maximum number of spheres in a volume
- From: Hatto von Aquitanien <abbot@xxxxxxxxxxxxxx>
- Date: Sun, 27 May 2007 14:21:41 -0400
"You are given a large number of steel balls of equal diameter d and a
container of known volume V. Every dimension of the container is much
greater than the diameter of a ball. What is the greatest number of balls
that can be placed in the container?"
My argument is this: The most efficient arrangement will be a lattice of
tetrahedrons. The spheres in one layer will be arranged so that their
centers form equilateral triangles with their neighbors. That means the
distance between adjacent rows will be Sqrt[3] d/2. The spheres in the
adjacent layer will be arranged over the centers of these triangles.
Suppose these are the vertices of the first triangle in the first layer
assuming d=1:
v1 = {0, 0, 0};
v2 = {1, 0, 0};
v3 = {1/2, Sqrt[3]/2, 0};
To find the fourth vertex in the tetrahedron, we first find the center of
the triangle:
vc=(v1+v2+v3)/3={1/2, 1/(2 Sqrt[3]), 0}
Then we find the vertical displacement that will result in an edge of unit
length.
vc.vc + z z= 1
z z=1-1/3
z=Sqrt[2/3]
v4={1/2, 1/(2 Sqrt[3]), Sqrt[2/3] }
We really didn't need v4 to solve the problem, but it helps in understanding
it. What we now have is the fact that there are X/d spheres in a row,
there are Y/(d Sqrt[3]/2) rows in a layer, and there are Z/(d Sqrt[2/3])
layers in a volume XYZ.
n = Sqrt[2] XYZ/(d^3)
Does that look right?
--
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