phase-packing

I just thought of a new kind of packing problem, a mutant extension of the Thomson problem. In this version, each particle has coordinates in two independent spaces; in each it is confined to a sphere (of some dimension). In each space, pairs of particles repel each other with a force inversely proportional to the square of their distance *in the other space*.

(I was imagining a head with striped hairs, as one does, and considered making each hair’s phase anti-correlated with those of its neighbors, and what that would mean.)

O fairest of randomizers

On most numbered dice, opposite sides are complementary; on a cube, for example, they add to 7. As a result, if you have the skill to throw a die so that the {1,2,3} corner lands on the table, the upward face must be at least 4.

I would prefer to design dice so that, if numbers are considered as masses, the center of mass coincides with the geometric center. I think this is equivalent to saying: the sum of the numbers in any hemisphere must be equal.

You can’t do that with a tetrahedron, cube, D10 or regular dodecahedron; but I found three solutions for the octahedron, and 876 for the *rhombic* dodecahedron. (At least I see no obvious way to reduce that number further with symmetries.)

For a cube the best you can do is put pairs of adjacent numbers on opposite faces.

The twelve best arrangements on the regular dodecahedron all have 0 and 11 opposite each other, 1–5 around 11, and 6–10 around 0.

The two best arrangements for D10 are 0285364179 and 0582367149 (reading around the fivefold axis, alternating upper and lower faces).

(Add 1 to each number if you don’t like zero-based indexing; it doesn’t affect the math.)

I’ll update here if I come up with an approach to the icosahedron problem that won’t take thousands of years.

witness on Whidbey

I watched *Behind the Curve* (2018), a documentary about the Flat Earth movement. In the beginning, Mark Sargent says (I paraphrase), “I know the Earth is not round because I can see Seattle from here [Whidbey Island].”

If I knew the distance from the Space Needle to Sargent’s house, the altitude of that house and the altitude of the lowest part of the Space Needle visible from there, I could put an upper bound on the curvature.

Scribbles: The Ensmoothening, Part III

Many of the curves in this chart have some unsightly wiggles. That’s because, when a function of degree 2 or higher tries to approximate a piecewise constant, it tends to go back and forth across the target. So here instead I fitted each such function not to the piecewise constant directly but to the fit of the next lower degree.

( . . more . . )

it’s in the literature

On a truncated icosahedron / buckyball / Telstar-style soccer ball, consider two adjacent hexagons and the two pentagons that are adjacent to both. These four faces can be removed, rotated by a right angle, and reattached, causing only a small change to the overall shape. Most fullerenes have at least one such patch.

If I ever get around to making more printable models of fullerenes, I would omit those that can be changed, by the above twist, into one of higher symmetry. I have a pretty good idea of how I’d go about listing the fullerenes and finding their siblings; but I do not have a grip on distinguishing symmetry groups of the same order – e.g., that of the regular tetrahedron versus that of a hexagonal prism – and a subgroup of one may not be a subgroup of the other.

So I got out *An Atlas of Fullerenes* in the hope of understanding how they did it – and happened to open to a chapter I had not looked at before, which covers the Stone-Wales transformation (for so it is named) and lists, up to C_{50} (15 hexagons), which fullerenes change with which.

The 812 smallest fullerenes are thus cut to 72 in 47 families. The biggest of these families has six remaining members, four with C2v symmetry (one axis of twofold rotation, and a reflection plane containing that axis) and two with C3 symmetry (chiral with one threefold axis). Their symmetry numbers are 4 and 3 respectively, but as C3 is not a subgroup of C2v I keep them all.

Surprisingly the ten families of C_{50} include two with no nontrivial symmetry at all.

another problem with my clothoids

I wrote:

each curve hits alternate dots: first exactly, then with offsets pushing it toward the other curve.

I don’t think I’ve mentioned here how the offsets work. ( . . more . . )

clothoid weekend update

For context, see past posts in the *curve-fitting* category that I just created. To recap:

The curves I’ve been drawing are the paths made by a point moving at constant speed at an angle which is a piecewise quadratic function of path length. Curvature, the first derivative of angle, is continuous.

Such a path that hits a given sequence of dots is fully determined if it loops, but otherwise it has two degrees of freedom. For any angle and curvature at the starting dot, there is a quadratic coefficient that lets the path reach the next node, and likewise for the next.

My current code starts with an estimate for the length of each segment (between two dots) and the angle at its midpoint, and uses these basis functions to fit those angles: a constant, a linear function, and a family of “solitons”: piecewise quadratics, zero outside a sequence of four dots, discontinuous in the second derivative at each of those dots. For n segments, there are n-2 solitons, so the constant and linear functions are needed to consume the last two degrees of freedom.

Eventually I noticed a flaw in this scheme: the curvature of the resulting path is the same at both ends, namely the slope of the linear component, because the solitons contribute nothing to it. That’s appropriate for ‘C’, but wrong for plenty of other strokes; in ‘S’ the end curvatures ought to have opposite sign.

The next thing I’ll try is a least-squares quadratic fit to the whole sequence, then fit the residues with solitons as before. That should be an improvement but it’s not ideal; curvature is a local feature. Perhaps I’ll think of something better later.