+- Southperry.net (https://www.southperry.net)
+-- Forum: Social (https://www.southperry.net/forumdisplay.php?fid=14)
+--- Forum: Rubik's Cube (https://www.southperry.net/forumdisplay.php?fid=58)
+--- Thread: Spheres in Cubes (/showthread.php?tid=36853)
Spheres in Cubes - 2147483647 - 2011-01-26
![[Image: bodycenteredcubic2.jpg]](http://img809.imageshack.us/img809/6396/bodycenteredcubic2.jpg)
In the above figure, the spheres are touching each other. I'm trying to derive a formula that tells me how much space of the cube the spheres occupy when the radius of the center sphere is varied. Unfortunately, I'm having trouble doing this...
To begin, each side of the cube is length s. Therefore, the volume of the cube is s^3. The volume of any sphere is 4/3 pi r^3.
Now let's define the outer semispheres as having radius 1/2, and the inner sphere having a radius of k/2. Now to relate s and k, the cube has to be divided down a diagonal, which looks like the following:
![[Image: imgc103.gif]](http://img692.imageshack.us/img692/2986/imgc103.gif)
One of the sides of this rectangle is sqrt(2) s, and the other side is s. The diagonal of this rectangle is therefore sqrt(3) s. Since the diagonal of this rectangle is the length of twice the inner radius plus twice the radius of the outer radius:
sqrt(3) s = 2 (k/2+ 1/2) = k+1
s = (k+1) / sqrt(3)
From here, the total volume of the two spheres divided by the total volume of the cube should give the formula. (Each of the 8 outer semispheres adds up to one sphere, and the inner sphere is one sphere.)
[4/3 pi (1/2)^3 + 4/3 pi (k/2)^3] / [(k+1) / sqrt(3)]^3
pi/6 (1+k^3) / [(k+1) / sqrt(3)]^3
[sqrt(3) pi /2] (1+k^3) / (k+1)^3
Unfortunately, this does not give me the formula I was looking for. When k = 0, the function should spit out p/6. Instead, it spits out sqrt(3) pi/2. Halp.
Spheres in Cubes - XTOTHEL - 2011-01-26
I might be wrong, but the diagonal in this cube is equal to 2® of the cut up spheres plus 2® of the whole one in the middle.
that big diagonal (D) is simply (sqrt ((s^2+s^2) + s^2))
then you vary the radius of the center sphere ®
the radius of the outside "sphere" ®
you get D - 2R = 2r, 0.5(D-2R) = r
then just multiply around using your formula for volume.
Simply put, you want d: because along that line is where your spheres are touching.
![[Image: Image13.gif]](http://mathforum.org/workshops/lucent/Image13.gif)
Spheres in Cubes - Cyadd - 2011-01-26
[COLOR="Red"]The easiest answer is going straight to body centered cubic from material science.
Body Centered cubic.
It's approx. 0.68[/COLOR]
Spheres in Cubes - 2147483647 - 2011-01-26
XTOTHEL Wrote:Simply put, you want d: because along that line is where your spheres are touching.
That's what I have in the first post.
XTOTHEL Wrote:you get D - 2R = 2r, 0.5(D-2R) = r
About this. The first time I did this problem, I set the sides of the cube to 1. That makes the diagonal of the diagonal of the cube (what you're talking about) sqrt(3).
2R + 2r = sqrt(3)
(sqrt(3) is what you have for D in the formula).
When you do it this way, you need to make R=kr so that the function can be written in terms of k, the ratio of the two radii.
2kr+2r = sqrt(3)
r = sqrt(3) / (2k+2)
2R +2R/k = sqrt(3)
R = sqrt(3) / (2+2/k)
4/3 pi [sqrt(3) / (2k+2)]^3 + 4/3 pi [sqrt(3) / (2+2k)]^3 / 1^3
That simplifies exactly to what I have in the first post, so it's still wrong. >_>"
Cyadd Wrote:[COLOR="Red"]The easiest answer is going straight to body centered cubic from material science.
Body Centered cubic.
It's approx. 0.68[/COLOR]
That's actually where I got this problem from. I want a formula that can tell me the efficiency of the packing for any k (ratio of the radii). Unfortunately, Wikipedia doesn't provide the formula. It only provides the formula for when the outer and the inner formulas are equal.
It also gives approximate solutions for k = 0 and k = 1, but I already know those. I'm just looking for a formula. By logic, the formula should spit out the same numbers for k and 1/k. In mathematical terms, f(k) = f(1/k). The formula I derived does that, but it's not giving me the efficiency anywhere except at k=1. I'm not sure why.
Spheres in Cubes - XTOTHEL - 2011-01-26
Let k = R/r
volume of the spheres should be (8pi/3)(s* sqrt(3)/(2+2k))^3 + 4pi/3(k*s* sqrt(3)/(2+2k))^3
so if you were to sub in 1 for s: (8pi/3)(sqrt(3)/(2+2k))^3 + 4pi/3(k sqrt(3)/(2+2k))^3
Spheres in Cubes - modular - 2011-01-26
After pondering this for a while and coming up with the same f(k) for the volume ratios, I questioned why you would ever input k = 0. That situation wouldn't make physical sense, because here k must be limited to sqrt(3) - 1 < k < 1/[sqrt(3)-1]. If k is outside this range, you end up implying the spheres can intersect each other.
However, for k = 1, you do get the same result stated for the APF on Wikipedia...
Spheres in Cubes - Russt - 2011-01-26
Yes. f(0) would imply that in this diagram:
the pairs of diagonal spheres would be touching each other, which would imply that they intersect the adjacent spheres.
Otherwise, I think the formula is valid.
Spheres in Cubes - 2147483647 - 2011-01-26
modular Wrote:After pondering this for a while and coming up with the same f(k) for the volume ratios, I questioned why you would ever input k = 0. That situation wouldn't make physical sense, because here k must be limited to sqrt(3) - 1 < k < 1/[sqrt(3)-1]. If k is outside this range, you end up implying the spheres can intersect each other.
Hmm. I didn't think of it that way. When I first posed the question, I wanted k = 0 to look like this:
![[Image: 92267200.jpg]](http://img522.imageshack.us/img522/4614/92267200.jpg)
Thanks. Now I know.
Spheres in Cubes - modular - 2011-01-26
Ooh, while in the shower...
You can write a piecewise continuous function to describe behavior for all positive values:
0 <= k <= sqrt(3) - 1
g(k) =
sqrt(3) - 1 <= k <= 1/[sqrt(3)-1]
f(k)
1/[sqrt(3)-1] <= k < inf
This function has the property of f(1/k) = f(k) like you want and properly describes it all without any nonphysical nonsense.
(I finally understood why pi/6 is the desired answer for f(0)... Just like you would expect from that picture)
edit: 1 final session of pondering ...
1/g approaches 0 at infinity and we want to construct it such that we approach pi/6. The concept remains that it's an inverse function to g, but I need to be more careful about dealing with the constant.
edit2: @ below
Ah yes indeed, see I was thinking in terms of efficiency decreasing linearly with volume, and skipping over the fact that k^3 is proportional to volume.
Spheres in Cubes - 2147483647 - 2011-01-29
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.