2011-01-29, 12:05 AM
I thought about this too. The way I wrote this problem,
k = R/r, where R is the inner sphere and r is the outer sphere.
So what happens if k < sqrt(3) -1? The outer radii can't expand beyond a radius of 1/2, so the inner sphere is slowly shrinking out of existence as k approaches 0. The volume of a sphere is 4/3 pi r^3, so that means at k = 0, the sphere is 4/3 pi (1/2)^3. Since k is the ratio between the two radii, we can redefine the volume of the inner sphere by 4/3 pi (k/2)^3. Thus, the f(k) on 0 < k < sqrt(3) -1 is:
f(k) = 4/3 pi (1/2)^3 + 4/3 pi (k/2)^3
f(k) = pi/6 (1+k^3)
Now we have to consider the opposite: what happens if k < 1 / (sqrt(3) -1)? As it turns out, the radius of the interior sphere can't expand past the walls of the box. Therefore, the radius of the inner sphere is a constant 1/2. However, our relation for the two radii stay the same:
2 (R+r) = sqrt(3)
Since k = R/r, 2 (R+ R/k) = sqrt(3)
We already know that R = 1/2, so:
2 (1/2 + 1/(2k)) = sqrt(3)
In other terms, the outer spheres are shrinking out of existence at a rate of 1/(2k). Plugging into the formula for the volume for a sphere:
f(k) = 4/3 pi (1/2)^3 + 4/3 pi (1/(2k))^3
f(k) = pi/6 (1+1/k^3)
The new function approaches pi/6 as k approaches infinity. The nice thing about this problem is that the unit cube is entirely symmetrical. By choosing a unit cube that's oriented in a different way, or "inverting the cube", or switching the positions of the inner spheres with the outer spheres, the efficiency isn't changed.
k = R/r, where R is the inner sphere and r is the outer sphere.
So what happens if k < sqrt(3) -1? The outer radii can't expand beyond a radius of 1/2, so the inner sphere is slowly shrinking out of existence as k approaches 0. The volume of a sphere is 4/3 pi r^3, so that means at k = 0, the sphere is 4/3 pi (1/2)^3. Since k is the ratio between the two radii, we can redefine the volume of the inner sphere by 4/3 pi (k/2)^3. Thus, the f(k) on 0 < k < sqrt(3) -1 is:
f(k) = 4/3 pi (1/2)^3 + 4/3 pi (k/2)^3
f(k) = pi/6 (1+k^3)
Now we have to consider the opposite: what happens if k < 1 / (sqrt(3) -1)? As it turns out, the radius of the interior sphere can't expand past the walls of the box. Therefore, the radius of the inner sphere is a constant 1/2. However, our relation for the two radii stay the same:
2 (R+r) = sqrt(3)
Since k = R/r, 2 (R+ R/k) = sqrt(3)
We already know that R = 1/2, so:
2 (1/2 + 1/(2k)) = sqrt(3)
In other terms, the outer spheres are shrinking out of existence at a rate of 1/(2k). Plugging into the formula for the volume for a sphere:
f(k) = 4/3 pi (1/2)^3 + 4/3 pi (1/(2k))^3
f(k) = pi/6 (1+1/k^3)
The new function approaches pi/6 as k approaches infinity. The nice thing about this problem is that the unit cube is entirely symmetrical. By choosing a unit cube that's oriented in a different way, or "inverting the cube", or switching the positions of the inner spheres with the outer spheres, the efficiency isn't changed.