[QUOTE=magellan01]
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.
[/QUOTE]
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.