Message #2303

From: Don Hatch <hatch@plunk.org>
Subject: Re: [MC4D] Re: Hyperbolic Honeycomb {7,3,3}
Date: Tue, 03 Jul 2012 15:51:50 -0400

Oh wait!
I realized I got last part wrong, just after I hit the "send" button :-)

The picture would start with an Apollonian gasket (see wikipedia article)
of circles on the sphere;
this is the intersection of the {7,3,3} with the sphere at infinity.
Then each circle in the gasket is filled with a {3,7},
the final result being the intersection of the {3,3,7} with the sphere
at infinity.
So, it isn’t true that there are isolated cluster points
in the *center* of each circle; the clustering is
towards the *boundary* of each circle. I think I have
a clear picture in my head of what this looks like now.

A cell of the {3,3,7} would touch the sphere
in 4 spherical triangles (its "feet"),
each foot in a different one of four mutually kissing circles
of the gasket, I think.

Don

On Tue, Jul 03, 2012 at 03:29:38PM -0400, Don Hatch wrote:
>
>
> Okay I think maybe I follow you now…
> But each face formed by truncation…
> it’s a triangle, not a hexagon, right?
> In fact it’s a spherical triangle, on the sphere at infinity, right?
> All of these spherical triangles, of different apparent sizes,
> would tile the sphere, 7 at each vertex…
> but with some "cluster points" which are the limit points
> of infinitely many of these triangles of decreasing size.
> I’d like to see a picture of this– it shouldn’t be too hard to generate
> (together with the spherical circles
> that are the intersection of the dual {7,3,3} with the sphere
> at infinity, in a different color… I think each cluster point
> would be the center of one of these circles).
>
> Don
>
> On Tue, Jul 03, 2012 at 02:43:52PM -0000, Andrey wrote:
> >
> >
> > 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/
> > >
> >
> >
>
> –
> Don Hatch
> hatch@plunk.org
> http://www.plunk.org/~hatch/
>
>

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