So this is on the behalf of another friend for a school assignment. The assignment is: Find how many soccer balls fit inside a cylindrical building. Obviously, since a soccer ball is a sphere, the resulting question is the title.
Initially my friend was working with a crate method, namely turning the spheres into cubes whose length was equal the sphere's diameter, and finding out how many cubes fit (with some estimates, this is easy enough). However, having my own doubts about the correctness of the method; I did some research on my own and came across close packing. From what I've gathered from previous Stack Exchange questions etc. the maximum density the sphere's can occupy in the close packing method is around 74%. However, I have yet to really understand how they derive this number. My own knowledge is only equivalent to an advanced class high schooler, so that's probably why this goes over my head. Would anyone mind clarifying why the maximum density is 74%?
Edit: I had some great responses, thank you very much for them! A few problems though:
- They just gave the formula. Finding the formula itself is easy enough to do, the real problem is finding out where the numbers come from ($N \, \le \, \frac{\pi}{3 \sqrt{2}} \frac{V_c}{V_s} \quad \approx \quad 0.74048 \, \frac{V_c}{V_s}$ is the formula that has been found). In particular, the $\frac{\pi}{3 \sqrt{2}}$.
- The formulas dealt with rectangular prisms / other containers, not necessarily cylinders.
I did find the following previous Stackexchange articles but did not understand them:
Problem of packing spheres of radius $\rho$ into a cylinder
Thanks again to everyone who has help me so far 🙂
Best Answer
The maximum density for packing equal spheres is about 74%, not 64%. This happens when the spheres form either a face-centered cubic lattice (FCC) or a hexagonal close packed lattice (HCP). Each sphere is then in contact with 12 other spheres.
This means that if the volume of one sphere is $V_s$, and you have a container with volume $V_c$, you can fit $$N \, \le \, \frac{\pi}{3 \sqrt{2}} \frac{V_c}{V_s} \quad \approx \quad 0.74048 \, \frac{V_c}{V_s}$$spheres in the container.
The rest of the volume is "wasted" in the small voids between the spheres, and in the voids between the spheres and the container walls.
The larger the container is compared to the size of the spheres, the closer you can get to the limit given above. When the container is not immense compared to the spheres, more volume (relatively speaking) is wasted between the spheres and the container walls, so the real $N$ you can achieve is less than what the above predicts. Down to zero, if the smallest size of a simple container is less than the diameter of the sphere; for example, even if you had cubic miles of space in a tube the diameter of five inches, you can't fit a single six-inch sphere in it.
Although each void is relatively small compared to the size of a sphere, there are lots of them – remember, each sphere touches 12 other spheres! –, totaling about 26% of the container volume.