Volume of air in the box (packing problem)

Question

Suppose I have a box with dimensions $L \times W \times H$.

What is the volume of air in the box, if I pack balls with radii $r$?

With increase of radius, does volume of air decrease?

Answer 1

Assuming that all spheres are of the same size, the maximum volume fraction that can be occupied by spheres is $\frac{\pi}{3 \sqrt{2}} \approx 0.74048$ (See close-packing of spheres). It was conjectured by Kepler and later proven by Hales, and includes regular as well as irregular arrangements. This result is NOT(!) dependent on the radius of the spheres.

The question here is a bit more complicated, as it does include a specific container. Both shape as well as dimensions are important, and in general lower and higher volume fractions can be obtained. It can range from 0%, when the radius of the spheres is larger than the smallest dimension of the box, upto 100% when using a spherical container of the same size.

In the case of a rectangular box and considering the occupied volume fraction as function of the inverse radius (so that we go from large to small spheres), this will be a function that consists of decreasing segments and discontinues jumps to higher values that converges to the closed packed limit $\pi/3 \sqrt{2}$. This can easily be understood, since for very large radii no sphere would fit in the box and the occupied volume fraction is 0. On decreasing the radius, at some size one (several) sphere(s) will fit in the box causing a discontinuous jump to a higher value in the function. Continuing to decrease the radius, initially no additional spheres can be added, resulting in a diminishing volume fraction, until one or more additional spheres could be added and consequentially another jump to higher value, after which the behaviour repeats it self.

For small spheres of radius $r$ the central part of the box of size $(L-2r)\times(W-2r)\times(H-2r)$ is not affected by the boundary of the box and can be filled to the maximum packing fraction. Since this will be the leading term in occupied volume for the whole box with respect to the boundary layer, it follows that in the limit of small spheres the volume faction converges to that value as well. (This can easily be shown with a more proper analysis).

So for the empty space as function of the radius we would have a function that at $r \rightarrow 0$ converges to $1 - \pi/3\sqrt{2}$ and consists of decreasing segments and discontinues jumps to higher values, ending when no sphere can fit in the box whereupon the box remains completely empty.

Here I only considered occupied vs empty space, ignoring the physics of the question.

Answer 2

Air is basically trapped in voids formed between balls, while packing them closely.

So, if you want to decrease the volume of trapped air, you need to decrease the radii of your balls.
When radius approaches $0$, volume of trapped air will also approach $0$;
whereas, if radius is increased, size of voids will also increase, leading to more air getting trapped between the balls, i.e. more overall volume of air inside the box.

So, it seems that volume of trapped air is directly proportional to radius of the balls.

But, depending upon specific conditions, it may or may not be possible;
because, if number of balls tends to $\infty$, the number of voids also tends to $\infty$, however small they might be becoming.

So, air trapped inside those tiny voids cant be neglected, until they become smaller than the size of an air molecule.
It may not be wrong to assume, without loss of generality, that radius factor is more dominating over their number.