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A Proof for Close Packing of Congruent Identical Spheres

  1. Jun 12, 2016 #1
    I developed two algorithms for calculating the density of close packed congruent identical spheres in two different arrangements:
    • A tetrahedron with four equilateral triangular faces, and
    • A square pyramid with a square base and four equilateral triangular faces, as shown below.

    tetrahedral ball stack w- n=3.jpg
    Figure 1. Tetrahedral ball stack.
    pyramidal ball stack.jpg
    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

    Dt equation.jpg (1)

    Density of Square Pyramidal Stack (Dp):

    Dp = ( n*(1+n)*(1+2*n)*Pi())/(3*((1+(2^0.5*n)))^3) or

    Dp equation.jpg (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:

    Dt limit.jpg = Dp limit.jpg (3)

    and got the following result:

    Wolfram Proof RTC = SPC.jpg (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."

    My question to you is this: can you provide a detailed proof that (1) = (2) as n→∞, i.e., that Dt equals Dp.jpg is true?

    Thanks for any input!

    Fizixfan.
     
  2. jcsd
  3. Jun 12, 2016 #2

    fresh_42

    User Avatar
    2017 Award

    Staff: Mentor

    You get the equality of the limits by calculating the coefficients of the highest power ##3## of ##n##. It is sufficient to see that
    $$\frac{2 \pi n^3}{3 \cdot \sqrt{2}^3 n^3} = \frac{4 \pi \sqrt{2} n^3}{3 \cdot 2^3 \cdot n^3}$$
    because all other terms don't grow as fast (and division of the complete fractions by ##n^3## (nominator and denominator) gives ##\frac{1}{n}## terms which become zero at infinity).
     
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