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Proof for Close Packing of Congruent Identical Spheres
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[QUOTE="fizixfan, post: 5496740, member: 506800"] I developed two algorithms for calculating the density of close packed congruent identical spheres in two different arrangements: [LIST] [*]A tetrahedron with four equilateral triangular faces, and [/LIST] [LIST] [*]A square pyramid with a square base and four equilateral triangular faces, as shown below. [/LIST] [ATTACH=full]101967[/ATTACH] Figure 1. Tetrahedral ball stack. [ATTACH=full]101968[/ATTACH] Figure 2. Square pyramidal ball stack. Here are the Excel-friendly algorithms (n = number of spheres along bottom row): Density of Tetrahedral Stack (Dt): Dt = (4*(2^0.5)*n*(n+1)*(n+2)*Pi())/(3*(2*n+2*3^0.5-2)^3 or [ATTACH=full]101969[/ATTACH] (1) Density of Square Pyramidal Stack (Dp): Dp = ( n*(1+n)*(1+2*n)*Pi())/(3*((1+(2^0.5*n)))^3) or [ATTACH=full]101970[/ATTACH] (2) I found that, as the number of spheres approaches infinity for both arrangements, that Dt = Dp = ≈ 0.74048... or π/√18. I resorted to Wolfram Alpha with the following query: [ATTACH=full]101971[/ATTACH] = [ATTACH=full]101972[/ATTACH] (3) and got the following result: [ATTACH=full]101973[/ATTACH] (4) True! I then forwarded my calculations to Dr. Thomas C. Hales, who proved the Kepler Conjecture, asking him if Dt = Dp was correct, and he responded, saying, "The reason for the equal densities in a tetrahedron and square pyramid is that they can both be viewed as part of the face-centered-cubic packing, each with a different set of exposed facets." [B]My question to you is this: can you provide a detailed proof that (1) = (2) as n→∞, i.e., that [ATTACH=full]101974[/ATTACH] is true?[/B] Thanks for any input! Fizixfan. [/QUOTE]
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Proof for Close Packing of Congruent Identical Spheres
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