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Please consider a central, ordinary 2-sphere $S_1$, of some radius $r_1$, and a second ordinary sphere, $S_2$, of radius $r_2$, where $r_2 \leq r_1$.

My question concerns optimal values for the number of spheres of type $S_2$ that can be packed in three-dimensional space so that they are non-overlapping and tangent to $S_1$. Is there an analytical result for optimal packing a function of the ${r_2 \over r_1}$, or are there subsets of cases that are solved with methods beyond something like simulating annealing? Does it simplify the problem to apply the further restriction that $r_2 << r_1$?

I've been having trouble finding answers with a literature search, particularly for the latter situation where $r_2 << r_1$, and I appreciate everyone's time.

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Even for disks in a circle this is quite hard, see for example this question mathoverflow.net/questions/24184/… –  j.c. Oct 12 '10 at 0:05
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up vote 4 down vote accepted

I believe what you are seeking is sometimes known as a spherical packing or a spherical code. Here is the MathWorld article on the topic. Here is Neil Sloane's webpage on the topic, including "putative optimal packings" up to $n=130$ spheres. See also integer sequence A126195 for the "Conjectured values for maximal number of solid spheres of radius 1 that can be rolled all in touch with and on the outside surface of a sphere of radius $n$."

There are bounds, but no "analytical result" of the type for which you might be hoping.

Tangential Addendum. There is an interesting variant of the sphere-packing kissing number that seems little known, and for which the bounds are wide enough to invite further work. It is called Hornich's Problem: What is the fewest number of non-overlapping closed unit balls that can radially hide a unit ball $B$, in the sense that every ray from the center of $B$ intersects at least one of the surrounding balls? In $\mathbb{R}^3$, it is known that at least 30 balls are needed, and 42 suffice. So rather different than the kissing number 12!

Described on p.117 of Research Problems in Discrete Geometry, Brass, Peter, Moser, William O. J., Pach, János, 2005. ISBN: 978-0-387-23815-9

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Thank you for the references, especially A126195! –  AfternoonCoffee Oct 12 '10 at 0:16
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This problem has been studied by many people in a general setting of convex bodies.

Given a convex body $K$ in $\mathbb R^d$ and an $\alpha > 0$, find the maximal number $H_{\alpha}(K)$ of nonoverlapping translates of $\alpha K$ touching K.

$H_{\alpha}(K)$ is often referred to as the generalized Hadwiger number (or the generalized kissing number).

L. Fejes Tóth studied the asymptotics of $H_{\alpha}(K)$ for polytopes in 1970s. His result was extended to the general case by K. Böröczky Jr., D. Larman, S. Sezgin, and C. Zong ("On Generalized Kissing Numbers and Blocking Numbers", Rend. Circ. Mat. Palermo, 65 (2000), pp. 39-57) who showed that $$H_{\alpha}(K)\sim \frac{C}{\alpha^{d-1}}$$ for any convex body $K\subset\mathbb R^d$. The constant $C$ can be calculated more or less explicitly in terms of $K$.

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Very cool, thanks Andrey! –  AfternoonCoffee Oct 12 '10 at 2:30
    
You're welcome. The asymptotics holds for small $\alpha>0$, of course. –  Andrey Rekalo Oct 12 '10 at 21:22
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Assuming w.l.o.g. that $r_1=1$, one can guess the asymptotic behaviour of the kissing number as $r_2\to 0$ as a function $f(r_2)$. Consider a radial projection of the smaller balls onto the surface of the large (closed) ball (where the projection is w.r.t. the centre of the large ball). As $r_2\to 0$, this projection will locally approximate a packing of disks in the plane, and the density of the projection of an optimal packing should approximate the density of an optimal packing of unit disks in the plane. This density is well known to be $\frac{\pi}{2\sqrt{3}}$. Since the projection of each small ball covers a surface area of approximately $\pi r_{\kern-1pt2}^2$ of the unit ball, and the unit ball has surface area $4\pi$, we should have for very small $r_2$ $$ f(r_2)\pi r_{\kern-1pt2}^2 \approx \frac{\pi}{2\sqrt{3}}4\pi~, $$ or $$ \lim_{r_2\to 0} ~f(r_2) = \frac{2\pi}{\sqrt{3}\cdot r_{\kern-1pt2}^{2}}~. $$

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