Ok, I finished the list of 7-limit ratios (including lower limits) with an exact Olympian representation:

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, 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.

The list only contains up-accidentals, and I omitted cent values, systematic comma names, and very large ratios (monzos are still listed). If the data is fine, I'll update the list in my opening post.## List of 7-prime limit accidentals

- Xen-Gedankenwelt
**Posts:**19**Joined:**Fri Sep 04, 2015 10:54 pm

- Dave Keenan
- Site Admin
**Posts:**239**Joined:**Tue Sep 01, 2015 2:59 pm**Location:**Brisbane, Queensland, Australia-
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### Re: List of 7-prime limit / 6125-odd limit accidentals

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

I note the potential for confusion of the vertical bar | with the digit 1. Some fonts are worse than others, for this. That's why I use square brackets for bra <...] and ket [...> vectors except when writing an inner product <... | ...>, in which case there will be a space on either side of the vertical bar. An alternative to using square brackets is to always include a space between the vertical bar and any digits.

Thanks for those. I'm pleased to see that we do have the reduced value 416/405 for in George's JI notation spreadsheet, the JI notation levels diagram and in footnote 19 on page 24 of http://sagittal.org/sagittal.pdf.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

I see that 576/572 was erroneously given for in sag_ji4.par. George's spreadsheet gives the 13-limit comma 567:572 for that symbol. So it seems "67" was accidentally transposed to "76" when making sag_ji4.par.

I have now corrected both of those in http://sagittal.org/sag_ji4.par. Thanks.

My apologies. You are correct that is not an exact representation of 9:10. That symbol represents a 19-limit ratio, as you say. When untempered they differ by only 0.003 cents. The only exact representations of 9:10 involve a change of nominal, e.g. C:D or C:E = C:EIn 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.

- Xen-Gedankenwelt
**Posts:**19**Joined:**Fri Sep 04, 2015 10:54 pm

### Re: List of 7-prime limit / 6125-odd limit accidentals

Thanks for your reply!

As a start, I separated the ratios into two lists, depending on whether they have an exact representation in the Olympian symbol set, or not.

Here is my current to-do list:

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?

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?

P.S.: Working with those lists is a great way for me to understand and learn Sagittal notation - thanks for your help so far!

As a start, I separated the ratios into two lists, depending on whether they have an exact representation in the Olympian symbol set, or not.

Here is my current to-do list:

- Complete the ratios and accidentals in the first list (exact / Olympian)
- Add accurate dual-accidentals and approximate single-accidentals in the second list
- Add systematic comma names

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?

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?

P.S.: Working with those lists is a great way for me to understand and learn Sagittal notation - thanks for your help so far!

- Dave Keenan
- Site Admin
**Posts:**239**Joined:**Tue Sep 01, 2015 2:59 pm**Location:**Brisbane, Queensland, Australia-
**Contact:**

### Re: List of 7-prime limit / 6125-odd limit accidentals

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?

I have an Excel spreadsheet that automatically names any ratio up to the double-apotome. I'd be happy to make it available after some tidying up. Would this be of any use to you? Do you have a recent version of Excel?

It presently uses:

complex (c)

supercomplex (sc)

hypercomplex (hc)

ultracomplex (uc)

5-complex (5c)

6-complex (6c)

...

But if you think the ordering of super, hyper, ultra is not sufficiently standardised, you could use 2c, 3c, 4c instead of sc, hc, uc.

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!