The closest possible packing for balls is shown in fig. 1. The balls form a regular tetrahedron.
fig 1.: closest packing for balls
The number m of balls in layer n is described by
m=binomial (n, n-2) for n>=2
m=1 for n=1
The total number M of balls up to layer n is
M=binomial (n, n-3) for n>=3
M=1 for n=1
M=4 for n=2
The volume of a single ball with the radius r is:
Vball=4/3*pi*r³
The volume of a regular tetrahedron with edge length a is given as (fig. 2):
Vtetra=a³*sqrt(2)/12
fig 2.: regular tetrahedron with edge length a
The edge length a for a pile with n layers can be substituted by r(2n-2)= 2r(n-1).
The percentage of the accumulated ball volume in the tetrahedron volume is:
M*Vball
p=------- (* see not below)
Vtetra
binomial(n,n-3)*4/3*pi*r³
p=-------------------------
8r³*(n-1)³*sqrt(2)/12
n*(n-1)*(n-2)
------------- *4/3*pi*r³
6
p=--------------------------------
8r³*(n-1)*(n-1)*(n-1)*sqrt(2)/12
After removing some terms:
n*(n-2)*pi
p=---------------------
(n-1)*(n-1)*3*sqrt(2)
If n goes to infinity we have to apply l'Hospital's rule twice which leads to:
pi
p=---------
3*sqrt(2)
------------------------------
(*) the accumulated ball volume in this formula is not exact, because the volume of the balls at the
edge of the tetrahedron are not complety part of the volume of the tetrahedron.
If n goes to infinity this error goes to 0, so that it is not necessary to calculate the exact volume
of the balls in the tetrahedron.
