7-limit planar temperaments

My understanding is that a 7-limit temperament family with one comma
(say 36/35) is a planar temperament. Per my discussions with the other
Paul, one can use determinants to find three temperaments that
determine this plane. (Or perhaps its going from the three
temperaments, to the comma). Could someone kindly show me the math for
taking three temperaments, in the 7-limit, that would generate a single
comma, or even better, how to take 36/35 and get some temperament
values.

I'm also open to learning more about 7-limit linear temperaments with
two commas or more.

Paul Hjelmstad

>My understanding is that a 7-limit temperament family with one comma
>(say 36/35) is a planar temperament. Per my discussions with the other
>Paul, one can use determinants to find three temperaments that
>determine this plane. (Or perhaps its going from the three
>temperaments, to the comma).

Something doesn't sound right here.

>Could someone kindly show me the math for taking three temperaments,
>in the 7-limit, that would generate a single comma, or even better,
>how to take 36/35 and get some temperament values.

I believe there is only one 7-limit planar temperament in which
36/35 vanishes. Go above the 7-limit or change "planar" to
"linear" and you get an infinite number of temperaments, I think.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >My understanding is that a 7-limit temperament family with one comma
> >(say 36/35) is a planar temperament. Per my discussions with the
other
> >Paul, one can use determinants to find three temperaments that
> >determine this plane. (Or perhaps its going from the three
> >temperaments, to the comma).
>
> Something doesn't sound right here.
>
> >Could someone kindly show me the math for taking three temperaments,
> >in the 7-limit, that would generate a single comma, or even better,
> >how to take 36/35 and get some temperament values.
>
> I believe there is only one 7-limit planar temperament in which
> 36/35 vanishes. Go above the 7-limit or change "planar" to
> "linear" and you get an infinite number of temperaments, I think.
>
There should be many temperaments on the plane that have 36/35 as the
comma in 7-limit (3-dimensional) space. I am basing this on an analogy
with the zoom diagrams in 5-limit space...

> -Carl

>> >My understanding is that a 7-limit temperament family with one comma
>> >(say 36/35) is a planar temperament. Per my discussions with the
>> >other Paul, one can use determinants to find three temperaments that
>> >determine this plane. (Or perhaps its going from the three
>> >temperaments, to the comma).
>>
>> Something doesn't sound right here.
>>
>> >Could someone kindly show me the math for taking three temperaments,
>> >in the 7-limit, that would generate a single comma, or even better,
>> >how to take 36/35 and get some temperament values.
>>
>> I believe there is only one 7-limit planar temperament in which
>> 36/35 vanishes. Go above the 7-limit or change "planar" to
>> "linear" and you get an infinite number of temperaments, I think.
>>
>There should be many temperaments on the plane that have 36/35 as the
>comma in 7-limit (3-dimensional) space. I am basing this on an analogy
>with the zoom diagrams in 5-limit space...

The *planar* temperaments are the planes in this space, each
defined by a single comma, just as the 5-limit linear temperament
were lines defined by single commas in the zoom diagrams.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> >My understanding is that a 7-limit temperament family with one
comma
> >> >(say 36/35) is a planar temperament. Per my discussions with the
> >> >other Paul, one can use determinants to find three temperaments
that
> >> >determine this plane. (Or perhaps its going from the three
> >> >temperaments, to the comma).
> >>
> >> Something doesn't sound right here.
> >>
> >> >Could someone kindly show me the math for taking three
temperaments,
> >> >in the 7-limit, that would generate a single comma, or even
better,
> >> >how to take 36/35 and get some temperament values.
> >>
> >> I believe there is only one 7-limit planar temperament in which
> >> 36/35 vanishes. Go above the 7-limit or change "planar" to
> >> "linear" and you get an infinite number of temperaments, I think.
> >>
> >There should be many temperaments on the plane that have 36/35 as
the
> >comma in 7-limit (3-dimensional) space. I am basing this on an
analogy
> >with the zoom diagrams in 5-limit space...
>
> The *planar* temperaments are the planes in this space, each
> defined by a single comma, just as the 5-limit linear temperament
> were lines defined by single commas in the zoom diagrams.
>
> -Carl

I know, that's what I meant. One can use Excel to calculate
determinants, inverses and adjoints so I would love to learn the math
for this...

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> My understanding is that a 7-limit temperament family with one comma
> (say 36/35) is a planar temperament. Per my discussions with the other
> Paul, one can use determinants to find three temperaments that
> determine this plane.

It's *called* planar, but it is actually three dimensions, so there
isn't a plane unless you octave reduce. I'm suggesting using "r3
temperament", short for "rank three temperament", for these. "r2"
would be what we call linear, and "r1" equal.

Given three r1 temperaments, we can find a single r3 temperament by
any of these methods:

(1) Wedging to get a trival, and taking the complement

(2) Taking the determinant of the three vals with indeterminates <a b
c d| for the fourth val

(3) Finding the kernel of the mapping defined by the three vals, if
Excel can do that

Excel should be able to handle (1), at least.

(Or perhaps its going from the three
> temperaments, to the comma). Could someone kindly show me the math for
> taking three temperaments, in the 7-limit, that would generate a single
> comma, or even better, how to take 36/35 and get some temperament
> values.

To get tuning values out of a single comma one easy method is to use
TOP tuning; however, this often retunes the octave and you may not
want that. For 36/35, octaves are tuned very flat.

36/35 is relatively easy to handle in that 2, 3, and 5 can be taken as
generators, and 7*(35/35) = 36/5 for the 7. This gives us the mapping
as [<1 0 0 2|, <0 1 0 2|, <0 0 1 -1|], and tunings can be optimized
from this; probably the easiest method to manage is rms, since you can
use least squares. If Excel can deal with linear programming, you can
also find minimax readily enough.

> I'm also open to learning more about 7-limit linear temperaments with
> two commas or more.

You mean in general, or particular examples? Most of our discussions
of 7-limit temperaments has been about these. Ones with three commas
are just r1 (or equal) temperaments. For example, combining 36/35 with
128/125 and 81/80 and then applying any of the above three methods
gives <12 19 28 34|, or septimal 12-equal.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>

Gene, Thanks!!! Please see QUESTION below...

> > My understanding is that a 7-limit temperament family with one
comma
> > (say 36/35) is a planar temperament. Per my discussions with the
other
> > Paul, one can use determinants to find three temperaments that
> > determine this plane.
>
> It's *called* planar, but it is actually three dimensions, so there
> isn't a plane unless you octave reduce. I'm suggesting using "r3
> temperament", short for "rank three temperament", for these. "r2"
> would be what we call linear, and "r1" equal.
>
> Given three r1 temperaments, we can find a single r3 temperament by
> any of these methods:
>
> (1) Wedging to get a trival, and taking the complement
>
> (2) Taking the determinant of the three vals with indeterminates <a
b
> c d| for the fourth val
>
> (3) Finding the kernel of the mapping defined by the three vals, if
> Excel can do that
>
> Excel should be able to handle (1), at least.
>
> (Or perhaps its going from the three
> > temperaments, to the comma). Could someone kindly show me the
math for
> > taking three temperaments, in the 7-limit, that would generate a
single
> > comma, or even better, how to take 36/35 and get some temperament
> > values.
>
> To get tuning values out of a single comma one easy method is to use
> TOP tuning; however, this often retunes the octave and you may not
> want that. For 36/35, octaves are tuned very flat.
>
> 36/35 is relatively easy to handle in that 2, 3, and 5 can be taken
as
> generators, and 7*(35/35) = 36/5 for the 7. This gives us the
mapping
> as [<1 0 0 2|, <0 1 0 2|, <0 0 1 -1|], and tunings can be optimized

QUESTION: So these are vals for 2 3 and 5? What do 2, 2 and -1
represent in the 7-slot?

> from this; probably the easiest method to manage is rms, since you
can
> use least squares. If Excel can deal with linear programming, you
can
> also find minimax readily enough.
>
> > I'm also open to learning more about 7-limit linear temperaments
with
> > two commas or more.
>
> You mean in general, or particular examples? Most of our discussions
> of 7-limit temperaments has been about these. Ones with three commas
> are just r1 (or equal) temperaments. For example, combining 36/35
with
> 128/125 and 81/80 and then applying any of the above three methods
> gives <12 19 28 34|, or septimal 12-equal.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul.hjelmstad@m...> wrote:
> >
>
> Gene, Thanks!!! Please see QUESTION below...
>
Actually, dumb question. I got it...its a matrix of course

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> My understanding is that a 7-limit temperament family with one comma
> (say 36/35) is a planar temperament. Per my discussions with the other
> Paul, one can use determinants to find three temperaments that
> determine this plane.

Perhaps you mean "find three generators".

> (Or perhaps its going from the three
> temperaments, to the comma). Could someone kindly show me the math for
> taking three temperaments,

"taking three generators"

> in the 7-limit, that would generate a single
> comma, or even better, how to take 36/35 and get some temperament
> values.

"and get some generator values".

> I'm also open to learning more about 7-limit linear temperaments with
> two commas or more.
>
> Paul Hjelmstad

> from this; probably the easiest method to manage is rms, since you
> can
> > use least squares. If Excel can deal with linear programming, you
> can
> > also find minimax readily enough.

One doesn't need linear programming to find minimax. One can get close
enough for all practical purposes by numerically minimising the sum of
a large even power of the errors.

Excel has a general-purpose numerical solver. You use ordinary
formulas to calculate the quantity to be minimised in one cell and you
tell the solver which other cells it is allowed to change, i.e. the
generators, in order to try to minimise this cell.

You may have to start off with a small even power, say the 2nd (which
will give the RMS optimum), to avoid IEEE-floating-point overflow, and
then when this is solved you can increase the power (to say the 10th
then 20th) and re-solve until the decimal places you care about no
longer change. This will be the minimax.

TOP is just minimax with the error in each prime multiplied by the log
of that prime before being raised to the large even power and added
together, and with all three generators (including the octave) being
allowed to change.

-- Dave

Oops!

That should have been

"TOP is just minimax with the error in each prime DIVIDED by the log
of that prime before being raised to the large even power and added
together, and with all three generators (including the octave) being
allowed to change."

-- Dave

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
wrote:

> > 36/35 is relatively easy to handle in that 2, 3, and 5 can be
taken
> as
> > generators, and 7*(35/35) = 36/5 for the 7. This gives us the
> mapping
> > as [<1 0 0 2|, <0 1 0 2|, <0 0 1 -1|], and tunings can be
optimized
>
> QUESTION: So these are vals for 2 3 and 5? What do 2, 2 and -1
> represent in the 7-slot?

2^2 3^2 5^(-1) = 36/5 ~ 7.

--- In tuning-math@yahoogroups.com, "Dave Keenan" <d.keenan@b...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
> > My understanding is that a 7-limit temperament family with one
comma
> > (say 36/35) is a planar temperament. Per my discussions with the
other
> > Paul, one can use determinants to find three temperaments that
> > determine this plane.
>
> Perhaps you mean "find three generators".

Could you point me to some examples? I've worked in Excel with
Adjoints, using Graham's method with (1 0 0 0) and three commas, is
that what you mean?

>
> > (Or perhaps its going from the three
> > temperaments, to the comma). Could someone kindly show me the
math for
> > taking three temperaments,
>
> "taking three generators"
>
> > in the 7-limit, that would generate a single
> > comma, or even better, how to take 36/35 and get some temperament
> > values.
>
> "and get some generator values".
>
> > I'm also open to learning more about 7-limit linear temperaments
with
> > two commas or more.
> >
> > Paul Hjelmstad

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> Given three r1 temperaments, we can find a single r3 temperament by
> any of these methods:
>
> (1) Wedging to get a trival, and taking the complement
>
> (2) Taking the determinant of the three vals with indeterminates <a b
> c d| for the fourth val

Could you show me an example of this one (2)?

>
> (3) Finding the kernel of the mapping defined by the three vals, if
> Excel can do that
>
> Excel should be able to handle (1), at least.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> > (2) Taking the determinant of the three vals with indeterminates <a b
> > c d| for the fourth val
>
> Could you show me an example of this one (2)?

For 81/80, 126/125 and 1029/1024, take

det([<a b c d|, |-4 4 -1 0>, |1 2 -3 1>, |-10 1 0 3>])
= -31a-49b-72c-87d, so the val is <31 49 72 87|.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > > (2) Taking the determinant of the three vals with indeterminates
<a b
> > > c d| for the fourth val
> >
> > Could you show me an example of this one (2)?
>
> For 81/80, 126/125 and 1029/1024, take
>
> det([<a b c d|, |-4 4 -1 0>, |1 2 -3 1>, |-10 1 0 3>])
> = -31a-49b-72c-87d, so the val is <31 49 72 87|.

Fantastic. I was doing the adjoint of 1 0 0 0 and three commas, but
this is cool too

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul.hjelmstad@m...> wrote:
> >
> > > > (2) Taking the determinant of the three vals with
indeterminates
> <a b
> > > > c d| for the fourth val
> > >
> > > Could you show me an example of this one (2)?
> >
> > For 81/80, 126/125 and 1029/1024, take
> >
> > det([<a b c d|, |-4 4 -1 0>, |1 2 -3 1>, |-10 1 0 3>])
> > = -31a-49b-72c-87d, so the val is <31 49 72 87|.
>
> Fantastic. I was doing the adjoint of 1 0 0 0 and three commas, but
> this is cool too

With the adjoint method, I get a first column of 31, 49, 72, 87 and
the second column is 0, 9, 5, -3. I see that 31+18=49, 62+10=72, and
93-6=87, so the generator is 2 steps and the period 31 steps. Why
does it work out to 31*1, 31*2 and 31*3 for the period values?
Also, if the generator is exactly 2 steps, what is the reasoning for
rms'ing it (or minimaxing it?) paul

--- In tuning-math@yahoogroups.com, "Dave Keenan" <d.keenan@b...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
> > My understanding is that a 7-limit temperament family with one
comma
> > (say 36/35) is a planar temperament. Per my discussions with the
other
> > Paul, one can use determinants to find three temperaments that
> > determine this plane.
>
> Perhaps you mean "find three generators".

I'm assuming you mean three generators of one temperament, like 0, 9,
5, -3 in 7-limit 31-et?

>
> > (Or perhaps its going from the three
> > temperaments, to the comma). Could someone kindly show me the
math for
> > taking three temperaments,
>
> "taking three generators"
>
> > in the 7-limit, that would generate a single
> > comma, or even better, how to take 36/35 and get some temperament
> > values.
>
> "and get some generator values".
>
> > I'm also open to learning more about 7-limit linear temperaments
with
> > two commas or more.
> >
> > Paul Hjelmstad

I was hoping, that by analogy with monzoom diagrams in the 5-limit
plane, you could take everything and extend it to a 7-limit 3D space.
But it looks like not everything carries over. I guess my question is:
How are temperament families determined in this space? Is there such
a thing, for example, as a family of _________ in 7-limit 3D space?

Paul

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> I'm assuming you mean three generators of one temperament, like 0, 9,
> 5, -3 in 7-limit 31-et?

Boy this is getting confusing. 0, 9, 5, -3 is the generator part of
the mapping to primes in the valentine, or 15&31 temperament, and so
indeed connected with 31-et:

[<1 1 2 3|, <0 9 5 -3|]

In a 7-limit "planar", or r3, temperament, there are three generators,
just as there are two (often given as period and "generator") for r2,
or "linear", temperaments. Possibly the question was how to find
these. My method works very well for Maple, but I like to use Hermite
reduction which Excel presumably doesn't have.

> I was hoping, that by analogy with monzoom diagrams in the 5-limit
> plane, you could take everything and extend it to a 7-limit 3D space.
> But it looks like not everything carries over. I guess my question is:
> How are temperament families determined in this space? Is there such
> a thing, for example, as a family of _________ in 7-limit 3D space?

How do you define a "temperament family"? To me, that term suggests
seven or higher limit temperaments which share the same five limit
comma. Also in the seven or higher limit, temperaments can be related
or unrelated, depending on whether they have nontrivial kernel
intersections or not. That is to say, meantone and miracle are
related, since they share the comma 225/224. Meantone and superpyth
are not related in this sense, even though they have the same generator.

If <<a1 a2 a3 a4 a5 a6|| and <<b1 b2 b3 b3 b5 b6|| are two 7-limit
wedgies, a quick test to see if the corresponding temperaments are
related is to take the pfaffian, a1b6-a2b5+a3b4+b1a6-b2a5+b3a4; if
this is zero then the temperaments are related, and if not, not. The
common comma can be found by taking the mapping to primes, putting it
together into a square matrix, and finding the kernel.

hi Paul,

> I was hoping, that by analogy with monzoom diagrams
> in the 5-limit plane, you could take everything and
> extend it to a 7-limit 3D space. But it looks like not
> everything carries over. I guess my question is:
> How are temperament families determined in this space?
> Is there such a thing, for example, as a family of
> _________ in 7-limit 3D space?

not sure how much illumination i can shed on this,
but here's my understanding of it:

in 2D space, any line represents a temperament family.
a particular temperament occupies a point which is the
intersection of an infinite number of lines, thus, that
particular temperament is a member of all the families
which are represented by those lines.

families grow as dimensions are added to the space
which represents tunings. thus, there are multiple
members of each new generation of the family.

(it's unfortuante that our geneology metaphor forces
us to use the word "generation", since "generator"
is already such an important word in tuning theory,
and the meaning here are quite different. oh well.)

in 3D space, any line is the intersection of an infinite
number of planes. thus, each of those planes is an
entire generation within the family, and each line
is a sibling within that generation.

if this isn't quite right, i'm sure Gene or someone
else who has a better mathematical grasp of how it
works will chime in and offer corrections. but i'm
pretty sure that this is the basic idea.

an example of what i'm talking about, the meantone
family tree, is here:

http://tonalsoft.com/enc/family.htm

-monz
http://tonalsoft.com
microtonal music software

--- In tuning-math@yahoogroups.com, "monz" <monz@t...> wrote:

> an example of what i'm talking about, the meantone
> family tree, is here:
>
> http://tonalsoft.com/enc/family.htm

Here's another relevant web page:

http://66.98.148.43/~xenharmo/commaseq.htm

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > I'm assuming you mean three generators of one temperament, like
0, 9,
> > 5, -3 in 7-limit 31-et?
>
> Boy this is getting confusing. 0, 9, 5, -3 is the generator part of
> the mapping to primes in the valentine, or 15&31 temperament, and so
> indeed connected with 31-et:
>
> [<1 1 2 3|, <0 9 5 -3|]
>

Could you remind me how to get the generator-to-primes from 15&31?
(I would check on Graham's website, but it appears to be down)

> In a 7-limit "planar", or r3, temperament, there are three
generators,
> just as there are two (often given as period and "generator") for
r2,
> or "linear", temperaments. Possibly the question was how to find
> these. My method works very well for Maple, but I like to use
Hermite
> reduction which Excel presumably doesn't have.

That was my question

> > I was hoping, that by analogy with monzoom diagrams in the 5-
limit
> > plane, you could take everything and extend it to a 7-limit 3D
space.
> > But it looks like not everything carries over. I guess my
question is:
> > How are temperament families determined in this space? Is there
such
> > a thing, for example, as a family of _________ in 7-limit 3D
space?
>
> How do you define a "temperament family"? To me, that term suggests
> seven or higher limit temperaments which share the same five limit
> comma. Also in the seven or higher limit, temperaments can be
related
> or unrelated, depending on whether they have nontrivial kernel
> intersections or not. That is to say, meantone and miracle are
> related, since they share the comma 225/224. Meantone and superpyth
> are not related in this sense, even though they have the same
generator.
>
> If <<a1 a2 a3 a4 a5 a6|| and <<b1 b2 b3 b3 b5 b6|| are two 7-limit
> wedgies, a quick test to see if the corresponding temperaments are
> related is to take the pfaffian, a1b6-a2b5+a3b4+b1a6-b2a5+b3a4; if
> this is zero then the temperaments are related, and if not, not. The
> common comma can be found by taking the mapping to primes, putting
it
> together into a square matrix, and finding the kernel.

The calculation with the Pfaffian is neat. Could you please show an
example of the last sentence (I know, I should be further along by
now..) Is the kernel when you set the matrix equal to zero and find
the solution?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> Could you remind me how to get the generator-to-primes from 15&31?
> (I would check on Graham's website, but it appears to be down)

Excel probably would not be able to deal with my method, but you can
certainly find the two vals: v15 = <15 24 35 42| and v31 = <31 49 72
87|. Then g = 31*v15 - 15*v31 = <0 9 5 -3| is also a val for the
temperament, and it is reduced and has a first coefficient of zero. It
is therefore the generator part of the mapping. Now take the
penultimate convergent to 31/15, which you can obtain by Euclidean
algorithm/continued fraction methods. You get 29/14; if you take
29*v15-14*v31 you get another val for the system, <1 10 7 0|. Wedging
<1 10 7 0| and <0 9 5 -3| gives <<9 5 -3 -13 -30 -21||, so these do
indeed generate valentine. However, using least squares shows the
generator is -1122 cents, so you may want to reduce it to a better
value. If instead we use <1 10 7 0|-<0 9 5 -3| = <1 1 2 3| for the
period part of the mapping, the generator is moved up an octave and
becomes 78 cents, which is probably what you want.

The
> > common comma can be found by taking the mapping to primes, putting
> it
> > together into a square matrix, and finding the kernel.
>
> The calculation with the Pfaffian is neat. Could you please show an
> example of the last sentence (I know, I should be further along by
> now..) Is the kernel when you set the matrix equal to zero and find
> the solution?

The kernel is what the matrix, considered as a linear transformation,
annihilates. Possibly Excel has something for this already, or if not
you could try solving vM = 0 or Mv = 0 (depending) for generic vectors v.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > Could you remind me how to get the generator-to-primes from 15&31?
> > (I would check on Graham's website, but it appears to be down)
>
> Excel probably would not be able to deal with my method, but you can
> certainly find the two vals: v15 = <15 24 35 42| and v31 = <31 49 72
> 87|. Then g = 31*v15 - 15*v31 = <0 9 5 -3| is also a val for the
> temperament, and it is reduced and has a first coefficient of zero.
It
> is therefore the generator part of the mapping. Now take the
> penultimate convergent to 31/15, which you can obtain by Euclidean
> algorithm/continued fraction methods. You get 29/14; if you take
> 29*v15-14*v31 you get another val for the system, <1 10 7 0|.
Wedging
> <1 10 7 0| and <0 9 5 -3| gives <<9 5 -3 -13 -30 -21||, so these do
> indeed generate valentine. However, using least squares shows the
> generator is -1122 cents, so you may want to reduce it to a better

Please show me, thanks! That is, least squares ....

> value. If instead we use <1 10 7 0|-<0 9 5 -3| = <1 1 2 3| for the
> period part of the mapping, the generator is moved up an octave and
> becomes 78 cents, which is probably what you want.
>
> The
> > > common comma can be found by taking the mapping to primes,
putting
> > it
> > > together into a square matrix, and finding the kernel.
> >

Don't you need more than one mapping to primes to fill out a square
matrix? Could you demonstrate this also? Thanks

On 5/17/05, Paul G Hjelmstad <paul.hjelmstad@medtronic.com> wrote:

> Could you remind me how to get the generator-to-primes from 15&31?
> (I would check on Graham's website, but it appears to be down)

This is on the Wiki now:

http://riters.com/microtonal/index.cgi/FindingLinearTemperaments

I seem to remember getting half way through revising it when the site
became unavailable.

Graham

--- In tuning-math@yahoogroups.com, Graham Breed <gbreed@g...> wrote:
> On 5/17/05, Paul G Hjelmstad <paul.hjelmstad@m...> wrote:
>
> > Could you remind me how to get the generator-to-primes from 15&31?
> > (I would check on Graham's website, but it appears to be down)
>
> This is on the Wiki now:
>
> http://riters.com/microtonal/index.cgi/FindingLinearTemperaments
>
> I seem to remember getting half way through revising it when the site
> became unavailable.
>
>
> Graham

Looks nice, but the link for online scripts is still broken...

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> Given three r1 temperaments, we can find a single r3 temperament by
> any of these methods:
>
> (1) Wedging to get a trival, and taking the complement

?Could you please demonstrate how one wedges to bet a tri-val out of
three temperaments? Do you start out with temperament vals or
generators?

> (2) Taking the determinant of the three vals with indeterminates <a
b
> c d| for the fourth val
>
> (3) Finding the kernel of the mapping defined by the three vals, if
> Excel can do that

?Any chance of demonstrating this also? Thanks! Paul

>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
> >
> > Given three r1 temperaments, we can find a single r3 temperament by
> > any of these methods:
> >
> > (1) Wedging to get a trival, and taking the complement
>
>
> ?Could you please demonstrate how one wedges to bet a tri-val out of
> three temperaments? Do you start out with temperament vals or
> generators?

It's just the triple wedge product. For 7-limit temperaments, this
tri-val is equivalent to a monzo--that is, it defines a 7-limit
interval, which you may find as the complement.

> > (3) Finding the kernel of the mapping defined by the three vals, if
> > Excel can do that
>
>
> ?Any chance of demonstrating this also? Thanks! Paul

It can be done by solving the system of three linear equations
v1(q)=0, v2(q)=0, v3(q)=0 where v1, v2, v3 are the three vals.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> > wrote:
> > >
> > > Given three r1 temperaments, we can find a single r3
temperament by
> > > any of these methods:
> > >
> > > (1) Wedging to get a trival, and taking the complement
> >
> >
> > ?Could you please demonstrate how one wedges to bet a tri-val out
of
> > three temperaments? Do you start out with temperament vals or
> > generators?
>
> It's just the triple wedge product. For 7-limit temperaments, this
> tri-val is equivalent to a monzo--that is, it defines a 7-limit
> interval, which you may find as the complement.
>
> > > (3) Finding the kernel of the mapping defined by the three
vals, if
> > > Excel can do that
> >
> >
> > ?Any chance of demonstrating this also? Thanks! Paul
>
> It can be done by solving the system of three linear equations
> v1(q)=0, v2(q)=0, v3(q)=0 where v1, v2, v3 are the three vals.

Thanx. I'll have to play with this a bit

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> > wrote:
> > >
> > > Given three r1 temperaments, we can find a single r3
temperament by
> > > any of these methods:
> > >
> > > (1) Wedging to get a trival, and taking the complement
> >
> >
> > ?Could you please demonstrate how one wedges to bet a tri-val out
of
> > three temperaments? Do you start out with temperament vals or
> > generators?
>
> It's just the triple wedge product. For 7-limit temperaments, this
> tri-val is equivalent to a monzo--that is, it defines a 7-limit
> interval, which you may find as the complement.

*Dumb question: How does one wedge three values? Is it similar to
taking the adjoint of a matrix of 1 0 0 0 and three commas? (or vals)

> > > (3) Finding the kernel of the mapping defined by the three
vals, if
> > > Excel can do that
> >
> >
> > ?Any chance of demonstrating this also? Thanks! Paul
>
> It can be done by solving the system of three linear equations
> v1(q)=0, v2(q)=0, v3(q)=0 where v1, v2, v3 are the three vals.

> > It's just the triple wedge product. For 7-limit temperaments, this
> > tri-val is equivalent to a monzo--that is, it defines a 7-limit
> > interval, which you may find as the complement.
>
> *Dumb question: How does one wedge three values? Is it similar to
> taking the adjoint of a matrix of 1 0 0 0 and three commas? (or vals)

If I take the adjoint of (1 0 0 0), (12 19 28 34), (19 30 44 54), (31
49 72 87) I get (4 -4 1 0) in the first column. Taking the negative
this is meantone... I am on the right track? How come it isn't
backwards?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> If I take the adjoint of (1 0 0 0), (12 19 28 34), (19 30 44 54), (31
> 49 72 87) I get (4 -4 1 0) in the first column. Taking the negative
> this is meantone... I am on the right track? How come it isn't
> backwards?

More to the point, the first column represents 2^4 3^(-4) 5^1, or
80/81. It is mapped to 4 by the octaves val and to 0 by the three
other vals, and represents their common comma. Of course, a 5-limit
comma complements immediately to a 5-limit mapping, so from the comma
you get the meantone mapping.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> *Dumb question: How does one wedge three values? Is it similar to
> taking the adjoint of a matrix of 1 0 0 0 and three commas? (or vals)

It's not the same thing, but in the 7-limit it ends up getting you to
the same place, which is finding a common comma. Note that the 1 0 0 0
could be replaced by 0 1 0 0, etc. If you don't have wedge product
routines available, I would suggest wedging three vals by taking the
determinant of a val with indeterminate values <v2 v3 v5 v7| with the
three other vals, or evaluating minors.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > *Dumb question: How does one wedge three values? Is it similar to
> > taking the adjoint of a matrix of 1 0 0 0 and three commas? (or
vals)
>
> It's not the same thing, but in the 7-limit it ends up getting you to
> the same place, which is finding a common comma. Note that the 1 0 0 0
> could be replaced by 0 1 0 0, etc. If you don't have wedge product
> routines available, I would suggest wedging three vals by taking the
> determinant of a val with indeterminate values <v2 v3 v5 v7| with the
> three other vals, or evaluating minors.

Thanks for your responses. Unfortunately Excel doesn't allow using
unknown variables (as far as I can tell).

I am currently really interested in the 7-limit. For example, take the
blues scale (0,3,5,6,7,10) Some interested facts: If C=0, then what you
have is a pentatonic scale on Eb with added minor third (Gb over Eb).
So we are working with C minor pentatonic or Eb pentatonic. Also,
(0,3,5 and 7,10,12) are the top halves of dominant seventh chords F7
and C7. So this relates to the 7-limit. Also, if you take the Z-related
complement of (0,3,5,6,7,10) you get (1,2,4,8,9,11) which is E7 and A
major triad superimposed. When one considers all the music based on the
blues scale and variants (blues, jazz, rock etc) I think this angle
might be worth pursuing. One gets (36, 42, 48, 49, 54, 63, 72) as one
possible just intonation version of the blues scale. You get a C minor
triad of (6,7,9) and an Eb minor triad of (6,7,9 - the simplest form for
a minor triad. Thoughts anyone?