# Message #2311

From: Andrey <andreyastrelin@yahoo.com>

Subject: Re: [MC4D] Hyperbolic Honeycomb {7,3,3}

Date: Wed, 04 Jul 2012 20:40:50 -0000

If we take trilateral based on {6,4} and use it as a face of infinite tetrahedron, what dihedral angle we get? If it’s not 2*pi/n, we’ll not get {3,3,n} honeycomb. But for the single H3 object it’s a good thing (with regular hexagons as faces of truncated tetrahedra it will be very tasty).

Andrey

— In 4D_Cubing@yahoogroups.com, Roice Nelson <roice3@…> 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

>