s + t, which is its staff position, and subtract the staff pitch. There remains a new pitch, which I also call x, but now with r + s + t = 0. As sharps/flats alter with a multiple of r - s, reduce using them so that only one of r, s is non-zero. Assume first that t = 1, i.e., one n. Then it must be either n - M or n - m. We have six microtonal symbols, sharp/natural/flat with up/down arrows, but it will, as we shall see, suffice with four. One way to make a choice is to conceptualize n as below or above (m + M)/2: if it is a small or large neutral. This choice is purely formal at this point, but will be of importance when plugging in values. If one thinks of n as between m and M, which is possible with actual values by reducing using sharps and flats, then n' := (m + M) - n (the minor third m3 complement) is also a neutral between m and M. If n is small, then n' is large, and vice versa. Returning to the situation above, assume n to be small. The up/down arrows will be thought of as changing with a small amount: n - m. There are two possibilities: n - m, and n - M. In the first case, n - m represents a small positive amount, so it is the natural with up arrow. In the second case, it is a large negative amount, so it is the flat with up arrow. Assume now that t = -1, so the two cases are m - n and M - n. The first case lowers with the small amount n - m, so it is a natural with a down arrow. And M - n raises with a large amount, so it must be a sharp with a down arrow. If the absolute value |t| of t is larger than 1, then one needs as many arrows as |t|: up if t is positive, and down if t is negative. Two symbols where not used: sharp with up arrow and flat with down arrow. But they conceptually fall without the region of raising a sharp M - m or lowering with a flat -(M - m), and can in fact be reduced using a natural with up/down arrow plus a sharp/flat. So here, one would need notation simplification algorithm. Hans _______________________________________________ lilypond-devel mailing list lilypond-devel@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-devel
