triangular vs. rectangular

Hi Paul,

For years you've backed triangular octave-equiv. lattices and
rectangular octave-spec. lattices. You wrote...

> The reason omitting the 2-axis forces one to make the lattice
> triangular is that typically many more powers of two will be
> needed to bring a product of prime factors into close position
> than to bring a ratio of prime factors into close position. So
> the latter should be represented by a shorter distance than the
> former. Simply ignoring distances along the 2-axis and sticking
> with a rectangular (or Monzo) lattice is throwing away
> information.
//
> ... a weight of log(axis) should be applied to all axes, and
> if a 2-axis is included, a rectangular lattice is OK. If a
> 2-axis is not included, a triangular lattice is better.
//
> ... in an octave-specific rectangular (or parallelogram)
> lattice, 7:1 and 5:1 are each one rung and 7:5 is two rungs. In
> an octave-specific sense, 7:1 and 5:1 really are simpler than
> 7:5; the former are more consonant. 7:4 and 5:4 are each three
> rungs in the rectangular lattice, but they still come out a little
> simpler than 7:5 since the rungs along the 2-axis are so short.
> If you can buy that 35:1 is as simple as 7:5, then the octave-
> specific lattice really should be rectangular, not triangular.
> 35:1 is really difficult to compare with 7:5 -- it's much less
> rough but also much harder to tune

...but I still don't understand what's wrong with an
octave-equiv. rect. lattice. Does your reasoning only apply
to weighted lattices (what about with unit taxicab lengths)?
Or, can you explain this another way? Thanks,

-Carl

>...but I still don't understand what's wrong with an
>octave-equiv. rect. lattice. Does your reasoning only apply
>to weighted lattices (what about with unit taxicab lengths)?
>Or, can you explain this another way?

I guess it depends on what you want to measure. So maybe
what I'm asking is, what do your two favored versions
measure (in musical terms), and why does oct-equiv-rect
fail?

-C.

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
>
> Hi Paul,
>
> For years you've backed triangular octave-equiv. lattices and
> rectangular octave-spec. lattices. You wrote...
>
> > The reason omitting the 2-axis forces one to make the lattice
> > triangular is that typically many more powers of two will be
> > needed to bring a product of prime factors into close position
> > than to bring a ratio of prime factors into close position. So
> > the latter should be represented by a shorter distance than the
> > former. Simply ignoring distances along the 2-axis and sticking
> > with a rectangular (or Monzo) lattice is throwing away
> > information.
> //
> > ... a weight of log(axis) should be applied to all axes, and
> > if a 2-axis is included, a rectangular lattice is OK. If a
> > 2-axis is not included, a triangular lattice is better.
> //
> > ... in an octave-specific rectangular (or parallelogram)
> > lattice, 7:1 and 5:1 are each one rung and 7:5 is two rungs. In
> > an octave-specific sense, 7:1 and 5:1 really are simpler than
> > 7:5; the former are more consonant. 7:4 and 5:4 are each three
> > rungs in the rectangular lattice, but they still come out a little
> > simpler than 7:5 since the rungs along the 2-axis are so short.
> > If you can buy that 35:1 is as simple as 7:5, then the octave-
> > specific lattice really should be rectangular, not triangular.
> > 35:1 is really difficult to compare with 7:5 -- it's much less
> > rough but also much harder to tune
>
> ...but I still don't understand what's wrong with an
> octave-equiv. rect. lattice.

Really? Reread the first paragraph above. For example, consider 5:3
vs. 15:8 in an octave-equivalent context. Following Partch and Kees,
we probably want to associate them with complexity values of log(5)
and log(15), respectively. But an octave-equivalent rectangular
lattice will give these the same length. That's means the lattice
isn't representing complexity by length, one of the things that you
want in order to make the lattices model some aspect of music.

> Does your reasoning only apply
> to weighted lattices (what about with unit taxicab lengths)?

It applies to both, clearly.

> Or, can you explain this another way?

Let me think about it. More questions will help . . .

>> Hi Paul,
>>
>> For years you've backed triangular octave-equiv. lattices and
>> rectangular octave-spec. lattices. You wrote...
>>
>> > The reason omitting the 2-axis forces one to make the lattice
>> > triangular is that typically many more powers of two will be
>> > needed to bring a product of prime factors into close position
>> > than to bring a ratio of prime factors into close position. So
>> > the latter should be represented by a shorter distance than the
>> > former. Simply ignoring distances along the 2-axis and sticking
>> > with a rectangular (or Monzo) lattice is throwing away
>> > information.
>> //
>> > ... a weight of log(axis) should be applied to all axes, and
>> > if a 2-axis is included, a rectangular lattice is OK. If a
>> > 2-axis is not included, a triangular lattice is better.
>> //
>> > ... in an octave-specific rectangular (or parallelogram)
>> > lattice, 7:1 and 5:1 are each one rung and 7:5 is two rungs. In
>> > an octave-specific sense, 7:1 and 5:1 really are simpler than
>> > 7:5; the former are more consonant. 7:4 and 5:4 are each three
>> > rungs in the rectangular lattice, but they still come out a little
>> > simpler than 7:5 since the rungs along the 2-axis are so short.
>> > If you can buy that 35:1 is as simple as 7:5, then the octave-
>> > specific lattice really should be rectangular, not triangular.
>> > 35:1 is really difficult to compare with 7:5 -- it's much less
>> > rough but also much harder to tune
>>
>> ...but I still don't understand what's wrong with an
>> octave-equiv. rect. lattice.
>
>Really? Reread the first paragraph above. For example, consider 5:3
>vs. 15:8 in an octave-equivalent context. Following Partch and Kees,
>we probably want to associate them with complexity values of log(5)
>and log(15), respectively. But an octave-equivalent rectangular
>lattice will give these the same length. That's means the lattice
>isn't representing complexity by length, one of the things that you
>want in order to make the lattices model some aspect of music.

In the 15-limit, I might consider 15 as consonant as 5:3 in some
sense. Or, maybe I'm just interested in 'chord modulation
distance'.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> In the 15-limit, I might consider 15 as consonant as 5:3 in some
> sense. Or, maybe I'm just interested in 'chord modulation
> distance'.

Right -- I'm waiting to see both how you propose to construct the 15-
limit lattice, and to explain what you meant about there being more
common-tone modulations via 15:8 than via 5:3.

But in a 15-limit lattice, we'd likely end up with a similar scenario,
comparing for example 15:13 vs. 195:128. Are you claiming these should
be the same length? Considerations of 'chord modulation distance'
certainly wouldn't support this, would it?