However, I noticed that there seem to be two errors in sag_ji4.par:
- 6656/6480 should be reduced to 416/405
- 576/572 doesn't make sense: If reduced, it becomes 144/143, but that doesn't match with the specified accidental / cent range
In addition,











Nicely done. You could include the proper Sagittal symbols in the PDF, at higher resolution than the forum smilies, if you install the Bravura font. Unfortunately, there isn't yet an easy way to type the symbols, as there is for the forum smiles. But it would be possible to make a Windows keyboard layout that would let us do so. Do you use Microsoft Windows?Xen-Gedankenwelt wrote:Ok, I finished the list of 7-limit ratios (including lower limits) with an exact Olympian representation:
7-limit Olympian set.pdf
Thanks for those. I'm pleased to see that we do have the reduced value 416/405 forHowever, I noticed that there seem to be two errors in sag_ji4.par:
- 6656/6480 should be reduced to 416/405
- 576/572 doesn't make sense: If reduced, it becomes 144/143, but that doesn't match with the specified accidental / cent range
My apologies. You are correct thatIn addition,and
don't seem to be an accurate representation of 9:10. A double-apotome
minus
is
which represents the 19-limit ratio 39/38, so
has to be a 19-limit ratio, too. Still, it seems to be a close approximation for 10/9, but not completely accurate.
Those are absolutely correct. Well done.Xen-Gedankenwelt wrote:Just to make sure that I understand systematic comma names correctly:
25:28 = +-[4 0 -2 1> (not in Olympian set) would be 7:25MS+A, and 1792:2025 = +-[-8 4 2 -1> (contained in Olympian set) would be c7:25MS+A, correct?
I think you have to check if there are more. I think it is guaranteed that all single-shaft symbols are the least complex in their size category. But when you add them to an apotome (3-shaft), or subtract them from one or two apotomes (2-shaft and X-shaft), you may end up with some that are not the least complex. It's possible that someone could prove there is at most one less complex ratio for these, but I don't know of such a proof.Some of the ratios in the Olympian set have a somewhat high absolute value of 3-exponent, and the highest I found is [-43 24 1 1>. Is it guaranteed that there is at most one less complex ratio (in terms of absolute value of 3-exponent), or do I have to check if there are more? In the latter case, would I prepend cc, ccc and so on?
My pleasure. It is good to have someone else who understands the mathematical details underlying Sagittal.P.S.: Working with those lists is a great way for me to understand and learn Sagittal notation - thanks for your help so far!