Packing part of a sphere with balls?

Question

A dish that is part of a uniform hollow sphere has the following dimensions:
36" in diameter at the top/brim
12" height (or depth of dish)
How many 1.5" balls can we pack in the dish?
The balls cannot rise above the mouth (top). They must be levelled at the top,

Answer 1

The general problem of how many small spheres one can pack into a larger sphere has not been solved, and is probably unsolvable except by numerical experiments. You've made it an even harder problem by making the volume to be filled a bowl! One can, however, put some limits on the number.

First, we need to determine the volume of your bowl. The bowl's shape would be called a "spherical cap" in geometry, and the volume of a spherical cap is given by:

pi* h^2 * (3*r^2 + h^2)/6

where h is the height of the "bowl", and r is the radius of the top of the bowl. In your case, this works out to be 7012.035 in^3.

It will also be useful to know the radius, R, of the sphere that the bowl is a portion of. We know from simple geometry that:

(R-h)^2 + r^2 = R^2

2*R*h = h^2 + r^2

R = (h^2 + r^2)/(2h)

In your case, R = 19.5 in, and the volume of the whole sphere would be 31,059.356 in^3. The volume of the bowl is 22.6% of the whole sphere.


If it were possible (but it's not) to pack the 1.5" balls in cubic or hexagonal closest packing in the bowl, they would occupy a fraction of the volume equal to pi/(3*sqrt(2)) ~ 74.05%.

Each 1.5" diameter ball has a volume of (4*pi*0.75^3)/4 = 1.767 in^3. 74.05 % of 7012.035 in^3 is equal to 5192.275 in^3. This is the same as the volume occupied by 2938.227 of the smaller balls, so an absolute maximum on the number of balls that could be contained in the bowl is 2938. Because of the shape of the bowl, and the fact that you can't allow any balls to stick above the rim, the actual number of balls that could be put in the bowl will be less.

If the balls are randomly packed (as opposed to closest packed), they will occupy about 64% of the available volume. This works out to be 2539 of the 1.5" balls. This is probably a reasonable estimate for how many you can get in the bowl in practice.


An empirical estimate, based mostly on the results of numerical experiments, for the maximum number of small spheres with uniform radius that can be packed into a larger sphere is given (by the 4th source below) as:

N = (1/d^3)*(0.7405*q^3 + 0.484*(1 - q^3)), where

N is the maximum number of small balls
d = r/R (the ratio of the radii of the balls and the sphere into which they are packed)
q = 1-2*d

In your case, d = 0.75/19.5 = 0.038, and q = 0.923, and the above equation predicts that one could pack 12,052 1.5"-diameter balls into a hollow sphere with radium 19.5". Your bowl is only 22.6% of the total volume of such a sphere, so a reasonable estimate for the maximum number of balls you could pack in the bowl, taking into account the geometry of the container, is 22.6% of 12,052 = 2721 balls. You would have to be very lucky to actually get this many in the bowl, and in most cases, you are more likely to end up with a number closer to the estimate given by the random packing density (2539).

Answer 2

In arithmetic, sphere packing complications are complications appropriate to preparations of non-overlapping same spheres which fill an section. in a lot of circumstances the gap in touch is 3-dimensional Euclidean area. in spite of the undeniable fact that, sphere packing complications might want to be generalised to 2 dimensional area (the position the "spheres" are circles), to n-dimensional area (the position the "spheres" are hyperspheres) and to non-Euclidean areas which comprise hyperbolic area. a conventional sphere packing difficulty is to discover an affiliation in which the spheres fill as large a percentage of the gap as achieveable. the percentage of area crammed by employing the spheres is termed the density of the affiliation. because the density of an affiliation can variety searching on the quantity over which it really is measured, the priority is in a lot of circumstances to maximize the perfect or asymptotic density, measured over a large adequate volume. a customary affiliation (called a periodic or lattice affiliation) is one in which the centres of the spheres form a really symmetric progression referred to as a lattice. preparations in which the spheres at the on the spot are not prepared in a lattice are referred to as abnormal or aperiodic preparations. customary preparations are extra accessible to handle than abnormal ones — their intense degree of symmetry makes it extra accessible to categorise them and to degree their densities.

Answer 3

An Interesting problem.

I seem to remember that there is no general solution to this problem. (Allthough that was some time ago)
Whatever solution you arrive at, it is not possible to proove that another arrangement does not have at least one more ball !

Good luck !