# Close Packing of Spheres in Regular Tetrahedral vs. Square Pyramidal C

#### fizixfan

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: (Figure 1) (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: (Figure 3) (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.

Fizixfan.

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#### Khashishi

Clearly, if the number of balls doesn't exactly fill a level, it will be less efficient to package.

#### fizixfan

Clearly, if the number of balls doesn't exactly fill a level, it will be less efficient to package.
It depends on what you mean by "exactly." All the balls in both stacks are touching all the adjacent balls. It's the size of the interstitial spaces that makes the difference. #### Andy Resnick

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:
<snip>

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.
I'm not exactly sure what your question is, but the packing of hard-sphere fluids is a long-studied problem. From experiments and simulations, we know:

1) the density (volume fraction) of random close-packed spheres in three dimensions is 0.638, compared with 0.7405 for FCC (face-centered cubic) or HCP (hexagonal close packed) arrangements.
2) a hard-sphere phase diagram is athermal, since the sphere-sphere interaction is either zero or infinite.
3) there is a solid-liquid phase transition as a function of volume fraction. On the phase boundary, a liquid phase with packing fraction 0.495 is at equilibrium with a FCC solid with a volume fraction of 0.545.

Further, by examining the lattice structures of close-packed arrangements, we find the primitive unit cell of FCC is not a cube but a (IIRC) cuboctohedron.

If we denote the possible 2-D triangular close-packed lattices A, B, and C (based on where the coordinate origin is), FCC is obtained by stacking ABCABCABC... while HCP is ABABABABAB..

#### fizixfan

I'm not exactly sure what your question is, but the packing of hard-sphere fluids is a long-studied problem. From experiments and simulations, we know:

1) the density (volume fraction) of random close-packed spheres in three dimensions is 0.638, compared with 0.7405 for FCC (face-centered cubic) or HCP (hexagonal close packed) arrangements.
2) a hard-sphere phase diagram is athermal, since the sphere-sphere interaction is either zero or infinite.
3) there is a solid-liquid phase transition as a function of volume fraction. On the phase boundary, a liquid phase with packing fraction 0.495 is at equilibrium with a FCC solid with a volume fraction of 0.545.

Further, by examining the lattice structures of close-packed arrangements, we find the primitive unit cell of FCC is not a cube but a (IIRC) cuboctohedron.

If we denote the possible 2-D triangular close-packed lattices A, B, and C (based on where the coordinate origin is), FCC is obtained by stacking ABCABCABC... while HCP is ABABABABAB..
I was trying to eliminate the square pyramidal arrangement as an efficient way of packing spheres. In one of the other forums, someone had said it was the same as a regular tetrahedral arrangement. I had to demonstrate, to myself at least, that it was not.   You can see that interstitial spaces in the "plan" of the square pyramid base (1st figure) are larger than those in the tetrahedron base (2nd figure). I also stacked some golf balls to hopefully give a real-world of demonstration of what I'm trying to do here (3rd and 4th figures). When a very large number of balls is used, the density of the pyramidal arrangement is ~0.6046, whereas the density of the tetrahedral arrangement is ~0.7048. I've shown this in the calculations listed in my initial post.

Perhaps Thomas Hales used the square pyramidal arrangement in his proof of the Kepler Conjecture (i.e., that this was NOT the most efficient, or densest, way of stacking equal-sized congruent spheres).

It may not be rocket science, but if you can't have fun solving geometric puzzles, what's the point?

#### fizixfan

...whereas the density of the tetrahedral arrangement is ~0.7048. I've shown this in the calculations listed in my initial post.
~0.7048 is incorrect. I meant ~0.74048.

#### M Quack

These are indeed the same structures, just oriented (and therefore terminated) differently.

Try to remove the 5 balls along the edge of the tetrahedral pyramid. You will see squares of 4 ball touching each other in a square arrangement, just as in the base of your square base pyramid.

The stacking is known as face-centered cubic (FCC) and occurs in many common metals such as copper.

#### fizixfan

These are indeed the same structures, just oriented (and therefore terminated) differently.

Try to remove the 5 balls along the edge of the tetrahedral pyramid. You will see squares of 4 ball touching each other in a square arrangement, just as in the base of your square base pyramid.

The stacking is known as face-centered cubic (FCC) and occurs in many common metals such as copper.
I tried removing the 5 balls along the edge of the tetrahedral pyramid, but I still did not see 4 balls touching each other in a square arrangement. In the illustration above, the balls in the square arrangement are show in the first figure. Notice that the ball in the center is touching eight other balls. In the second figure, the base layer of the tetrahedral arrangement, the center ball is touching only six other balls, making it denser. Not to complicate things too much, but the second figure is also a cross-sectional view (looking horizontally at the stack) of both the square pyramid and the tetrahedron. But it's the way the balls are arranged in each horizontal layer of the stack that makes the difference in the density.

#### fizixfan 