Message #2300

From: Andrey <andreyastrelin@yahoo.com>
Subject: [MC4D] Re: Hyperbolic Honeycomb {7,3,3}
Date: Tue, 03 Jul 2012 14:43:52 -0000

Yes, dihedral… I mean angle between hexagonal and triangular faces of truncated tetrahedron. By the selection of truncating planes it will be pi/2. And angles between hexagonal faces are 2*pi/7.

Andrey

— In 4D_Cubing@yahoogroups.com, Don Hatch <hatch@…> wrote:
>
> Hi Andrey,
>
> I’m not sure if I’m understanding correctly…
> is "behedral angle" the same as "dihedral angle"?
> If so, isn’t the dihedral angle going to be 2*pi/7,
> since, by definition, 7 tetrahedra surround each edge?
>
> Don
>
>
> On Tue, Jul 03, 2012 at 05:14:47AM -0000, Andrey wrote:
> >
> >
> > Hi all,
> > About {3,3,7} I had some idea (but it was long ago…). We know that its
> > cell is a tetragedron that expands infinitely beyond "vertices". For each
> > 3 its faces we have a plane perpendicular to them (that cuts them is the
> > narrowest place). It we cut {3,3,7} cell by these planes, we get truncated
> > tetrahedron with behedral angles = 180 deg. What happens if we reflect it
> > about triangle faces, and continue this process to infinity? It will be
> > some "fractal-like" network inscribed in the cell of {3,3,7} - regular
> > polyhedron with infinite numer of infinite faces but with no vertices. I’m
> > sure that it has enough regular patterns of face coloring, and it may be a
> > good base for 3D puzzle.
> >
> > Andrey
> >
> > — In 4D_Cubing@yahoogroups.com, Don Hatch <hatch@> wrote:
> > >
> > > Hi Nan,
> > >
> > > Heh, I try not to make judgements…
> > > I think {3,3,7} is as legit as {7,3,3}
> > > (in fact I’d go so far as to say they are the same object,
> > > with different names given to the components).
> > > But pragmatically, {7,3,3} seems easier to get a grip on,
> > > in a viewer program such as yours which focuses naturally
> > > on the vertices and edges.
> > >
> > > Perhaps the best way to get a feeling for {3,3,7}
> > > would be to view it together with the {7,3,3}?
> > > Maybe one color for the {7,3,3} edges,
> > > another color for the {3,3,7} edges,
> > > and a third color for the edges formed where
> > > the faces of one intersect the faces of the other.
> > > And then perhaps, optionally,
> > > the full outlines of the characteristic tetrahedra?
> > > There are 6 types of edges in all (6 edges of a characteristic tet);
> > > I wonder if there’s a natural coloring scheme
> > > using the 6 primary and secondary colors.
> > >
> > > Don
> > >
> > >
> > > On Sat, Jun 30, 2012 at 11:29:32PM -0000, schuma wrote:
> > > >
> > > >
> > > > Hi Don,
> > > >
> > > > Nice to see you here. Here are my thoughts about {3,3,7} and the
> > things
> > > > similar to it.
> > > >
> > > > Just like when we constructed {7,3,3} we were not able locate the cell
> > > > centers, when we consider {3,3,7} we have to sacrifice the vertices.
> > Let’s
> > > > start by considering something simpler in lower dimensions.
> > > >
> > > > For example, in 2D, we could consider a hyperbolic "triangle" for
> > which
> > > > the sides don’t meet even at the circle of infinity. The sides are
> > > > ultraparallel. Since there’s no "angle", the name "triangle" is not
> > > > appropriate any more. I’ll call it a "trilateral", because it does
> > have
> > > > three sides (the common triangle is also a trilateral in my notation).
> > > > Here’s a tessellation of H2 using trilaterals, in which different
> > colors
> > > > indicate different trilaterals.
> > > >
> > > >
> > http://games.groups.yahoo.com/group/4D_Cubing/files/Nan%20Ma/figure3.gif
> > > >
> > > > I constructed it as follows:
> > > >
> > > > In R2, a hexagon can be regarded as a truncated triangle, that is,
> > when
> > > > you extend the first, third, and fifth side of a hexagon, you get a
> > > > triangle. In H3, when you extend the sides of a hexagon, you don’t
> > always
> > > > get a triangle in the common sense: sometimes the extensions don’t
> > meet.
> > > > But I claim you always get a trilateral. So I started with a regular
> > {6,4}
> > > > tiling, and applied the extensions to get the tiling of trilaterals.
> > > >
> > > > And I believe we can do similar things in H3: extend a properly scaled
> > > > truncated tetrahedron to construct a "tetrahedron" with no vertices.
> > > > Fortunately the name "tetrahedron" remains valid because hedron means
> > face
> > > > rather than vertices. But I have never done an illustration of it yet.
> > > > Then, maybe we can go ahead and put seven of them around an edge and
> > make
> > > > a {3,3,7}.
> > > >
> > > > I agree that these objects are not conventional at all. We lost
> > something
> > > > like the vertices. But just like the above image, they do have nice
> > > > patterns and are something worth considering.
> > > >
> > > > So, what do you think about them? Do they sound more legit now?
> > > >
> > > > Nan
> > > >
> > > > — In 4D_Cubing@yahoogroups.com, Don Hatch <hatch@> wrote:
> > > > > {3,3,7} less so… its vertices are not simply at infinity (as in
> > > > {3,3,6}),
> > > > > they are "beyond infinity"…
> > > > > If you try to draw this one, none of the edges will meet at all (not
> > > > even at
> > > > > infinity)… they all diverge! You’ll see each edge
> > > > > emerging from somewhere on the horizon (although there’s no vertex
> > > > > there) and leaving somewhere else on the horizon…
> > > > > so nothing meets up, which kind of makes the picture less
> > satisfying.
> > > > > If you run the formula for edge length or cell circumradius, you’ll
> > get,
> > > > not infinity,
> > > > > but an imaginary or complex number (although the cell in-radius is
> > > > finite, of
> > > > > course, being equal to the half-edge-length of the dual {7,3,3}).
> > > >
> > > >
> > >
> > > –
> > > Don Hatch
> > > hatch@
> > > http://www.plunk.org/~hatch/
> > >
> >
> >
>
> –
> Don Hatch
> hatch@…
> http://www.plunk.org/~hatch/
>