Tetrad geometry of planar temperaments

These are seven limit temperaments from a single comma, thus rank 3,
and the ones considered here are planar in the strict sense--the
octave is not divided into the equivalent of periods, and pitch
classes appear on one plane, not layered into several. Using Hermite
reduction, you can get a 4x3 matrix, the first column of which is the
octaves column which we ignore. The remaining two columns can be
regarded as a stack of four two-element vectors, and these vectors as
representing points in the generator plane, one of which is [0 0]. We
thereby get either a quadrilateral or a triangle. I list below the
comma, the number of boundry points B, the number of interior points
I, the number of verticies V, and the area A. By Pick's theorem,
A = B/2 + I - 1.

The only two temperaments on the list to have all of these invariants
the same are 81/80 and 1029/1024; however these are transformable one
to the other and therefore ought to be the same. The area works as a
complexity measure, so you can consider the list as ordered from low
to high complexity. Note that the complexity of 2401/2400, despite
being nanotempering, is low! It actually has the same number of
boundry and interior points as 225/224, and differs by being a
triangle rather than a quadrilateral. An extraneous note to the chord
appears on the boundry rather than in the interior, and the 7 is an
interior point, not a boundry point.

126/125
B = 5
I = 0
V = 3
A = 1.5

225/224
B = 4
I = 1
V = 4
A = 2

2401/2400
B = 4
I = 1
V = 3
A = 2

81/80
B = 6
I = 0
V = 3
A = 2

1029/1024
B = 6
I = 0
V = 3
A = 2

1728/1715
B = 6
I = 0
V = 4
A = 2

245/243
B = 3
I = 2
V = 3
A = 2.5

6144/6125
B = 5
I = 1
V = 4
A = 2.5

3136/3125
B = 7
I = 0
V = 3
A = 2.5

4375/4374
B = 3
I = 2
V = 3
A = 3.5

5120/5103
B = 9
I = 0
V = 4
A = 3.5

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> These are seven limit temperaments from a single comma, thus rank 3,
> and the ones considered here are planar in the strict sense--the
> octave is not divided into the equivalent of periods, and pitch
> classes appear on one plane, not layered into several. Using Hermite
> reduction, you can get a 4x3 matrix, the first column of which is
the
> octaves column which we ignore. The remaining two columns can be
> regarded as a stack of four two-element vectors, and these vectors
as
> representing points in the generator plane, one of which is [0 0].
We
> thereby get either a quadrilateral or a triangle. I list below the
> comma, the number of boundry points B, the number of interior points
> I, the number of verticies V, and the area A. By Pick's theorem,
> A = B/2 + I - 1.

Question: What exactly is represented by lines in these
quadrilaterals and triangles?
>

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

>
> Question: What exactly is represented by lines in these
> quadrilaterals and triangles?

The figures in question are either quadilaterals, in which case
{1,5/4,3/2,7/4} are the verticies and the sides are the line segments
between them, or three out of the above set are verticies, with the
remaining one (which could be 1) either an interior or a boundry
point. The three sides of the triangle then connect the three vertex
subset of {1,5/4,3/2,7/4}.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> >
> > Question: What exactly is represented by lines in these
> > quadrilaterals and triangles?
>
> The figures in question are either quadilaterals, in which case
> {1,5/4,3/2,7/4} are the verticies and the sides are the line segments
> between them, or three out of the above set are verticies, with the
> remaining one (which could be 1) either an interior or a boundry
> point. The three sides of the triangle then connect the three vertex
> subset of {1,5/4,3/2,7/4}.

Thanks! I'm really starting to love the stuff that can be visualized
geometrically. (Stuff you can even dream about)

Paul