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RE: Fokker periodicity blocks

🔗Carl Lumma <clumma@xxx.xxxx>

9/29/1999 10:31:57 PM

>>That would seem to make sense. I just wondered if the larger tuning could
>>have a vector in addition to the required one(s) that would destroy the
>>scale. Is that possible?
>
>Only if it causes different notes in the scale to become equivalent to one
>another, reducing the number of notes it contains.

That's what I was worried about. So it is possible.

>>And exactly what properties are specified by the unison vector (which
ones >>are preserved across different embeddings)?
>
>I'm sure you know what a unison vector means, so I'm not sure what you're
>asking.

Poor wording on my part. I was asking what properties two scales must
share if they share one unison vector. We agree that cardinality isn't
one, which should pretty much diss everything else.

>Matrix multiplication is not commutative. The definition is, if A*B=C, then
>
>C(i,j) = A(i,1)*B(1,j) + A(i,2)*B(2,j) + A(i,3)*B(3,j) . . .

So C(i,j) is a matrix? And a matrix is a set of vectors? What are i and j?

-C.

🔗Paul H. Erlich <PErlich@xxxxxxxxxxxxx.xxxx>

9/30/1999 11:10:02 AM

Carl Lumma wrote,

>Poor wording on my part. I was asking what properties two scales must
>share if they share one unison vector. We agree that cardinality isn't
>one, which should pretty much diss everything else.

I don't know what you mean. Clearly scales which share the syntonic comma as
unison vector are meantone scales, what's to diss about that?

I wrote,

>>Matrix multiplication is not commutative. The definition is, if A*B=C,
then
>
>>C(i,j) = A(i,1)*B(1,j) + A(i,2)*B(2,j) + A(i,3)*B(3,j) . . .

>So C(i,j) is a matrix? And a matrix is a set of vectors? What are i and
j?

C(i,j) is the entry in the i_th row and j_th column of matrix C.

🔗Carl Lumma <clumma@xxx.xxxx>

10/2/1999 7:00:42 AM

>>Poor wording on my part. I was asking what properties two scales must
>>share if they share one unison vector. We agree that cardinality isn't
>>one, which should pretty much diss everything else.
>
>I don't know what you mean. Clearly scales which share the syntonic comma as
>unison vector are meantone scales, what's to diss about that?

Didn't you just say,

>Only if it causes different notes in the scale to become equivalent to one
>another, reducing the number of notes it contains.

So what exactly are some properties that meantone scales share, even when
they don't share cardinality (other than the definition of the syntonic
comma unison vector itself- actually, can't even that be destroyed, say, by
making another unison vector at +2 0)?

>>>Matrix multiplication is not commutative. The definition is, if A*B=C,
>>>then C(i,j) = A(i,1)*B(1,j) + A(i,2)*B(2,j) + A(i,3)*B(3,j) . . .
>>
>>So C(i,j) is a matrix? And a matrix is a set of vectors? What are i and
>>j?
>
>C(i,j) is the entry in the i_th row and j_th column of matrix C.

Okay. Hold on, that's heavy...

-C.

🔗Paul H. Erlich <PErlich@xxxxxxxxxxxxx.xxxx>

10/2/1999 7:23:04 AM

>Didn't you just say,

>>Only if it causes different notes in the scale to become equivalent to one
>>another, reducing the number of notes it contains.

Whoops -- I must clarify. I meant only if it causes different notes in the
_diatonic_ scale to become equivalent to one another, leaving it with less
than 7 notes.

>So what exactly are some properties that meantone scales share, even when
>they don't share cardinality (other than the definition of the syntonic
>comma unison vector itself

Well, they're all logically equivalent to that, but one is that meantone
scales can be notated conventionally, with the approximations to 3- and
5-limit intervals occuring as normally expected.

>- actually, can't even that be destroyed, say, by
>making another unison vector at +2 0)?

Well, then you get a 2-tone scale, but strictly speaking the syntonic comma
still vanishes.

>>C(i,j) is the entry in the i_th row and j_th column of matrix C.

>Okay. Hold on, that's heavy...

Not too bad once you get the hang of it. We had to do lots of these in 9th
grade, before learning a single application for matrix multiplication.

🔗Carl Lumma <clumma@xxx.xxxx>

10/3/1999 7:47:18 AM

>>So what exactly are some properties that meantone scales share, even when
>>they don't share cardinality (other than the definition of the syntonic
>>comma unison vector itself
>
>Well, they're all logically equivalent to that, but one is that meantone
>scales can be notated conventionally, with the approximations to 3- and
>5-limit intervals occuring as normally expected.

Paul, all I was asking: is there any property, say something that could be
defined on the interval matrix, that scales would share if they shared a
unison vector (they would all be Erlich-CS or not, to pull an example out
of the air). I suspected and you confirmed that cardinality is not
necessarily shared, so it's doubtful that anything else is. No big deal; I
was more trying to smooth over my understanding of periodicity blocks than
to uncover any great property of them.

-C.

🔗Paul H. Erlich <PErlich@xxxxxxxxxxxxx.xxxx>

10/3/1999 9:57:42 AM

>Paul, all I was asking: is there any property, say something that could be
>defined on the interval matrix, that scales would share if they shared a
>unison vector (they would all be Erlich-CS or not, to pull an example out
>of the air).

I'm sure there are many such properties -- the meantone one, that four
fifths equal a major third, is certainly defined on the interval matrix. As
I said before, I believe all periodicity blocks are CS, and I'm using
Kraig's old definition, which there's no reason to attach my name to.