Yahoo Tuning Groups Ultimate Backup tuning Fine tuning 5-limit 2D temps -- has this been discussed?

Hello tuners.

Although I've noticed that much of the discussion has moved over to XA, I think this topic deserves being brought up here.

Suppose we have three different versions of meantone: 1/4-comma, 1/3-comma, 2/7-comma. Let's find the mistuning in the following factors: 2/1, 3/1, 5/1, 3/2, 5/2, 5/3.

In the first case, we have the following comma fractions:

0, -1/4, 0, -1/4, 0, 1/4

For our second example, we get:

0, -1/3, -1/3, -1/3, -1/3, 0

And finally, we have:

0, -2/7, -1/7, -2/7, -1/7, 1/7

The sum of absolute errors is therefore 3/4, 4/3, or 1 comma, respectively.

According to this type of measurement, 1/4-comma meantone should be more "in tune" than 2/7-comma. But this omits the fact that only 2 of the 6 factors have the highest error in 2/7-comma meantone while in 1/4-comma meantone there are 3 of them. This becomes even more noticeable when I compare just 3 factors (3/1, 5/1, 5/3), only one of which has the highest error in 2/7-comma meantone.

This raises several questions:

1) What other procedure for finding a generator should we choose (I've excluded minimax tuning right now) if we want the lowest non-zero absolute error to be the lowest possible and also to appear in as many intervals as possible?

2) How should we apply this consideration in the case of tempered octaves?

Any suggestions or comments will be highly appreciated.

Petr

🔗Mike Battaglia <battaglia01@...>

🔗Petr Pařízek <petrparizek2000@...>

Hi Mike.

Well, probably both of these, provided I understand the termonilogy correctly. Hope my terminology won't be too confusing for you in what I'm going to say next.

Honestly, I'm not sure what would be the best way to go if octaves were tempered as well. But for non-tempered octaves, the lowest non-zero error always appears more than once if I compare 3/1, 5/1 and 5/3. And I think this should be one of the criteria. Of course, the error appears twiceeven in 1/4-comma meantone, but it isn't the lowest possible one (i.e. 1/4-comma) compared to the lowest error in 2/7-comma meantone (only 1/7-comma). This indicates to me that now we have one more requirement -- that one of the errors appearing more than once should be the lowest possible non-zero value.

It seems to me that both of these requirements should be met. Because if I went for, let's say, 1/11-comma meantone, then the error in the 3/1 is tiny but it appears just once, which makes the other two errors (in 5/1 and 5/3) meaninglessly high.

Hope my words are understandable.

Petr

🔗Carl Lumma <carl@...>

Hi Petr,

Not sure I'm following your inquiry. It looks like you are attempting some sort of tuning optimization, where the errors of 3/1, 5/1, and 5/3 are given special importance.

A lot of time has been spent on tuning optimizations on these lists over the years, as you may be aware. A few interesting discoveries were made, having to do with the relationship between errors of prime intervals (2/1, 3/1, 5/1) and all other intervals (3/2, 5/3 etc) in regular tuning systems. These are described in some detail on the wiki, e.g.

http://xenharmonic.wikispaces.com/TOP+tuning
http://xenharmonic.wikispaces.com/Tenney-Euclidean+Tuning

Or maybe I have misunderstood your point/question,

-Carl

--- In tuning@yahoogroups.com, Petr PaÅÃzek <petrparizek2000@...> wrote:
>
> Hi Mike.
>
> Well, probably both of these, provided I understand the termonilogy
> correctly. Hope my terminology won't be too confusing for you in
> what I'm going to say next. [snip]

🔗Petr Pařízek <petrparizek2000@...>

Carl wrote:

> Not sure I'm following your inquiry. It looks like you are attempting some > sort of tuning optimization,
> where the errors of 3/1, 5/1, and 5/3 are given special importance.

Well, I'm trying to take a concept which I more or less know how to apply to 2D temperaments with pure octaves and find a way of generalizing the concept also for the case of tempered octaves.

Let me give you three examples, I hope you'll get the idea then.

In quarter-comma meantone, the lowest non-zero absolute error in the three primary ratios (3/1, 5/1, 5/3) is 1/4-comma and it appears in two of them (since 5/1 is pure). However, in 2/7-comma meantone, although 3/1 is a bit more mistuned, the error which appears more than once (this time in 5/1 and 5/3) is even lower (only 1/7-comma). If we leave the octaves pure, then this is the configuration where the error appearing more than once is the lowest possible.

In negri, if the generator is the 7th root of 5/3, then both 3/1 and 5/1 are mistuned by 1/7 of the vanishing interval (let's call that a wide diesis). However, if the generator is the 11th root of 20/9, then the error appearing more than once (both in 3/1 and in 5/3) is only 1/11 of the wide diesis.

In semisixths, if the generator is the 9th root of 10, then both 3/1 and 5/3 are mistuned by 1/9 of the vanishing "semi-diesis". However, if the generator is the 16th root of 60, then both 3/1 and 5/1 are mistuned by only its 1/16.

Therefore, if the generator counts can easily be found and if octaves are untempered, then the procedure is simple:

- Find the three generator counts needed for 3/1, 5/1, 5/3.

- Pick the two that have the highest absolute values and take their sum.

- Find the JI ratio which is approximated by this number of generators (with no periods), take the exact JI interval size, and split it into that number of equal parts. The result is the desired generator.

What I'm not sure is 1) how I should proceed if only MOS step sizes are known but generator counts aren't, and 2) how I can find a tuning where the error appearing more than once is the lowest possible but pure octaves aren't required.

Petr

🔗petrparizek2000 <petrparizek2000@...>

🔗Mike Battaglia <battaglia01@...>

You should look at the TE tuning. TE tuning, according to Graham's
research, is indistinguishably close from the optimal tuning that you get
if you minimize RMS error of Tenney-bounded subsets of the lattice as the
bound increases to infinity. If you want pure octaves, go with POTE, and
that gives you something indistinguishably close from what you get if you
minimize the RMS error of odd-limit-bounded subsets of the lattice as the
bound increases to infinity.

-Mike

On Sun, Apr 7, 2013 at 5:29 PM, Petr Pařízek <petrparizek2000@...>wrote:

> **
>
>
> Carl wrote:
>
> > Not sure I'm following your inquiry. It looks like you are attempting
> some
> > sort of tuning optimization,
> > where the errors of 3/1, 5/1, and 5/3 are given special importance.
>
> Well, I'm trying to take a concept which I more or less know how to apply
> to
> 2D temperaments with pure octaves and find a way of generalizing the
> concept
> also for the case of tempered octaves.
>
> Let me give you three examples, I hope you'll get the idea then.
>
> In quarter-comma meantone, the lowest non-zero absolute error in the three
> primary ratios (3/1, 5/1, 5/3) is 1/4-comma and it appears in two of them
> (since 5/1 is pure). However, in 2/7-comma meantone, although 3/1 is a bit
> more mistuned, the error which appears more than once (this time in 5/1
> and
> 5/3) is even lower (only 1/7-comma). If we leave the octaves pure, then
> this
> is the configuration where the error appearing more than once is the
> lowest
> possible.
>
> In negri, if the generator is the 7th root of 5/3, then both 3/1 and 5/1
> are
> mistuned by 1/7 of the vanishing interval (let's call that a wide diesis).
> However, if the generator is the 11th root of 20/9, then the error
> appearing
> more than once (both in 3/1 and in 5/3) is only 1/11 of the wide diesis.
>
> In semisixths, if the generator is the 9th root of 10, then both 3/1 and
> 5/3
> are mistuned by 1/9 of the vanishing "semi-diesis". However, if the
> generator is the 16th root of 60, then both 3/1 and 5/1 are mistuned by
> only
> its 1/16.
>
> Therefore, if the generator counts can easily be found and if octaves are
> untempered, then the procedure is simple:
>
> - Find the three generator counts needed for 3/1, 5/1, 5/3.
>
> - Pick the two that have the highest absolute values and take their sum.
>
> - Find the JI ratio which is approximated by this number of generators
> (with
> no periods), take the exact JI interval size, and split it into that
> number
> of equal parts. The result is the desired generator.
>
> What I'm not sure is 1) how I should proceed if only MOS step sizes are
> known but generator counts aren't, and 2) how I can find a tuning where
> the
> error appearing more than once is the lowest possible but pure octaves
> aren't required.
>
> Petr
>
>
>
>
>

🔗Petr Pařízek <petrparizek2000@...>

🔗Mike Battaglia <battaglia01@...>

Computing it is nontrivial and requires a decent amount of linear
algebra; you'll specifically need a routine to compute the
Moore-Penrose pseudoinverse. Graham wrote the original paper on it
here, calling it TOP-RMS: http://x31eq.com/primerr.pdf . The Wiki has
good information as well, under the "Tenney-Euclidean tuning"
sections.

It's all quite abstract and mathematical, and Graham's done the work
by coding it up on his temperament finder anyway. So, you need to ask
yourself: is your goal really to understand and code up an
implementation of the algorithm yourself? If so, I can help walk you
through it, but if not, you might be happy just to use what the
temperament finder spits out.

So you at least understand the basic goal of the whole procedure, for
prime-limits, the approach is essentially as follows:
1) For any tuning of your temperament, compute the tuning of each
prime. This set of prime tunings is called the "tuning map" for the
temperament.
2) Subtract the tempered primes from their idealized JI equivalents
and take the absolute value to come up with the "error" for that
prime.
3) Take the root-mean-squared of all of the prime errors. This is the
"RMS error" for that tuning.
4) Find the tuning map supporting your temperament which has RMS error
as low as mathematically possible.
5) This tuning map has the magical "minimizing average error over all
intervals" properties that I mentioned before, where "all intervals"
is formulated as that one specific limit I mentioned.

#4 is the nontrivial step which is made easy if you have a good
pseudoinverse routine. But again, you don't need to reinvent the wheel
because Graham's already done the work on his temperament finder. For
instance, here's 5-limit meantone:
http://x31eq.com/cgi-bin/rt.cgi?ets=12_19&limit=5 . Note that it says

Tuning Map (cents)
<1201.397, 1898.446, 2788.196]

This is, in fact, the TE tuning for meantone. Done! You read this
tuning map as though it were a val with real coefficients, so in this
case the first one is the tuning for 2/1, the second is the tuning for
3/1, and the last is the tuning for 5/1. Note that octaves are detuned
as well, so that error is spread out along the octaves too.

To compute the POTE tuning instead of the TE tuning, you want to scale
the whole thing linearly so that the first coefficient becomes
1200.000. So in this case, multiply the whole map by
1200.000/1201.397, and you get <1200.000 1896.238 2784.954|. This is
the POTE tuning, easy.

TOP is similar, except you can think of it as minimizing maximum error
rather than average error. It strives to make the worst-case detuning
as palatable as possible. TE, in contrast, is better at spreading the
error out as evenly as possible. In practice they both almost always
agree very closely, to within a few cents, so it's not worth really
agonizing over.

A final caveat: Graham's code currently doesn't compute the TE tuning
properly for subgroup temperaments. If you want to compute it for
subgroup temperaments, just compute it for the full-limit version
instead with the same kernel and then throw the extra intervals away.
So for instance, if you want to work out the TE tuning for the
2.9/7.5/3 245/243 temperament, which is basically a subgroup version
of sensi, just use the temperament finder to compute it for the
2.3.5.7 245/243 temperament instead, and then throw away the intervals
you don't want.

-Mike

On Mon, Apr 8, 2013 at 5:15 AM, Petr Pařízek <petrparizek2000@...> wrote:
>
>
>
> Hi Mike,
>
> can you give an example of the procedure, one with pure octaves and one
> without?
>
> Petr

🔗petrparizek2000 <petrparizek2000@...>

🔗Marcel de Velde <marcel@...>

Assuming that we make the octave and 5/1 a little wider and the 3/1 a little narrower:
4 * (3/1 - x) - 4 * (2/1 + x) = 5/1 + x

x = ?

And if I'm judging this wrong and the octave need to be narrower then it is:
4 * (3/1 - x) - 4 * (2/1 - x) = 5/1 + x

I'll let someone who's not a highschool dropout like me solve the algebra ;-p

> Hmmm, I was wrong again. I wanted to say: If the three ratios of 2/1 > and 3/1 and 5/1 could have the same error and that would be the lowest > possible one. Yes, that's it. Finally I formulated my wish into words > ... Never thought I would have so much trouble with that. :-D
>
> Petr
>
>

🔗Graham Breed <gbreed@...>

On Monday 08 April 2013 11:00:19 you wrote:
> Computing it is nontrivial and requires a decent amount
> of linear algebra; you'll specifically need a routine to
> compute the Moore-Penrose pseudoinverse. Graham wrote
> the original paper on it here, calling it TOP-RMS:
> http://x31eq.com/primerr.pdf . The Wiki has good
> information as well, under the "Tenney-Euclidean tuning"
> sections.

No, it's not that difficult, not for rank 2. There's a worked
example in the PDF (p.11 and neighbors) and you can follow
it with a pocket calculator. Ideally a scientific calculator
with statistical features to calculate the sums and sum-
squareds.

There's also the standard deviation algorithm, which is a
good approximation the the POTE tuning and error. It's on
page 14 of the PDF. The optimal generator as a fraction of
the period is

-cov(M0, M1)/var(M1)

where:

cov() is the covariance.
M0 is the weighted period mapping.
M1 is the weighted octave-equivalent generator mapping.
var() is the variance (square of the standard deviation).

Scientific calculators may calculate the covariance and
standard deviation directly.

For higher ranks, you don't need a Moore-Penrose
pseudoinverse. What you need is a way of solving a linear
least squares problem. The pseudoinverse will do that, and
it may be simple to find the pseudoinverse once you've solved
the least squares problem, but people managed to solve such
problems before Moore and Penrose came along.

Graham

🔗Marcel de Velde <marcel@...>

Oh I just found an online math solver :)
http://www.myalgebra.com/

And of course I made a stupid errors in my formula in my previous message, I wrote it mathematically all wrong. I used - and + where I should have written / or *, and should use "powers of" in some places where I used * and /.

So when I input the correct formula:
(3/x * 3/x * 3/x * 3/x) / (2*x * 2*x * 2*x * 2*x) = 5*x
There is "no solution"..
I'm taking this means there is an infinite amount of x that will work.
I don't know how to input in a calculator how to give the lowest value of x.
Can someone with math experience do this? Then you'll have your answer.

For
(3/x * 3/x * 3/x * 3/x) / (2/x * 2/x * 2/x * 2/x) = 5*x
The answer is x = 81/80
Which is true of course as 4 * (3/1 lowered by a syntonic comma) "minus" 4 * (2/1 lowered by a syntonic comma) = 5/1 raised by a syntonic comma (or better said a Pythagorean major third of 81/16)

For (3/x * 3/x * 3/x * 3/x) / (2/x * 2/x * 2/x * 2/x) = 5/x
x = 80/81

For (3*x * 3*x * 3*x * 3*x) / (2/x * 2/x * 2/x * 2/x) = 5/x
x = "no solution"

For (3*x * 3*x * 3*x * 3*x) / (2*x * 2*x * 2*x * 2*x) = 5/x
x = 80/81

For (3/x * 3/x * 3/x * 3/x) / (2*x * 2*x * 2*x * 2*x) = 5/x
x = "no algebraic solution". (I'm taking this means that there is no possible answer for x)

For (3*x * 3*x * 3*x * 3*x) / (2/x * 2/x * 2/x * 2/x) = 5*x
x = "no algebraic solution"

For (3*x * 3*x * 3*x * 3*x) / (2/x * 2/x * 2/x * 2/x) = 5*x
x = 81/80

These are all 8 possibilities for the formula.

-Marcel

> Assuming that we make the octave and 5/1 a little wider and the 3/1 a > little narrower:
> 4 * (3/1 - x) - 4 * (2/1 + x) = 5/1 + x
>
> x = ?
>
> And if I'm judging this wrong and the octave need to be narrower then > it is:
> 4 * (3/1 - x) - 4 * (2/1 - x) = 5/1 + x
>
> I'll let someone who's not a highschool dropout like me solve the > algebra ;-p
>
>
>
>> Hmmm, I was wrong again. I wanted to say: If the three ratios of 2/1 >> and 3/1 and 5/1 could have the same error and that would be the >> lowest possible one. Yes, that's it. Finally I formulated my wish >> into words ... Never thought I would have so much trouble with that. :-D
>>
>> Petr
>>
>>
> __._,_.__

🔗Marcel de Velde <marcel@...>

Sorry for yet another email on this subject.
But just input the formula in wolfram alpha and it does give the solutions:
http://www.wolframalpha.com/input/?i=%283%2Fx+*+3%2Fx+*+3%2Fx+*+3%2Fx%29+%2F+%282*x+*+2*x+*+2*x+*+2*x%29+%3D+5*x
Indeed it is 1/9 comma :)

And here are the other ones:
http://www.wolframalpha.com/input/?i=%283%2Fx+*+3%2Fx+*+3%2Fx+*+3%2Fx%29+%2F+%282%2Fx+*+2%2Fx+*+2%2Fx+*+2%2Fx%29+%3D+5*x

http://www.wolframalpha.com/input/?i=%283%2Fx+*+3%2Fx+*+3%2Fx+*+3%2Fx%29+%2F+%282%2Fx+*+2%2Fx+*+2%2Fx+*+2%2Fx%29+%3D+5%2Fx

http://www.wolframalpha.com/input/?i=%283*x+*+3*x+*+3*x+*+3*x%29+%2F+%282%2Fx+*+2%2Fx+*+2%2Fx+*+2%2Fx%29+%3D+5%2Fx

http://www.wolframalpha.com/input/?i=%283*x+*+3*x+*+3*x+*+3*x%29+%2F+%282*x+*+2*x+*+2*x+*+2*x%29+%3D+5%2Fx

http://www.wolframalpha.com/input/?i=%283%2Fx+*+3%2Fx+*+3%2Fx+*+3%2Fx%29+%2F+%282*x+*+2*x+*+2*x+*+2*x%29+%3D+5%2Fx

http://www.wolframalpha.com/input/?i=%283*x+*+3*x+*+3*x+*+3*x%29+%2F+%282%2Fx+*+2%2Fx+*+2%2Fx+*+2%2Fx%29+%3D+5*x

http://www.wolframalpha.com/input/?i=%283*x+*+3*x+*+3*x+*+3*x%29+%2F+%282*x+*+2*x+*+2*x+*+2*x%29+%3D+5*x
(btw I wrote the formula wrong for this last one in my previous email, but correct here)

-Marcel

> Oh I just found an online math solver :)
> http://www.myalgebra.com/
>
> And of course I made a stupid errors in my formula in my previous > message, I wrote it mathematically all wrong. I used - and + where I > should have written / or *, and should use "powers of" in some places > where I used * and /.
>
> So when I input the correct formula:
> (3/x * 3/x * 3/x * 3/x) / (2*x * 2*x * 2*x * 2*x) = 5*x
> There is "no solution"..
> I'm taking this means there is an infinite amount of x that will work.
> I don't know how to input in a calculator how to give the lowest value > of x.
> Can someone with math experience do this? Then you'll have your answer.
>
> For
> (3/x * 3/x * 3/x * 3/x) / (2/x * 2/x * 2/x * 2/x) = 5*x
> The answer is x = 81/80
> Which is true of course as 4 * (3/1 lowered by a syntonic comma) > "minus" 4 * (2/1 lowered by a syntonic comma) = 5/1 raised by a > syntonic comma (or better said a Pythagorean major third of 81/16)
>
> For (3/x * 3/x * 3/x * 3/x) / (2/x * 2/x * 2/x * 2/x) = 5/x
> x = 80/81
>
> For (3*x * 3*x * 3*x * 3*x) / (2/x * 2/x * 2/x * 2/x) = 5/x
> x = "no solution"
>
> For (3*x * 3*x * 3*x * 3*x) / (2*x * 2*x * 2*x * 2*x) = 5/x
> x = 80/81
>
> For (3/x * 3/x * 3/x * 3/x) / (2*x * 2*x * 2*x * 2*x) = 5/x
> x = "no algebraic solution". (I'm taking this means that there is no > possible answer for x)
>
> For (3*x * 3*x * 3*x * 3*x) / (2/x * 2/x * 2/x * 2/x) = 5*x
> x = "no algebraic solution"
>
> For (3*x * 3*x * 3*x * 3*x) / (2/x * 2/x * 2/x * 2/x) = 5*x
> x = 81/80
>
> These are all 8 possibilities for the formula.
>
> -Marcel
>
>
>> Assuming that we make the octave and 5/1 a little wider and the 3/1 a >> little narrower:
>> 4 * (3/1 - x) - 4 * (2/1 + x) = 5/1 + x
>>
>> x = ?
>>
>> And if I'm judging this wrong and the octave need to be narrower then >> it is:
>> 4 * (3/1 - x) - 4 * (2/1 - x) = 5/1 + x
>>
>> I'll let someone who's not a highschool dropout like me solve the >> algebra ;-p
>>
>>
>>
>>> Hmmm, I was wrong again. I wanted to say: If the three ratios of 2/1 >>> and 3/1 and 5/1 could have the same error and that would be the >>> lowest possible one. Yes, that's it. Finally I formulated my wish >>> into words ... Never thought I would have so much trouble with that. :-D
>>>
>>> Petr
>>>
>>>
>> __._,_.__
>

🔗Mike Battaglia <battaglia01@...>

🔗Marcel de Velde <marcel@...>

Did I misunderstand Petr's request?
I thought I solved it already?

http://www.wolframalpha.com/input/?i=%283%2Fx%29%5E4+%2F+%282*x%29%5E4+%3D+5*x
This is the lowest possible error for combination for 2/1 3/1 and 5/1 (where 4 fifths make 5/1 +-error of course).
It is indeed 1/9 comma error as Petr already found himself.

The next solution is:
http://www.wolframalpha.com/input/?i=%283%2Fx%29%5E4+%2F+%282*x%29%5E4+%3D+5%2Fx
Which has 1/7 comma error.

Next come 2 solutions with Syntonic comma errors.
And those are all the possible solutions to what I understood to be what Petr was asking for.

-Marcel

> On Mon, Apr 8, 2013 at 3:31 PM, Graham Breed <gbreed@... > <mailto:gbreed%40gmail.com>> wrote:
> >
> > No, it's not that difficult, not for rank 2. There's a worked
> > example in the PDF (p.11 and neighbors) and you can follow
> > it with a pocket calculator. Ideally a scientific calculator
> > with statistical features to calculate the sums and sum-
> > squareds.
>
> What you and I think is difficult isn't the same thing that other
> people think is difficult. If Petr already knows the basics of solving
> least-squares problems then he'll know how to do it already, but if he
> just wants to code something up, the pseudoinverse is the way I'd
> recommend doing it.
>
> Mike
>
>

🔗petrparizek2000 <petrparizek2000@...>

@Marcel,
Mike was answering Graham as they were mentioning a slightly different topic previously.
Your idea is ingenious, thank you for the contribution. :-)
I was also thinking of the possibility where the coefficients, instead of being 2 and 3 and 5, would be 1200 and ~1901.955 and ~2786.314 cents, which would allow to add and multiply interval sizes the way you were doing in the "wrong" equations.
In general, we then need to find the zero points of the following polynomials:
(b-x)*d - (a+x)*d - c+x
or:
(b-x)*d - (a+x)*d - c-x
I've just entered the following statement into WA's edit box:
find roots of x for (b-x)*d - (a+x)*d - c+x
Currently, I'm unable to figure out which of the suggested roots would be the one we're looking for.
But this begins to sound interesting.
Petr

🔗Freeman Gilmore <freeman.gilmore@...>

That should be x = - 1

On Mon, Apr 8, 2013 at 2:44 PM, Marcel de Velde
<marcel@...>wrote:

> **
>
>
> Assuming that we make the octave and 5/1 a little wider and the 3/1 a
> little narrower:
> 4 * (3/1 - x) - 4 * (2/1 + x) = 5/1 + x
>
> x = ?
>
> And if I'm judging this wrong and the octave need to be narrower then it
> is:
> 4 * (3/1 - x) - 4 * (2/1 - x) = 5/1 + x
>
> I'll let someone who's not a highschool dropout like me solve the algebra
> ;-p
>
>
>
>
>
> Hmmm, I was wrong again. I wanted to say: If the three ratios of 2/1 and
> 3/1 and 5/1 could have the same error and that would be the lowest possible
> one. Yes, that's it. Finally I formulated my wish into words ... Never
> thought I would have so much trouble with that. :-D
>
> Petr
>
>
>
>

🔗Freeman Gilmore <freeman.gilmore@...>

4 * (3/1 - x) - 4 * (2/1 + x) = 5/1 + x

x = 1

On Mon, Apr 8, 2013 at 2:44 PM, Marcel de Velde
<marcel@...>wrote:

🔗Graham Breed <gbreed@...>

On Monday 08 April 2013 18:39:29 you wrote:
> Hmmm, I was wrong again. I wanted to say: If the three
> ratios of 2/1 and 3/1 and 5/1 could have the same error
> and that would be the lowest possible one. Yes, that's
> it. Finally I formulated my wish into words ... Never
> thought I would have so much trouble with that. :-D

Having the same error is a minimax thing, so you could look
at classic TOP. A variant of that is the worst-Kees error,
where you take the difference between the highest and lowest
weighted errors in those prime intervals.

These things give 1/4-comma for meantone, though. It's RMS
functions that tend to push it to around 2/7-comma.

Graham

🔗petrparizek2000 <petrparizek2000@...>

🔗Marcel de Velde <marcel@...>

Btw, you can then click on "approximate form" under "real solutions" at the wolfram alpha results.
And then click "more digits" and then "A -> copyable plaintext" and then copy the text and input it in a ratio to cents converter like http://www.sengpielaudio.com/calculator-centsratio.htm and then you'll have your value in cents.
Which is only 0.139551484857567 cents in the below example. (result is 3/2 fifths smaller by this amount, 2/1 octaves smaller by this amount, and 5/4 smaller by this amount).

-Marcel

> Hi Petr,
>
> Yes that would be this:
> http://www.wolframalpha.com/input/?i=%282%2F3x%29%5E8+*+%282x%29%5E5+%3D+5%2F4%2Fx
> 8 fifths down (3/4)^8, times 5 octaves up 2^5 = ~5/4
> I think this is the smallest error, but haven't tried all 8 (actually > 4 since they are mirrors of eachother) possibilities where we divide > or multiply by x.
>
> -Marcel
>
>> @Marcel,
>>
>> do you have an idea how you would do something similar for the >> Helmholtz/Groven temperament where 5/4 is approximated by a >> diminished fourth?
>>
>> Petr
>>
>> >

🔗Marcel de Velde <marcel@...>

Hi Petr,

Bit late reply, I'm behind with my mail.
I didn't follow the whole thread, thought I could extract what it was about from your post where you stated what you were looking for, but apparently that didn't work haha.
But glad my equation is still of use :)

About the cents version, it works like this: http://www.wolframalpha.com/input/?i=%281902-x%29*4+-+%281200%2Bx%29*4+%3D+2786%2Bx
The answer is also in cents. (2.444... cents).

I really have no math knowledge, so I'm not sure what your equation does with finding roots.
Is it the same result you're after as the cents version above? Or is it something different?

-Marcel

> @Marcel,
> Mike was answering Graham as they were mentioning a slightly different > topic previously.
> Your idea is ingenious, thank you for the contribution. :-)
> I was also thinking of the possibility where the coefficients, instead > of being 2 and 3 and 5, would be 1200 and ~1901.955 and ~2786.314 > cents, which would allow to add and multiply interval sizes the way > you were doing in the "wrong" equations.
> In general, we then need to find the zero points of the following > polynomials:
> (b-x)*d - (a+x)*d - c+x
> or:
> (b-x)*d - (a+x)*d - c-x
> I've just entered the following statement into WA's edit box:
> find roots of x for (b-x)*d - (a+x)*d - c+x
> Currently, I'm unable to figure out which of the suggested roots would > be the one we're looking for.
> But this begins to sound interesting.
> Petr
>
>

🔗Petr Pařízek <petrparizek2000@...>

Hi Marcel.

I was thinking about it more and more and I realized that i must already have known the answer back in, believe it or not, July 2011. :-D Right, this is probably the first time that I've asked about something which I later find I already knew the answer to. Yes, I was making a small utility for Microsoft QuickBasic (still under MSDos) that could generate pitch sequences for triadic comma pumps (you just type in the ratio and it does all the rest for you). And if the tempering is to be the least possible, then the steps are as follows:

#1. Find the logarithmic size of the interval which will be tempered out (i.e. for meantone, it's ~21.506 cents).

#2. Find the exponents of the 3 ratios in question. If the ratios are 2/1, 3/1, 5/1, then the exponents are -4, 4, -1.

#3. Find their absolute values and add them up. You get 9. Divide the comma size by 9 and you get the required amount of tempering.

#4. Make a sequence of 4 falling octaves, 4 rising 12ths, 1 major 17th, in an order of your choice (I usually try to minimize the repetitions of one ratio more times in a row so I would probably alternate between 12ths and octaves). You end up a comma higher.

#5. Shift each consecutive pitch down by the resulting tempering amount. You end up where you started. If you add or subtract a proper number of octaves for each of the 9 pitches, you end up with a pentatonic scale because the 4 other pitches are octave equivalents of some other ones.

I'm sending a link to one of my experimental results from 2011. Here the vanishing interval is the schisma but the 3 ratios change from case to case. They are labeled by letters from A to F, meaning 2/1, 3/1, 3/2, 5/1, 5/2, 5/3, respectively.

BTW: Sorry for the "0" before the prime exponents; the program was originally intended for something else and I didn't want to rewrite too much of it.

https://dl.dropboxusercontent.com/u/8497979/CPLIST_SCHISMA.TXT

Petr

🔗Glenn <glenn.leider@...>

--- In tuning@yahoogroups.com, Marcel de Velde <marcel@...> wrote:
>
> Btw, you can then click on "approximate form" under "real solutions" at
> the wolfram alpha results.
> And then click "more digits" and then "A -> copyable plaintext" and then
> copy the text and input it in a ratio to cents converter like
> http://www.sengpielaudio.com/calculator-centsratio.htm and then you'll
> have your value in cents.
> Which is only 0.139551484857567 cents in the below example. (result is
> 3/2 fifths smaller by this amount, 2/1 octaves smaller by this amount,
> and 5/4 smaller by this amount).
>
> -Marcel
>
> > Hi Petr,
> >
> > Yes that would be this:
> > http://www.wolframalpha.com/input/?i=%282%2F3x%29%5E8+*+%282x%29%5E5+%3D+5%2F4%2Fx
> > 8 fifths down (3/4)^8, times 5 octaves up 2^5 = ~5/4
> > I think this is the smallest error, but haven't tried all 8 (actually
> > 4 since they are mirrors of eachother) possibilities where we divide
> > or multiply by x.
> >
> > -Marcel
> >
> >> @Marcel,
> >>
> >> do you have an idea how you would do something similar for the
> >> Helmholtz/Groven temperament where 5/4 is approximated by a
> >> diminished fourth?
> >>
> >> Petr
> >>
> >>
> >
>
This may be easier to visualize when you realize that this ratio is merely the 14th root of the schisma. The schisma's ratio is 32805/32768, or 1.95372078793416 cents (using Windows Calculator and rounding to 15 sig figs), so its 14th root is 1.95372078793416 / 14 or 0.13955148485244 cents.

🔗Glenn <glenn.leider@...>

--- In tuning@yahoogroups.com, Marcel de Velde <marcel@...> wrote:
>
> Hi Petr,
>
> Bit late reply, I'm behind with my mail.
> I didn't follow the whole thread, thought I could extract what it was
> about from your post where you stated what you were looking for, but
> apparently that didn't work haha.
> But glad my equation is still of use :)
>
> About the cents version, it works like this:
> http://www.wolframalpha.com/input/?i=%281902-x%29*4+-+%281200%2Bx%29*4+%3D+2786%2Bx
> The answer is also in cents. (2.444... cents).
>
> I really have no math knowledge, so I'm not sure what your equation does
> with finding roots.
> Is it the same result you're after as the cents version above? Or is it
> something different?
>
> -Marcel
>
> > @Marcel,
> > Mike was answering Graham as they were mentioning a slightly different
> > topic previously.
> > Your idea is ingenious, thank you for the contribution. :-)
> > I was also thinking of the possibility where the coefficients, instead
> > of being 2 and 3 and 5, would be 1200 and ~1901.955 and ~2786.314
> > cents, which would allow to add and multiply interval sizes the way
> > you were doing in the "wrong" equations.
> > In general, we then need to find the zero points of the following
> > polynomials:
> > (b-x)*d - (a+x)*d - c+x
> > or:
> > (b-x)*d - (a+x)*d - c-x
> > I've just entered the following statement into WA's edit box:
> > find roots of x for (b-x)*d - (a+x)*d - c+x
> > Currently, I'm unable to figure out which of the suggested roots would
> > be the one we're looking for.
> > But this begins to sound interesting.
> > Petr
> >
> >
>
I'm not sure which is more correct, but your cents would be a little more accurate with a more precise equation:
http://www.wolframalpha.com/input/?i=%281901.955-x%29*4+-+%281200%2Bx%29*4+%3D+2786.3137%2Bx
which yields a solution of
9x = 21.5063 (the familiar syntonic comma), or
x = 2.38959 cents approximately.