### Re: Magrathean diacritics

Posted:

**Sat May 23, 2020 10:14 pm**Your coin-counting idea at least tells us that we don't need to go beyond 9.5 tinas additive. And if we want to allow users to make the choice between additive and subtractive in the range from 7.5 to 9.5 then we need to go up to 9.5. For your example that needs no tinas, but up to 4 minas, I'm happy to just use two 2 t symbols.

Thanks for giving the fractional tinas for the commas. They led to the thought that the dot up and dot down don't have to be ±¹⁄₂ t. They could be ±¹⁄₃ t. Is there an accurate EDO that's approximately 3 × 8539edo? That would also mean that the maximum error allowed for the various commas would be ±¹⁄₆ t.

And if we impose a requirement that n>7 tina commas must be 5-schisma complements of n<7 tina commas, then the fact that there are 13.90 tinas in the 5-schisma, not 14, implies even tighter constraints on the errors.

Those below 7 tinas can have an error of -0.167 t but only +0.067 t. And vice versa for those above 7 tinas.

It seems your list is still not complete, because when I subtract your 9.21 tina comma from the 5-schisma, I obtain the following 4.69 tina comma that isn't in your list.

But these do not make an acceptable pair for 5 and 9 tinas, because they are more than a sixth of a tina away from 5 and 9.

The obvious 9 tina comma is:

It's annoying that its 5-schisma complement (a candidate for 5 tinas) has a 3-exponent of 13:

There is a clear winner for 8 tinas. It has the best popularity ranking by far, and it is the 5-schisma complement of the 6 tina (2 mina) comma.

The idea occurs, to obtain the 2 tina comma as the difference between the 8 and 6 tina commas, but no, it's complete rubbish, with a 3-exponent of 14 and a popularity rank of 67.

So what about the 5 tina as the difference between the 8 and 3 tinas? Perhaps unsurprisingly, it gives us the same comma we found for 5 tinas as 14 - 9 tinas. i.e. the one with a 3-exponent of 13 (popularity 30). Maybe we should use it anyway.

It seems, at first, that there are two potential complementary 4 tina/10 tina pairs:

But that 9.83 tinas is actually 9.830 or 9.829 and so is more than 1/6th of a tina away from 10 tinas. So it has to be the 17-limit pair.

Unfortunately the 5 tina comma obtained as 9 tinas minus 4 tinas has a 3-exponent of -15 (popularity 33).

Thanks for giving the fractional tinas for the commas. They led to the thought that the dot up and dot down don't have to be ±¹⁄₂ t. They could be ±¹⁄₃ t. Is there an accurate EDO that's approximately 3 × 8539edo? That would also mean that the maximum error allowed for the various commas would be ±¹⁄₆ t.

And if we impose a requirement that n>7 tina commas must be 5-schisma complements of n<7 tina commas, then the fact that there are 13.90 tinas in the 5-schisma, not 14, implies even tighter constraints on the errors.

Those below 7 tinas can have an error of -0.167 t but only +0.067 t. And vice versa for those above 7 tinas.

It seems your list is still not complete, because when I subtract your 9.21 tina comma from the 5-schisma, I obtain the following 4.69 tina comma that isn't in your list.

tinas comma monzo interval cents limit pop 13.90 5s [-15 8 1 0 0 0 0 0 ⟩ 32805/32768 1.953721 5 5 9.21 41503n [ -9 -4 0 3 2 0 0 0 ⟩ 41503/41472 1.293601 11 43 4.69 ?????n [ -6 12 1 -3 -2 0 0 0 ⟩ ?????/????? ?.?????? 11 48

But these do not make an acceptable pair for 5 and 9 tinas, because they are more than a sixth of a tina away from 5 and 9.

The obvious 9 tina comma is:

8.93 539n [ 17 -5 0 -2 -1 0 0 0 ⟩ 131072/130977 1.255240 11 25

It's annoying that its 5-schisma complement (a candidate for 5 tinas) has a 3-exponent of 13:

4.97 ???n [-32 13 1 2 1 0 0 0 ⟩ ??????/?????? ?.?????? 11 30

There is a clear winner for 8 tinas. It has the best popularity ranking by far, and it is the 5-schisma complement of the 6 tina (2 mina) comma.

7.98 13:77n [-20 11 0 1 1 -1 0 0 ⟩ 13640319/13631488 1.121197 13 31

The idea occurs, to obtain the 2 tina comma as the difference between the 8 and 6 tina commas, but no, it's complete rubbish, with a 3-exponent of 14 and a popularity rank of 67.

So what about the 5 tina as the difference between the 8 and 3 tinas? Perhaps unsurprisingly, it gives us the same comma we found for 5 tinas as 14 - 9 tinas. i.e. the one with a 3-exponent of 13 (popularity 30). Maybe we should use it anyway.

It seems, at first, that there are two potential complementary 4 tina/10 tina pairs:

3.84 17:245n [-12 10 -1 -2 0 0 1 0 ⟩ 1003833/1003520 0.539891 17 36 10.06 17:1225n [ -3 -2 2 2 0 0 -1 0 ⟩ 1225/1224 1.413829 17 41 and 4.07 7:3025n [ -4 -3 2 -1 2 0 0 0 ⟩ 3025/3024 0.572403 11 39 9.83 7:605n [-11 11 -1 1 -2 0 0 0 ⟩ 1240029/1239040 1.381317 11 34

But that 9.83 tinas is actually 9.830 or 9.829 and so is more than 1/6th of a tina away from 10 tinas. So it has to be the 17-limit pair.

Unfortunately the 5 tina comma obtained as 9 tinas minus 4 tinas has a 3-exponent of -15 (popularity 33).