Re: rtomes@kcbbs.gen.nz (Ray Tomes)

Ray Tomes wrote,

>>>This shape is interesting because it is the same as Kepler's sphere
>>>stacking problem. It can be represented as a triangular lattice with
>>>another one on top or a square lattice with another one on top.

I wrote,

>>That is, one can represent it by stacking triagular lattices in _three_
>>staggered positions, or square lattices in two staggered positions. For
>>viewing more than three layers of this lattice at once, an oblique view is
>>necessary.

>That is true in the position with the square base, but in the triangular
>view that you drew multiple layers can be seen at once.

>$ A---------E---------B
> /|\ /|\ / \
> B--/-|-\--F#-/-|-\--C# / \
> | / Eb--------Bb \ | / * \
> |/,' `.\|/,' `.\|/ \
> F---------C---------G---------D

Ray, that is only three layers, corresponding to the three staggered
positions. Try to add a fourth layer, and you will have overlaps. That is
why an oblique view would be necessary.

>Has any consensus been reached on the relative sizes of these?
>I stated earlier that p*ln(p) was the relative distance for prime p but
>later realised that this ratio refers to something slightly different,
>namely how often each prime should be present (inversely).

In the case where there is one axis for each _odd_ rather than for each
_prime_ , I have given my reasons for making the length of each connection
proportional to the log of the odd limit of the ratio. Erv Wilson, the
undisputed lattice king, uses one axis for each odd, and the unfortunate
result is that every note is duplicated an infinite number of times in the
lattice. However, using the shortest path in this lattice (with the lengths
I suggest) gives you a good measure of consonance. This was discussed very
extensively before you joined -- check the archives.