Yahoo Tuning Groups Ultimate Backup tuning Fokker tunes

I was pondering the question of whether and to what extent being a Fokker block is an audible property, and one idea I came up with was to construct and listen to what might be called the "Fokker tune" for the block. A Fokker tune is constructed as follows:

(1) Take all the rank two temperaments which temper the block to a MOS (for a given val making the block weakly epimorphic) and order them in terms of complexity from least to greatest.

(2) Take the scale in Scala form, so having 2/1 but not 1/1, and order it for each temperament in terms of that temperament. By this I mean if [periods, generators] is the Hermite-reduced mapping for the temperament, and if a and b are two intervals, then a<b if generators(a)<generators(b), and a>b if generators(a)>generators(b). If generators(a)=generators(b), then a<b if periods(a)<periods(b), and a>b if periods(a)>periods(b). If both periods and generators are equal you don't have a temperament for the block, but in general you can say a<b if a<b numerically, etc.

(3) Now take these orderings and turn them into a tune by traversing the orderings from least to most complex; I add timing information by using a longer time value when the next note moves by a generator rather than a period.

Here's an example: Carl Lumma's prism scale is an august-pajara-hexe-augene wakalix. Below is a link to a version of the Fokker tune, and the seq file used to create it:

http://micro.soonlabel.com/gene_ward_smith/Fokker%20tunes/prism.mp3

0 exclude 10
0 division 16
0 frequency 261.6255653
0 tempo 3000000
! sorting of prism.scl

0 track 1
! august sort
0 program 66
0 note (16/15) 4
4 note (4/3) 4
8 note (5/3) 8
16 note (5/4) 4
20 note (8/5) 4
24 note (2) 8
32 note (7/6) 4
36 note (112/75) 4
40 note (28/15) 8
48 note (28/25) 4
52 note (7/5) 4
56 note (7/4) 8
! pajara sort
64 program 69
64 note (7/6) 5
69 note (5/3) 5
74 note (5/4) 5
79 note (7/4) 5
84 note (4/3) 5
89 note (28/15) 7
96 note (7/5) 5
101 note (2) 5
106 note (16/15) 5
111 note (112/75) 5
116 note (28/25) 5
121 note (8/5) 7
! hexe sort
128 program 80
128 note (16/15) 5
133 note (7/6) 5
138 note (4/3) 5
143 note (112/75) 5
148 note (5/3) 5
153 note (28/15) 7
160 note (28/25) 5
165 note (5/4) 5
170 note (7/5) 5
175 note (8/5) 5
180 note (7/4) 5
185 note (2) 7
! augene sort
192 program 61
192 note (7/6) 4
196 note (112/75) 4
200 note (28/15) 8
208 note (28/25) 4
212 note (7/5) 4
216 note (7/4) 8
224 note (16/15) 4
228 note (4/3) 4
232 note (5/3) 8
240 note (5/4) 4
244 note (8/5) 4
248 note (2) 12

🔗Mike Battaglia <battaglia01@...>

On Mon, Apr 16, 2012 at 12:31 AM, genewardsmith
<genewardsmith@...> wrote:
>
> I was pondering the question of whether and to what extent being a Fokker
> block is an audible property, and one idea I came up with was to construct
> and listen to what might be called the "Fokker tune" for the block. A Fokker
> tune is constructed as follows:

Well, it just sounds like 12-EDO to me, although I doubt that'll
surprise anyone. This is an interesting idea, though.

Keenan made the point that Fokker blocks don't have to have any
particularly special harmonic properties, though. For instance, you
could create a 7-note Fokker block in the {81/80, 25/24, 128/125}
basis of 5-limit JI. This will be a Fokker block, but it won't be too
spectacular harmonically. Once you transform the lattice back to {2/1,
3/1, 5/1}, it'll still be a Fokker block, and still not be too
spectacular harmonically.

However, what does seem to matter are the following three properties:
1) Epimorphicity
2) Being a (d-1)-connected set, where d is the number of dimensions in
the JI system you're working in (not counting the interval of
equivalence), -and where the basis is the primes-
3) Convexity

Parallelograms and hexagons (and their higher-dimensional
counterparts) are two shapes I can think of which satisfy all of these
requirements. Are there any others?

It might also be interesting to reimagine #2 on a triangular lattice.

-Mike

🔗Mike Battaglia <battaglia01@...>

On Mon, Apr 16, 2012 at 5:03 PM, Mike Battaglia <battaglia01@...> wrote:
>
> However, what does seem to matter are the following three properties:
> 1) Epimorphicity
> 2) Being a (d-1)-connected set, where d is the number of dimensions in
> the JI system you're working in (not counting the interval of
> equivalence), -and where the basis is the primes-
> 3) Convexity
>
> Parallelograms and hexagons (and their higher-dimensional
> counterparts) are two shapes I can think of which satisfy all of these
> requirements. Are there any others?
>
> It might also be interesting to reimagine #2 on a triangular lattice.

In case that wasn't clear, I suggest that what's really fundamental is
the notion that everything be related to everything else by as many
p-limit consonances as possible, which is why I suggested
(d-1)-connectedness in the lattice in which the primes are the basis.
If you couple that condition with epimorphicity and convexity, you'll
find that things like parallelograms and hexagons and possibly other
sorts of things turn up.

All of these parallelograms will be Fokker blocks, but the converse is
not true. I suggest that Fokker blocks which meet this special
property are the ones which are really of interest. I'll refrain from
suggesting the name "Mother-Fokker blocks" for such constructions.
Additionally, this suggests that hexagons might be of interest.

-Mike

🔗genewardsmith <genewardsmith@...>

🔗Mike Battaglia <battaglia01@...>

On Mon, Apr 16, 2012 at 6:30 PM, genewardsmith <genewardsmith@...>
wrote:
>
> --- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > Keenan made the point that Fokker blocks don't have to have any
> > particularly special harmonic properties, though. For instance, you
> > could create a 7-note Fokker block in the {81/80, 25/24, 128/125}
> > basis of 5-limit JI. This will be a Fokker block, but it won't be too
> > spectacular harmonically.
>
> Show it to us.

The same set of points on the lattice will be a Fokker block no matter
what JI intervals you say that |1 0 0>, |0 1 0>, and |0 0 1>
represent. Put another way, if you apply any unimodular transformation
to a Fokker block, it remains a Fokker block. For instance, consider
this set of monzos:

|0 0 0>
|-3 2 0>
|-2 0 1>
|2 -1 0>
|-1 1 0>
|0 -1 1>
|-3 1 1>

This is the Zarlino major scale, which has unison vectors |-4 4 -1>
and |-3 -1 2>. Now let's multiply by the matrix with rows <612 970
1421|, <118 187 274|, <-171 -271 -397|, which is unimodular. You then
get

|0 0 0>
|104 20 -29>
|197 38 -55>
|254 49 -71>
|358 69 -100>
|451 87 -126>
|555 107 -155>

This is still a Fokker block, and it has unison vectors |36 7 -10> and
|11 2 -3>. It's just that it's now a very long and very thin
parallelogram.

-Mike

🔗Mike Battaglia <battaglia01@...>

On Mon, Apr 16, 2012 at 6:33 PM, genewardsmith <genewardsmith@...>
wrote:
>
> --- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > All of these parallelograms will be Fokker blocks, but the converse is
> > not true.
>
> What are "these parallelograms" and why isn't the converse true? It's
> seems to me it's only the converse which is true.

"These parallelograms" are parallelograms which meet the three
criteria that I outlined.

How could only the converse be true? If the statement "all of these
parallelograms will be Fokker blocks" is false, then which
parallelograms are there that aren't Fokker blocks?

-Mike

🔗genewardsmith <genewardsmith@...>

🔗Mike Battaglia <battaglia01@...>

On Mon, Apr 16, 2012 at 10:23 PM, genewardsmith
<genewardsmith@...> wrote:
>
> --- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > This is still a Fokker block, and it has unison vectors |36 7 -10> and
> > |11 2 -3>. It's just that it's now a very long and very thin
> > parallelogram.
>
> By my calculation, it isn't a Fokker block. Can you recheck the values?

Yes, they're right. You can confirm them yourself by multiplying the
monzos in the Zarlino major scale by the transformation matrix I
specified. And you know that a Fokker block ought to remain a Fokker
block when a unimodular transformation is applied to the lattice, so
it has to be a Fokker block.

-Mike

🔗genewardsmith <genewardsmith@...>

🔗Mike Battaglia <battaglia01@...>

On Mon, Apr 16, 2012 at 11:43 PM, genewardsmith
<genewardsmith@...> wrote:
>
> --- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> > Yes, they're right. You can confirm them yourself by multiplying the
> > monzos in the Zarlino major scale by the transformation matrix I
> > specified.
>
> That seemed wrong also. Do you really want to send 2 to 2^612 * 3^970 *
> 5^1421?

Yes, that's the point. I'm sending normal intervals to ridiculous ones
to demonstrate a scale which is a Fokker block, but in which the notes
are all related to each other by absurd harmonic relationships.

-Mike

🔗genewardsmith <genewardsmith@...>

--- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> > That seemed wrong also. Do you really want to send 2 to 2^612 * 3^970 *
> > 5^1421?
>
> Yes, that's the point. I'm sending normal intervals to ridiculous ones
> to demonstrate a scale which is a Fokker block, but in which the notes
> are all related to each other by absurd harmonic relationships.

OK. Now the val <-942 -81 461| sends your first row of 2^612 * 3^970 * 5^1421 to 7, the next row to 11, and the last to 16. If I apply your transformation to Zarlino, I don't get what you got, but I do get something which this val sorts epimorphically. I think you screwed up somewhere, but now I've got a val which sends 2 to -942 so it can send your humongous replacement to 7, and use it in all of its queasy glory as an interval of repetition.

🔗Mike Battaglia <battaglia01@...>

2012/4/17 genewardsmith <genewardsmith@...>:
>
> --- In tuning@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
>> > That seemed wrong also. Do you really want to send 2 to 2^612 * 3^970 *
>> > 5^1421?
>>
>> Yes, that's the point. I'm sending normal intervals to ridiculous ones
>> to demonstrate a scale which is a Fokker block, but in which the notes
>> are all related to each other by absurd harmonic relationships.
>
> OK. Now the val <-942 -81 461| sends your first row of 2^612 * 3^970 * 5^1421 to 7, the next row to 11, and the last to 16. If I apply your transformation to Zarlino, I don't get what you got, but I do get something which this val sorts epimorphically. I think you screwed up somewhere, but now I've got a val which sends 2 to -942 so it can send your humongous replacement to 7, and use it in all of its queasy glory as an interval of repetition.

Oh, I I think I know what happened - we're screwed up on eigenspaces
again. I was assuming that monzos were column vectors here. So take
the transpose of the matrix I gave if you want to do it that way. But
since the transpose of a unimodular matrix is another unimodular
matrix, the example still works.

Either way, you can still keep 2/1 in the basis if you want to keep it
as the equivalence interval.

The point of this exercise was to demonstrate that, in response to
your question of what the "audible sound" of Fokker blocks is: from a
harmonic standpoint, without any additional restrictions placed, it's
possible for me to come up with arbitrarily absurd Fokker blocks, such
as the one I gave. So, harmonically, Fokker blocks don't have to sound
like much at all.

On the other hand, from a melodic scalar standpoint, what I find
interesting is that they generalize MOS's in being melodically even,
and in that you can move around from mode to mode (and from dome to
dome) by displacing notes by a chroma, ending up with another Fokker
block each time. Except in the rank > 2 case, you have more than one
chroma to deal with, which makes it more complex and beautiful in a
certain way. That's just me, anyways.

-Mike