Message #2338

From: Melinda Green <melinda@superliminal.com>
Subject: Re: [MC4D] Re: Hyperbolic Honeycomb {7,3,3}
Date: Tue, 17 Jul 2012 19:44:14 -0700

I have no idea but the pattern reminds me a lot of the background image
<http://draves.org/blog/206.jpg> of Scott Drave’s Blog. He’s the
inventor of the Electric Sheep, BTW.

-Melinda

On 7/17/2012 3:43 PM, Roice Nelson wrote:
>
>
> I made a little progress with this, using info from Don’s email, but
> working in the plane rather than with the sphere. My best image so
> far, with 20k tetrahedra (80k triangles), is here:
>
> http://www.gravitation3d.com/roice/math/%7B3,3,7%7D_sphere_at_inf.png
>
> For reference, the blue circle is the sphere equator, and the green
> arcs are a tetrahedral tiling on the sphere. (The incircles of that
> tiling were the 4 starting circles for the previous {3,3,inf} image.)
>
> Clearly, much improvement could be made as far as filling things in.
> However, I still wanted to post because there are some areas which
> genuinely look like they will NEVER get filled in, no matter how much
> one recursed. These "holes" in the picture seem to be the dual areas
> of what is getting filled in (notice the 3 holes at the vertices of
> the tetrahedral tiling). Does this make sense to anyone? I’m
> wondering if it is possible that reflections of the infinite
> fundamental tetrahedron never reach certain portions of H3. The more
> likely scenario is that I’m doing something incorrectly :)
>
> Roice
>
>
> On Mon, Jul 16, 2012 at 9:35 AM, Andrey <andreyastrelin@yahoo.com
> <mailto:andreyastrelin@yahoo.com>> wrote:
>
> Don,
> I think it’s much more convenient to work not with sphere, but
> with boundary plane in half-space Poincare model. Every plane in
> H3 inersects with this plane by circle, and transformations of H3
> are equivalent to Moebius transformations of boundary plane. And
> probably {3,3,8} will give better formulae for coordinates and
> movements.
> Andrey
>
> — In 4D_Cubing@yahoogroups.com
> <mailto:4D_Cubing@yahoogroups.com>, Don Hatch <hatch@…> wrote:
> >
> > Here’s a more detailed outline of how I’d go about drawing
> > the intersection of {3,3,7} with the sphere at infinity.
> > I’ll probably do this in a few days if no one else does it first.
> >
> > To get started, consider the center cell
> > of a cell-centered {3,3,7}.
> > The intersection of that cell with the sphere at infinity
> > consists of three little regular spherical triangles;
> > we need to know the euclidean coords of one of these
> > little spherical triangles.
> >
> > We can find that in 4 steps:
> >
> > Step 1: compute the cell mid-radius r31{3,3,7}.
> >
> > The reference paper "Regular Honeycombs In Hyperbolic Space"
> by HSM Coxeter
> > doesn’t give a direct formula for cell mid-radius r31,
> > but it gives a formula for the cell in-radius r32.
> > From that, we can use the identity
> > sin(A)/sinh(a) = sin(B)/sinh(b) = sin(C)/sinh(c)
> > on the right hyperbolic triangle
> > formed by the cell center, face center, and edge center
> (draw it):
> > sin(pi/2)/sinh(r31) = sin(pi/r)/sinh(r32)
> > i.e. 1/sinh(r31) = sin(pi/r)/sinh(r32)
> > i.e. r31 = asinh(sinh(r32)/sin(pi/r))
> > So, plug in the formula for r32 from the paper:
> > r32{p,q,r} =
> acosh(sin(pi/p)*cos(pi/r)/sqrt(1-cos(pi/p)^2-cos(pi/q)^2))
> > and now we have a formula for r31{p,q,r}, the cell mid-radius:
> > r31{p,q,r} =
> asinh(sinh(acosh(sin(pi/p)*cos(pi/r)/sqrt(1-cos(pi/p)^2-cos(pi/q)^2)))/sin(pi/r))
> > (This can certainly be simplified, using the identity
> > sinh(acosh(x)) = sqrt(x^2-1), and probably further
> simplifications,
> > but I’m not bothering here.)
> > (And note, I’m not positive I got all of the above exactly
> right,
> > but the method should be sound.)
> > Substitute p=3,q=3,r=7 to get the desired cell-mid-radius of
> {3,3,7}.
> >
> > Step 2: from r31, compute the euclidean distance
> > from the origin to an edge of the center cell in the
> poincare ball model.
> >
> > If I recall correctly, that will be tanh(r31/2).
> >
> > Step 3: from that, compute the actual coords of the two
> endpoints-at-infinity
> > of one edge of the center cell.
> >
> > For definiteness, align the center cell
> > with the regular euclidean tetrahedron
> > with verts:
> > (-1,-1,1)
> > (-1,1,-1)
> > (1,-1,-1)
> > (1,1,1)
> > The center of the cell’s edge closest to joining -1,-1,1 to
> 1,1,1
> > lies on the +z axis, so by Step 2 this edge center is:
> > (0,0,tanh(r31/2))
> > The two endpoints-at-infinity of that edge
> > will be (-sqrt(1/2),-sqrt(1/2),0) and (sqrt(1/2),sqrt(1/2),0)
> > "translated by (0,0,tanh(r31/2))",
> > i.e. transformed by the translation
> > that takes the origin to that edge center (0,0,tanh(r31/2)).
> >
> > Recall that for any points p,t in the poincare ball
> > (of any number of dimensions), p translated by t is
> > given by the magic formula:
> > ((1 + 2*p.t + p.p)*t + (1-t.t)*p) / (1 + 2*p.t + p.p * t.t)
> > where "." denotes dot product. (Hope I wrote that down right.)
> > So plug in:
> > t = (0,0,tanh(r31/2))
> > p = (sqrt(1/2),sqrt(1/2),0)
> > (a bunch of terms simplify and drop out since p.p=1 and
> p.t=0, but whatever);
> > The resulting endpoint coords are (a,a,b) for some a,b
> > (then the other endpoint is (-a,-a,b), but we don’t need
> that at this point).
> >
> > Step 4: rotate one of those endpoints-at-infinity
> > around the appropriate axis
> > to get the other two vertices of the little spherical triangle.
> > The three spherical triangle vertices are:
> > (a,a,b)
> > (b,a,a)
> > (a,b,a)
> > (where a,b are the result of step 3).
> >
> >
> =========================================================================
> > So now we have one little spherical triangle.
> > Now, choose a set of 3 generators for the rotational symmetry group
> > of {3,3,7}, and use them repeatedly to send the triangle everywhere.
> > There are lots of choices of 3 generators; here’s one:
> > A: plain old euclidean rotation by 120 degrees about the
> vector (1,1,1)
> > B: plain old euclidean rotation by 120 degrees about the
> vector (1,1,-1)
> > C: rotation by 2*pi/7 about an edge of {3,3,7}.
> > for this, we can use the composition of:
> > translate the edge center to the origin
> > (i.e. translate the origin to minus the edge center)
> > followed by plain old euclidean rotation of 2*pi/7
> about this edge-through-the-origin
> > followed by translating the origin back to the
> original edge center
> > A specific edge center, and the translation formula,
> > can be found in Step 3 above.
> >
> > Don
>
>
>
>
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