Question for Topologists

(No, no…not the smoker’s toothpaste)

I’m having trouble envisioning something in my head, and was wondering if anyone could help.

If you take a set of circles of equal radius that do not overlap, and group them as closely as possible to each other, you get an arrangement with one in the center and six surrounding it with all borders touching. If you then “squish” them together along the axes defined by their points of contact so that the curved borders become flat faces, you get tessellating hexagons (like a honeycomb).

With me so far?

Okay. My question is: what happens when you use spheres? Spheres (on a single level) group like circles (for obvious reasons), but when you stack the layers, they are offset from each other, with each layer sitting in the “valleys” produced by the previous one.

If you again “squish” them together along all axes defined by their points of contact, what shape do they become? It seems that they should become regular polyhaedra, but there isn’t an obviously hexagonal one of those among the five Pythagoreans. Since those five are the only regular ones there are (so we’re told…hmmm :dubious: :slight_smile: ), what am I missing?

Thanks and all.

Dodecahedrons. Find a friend with gaming dice and you’ll see why.

Hmm…maybe not. It’s going to have to be something that can tessellate 3-space, and dodecahedrons can’t.

The simplest shape you can find in the planar case is the equilateral triangle; the 3-d analog of this, the tetrahedron, is the only Platonic solid you’ll get in this way. There are other somewhat-regular solids you can make by joining larger clusters of adjacent-sphere centers; for example the Archimedian solid known as the cuboctahedron (link contains useful illustration)–this is the one that seems to me most analogous to the hexagon.

The answer is complicated somewhat by the nonuniqueness of closest-packed arrangements in three dimensions. After arranging one layer of spheres in the planar hexagonal arrangement, there are two possible closest-packed places to put the next layer. Call these offsets “B” and “C” (where “A” indicates the zero offset relative to the first layer). If you arrange the first two layers as AB, then the third layer can be either A or C. The arrangement ABABAB… is the “hexagonal closest-packed” (hcp) structure; ABCABCABC… is the “face-centered cubic closest-packed” (fcc) structure. (You may remember these from chemistry; they are common crystal structures.) You could make more complicated lattices: for example, ABABCABABC… (repeating every five layers), and even quasiperiodic structures. Different lattices will obviously have different solids “embedded” in them; the cuboctahedron, for example, is present in the fcc arrangement, but not in the hcp arrangement.

I thought of dodecahedrons, but they have pentagonal surfaces, and the packing of circles (and thus, I’m thinking, the packing of spheres) would result in hexagonal surfaces.

The problem is that there’s no “hexahecahedron,” which is what seems to be called for.

Yeah, but the equilateral triangle isn’t the basis that I’m shooting for. Perhaps if you compress circles far enough you will get an ET, but I’m looking for (for want of a better terminology) the 3D version of a hexagon.

Ouch. No wonder I was having some difficulty. The bottom illustration on your link seems closest to what I’m shooting for, but I’ll admit, I didn’t realize that there were more than one “closest-packed place” to put the next layer. That seems counter-intuitive to me, but obviously my intuition is wrong…again.

I’m going to have to buy a box of marbles, or something. Once again, the world is stranger than I can imagine.

Great link, btw. I’m going to be digging around that all night, for sure. Thanks again.

Nitpick: Packing cubes tessellate 3-d space quite nicely. Tetrahedra don’t.

The cuboctahedron is “embedded” in the structure by connecting the centers of the spheres. But “squishing” them together forms a tessellation of rhombic dodecahedra. Squishing the spheres together in the hcp arrangement will give you another tessellation of different irregular dodecahedra.

For what it’s worth, the subject of sphere packing is not part of topology. Here’s the Wikipedia entry on sphere packing:

The reference at the end says that it’s part of discrete geometry. Remember, in topology, you’re able to distort distances as long as you do it continuously. The question you’ve asked is irrelevant to topology.

Yes, but cubes don’t appear in the 3-d closest-packed lattices in the way that tetrahedra do (as centers of adjacent spheres). The closest-packed structures don’t tesselate the space with tetrahedra; they leave lots of gaps.

I misunderstood the construction that Dijon Warlock was describing, though; I think you got it right. I thought he was talking about the polyhedra you can get by connecting adjacent sphere centers, but now I think he’s talking about the polyhedron you get by considering all points closest to a given point in the closest-packed arrangement, the Voronoi cell for the lattice–which, as you say, is (for the fcc lattice) the dual of the cuboctahedron, the rhombic dodecahedron.

The question’s been more or less answered, but I wanted to point out the nitpick that this is a question for geometers, not topologists.

Oh, man…I think we may have a winner here.

This is sounding exactly like what I’m looking for. Doesn’t look as I expected, but sounds right.

Ah, well then. I always understood topology as the study of distortable surfaces, which is why I thought this question would be addressed by that discipline. I stand happily corrected.

That point has already been made, but thanks (once again) for snarking in one of my GQ threads. Yer a joy, you is.

Well, you were on the right track. Just as six circles will enclose a seventh (and squishing the structure turns the middle circle into a hexagon), twelve spheres will enclose a thirteenth, so squishing the structure will turn the center sphere into a dodecahedron. It just won’t be a regular dodecahedron, since that would mean wasted space.

I don’t think it’s possible to perfectly stack polyhedrons with obtuse angles (which disqualifies any regular polyhedron of higher order than a cube), but at the moment, the proof escapes me.

Well, I take that back, since the rhombicdodecahedron clearly does have obtuse angles. Hmmm…

Topology is the study of properties of sets that are preserved by continuous maps. The distance between points is not one of those.

Again, I learn.

Now that I’ve posted that…

ultrafilter: Is your rendition of topology fundamentally different than mine, and if so, why? It sounds the same, only in different terms.

Granted, “sets” are more general than “surfaces”, but is the concept different otherwise?

The study of deformable surfaces is a subset of topology, but it’s only a very small subset. You can also have a topology consisting of the sets {{}, {1}, {2,3}, {1,2,3}}, for instance (admittedly that’s not a very interesting or useful topology, but it’s a perfectly valid one nonetheless). No surfaces anywhere in sight.

The rhombic dodecahedron shown in the Wiki link doesn’t look how you’d expect, because they chose a funny angle to look at it from. The simplest way to construct a rhombic dodecahedron is to start with a cube, and stick identical square-based pyramids on all of the faces so that the faces of the pyramids match up. Viewed that way, it’s perhaps a bit easier to see how they arise from face-centered cubic packing.

Not only continuous maps but homeomorphisms[1]. The maps have to form a group, following the Erlangen Program.

To make it extremely clear to the non-mathematicians: in the OP the “spheres” are actually balls (a sphere is just the surface of a ball), and the polyhedra the OP is asking for are still balls, since they’re homeomorphic (the topology notion of equivalent) to the original balls. If you’re thinking of such a deformation as changing anything, it’s not topology.

[1] continuous with a continuous inverse. Yes, all parts of that condition are required.

Actually, that makes it much harder for me. (I know :rolleyes: ) I’ll admit, I got math-lost somewhere in the middle of pre-calc, and have never quite recovered, so I’m trying to look at this question purely from an intuitional standpoint. However, while sticking identical square-based pyramids on all of the faces of a cube makes sense, doing it so that “the faces of the pyramids match up” doesn’t work for me. It seems they would be all separate from each other. Are you talking about more than one of these figures tessellating 3-space?

The reason this doesn’t work for me is that in 2-space, the geometry of the circle/hexagon process is very 6-fold (or 3-fold in multiples). I would therefore expect the same out of the 3-space version of same. Cubic geometry doesn’t make sense to my brain here. Sure, it’s 6-fold counting all faces, but it’s square. 90 degree angles, and all that. What I’m shooting for should be 60 degree angles (just in all three dimensions). At least, it seems so to me.

Omphaloskeptic: Sorry I don’t have the terminology to be more clear with my question. I’ve tried to figure which of your renditions of my question was the one I was asking…and finally gave up. Higher mathematics hurts my head. It intrigues, but it hurts. The best I can render this is the illustration I gave about the closest-packed circles turning into hexagons when compressed, and what is the 3D equivalent.

I’m still not seeing how there is more than one way to most closely pack spheres, but that seems to be the case, anyway. I am going to have to buy a box of marbles.

Google ad at the bottom of this thread currently:

“Easy Raccoon Deterrent”

Da hell?

Think of it like this: There’s six faces on a cube, and five on each pyramid. Gluing the square base of a pyramid to the square face of the cube replaces one face (of the cube) with four (of the new polyhedron). So after gluing on all the pyramids there are 24 faces.

But we want 12 (dodecahedron). If we choose the angles of the pyramids properly, the top face of the pyramid on the front of the cube will be in the same plane as the front face of the pyramid at the top of the cube. Instead of 24 triangular faces, we’ll have 24 rhombi. In particular, since the cube faces meet at 90 degrees, we should choose the pyramids so that the triangular faces meet the base at 45 degrees. Then from one face to the base of its pyramid (45) plus from one face to another of the cube (90) plus from the base of the new pyramid to its face (45) is 180 degrees, and the two triangles lay flat.