The full title of this post is "Close Packing of Spheres in Regular Tetrahedral vs. Square Pyramidal Container"

I'm not sure where this post belongs, but Greg Bernhardt suggested I just post it where I thought best, and he would find a place for it. So here goes:

In 1611, Johannes Kepler proposed that face-centered cubic packing achieves the greatest density. In 1998, Thomas Hales proved he was right.

I took a different approach on this and compared the packing density of equal-sized congruent spheres within a regular tetrahedral container (four similar equilateral triangular faces) vs. a square pyramidal container (four similar equilateral triangular faces with a square base), as shown below:

Attachment 73250 (Figure 1)

Attachment 73251 (Figure 2)

I found that, with a very large number of spheres (>100,000), the "packing efficiency" of spheres (volume of spheres/volume of container) within a regular tetrahedral container approached ≈74.048% ("Hales' density" or ∏/√18 ), whereas the packing efficiency of spheres within a square pyramidal container approached only ≈60.460%.

Oddly enough, the space-efficiency of a tetrahedral ball stack decreases in proportion to the number of balls it contains, whereas the space-efficiency of a pyramidal ball stack increases with the number of balls, as shown below:

Attachment 73252 (Figure 3)

Attachment 73253 (Figure 4)

Here are the formulas I used in my calculations (note that I let the radius of each sphere equal 1).

For a regular tetrahedral container:

Volume of each sphere = 4/3∏r³ (r = 1) = 4.1888

Number of spheres in a regular tetrahedron Tt = n(n+1)(n+2)/6 where n = number of spheres along base (n=5 in Figure 1), so the number of spheres in a tetrahedron with 5 spheres along the base would equal 35.

Volume of spheres in a regular tetrahedron = Vst = n(n+1)(n+2)/6*4/3*∏r^3 ≈ 146.608 (for n=5)

Length of side of regular tetrahedral container = b = ((2n)-2)+2*3^0.5 = 11.464 (for n=5)

Volume of regular tetrahedral container = Vt = b^3/6*2^0.5 = (((2n)-2)+2*√3)^3/(6*(2^0.5)) = 177.564 (for n=5)

"Packing efficiency" = Vst/Vt ≈ 82.566% (for n=5)

The packing efficiency of a regular tetrahedral container decreases with the number of spheres until it reaches a limit of ≈74.048% (Figure 3). Note that the packing efficiency actually

increases between 1 sphere to 4 spheres. This is unexpected, and could be an error in my calculations, or it could be accurate. I've checked it numerous times. Feel free to check it for yourself.

For a square pyramidal container:

Volume of each sphere = 4/3∏r³ (r = 1) ≈ 4.189

Number of spheres in a square pyramid = Tp = n(n+1)(2n+1)/6 where n = number of spheres along base (n=4 in Figure 2), so the number of spheres in a square pyramid with 4 spheres along the base would equal 30. With 5 spheres along the base, the total number of spheres would equal 55. (I couldn't find an illustration that showed 5 spheres along the base, so I had to use the one with 4 spheres along the base).

Volume of spheres in a square pyramid = Vsp = n(n+1)(2n+1)/6*4/3*∏r^3 ≈ 230.383 (for n=5)

Length of side of square pyramidal container = b = ((2n)-2)+2*3^0.5 ≈ 11.464 (for n=5)

Height of square pyramidal container = h = (((n*2-2)+(2*(3^0.5)))*3^0.5/2) ≈ 9.928 (for n=5)

Volume of square pyramidal container = Vp = b^2*h/3 = ((((2*n)-2)+2*3^0.5)^2*(2n)-2) + 2*(3^0.5)*2/3^0.5)/3 ≈ 434.94 (for n=5)

"Packing efficiency" = Vsp/Vp ≈ 52.969% (for n=5)

The packing efficiency of a square pyramidal container increases with the number of spheres until it reaches a limit of ≈60.460% (Figure 4).

There may be practical applications for this (if my calculations are right) for the packing of spherical objects (oranges, tennis balls, baseballs, cannonballs, etc.). If so, it would mean that the more space efficient container for packing equal-sized spheres would be a regular tetradhedral container rather than a square pyramidal container.

Your comments are welcome!

Fizixfan.