Message #2318

From: Don Hatch <hatch@plunk.org>
Subject: Re: [MC4D] Hyperbolic Honeycomb {7,3,3}
Date: Fri, 06 Jul 2012 05:05:42 -0400

The formula for cell in-radius is cosh(inradius({p,q})) = cos(pi/q)/sin(pi/p).
For the {n,ultrainf} based on {6,4},
I think we want n such that cosh(inradius({3,n})) == cosh(inradius(6,4)),
that is:
cos(pi/n)/sin(pi/3) = cos(pi/4)/sin(pi/6)
=> cos(pi/n)/(sqrt(3)/2) = (sqrt(2)/2)/(1/2)
=> n = +-pi/acos(sqrt(3/2))
= 4.770984191560898 i
So the {3,ultrainf} based on {6,4}
is {3, 4.770984191560898 i}.
Not a nice number like I had hoped.
Someone check my math?

Don

On Fri, Jul 06, 2012 at 04:16:04AM -0400, Don Hatch wrote:
>
>
> Thinking more about the parametrization…
> as we increase n to infinity, the cell in-radius of {3,n}
> increases, approaching a finite limit (the in-radius of {3,inf}), right?
> Then you can keep increasing the in-radius towards infinity,
> resulting in various kinds of what we’ve been calling {3,ultrainf}.
>
> I wonder if we can invert the formula for in-radius in terms of n,
> giving n in terms of in-radius? I guess the n’s
> for various kinds of {3,ultrainf} would be imaginary or complex?
>
> So maybe n is still a natural parameter for all of these,
> as an alternative to in-radius or distance-between-pairs-of-edges.
>
> In particular, I wonder what the parameter n is
> for the picture derived from the {6,4}?
>
> Let me find that formula again…
>
> Don
>
> On Wed, Jul 04, 2012 at 02:33:55PM -0500, Roice Nelson wrote:
> >
> >
> > Sure, I’m interested in what you guys came up with
> > along the lines of a {3,ultrainfinity}…
> > I guess it would look like the picture Nan included in his previous
> > e-mail
> > (obtained by erasing some edges of the {6,4})
> > however you’re free to choose any triangle in-radius
> > in the range (in-radius of {3,infinity}, infinity], right?
> >
> > yep, our discussion finished on that picture, so Nan already shared most
> > of what we talked about. I like your thought to use the inradius as the
> > parameter for {3,ultrainf}, and that range sounds right to me.
> >
> >
> > Is there a nicer parametrization of that one degree of freedom?
> > Or is there some special value which could be regarded as the canonical
> > one?
> >
> > Nan and I had discussed the parametrization Andrey mentions, the
> (closest)
> > perpendicular distance between pairs of the the 3 ultraparallel lines.
> > Since "trilaterals" have no vertices, these distances can somewhat play
> > the role of angle - if they are all the same you have a
> > regular trilateral. The trilateral derived from the {6,4} tiling that
> > Nan shared is even more regular in a sense. Even though trilaterals have
> > infinite edge length, we can consider the edge lengths between the
> > perpendicular lines above. Only for the trilateral based on the {6,4}
> are
> > those lengths equal to the "angles". So perhaps it is the best canonical
> > example for {3,ultrainf}.
> > seeya,
> > Roice
> >
> >
> >
>
> –
> Don Hatch
> hatch@plunk.org
> http://www.plunk.org/~hatch/
>
>

–
Don Hatch
hatch@plunk.org
http://www.plunk.org/~hatch/