# 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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