Spheres within a sphere...

The stacking gumballs in a sphere....Well my first observation is that there is an associated void volume.

A place to start is at the middle. I'll start with a tetrahedron and four gumballs. A tetrahedron is a polyhedron with four vertices, four faces, and six edges.

I'll orient four gumballs with one at each vertex. Each tetrahedron edge is the length of a gumball diameter (two radiuses). From solid geometry, each portion of a sphere is (I'm dealing with memory) one twenty-seventh of that sphere, and four spheres represent 4/27 a total sphere. So the volume of a sphere inside the tetrahedron is 4/27 the volume of a gumball [Vgb=(4/3*pi*R^3)*4/27]=0.6256R^3

The volume of a regular tetrahedron with an edge equal to 2R is given by Vt=0.11785(2R)^3 (I looked this one up)=0.9428R^3

So the void volume for a tetrahedral stacking of gumballs is
Vf=(0.9428-0.6256)/0.9428=0.3364.

So at best 1 third of the gumball machine is filled with air, even when it's full of gumballs.


So the ratio of the maximum possible (presmushed gumballs) to a tetrahedral close packing is 2/3 (about 1-.3364) and the number of close packed gumballs is about 1896.

Or maybe knot.

Rap

I believe the edge effects will skew the tetrahedron. So, heres a bit of applied. When the machine is half full. pour in as much water as it takes to just top the gumballs. Remove the water and make note of its volume. Compute the inside volume of the gumball machine/2=Sig { ofall the gumballs + volume of water}.
its a porosity problem wherein all the grains of sand are quite the same.

1/2(M/g) is 1/2(9/.5)^3 = 2916 gumballs (the other terms cancel out).

A quick search for sphere packing indicates that the density for random packing is .64 which would give you 1866 gumballs. Random loose packing density is .55, or 1604 gumballs. Close packing density is .74, or 2158 gumballs. If you shake the machine, you should get better packing and could estimate the number more accurately.

More info on sphere packing

Farmerman, you wouldn't have to damage the gumballs with water, you could assume that the gumballs weren't gas permeable and introduce a metered amount of nitrogen into the gumball machine (after it was filled to the max) and measure the change in pressure. It would then be easy to use the ideal gas law to determine the void volume (at a constant temperature).

On the other hand, if the tetrahedrons were arranged to form a dodecahedron the edge effects would be minimized.

In reality, I remember Terry's given parameters from a long gone project concerning packed beds and Dowex ion exchange resins.

Rap

hee hee.

Alternate solution

The synergistic close packing of uniform spheres problem resulted in the discovery of a polyhedron that maximizes spheres around a central sphere.

If you place one sphere at the middle, a maximum of 12 spheres can be arranged around that central sphere. If you take a vector from the center of each of the spheres arranged around the middle sphere you'll get a dodecahedron. But this dodecahedron is not a platonic solid. Instead of getting a dodecahedron of 12 pentagons, you get one with 6 squares and 6 equilateral triangles that is based upon an arrangement of 12 internal tetrahedrons. A common name for this polyhedron is the quadray.

Sphere packing, using quadray can be used to kinda find a solution for your problem if you make a "assuming a square hole" conjecture.

Anywho's here it goes.

A quadray is roughly spherical, so make the assumption that a spherical gumball machine is close enough. Now the gumball is 9 inches in radius and a gumball is 1 inch in diameter (R=1/2 inch), so a radial vector has a length of 9 gumballs plus ½ inch for the central gumball and the radius is 9.5 inches. But this won't fit in a 9 inch machine so the quadray consists of 8 layers of close packed gumballs around a central gumball.

Counting the gumballs

The zeroth layer consists of 1 gumball; the first layer consists of twelve gumballs, the second by 42, and the third by 92 and so on. Further study indicated a pattern (one that is supported by geometry and counting) that the number of gumballs in a layer (except for the zeroth) can be given by 10F^2+2 where F is layer. The total number of balls in a quadray of F layers can be determined by summing up the layers. A little algebra, gives this formula.

Tb=10n(n+1)(2n+1)/6+2n+1
Which rearranges to
Tb=10/3n^3+5n^2+11/3n+1

So if there are 8 layers of gumballs in the quadray that makes the spherical gumball machine, there are 642 gumballs in the outermost layer (F=8) and a total of 2057 gumballs.

OK now you say, that's real nice, but a hole may not be square and a sphere is not a polyhedron called a quadray. As a matter of fact if a sphere is going to be a polyhedron it would be a platonic one.

My answer---Oh so now you want a measurement! Wait a minute. I'll go get the technician.

This is what the tech does
Vs=4/3pi9^3=3053.63 in^3
Vgb=4/3pi(1/2)^3=0.5236 in^3
Vv=(3053.65-0.5236n) in^3 where n is the number of gumballs

Using a gas (nitrogen is good) and assuming the gumballs are gas impermeable introduce a known amount of nitrogen (m grams) into the gumball machine and measure the pressure change.
Assuming the ideal gas law then
Vv= (m/DelP)RT
The number of gumballs (n) is measured to be
n=[3056.63-(m/DelP)RT]/0.5236

Rap

Calling all history buffs.

This gumball problem is a variation of the so-called cannonball problem.

It seems that Johann Kepler developed this conjecture in 1606 in correspondence to a Thomas Heriot who was an assistant of Sir Francis Drake. Drake wanted to know how to stack cannonballs on the deck of a pitching ship. Kepler's Conjecture is that the cuboctahedron provided the smallest space and the cannonballs would occupy approximately 75% of the total volume.

I have also seen pictures of artillery ordinance stacked during the civil war. The octahedron arrangement can be clearly seen as the most common way to stack cannonballs (answering that old time question).

Quadray is a Fullerism. When Fuller was investigating dymaxion geometries he rediscovered this structure, but in his own buckyway he gave its own name.

For chemists and crystallographers, this is old news--its face centered.

Rap