Question for math brains

Factual Questions
1

Take a hollow sphere, a basketball, for instance. Next, take smaller spheres: either softballs, ping pong balls and bbs, etc. The question is how much of the volume of the sphere is taken up by these other spheres if the basketball is filled to maximum capacity? Scenario A would fit as many softballs into the basketball, Scanario B would fit as many ping pong balls, etc.

Intuition seems to say that the smaller the spheres, the more of the volume that would be taken up. On the other hand, the sphere that would take up the volume most efficeiintly would be a single sphere that is just slightly smaller than a basketball.

I guess the questions are: is my intuition misleading me? Meaning that although it appears that the bbs take up more of the space, that they are taking up no more space than the few softballs that would fit inside the basketball

Also, what would an equation look like for this? And if plotted on a graph (amount of space taken up on one axis and diameter of spheres used to fill the basketball on the other, is the line straight ? Curved? Does it form a line then break when we use a sphere just slightly stronger than a basketball?

Feel free to offer an answer in a way other than what I described. Obviously, math is not my first language, so take it easy. Thanks.

2

Your question is wrapped around packing theory, which I took a very casual look at when I was working for an educational multimedia company… quite a long time ago. Packing theory is by no means simple, and not for dabblers like myself, but that said… here’s some thinkage you might at least find helpful.

  1. Forget spheres for the time being and simplify the problem. Easier to draw, visualize, and do computations with circles.
  2. Start with a circle A, with a radius 1. With our familar Area = pi(r^2), the area of the circle is pi.
  3. Let’s say that we throw another circle in there - circle B - that has half the radius of circle A. The area of this circle is going to be pi(1/2)(1/2), or pi/4. Circle B fills 1/4 of circle A.
  4. You actually have room for two circle B’s in there - and no more - so with two circles of half the radius of circle A, you can fill only half of circle A.
  5. You can get the same circle filled more efficiently with a larger circle. A bit of algebra tells us that a single circle with radius 1/sqrt 2 will do just as well as two circles with radius 1/2.

What this tells us:
With a circle of radius between 1 and 1/sqrt 2, you fill circle A less and less efficiently until you hit the radius of exactly 1/2, when your efficiency jumps back up to filling half the circle. If you continue to shrink the smaller circle, you will lose efficiency again until you can squeeze three circles inside - and, for the first time since 1/sqrt 2, an efficiency better than 1/2.

My hunch: Efficiency fluctuates. With many small circles, there are “optimum” sizes that are closer to 1. As you shrink the circles from these “optimum” sizes, you lose some efficiency (although less the smaller the circles are) until you reach another optimum size that brings you closer to 1 than you’ve seen yet. Calculation of optimum sizes becomes more complicated with larger number of circles.

Hope this makes sense without drawings… I’ve made a few sketches over here while I was writing this, wish I could share.

3

This is actually a really interesting question that I don’t have a good answer to. But your intuition is correct. This is what’s called a “packing problem”, where the goal is to utilise space as efficiently as possible. Unfortunately, packing problems are notoriously hard to solve perfectly. Here’s an example that I hope I’m remebering correctly (sorry, no cite):

You have a square that is 1,000.1 inches x 1,000.1 inches. How many 1 inch x 1 inch tiles can fit inside the square? The intuitive answer is 1,000 * 1,000 = 1,000,000. However, someone (Paul Erdos, maybe?) proved that if you twisted the tiles slightly, you could actually fit more than 1 million tiles in the square. But it remains an open question whether or not a more efficient packing is possible. And this is for a simple, 2-d question. It gets harder in three dimensions (or more!) like your question.

Back to the question at hand: Let’s define E(r) to be our efficiency function (the ratio of the volume of inside spheres o radius “r” to the volume of a sphere of radius 1). The domain is 0 < r <= 1. (Since you can’t have a sphere of radius 0, but radius 1 is possble). We can rewrite E(r) as (S(r)4/3pir^3)/(4/3pi*1^3) where S(r) is the number of spheres of radius r that can fit into a sphere of radius 1, which is always a whole number. Simplified, we get E(r) = S(r)*r^3. So to create our graph, we need to find S(r). That’s the real sticky wicket here. Here’s what I’ve gotten:

for 1 < r, S(r) = 0
for 1/2 < S(r) <= 1, S(r) = 1
for .464… < S(r) <= 1/2, S(r) = 2* .464… is 3/(2sqrt(3) + 3) [I’m glad to post how I got this, but it’s from three equal circles packed into a larger circle.]
for (something) < r <= .464…, S(r) = 3

As r gets closer to zero, I have no idea as to how to proceed, but you pretty quickly need to start looking for approximate solutions. But your graph of E(r) will look like a cubic function that has breaks in it where S(r) jumps from one integer to the next.

4

And Engywook’s suggestion to reduce the question to two dimensions is a good one. Everything in my above post would still apply, except now E® = S®*r^2. And it is much easier to visualize

5

If your smaller spheres are significantly smaller than the big one, then it doesn’t much matter what exactly the size is. It’ll fluctuate, of course, but the fluctuations will be very, very small. So spheres the size marbles and spheres the size of atoms will both fill the basketball with 74.05…% efficiency. Maybe you can shift things around such that you can squeeze in one more marble, but one more marble out of thousands is insignificant. For spheres comparable to the size of the container, though, the edges of the container become significant, and one filling sphere more or less becomes very significant. It can make a huge difference to go from one packing scheme to a completely different one, if that new packing scheme can let you get one more ball in. So for this case, the calculation is (or can be) horrendously ugly, and there are no straightforward formulae.

6

Interesting question. What is the shape of this function? I’ll have a partial stab at it.

Let the horizontal axis, r, represent the radius of the packing-spheres as a fraction of the sphere to be packed. So r takes on positive real values greater than 0 and less than or equal to 1. Let the vertical axis, e, be the efficiency of the best possible packing for a given r, as a fraction from 0 to 1.

Near the left edge (r tiny), the function will be constrained to hover very close to 74.05%. As Chronos says, the efficiency isn’t going to change much when the spheres are small.

On the high end, any r greater than 0.5 will give an e that’s simply the ratio of the two spherical volumes. (You can only fit one such sphere into the larger one.) So for r > 0.5 and <= 1, e(r) will go as r[sup]3[/sup].

Below that, there’s an interval where exactly two spheres is all you can put into the larger sphere. For that interval, ending at r = 0.5 (but starting I know not exactly where), e(r) will go as 2r[sup]3[/sup].

Below that, there’s a smaller interval where exactly 3 spheres is the most you can put in the larger sphere. Or, does it leap up to 4? As you shrink the packing-sphere radius, does the best packing suddenly become a “tetrahedron”, or is it an “equalateral triangle” for a while? In general, does every integer get touched on the way down, or are there jumps?

In any case, e(r) is going to have the form k(r) x r[sup]3[/sup], where k is a discrete integer function of r: the maximum number of spheres you can pack for the given radius. When r is small, the intervals will be tiny, and k will go approximately as 1/r[sup]3[/sup]. So, it’s getting astronomically larger, but the intervals are smaller and smaller, and the function e(r) itself still hovers around 0.74. As r gets larger, the intervals get larger, and the range covered by e(r) within those intervals also gets larger.

In short, this is not a simple function with a pre-defined name. At least not so far as I’m aware.

7

This is what I was trying to communicate towards the bottom of post #3. And there is an interval where three will fit, but four will not. (The three are coplanar; picture three circles shoved into a bigger circle. Now imagine all of the circles are spheres. There’s no room for another sphere inside.)

But where are you and Chronos arriving at 74.05…% ? Is that the percentage of the volume of a tetrahedron taken up by spheres centered at each vertex with a radius equal to half the length of a side?

8

Right, I see that now. (I had only skimmed your post before. Sorry.)

That’s the geometry I had in mind, but it wasn’t immediately obvious to me (as I was writing earlier) that spheres of the same radius couldn’t also be arranged into a tetrahedron, if they were just small enough to make a triangle. But I agree with you now.

The exact value is pi / (3 x sqrt(2)) = 0.74045865… It’s the portion of infinite space that an infinite number of spheres would occupy, if they were packed in the “obvious” arrangment. Here’s a Mathworld page on the topic.

9

Thanks for the link. I always forget to check MathWorld.

10

Well, the bad news is that my question was not as simple to answer as I thought. The good news is that I don’t feel as dumb as I thought I might. Bad news again, clarity (for me) is still a ways off.

Thanks to all for you answers. I’ve done a lot of searching and that 74% number seems to be magic, like a universal constant. One math brain that I emailed was kind enough to reply with this below. I’m sure some of you will get more out of it than I. I’m still seeking an equation that would allow me to calculate how many spheres of radius “r” would fit into a large sphere with radius “R”. Maybe it’s here and I’m just not seeing it.

Here’s the link: http://groups.google.com/group/sci.math/msg/05dda67e4fb7c0e2. The first link in the quote box above looks like interesting fun, but it doesn’t seem to work on my computer (Mac).

Thanks again.