How many spheres of radius $R/2$ can be packed into a can having depth $R$, height $H$, with a hemispherical top

Question

The hemispherical "cap" is on the top of the cylindrical can. So, in the side view, the total height of the container will be $R+H$.

Answer 1

Let $N$ be the number of spheres one can pack into the can.

As a function of $H$, it is not hard to see:

• when $0 \le H < \frac{R}{2}$, the can can hold only 1 sphere $\implies N = 1$.

• when we increase $H$ to $\frac{R}{2}$, we can start to put 2 spheres at the bottom of can $\implies N = 2$. For this $H$, there are some free spaces between these two spheres and the hemisphere cap. However, it is not large enough to hold another sphere.

• if we further increase $H$ by an amount $\frac{\sqrt{3}-1}{2}R$, i.e. when $H$ becomes $\frac{\sqrt{3}}{2}R$, the free space becomes large enough to hold another sphere $\implies N = 2 + 1 = 3$.

• when we further increase $H$ to $(\frac12 + \frac{1}{\sqrt{2}})R$, there are enough space in the base cylinder part of the can to put another layer of spheres there. The center of spheres in these two layers form a regular tetrahedron of side $R$. The free space above the spheres get reduced, no longer able to hold the extra sphere. As a result, $N = 3 + 2 - 1 = 4$.

If we keep on increasing $H$, these pattern will repeat at

$$H = \left(\frac{\sqrt{3}}{2} + \frac{n-1}{\sqrt{2}}\right)R \quad\text{ and }\quad \left( \frac12 + \frac{n}{\sqrt{2}} \right)R$$

for every positive integer n. So the general formula for $N$ when $H \ge \frac{R}{2}$ is:

$$N = 1 + \left\lfloor 1 + \sqrt{2}( \frac{H}{R} - \frac12 )\right\rfloor + \left\lfloor 1 + \sqrt{2}( \frac{H}{R} - \frac{\sqrt{3}}{2} )\right\rfloor$$

Below is an illustration of the arrangement of 7 spheres when $\displaystyle H = (\frac{\sqrt{3}}{2} + \frac{2}{\sqrt{2}})R \sim 2.28 R$.