# How many spheres can fit inside this larger sphere?

I would like to know if there is a way to do the following: calculate the maximal number of spheres of unit radius that can fit inside a sphere of radius 200 times the unit radius.

This is a generalisation of a question that was asked in a biology class. I was wondering if there exist some theorems on this, since I don't know how to start on it.

• You could estimate by taking the volume of the big sphere divided by the packing efficiency of the small spheres. – Cbjork Sep 30 '19 at 19:37
• Even in 2D there is no easy solution. I would not expect the case of spheres of radius ratio 200 to be tractable (for the exact solution). – Yves Daoust Sep 30 '19 at 19:40
• A good approximation should be $\frac{\pi}{3\sqrt{2}}20^3 \approx 5923$, see here, but the actual maximum number might be considerably smaller (closer to $5200$), since packing spheres near the sphere surface is harder then packing them in the interior. – Jack D'Aurizio Sep 30 '19 at 19:46
• OEIS A084828 suggests that exact numbers are only known up to $r=5$ and even that is experimental – Henry Sep 30 '19 at 19:56

A simple approach for producing reasonable lower bounds is to use a face-centered cubic packing or a hexagonal packing (both have the optimal density, $$\frac{\pi}{2\sqrt{3}}\approx 74\%$$, in the unconstrained space) and to count the number of spheres met by $$x^2+y^2+z^2=(20)^2$$. Recalling that the optimal packing density in the plane is $$\frac{\pi\sqrt{3}}{6}$$, in a sphere with radius $$20$$ it should be possible to pack around
$$\frac{\pi}{2\sqrt{3}}\cdot 20^3 - \frac{\pi\sqrt{3}}{6}\cdot 4(20)^2 \approx\color{red}{5804}$$ spheres, but not many more. The estimated density is so $$\approx 72.5\%$$.
There is also a packing arrangement known as Random Close Pack. RCP depends on the object shape - for spheres it is 0.64 meaning that the packing efficiency is 64% (as you can also see in Jack D'Aurizio's link). Therefore, if the balls are randomly distributed, then you can fit approximately $$0.64 \cdot \frac{\frac{4}{3}\pi (20r)^3}{\frac{4}{3}\pi r^3} \approx 5120$$, which is far from the highest estimation but still pretty good.