packing S3 and S4

[Tamfang]

What do you get if you pack N>>5! balls in S3?
In S2 you get "kissing pennies" with twelve flaws, where local symmetry is 5fold rather than 6fold; so in S3 I would expect a fcc lattice (the familiar "stack of oranges") with flaws, but what sort of flaws?

My casual curiosity about this and related questions was aroused by my latest rereading of Greg Egan's Diaspora which obliquely mentions "the five-dimensional version of" a geodesic dome.

[wendy]

In four dimensions, one can pack at least 24, and prehaps 25 spheres around a sphere.

In five, six and seven dimensions, it is believed, but not proven that 40, 72 and 126 spheres go around a central sphere. There is certianly more than one solution. In eight dimensions, 240 is the best known result, is quite a dense packing.

The idea of the geodesic sphere is to reduce excess curvature evenly over the sphere, by making the faces very much run as the sphere does. One starts off with a polytope with lots of small faces, eg 20 triangles, or 600 tetrahedra.

[anderscolingustafson]

In four dimensions, one can pack at least 24, and prehaps 25 spheres around a sphere.

In four dimensions you might even be able to pack 120 spheres around a sphere since the 120 cell is the 4d equivalent of the dodecahedron and in 3d the number of spheres that can fit around a sphere is the same as the number of sides a dodecahedron has. So the maximum number of spheres you can pack around a sphere in 4d might just be the same as the number of sides to the 120 cell.

[Tamfang]

So the maximum number of spheres you can pack around a sphere in 4d might just be the same as the number of sides to the 120 cell.

Might be, but isn't. You need to restrain this impulse to apply analogies blindly.

[Keiji]

So the maximum number of spheres you can pack around a sphere in 4d might just be the same as the number of sides to the 120 cell.

Might be, but isn't. You need to restrain this impulse to apply analogies blindly.

This.

[wendy]

In practice, you are hard pressed to pack 25 spheres around a given sphere. You certianly can get 24.

Google for 'kissing numbers' 'spheres' or see, http://en.wikipedia.org/wiki/Kissing_number_problem

The case for 120 spheres is a hyperbolic case, since {3,3,3,5} is hyperbolic.

[pat]

I used my raytracer to draw 24 3-spheres touching a center sphere in 4-D. Unfortunately, I can't find the image or input file for it right now. There is small picture from outside the kissing configuration on page 35 of these slide: http://old.nklein.com/products/rt/rt2.5 ... -guide.pdf

The one that I printed out really huge was a fish-eye view from the origin of the 24 surrounding spheres. I'll have to try to dig through old backups somewhere.

[anderscolingustafson]

I wander how many spheres could be packed around a sphere in five dimensions.

[Keiji]

I wander how many spheres could be packed around a sphere in five dimensions.

Don't we all?

IIRC the value for 8 dimensions was found to be 240, so it can't be higher than that

[Tamfang]

See http://en.wikipedia.org/wiki/Kissing_number_problem
Apparently in E5 it's known that you can't get better than 44, but the best known is 40 (vertices of a rectified cross-polytope, as in E4 and E3).

[Tamfang]

It's interesting that not a single reply has even acknowledged the original question. :\

[Tamfang]

... but hey, don't go away mad ...

[wendy]

Basicly, the kissing pennies and the geodesic dome are two different issues.

Kissing pennies are about spheres of equal size around a centre one. This is known exactly in 2 and 8 dimensions (6 and 240), while in other dimensions, follow the series 3d=12, 4d=24, 5d=40, 6d=72, 7d=126.

The geodesic dome issue is about finding rigid polytopes that have lots of vertices. The number of vertices do not equate to the kissing pennies, since there is no requirement that the edge equals the diameter (in 3d, it is the icosahedron, where the edge is longer than the radius. In 4d, the 600ch gives the opposite effect, the radius is 1.618 of the edge). Based on using simple faces, the 5d version would have 16, 6d has 27, 7d has 56 and 8d 240, these being thorald gosset's series of polytope.

[Tamfang]

Geodesic domes are closely related to the problem of packing many smaller balls around one big ball, which is the topic that I tried to raise here.

[wendy]

That's why the kissing pennies trick doesn't work. The case in 4d is for the coins around the sphere to be 0.61803398875 = 0:74.19 of the central size. In no other dimension it gets so small. Even in 24 dimensions, the kissing number is 13.7800 (twelfty), but i don't think it is any better in chords than the cuboctahedron or 24 cell.

There are not a lot of polytopes of high symmetry that equate to packing small spheres around a big one. In any case, the three-dimensional case uses larger spheres than the pennies (the edges of 3,5 are bigger than the radius). The kissing sphere of some size is the right approach, but the range of polytopes runs out sooner rather than later for an authoritive answer in this manner.

From what i see of the geodesic dome, one uses simplex-tilings, with lots of vertices to reduce the excess deficit at each corner. By making all faces of as few types as possible the strutt-set is kept to a minimum. In five to eight dimensions, the polytopes in question are gosset's semiregular figures. Beyond eight, one starts getting into the assorted litroforms.

[Tamfang]

That's why the kissing pennies trick doesn't work.

For what?

The case in 4d is for the coins around the sphere to be 0.61803398875 = 0:74.19 of the central size.

For what? (By the way, that number has names)

... There are not a lot of polytopes of high symmetry that equate to packing small spheres around a big one.

I didn't ask about high symmetry ...

In any case, the three-dimensional case uses larger spheres than the pennies (the edges of 3,5 are bigger than the radius). The kissing sphere of some size is the right approach, ...

I'm still not convinced that we're anywhere near the same page, so I'll rephrase my OP.

If you close-pack a VERY LARGE number of dots on a 2-sphere (the surface of a 3-ball), enough so that on the scale of a small cluster of dots the curvature of the substrate is small, most of their Voronoi cells are approximately regular hexagons. But because of the curvature, the pattern has twelve flaw-points whose neighborhoods have (rough) pentagonal symmetry.

I'm not interested here in symmetries of the arrangement as a whole.

If you close-pack a VERY LARGE number of dots on a 3-sphere (the surface of a 4-ball), enough so that on the scale of a small cluster of dots the curvature of the substrate is small, presumably most of their Voronoi cells are approximately rhombic dodecahedra (with random small truncation of the 'outer' vertices). And again, because of the curvature there must be flaws. But what kinds of flaws? Are the flaws concentrated around 120 points, or do they extend along 720 edges? (Or have I guessed wrong about these numbers? Since the lattice is face-centred cubic, maybe its flaws belong to the vertices and edges of a tesseract.) What does the Voronoi map look like in the neighborhood of a flaw?

I'm not interested here in symmetries of the arrangement as a whole.

[wendy]

The idea of the geodesic dome is to use a minimum number of strut sizes to get a maximum dome. That's the point. Using highly symmetric figures made of triangles etc does the job.

The 'kissing pennies' trick supposes that the coin in the middle is the same size as the external ones. A soluion to the kissing pennies supposes a maximum strutt size. However, while sphere-packing in spherical space is of some help. But there are solutions to the geodesic dome that are unrelated to the kissing-sphere issue.

In practice, the kissing-number at Sloane's site solves the case for spheres of the same size (irrespective of any lattice), these only diverge after something like 9 or 10 dimensions: the kissing number and lattice contact number are identical up to that point.

The sphere in 4d can be solved by the edges of {3,3,5}, where the edge is 0.618 of the radius. However, while this is a packing of spheres of high efficiency, it is not a generic 'kissing number' solution (where all spheres are the same as the inner one). In three dimensions, the {3,5} is used for geodesic domes, although the edge of this is longer than the radius.

One could use several different approaches, some of which have been solved in three dimensions: the largest circle that N will pack into S2. In some cases, a packing of N spheres might allow more spheres of the same size to join, such as one might have the 16 vertices of {4,3,3} vs the 24 spheres of {3,4,3}. One can pack 20 spheres in the vertices of x3o3o3x, but this is identical in size to 20 vertices of 3,4,3, so some rearrangement is still possible there. However, i don't think that outside of the regular cases, anyone has brooked a solution. The {4,3,3} and {5,3,3} are probably not maximal.

However, packing spheres of a given size is hardly the 'geodesic dome', which implies minimising the lengths of struts.

[wendy]

We now turn to the flaws.

It should be understood that the flaws are a result of the large structure. One replaces the faces of some target polytope by a surface tiling that fits without cell-loss into that shape.

The high symmetry does reflect the flaws. The idea is to replace the faces of a polytope by a sheet of tiles that generally fit. In 3d, the faces of an icosahedron are replaced by sheets of triangles. You get flaws from levels below the margins between the faces. In three dimensions, you get flaws at the vertices of the faces of x3o5o, which give 5 instead of 6 pentagons.

In four dimensions, the optimal solution is the {3,3,5}, with walls replaced by the oct-tet truss. At the vertex, one gets an icosahedra of tetrahedra. The edges give rise to new vertices, with five copies of tetra-octa-tetra (that forms the tetrahedral edge). You get some margin disconuity caused by the change from alternating octahedra/tetra on the faces, to matching octahedra and matching tetra at the triangles of {3,3,5}.

In five dimensions, one might suppose the optimal solution is oEx3o3o3o. Here, the cells reduce to tilings of A4 (x3o3o3o3o3z), and {3,3,4,3}. This means that even in the simplest form, the cells are alternately tilings of x3o3o3o and o3x3o3o (penta and rect penta), which change at places to x3o3o4o, of which some are halved. The corners of oEx3o3o3o are o3x3o3o, which have 5 octa and 5 tetra faces, becoming x3o3o4o and x3o3o3o.

[Tamfang]

The idea of the geodesic dome is to use a minimum number of strut sizes to get a maximum dome. That's the point.

The idea of the geodesic dome is to distribute forces efficiently; that's the point of making it round. If you really want to minimize the number of kinds of elements, use flat facets.

The 'kissing pennies' trick supposes that the coin in the middle is the same size as the external ones.

What would you call a close-packed array of (many more than seven) discs in E2? That's what I meant when I used the phrase "kissing pennies" — and **** am I sorry I did!