Estimate the number of Ping‐Pong balls that would fit into a typical‐size room (without being crushed).?

Question

Explain fully, including all assumed values

Answer 1

A regulation ping-pong has a diameter of 40 millimeters. The best packing efficiency is achieved with face-centered cubic packing:
efficiency = π / (2*sqrt(3))
efficiency = 0.740480

Let's assume our room is 3 meters tall, 4 meters wide, and 6 meters long. It has a volume of
V = l*w*h
V = (6 m)*(4 m)*(3 m)
V = 72 m^3

Our packing efficiency of 0.740480 means that the amount of space that can actually be occupied with ping-pong balls is
Veffective = 0.740480 * V
Veffective = 0.740480 * (72 m^3)
Veffective = 53.3146 m^3

This of course ignores the poorer packing efficiency at the edges of the room, but we will assume that this error is small, because only a small fraction of the balls will be adjacent to the wall. Now, the volume of a single ping-pong ball is
Vball = 4/3*π*R^3
Vball = 4/3*π*(20 mm)^3
Vball = 4/3*π*(0.02 m)^3
Vball = 0.0000335103 m^3

So, the number of balls that fit into the effective volume of the room (which takes packing efficiency into account) is

N = Veffective / Vball
N = (53.3146 m^3) / (0.0000335103 m^3)
N = 1.591 * 10^6

So, about 1,591,000 balls will fit into a 3x4x6 meter room if packed in the most efficient manner (face-centered cubic packing).

Answer 2

Assuming a room of a rectangular brick formation.
The quickest and somewhat accurate formulate would
simply include the diameters as if the balls were cubes:
area=4/3(pi)r^3 becomes area=2r*2r*2r. Measure
the distances of each corners edge on the x, y, and z
axis and divide by the length 2r then use the floor function
to round to whole balls/cubes( floor(1.3)=1, floor(2.7)=2, etc).

Although such an estimation would be somewhat accurate
consider that each ball can sit within the crevice of the
curved balls underneath it. By treating the balls as
cubes we neglected the space around the curved
edges. Imagine that for every 4 balls a ball can sit
on top of them inbetween the four and conserve more
space. However, it isn't that simple, consider that the
walls will prevent some of the balls around the edges
of the room from being seated on top. Furthermore,
consider that for all of the middle ball alignments
some of the balls which sit on top share bunkers
provided by the lower layer with the balls on the same
layer. Therefore, we don't get a constant 1 ball for
every 4 underneath because 2 balls on the same
layer might share 2 balls from the layer beneath.
If all of the balls were aligned then the formula
might look something like:
bottom layer: same as we calculated above(treat as cubes)
middle layers:
[(area)-(edge area) ] / 2r; with height being less than
2r because the balls are seated in crevices perhaps
(4/5)*2r would be a good approximation.
top layer: allow space which isn't large enough to fit balls