5-limit, 12-note Fokker blocks

The Duodene is well-known as a 5-limit, 12-note Fokker block; I
decided to check on whether Scala knows of four other examples, and
got this. (The scales can be obtained by letting i range from 1 to
12.)

(16/15)^i (2048/2025)^(-5i/12) (81/80)^(-2i/12)
No Scala name

(16/15)^i (81/80)^(8i/12) (648/625)^(-5i/12)
No Scala name

(16/15)^i (128/125)^(-8i/12) (648/625)^(3i/12)
No Scala name, a mode of what I've called the thirds scale

(16/15)^i (2048/2025)^(-3i/12) (128/125)^(-2i/12)
Scala identifies it as lumma5r.scl

! lumma5r.scl
!
Carl Lumma's scale, 5-limit just version, TL 19-2-
99
12
!
16/15
9/8
75/64
5/4
4/3
45/32
3/2
8/5
5/3
225/128
15/8
2/1

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> (The scales can be obtained by letting i range from 1 to
> 12.)

I forgot to put in that you must round the exponents to integers; if
you don't you'll merely get 12-equal.

> (16/15)^i (2048/2025)^round(-5i/12) (81/80)^round(-2i/12)
> No Scala name
>
>
> (16/15)^i (81/80)^rounnd(8i/12) (648/625)^round(-5i/12)
> No Scala name
>
>
> (16/15)^i (128/125)^round(-8i/12) (648/625)^round(3i/12)
> No Scala name, a mode of what I've called the thirds scale
>
>
> (16/15)^i (2048/2025)^round(-3i/12) (128/125)^round(-2i/12)
> Scala identifies it as lumma5r.scl
>
> ! lumma5r.scl
> !
> Carl Lumma's scale, 5-limit just version, TL 19-2-
> 99

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> The Duodene is well-known as a 5-limit, 12-note Fokker block;
> I
> decided to check on whether Scala knows of four other examples, and
> got this.

Scala has many others, including Marpurg's Monochord #1, Ramos/Ramis,
etc . . . see the Gentle Introduction to Fokker Periodicity Blocks.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:

> Scala has many others, including Marpurg's Monochord #1,
Ramos/Ramis,
> etc . . . see the Gentle Introduction to Fokker Periodicity Blocks.

"Many" is a little vague, though I see tamil_vi.scl is a mode of
ramis.scl. Can you tell me what two commas give Marpurg #1 and Ramis?
Do you know of any other specific examples?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> "Many" is a little vague, though I see tamil_vi.scl is a mode of
> ramis.scl. Can you tell me what two commas give Marpurg #1 and Ramis?
> Do you know of any other specific examples?

I'm going to answer part of my own question. Determination of whether
or not something was a Fokker block might be considered as an addition
to Scala.

Consider the intervals of ramis.scl. These go

135/126, 256/243, 16/15, 135/128, 16/15, 135/128, 16/15, 256/243,
135/128, 16/15, 135/128, 16/15

There are only three sizes (in the 5-limit, there can be as many as
four) and none is a singleton, which we should avoid using as a basis.
Pick 16/15 as the "basis", and take ratios with the other two step
sizes to find the commas: (16/15)/(135/128) = 2048/2025, and
(16/15)/(256/243) = 81/80. Now invert the matrix whose rows are the
monzos for 16/15, 81/80, 2048/2025 and get the matrix whose columns
are the standard vals h12, -h2, -h5. If we take the triples
[h12(q), h2(q), h5(q)] for q in the Ramis scale, we get

1 [0, 0, 0]
135/128 [1, 0, 1]
10/9 [2, 1, 1]
32/27 [3, 1, 1]
5/4 [4, 1, 2]
4/3 [5, 1, 2]
45/32 [6, 1, 3]
3/2 [7, 1, 3]
128/81 [8, 2, 3]
5/3 [9, 2, 4]
16/9 [10, 2, 4]
15/8 [11, 2, 5]
2 [12, 2, 5]

The values for h2 and h5 are nondecreasing, and for h12 they go up one
at a time (the scale is h12-epimorphic.) This tells us we have a
Fokker block.

Doing the same kind of analysis for Marpurg Monochord 1, we get
16/15, 128/125 and 81/80, and h12, h5, and h3 as the vals to check it
with. This time we get

1 [0, 0, 0]
25/24 [1, 1, 0]
9/8 [2, 1, 1]
6/5 [3, 1, 1]
5/4 [4, 2, 1]
4/3 [5, 2, 1]
45/32 [6, 3, 2]
3/2 [7, 3, 2]
25/16 [8, 4, 2]
5/3 [9, 4, 2]
9/5 [10, 4, 3]
15/8 [11, 5, 3]
2 [12, 5, 3]

Once again, we have the correct monotonic and epimorphic properties,
and Marpurg Monochord 1 is a Fokker block.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> Consider the intervals of ramis.scl. These go
>
> 135/126, 256/243, 16/15, 135/128, 16/15, 135/128, 16/15, 256/243,
> 135/128, 16/15, 135/128, 16/15
>
> There are only three sizes (in the 5-limit, there can be as many as
> four) and none is a singleton, which we should avoid using as a basis.
> Pick 16/15 as the "basis", and take ratios with the other two step
> sizes to find the commas: (16/15)/(135/128) = 2048/2025, and
> (16/15)/(256/243) = 81/80. Now invert the matrix whose rows are the
> monzos for 16/15, 81/80, 2048/2025 and get the matrix whose columns
> are the standard vals h12, -h2, -h5. If we take the triples
> [h12(q), h2(q), h5(q)] for q in the Ramis scale, we get
>
> 1 [0, 0, 0]
> 135/128 [1, 0, 1]
> 10/9 [2, 1, 1]
> 32/27 [3, 1, 1]
> 5/4 [4, 1, 2]
> 4/3 [5, 1, 2]
> 45/32 [6, 1, 3]
> 3/2 [7, 1, 3]
> 128/81 [8, 2, 3]
> 5/3 [9, 2, 4]
> 16/9 [10, 2, 4]
> 15/8 [11, 2, 5]
> 2 [12, 2, 5]

If now we take [i-h12(q), i/6-h2(q), 5i/12-h5(q)] we get

0 [0, 0, 0]
1 [0, 1/6, -7/12]
2 [0, -2/3, -1/6]
3 [0, -1/2, 1/4]
4 [0, -1/3, -1/3]
5 [0, -1/6, 1/12]
6 [0, 0, -1/2]
7 [0, 1/6, -1/12]
8 [0, -2/3, 1/3]
9 [0, -1/2, -1/4]
10 [0, -1/3, 1/6]
11 [0, -1/6, -5/12]

The midpoint of values of the second column is (-2/3 +1/6)/2 = -1/4,
and the midpoint of the third column is (-7/12 + 1/3)/2 = -1/8. Hence
we can express the exponents in terms of round(-5i/12 - 1/8) and
round(-i/6 - 1/4), and get

ramis[i] = (16/15)^i * (2048/2025)^round(-5i/12-1/8) *
(81/80)^round(-i/6-/14)

We can automate this process with the following steps

1. Find the set of intervals between steps. If there are more of these
than there are primes involved in the factorization of the scale
degrees, return false.

2. Check if the scale is epimorphic. If not, return false.

3. Pick a "base" element from the scale steps, by any means (first
step, smallest step, most common step, least Tenney height, etc.) Now
take ratios of the basis element with the other scale steps, finding
(n-1) other ratios (where n is the number of primes involved) which
together with the basis element gives a matrix of row vectors whose
determinant is +-1 (unimodular matrix.) If this is impossible, return
false.

4. Invert the above unimodular matrix, and obtain vals from the
columns of the inverted matrix.

5. For each of these vals aside from the first, take the maximum and
minimum values of v(2)*i/N, where N is the number of scale steps and i
goes from 0 to N-1, and find the average of the maximum and minimum
(the midpoint value.)

6. Using the midpoint value as an offset, calculate a scale by the
above method. If it gives the correct result, return true and the data
defining the means of computing the scale (v(2) for the various vals,
corresponding commas, and offsets. Just giving the formula would be good.)

Can anyone improve on this?

* Please see question below Thanks
--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
>
> > "Many" is a little vague, though I see tamil_vi.scl is a mode of
> > ramis.scl. Can you tell me what two commas give Marpurg #1 and
Ramis?
> > Do you know of any other specific examples?
>
> I'm going to answer part of my own question. Determination of
whether
> or not something was a Fokker block might be considered as an
addition
> to Scala.
>
> Consider the intervals of ramis.scl. These go
>
> 135/126, 256/243, 16/15, 135/128, 16/15, 135/128, 16/15, 256/243,
> 135/128, 16/15, 135/128, 16/15
>
> There are only three sizes (in the 5-limit, there can be as many as
> four) and none is a singleton, which we should avoid using as a
basis.
> Pick 16/15 as the "basis", and take ratios with the other two step
> sizes to find the commas: (16/15)/(135/128) = 2048/2025, and
> (16/15)/(256/243) = 81/80. Now invert the matrix whose rows are the
> monzos for 16/15, 81/80, 2048/2025 and get the matrix whose columns
> are the standard vals h12, -h2, -h5. If we take the triples
> [h12(q), h2(q), h5(q)] for q in the Ramis scale, we get
>
> 1 [0, 0, 0]
> 135/128 [1, 0, 1]
> 10/9 [2, 1, 1]
> 32/27 [3, 1, 1]
> 5/4 [4, 1, 2]
> 4/3 [5, 1, 2]
> 45/32 [6, 1, 3]
> 3/2 [7, 1, 3]
> 128/81 [8, 2, 3]
> 5/3 [9, 2, 4]
> 16/9 [10, 2, 4]
> 15/8 [11, 2, 5]
> 2 [12, 2, 5]
>
> The values for h2 and h5 are nondecreasing, and for h12 they go up
one
> at a time (the scale is h12-epimorphic.) This tells us we have a
> Fokker block.

*Can someone tell me what determines the breakpoints for h2 and h5?
(Why for example does h2 go to 2 when i=8.)
>
> Doing the same kind of analysis for Marpurg Monochord 1, we get
> 16/15, 128/125 and 81/80, and h12, h5, and h3 as the vals to check
it
> with. This time we get
>
> 1 [0, 0, 0]
> 25/24 [1, 1, 0]
> 9/8 [2, 1, 1]
> 6/5 [3, 1, 1]
> 5/4 [4, 2, 1]
> 4/3 [5, 2, 1]
> 45/32 [6, 3, 2]
> 3/2 [7, 3, 2]
> 25/16 [8, 4, 2]
> 5/3 [9, 4, 2]
> 9/5 [10, 4, 3]
> 15/8 [11, 5, 3]
> 2 [12, 5, 3]
>
> Once again, we have the correct monotonic and epimorphic properties,
> and Marpurg Monochord 1 is a Fokker block.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> * Please see question below Thanks
No reply neccessary. I figured it out. I get the same results for
everything except h2 in Ramis scale. h5 checks out, and so does
h5 and h3 in Marpung.

> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> > wrote:
> >
> > > "Many" is a little vague, though I see tamil_vi.scl is a mode
of
> > > ramis.scl. Can you tell me what two commas give Marpurg #1 and
> Ramis?
> > > Do you know of any other specific examples?
> >
> > I'm going to answer part of my own question. Determination of
> whether
> > or not something was a Fokker block might be considered as an
> addition
> > to Scala.
> >
> > Consider the intervals of ramis.scl. These go
> >
> > 135/126, 256/243, 16/15, 135/128, 16/15, 135/128, 16/15, 256/243,
> > 135/128, 16/15, 135/128, 16/15
> >
> > There are only three sizes (in the 5-limit, there can be as many
as
> > four) and none is a singleton, which we should avoid using as a
> basis.
> > Pick 16/15 as the "basis", and take ratios with the other two step
> > sizes to find the commas: (16/15)/(135/128) = 2048/2025, and
> > (16/15)/(256/243) = 81/80. Now invert the matrix whose rows are
the
> > monzos for 16/15, 81/80, 2048/2025 and get the matrix whose
columns
> > are the standard vals h12, -h2, -h5. If we take the triples
> > [h12(q), h2(q), h5(q)] for q in the Ramis scale, we get
> >
> > 1 [0, 0, 0]
> > 135/128 [1, 0, 1]
> > 10/9 [2, 1, 1]
> > 32/27 [3, 1, 1]
> > 5/4 [4, 1, 2]
> > 4/3 [5, 1, 2]
> > 45/32 [6, 1, 3]
> > 3/2 [7, 1, 3]
> > 128/81 [8, 2, 3]
> > 5/3 [9, 2, 4]
> > 16/9 [10, 2, 4]
> > 15/8 [11, 2, 5]
> > 2 [12, 2, 5]
> >
> > The values for h2 and h5 are nondecreasing, and for h12 they go
up
> one
> > at a time (the scale is h12-epimorphic.) This tells us we have a
> > Fokker block.
>
> *Can someone tell me what determines the breakpoints for h2 and h5?
> (Why for example does h2 go to 2 when i=8.)
> >
> > Doing the same kind of analysis for Marpurg Monochord 1, we get
> > 16/15, 128/125 and 81/80, and h12, h5, and h3 as the vals to
check
> it
> > with. This time we get
> >
> > 1 [0, 0, 0]
> > 25/24 [1, 1, 0]
> > 9/8 [2, 1, 1]
> > 6/5 [3, 1, 1]
> > 5/4 [4, 2, 1]
> > 4/3 [5, 2, 1]
> > 45/32 [6, 3, 2]
> > 3/2 [7, 3, 2]
> > 25/16 [8, 4, 2]
> > 5/3 [9, 4, 2]
> > 9/5 [10, 4, 3]
> > 15/8 [11, 5, 3]
> > 2 [12, 5, 3]
> >
> > Once again, we have the correct monotonic and epimorphic
properties,
> > and Marpurg Monochord 1 is a Fokker block.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> We can automate this process with the following steps
>
> 1. Find the set of intervals between steps. If there are more of these
> than there are primes involved in the factorization of the scale
> degrees, return false.

Sorry, I forgot my own analysis. Skip this step.

> 2. Check if the scale is epimorphic. If not, return false.
>
> 3. Pick a "base" element from the scale steps, by any means (first
> step, smallest step, most common step, least Tenney height, etc.) Now
> take ratios of the basis element with the other scale steps, finding
> (n-1) other ratios (where n is the number of primes involved) which
> together with the basis element gives a matrix of row vectors whose
> determinant is +-1 (unimodular matrix.) If this is impossible, return
> false.
>
> 4. Invert the above unimodular matrix, and obtain vals from the
> columns of the inverted matrix.
>
> 5. For each of these vals aside from the first, take the maximum and
> minimum values of v(2)*i/N, where N is the number of scale steps and i
> goes from 0 to N-1, and find the average of the maximum and minimum
> (the midpoint value.)
>
> 6. Using the midpoint value as an offset, calculate a scale by the
> above method. If it gives the correct result, return true and the data
> defining the means of computing the scale (v(2) for the various vals,
> corresponding commas, and offsets. Just giving the formula would be
good.)
>
> Can anyone improve on this?

I've paid attention. It will be a useful addition, so I've put it
on the todo list.

Manuel