Okay letsrun brainiacs, I have a tough question and I want to see who can solve it:
I want to know how many ping pong balls you can fit within a much larger sphere.
The larger sphere has a circumference of ~96in.
A ping pong ball has a circumference of ~4.9634in.
You do the math and as always, show your work.
Anyone who comes up with the correct answer is up for a prize.
VERY HARD math problem, winner gets a special prize: TBD
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What I know:
This is a calculus 3 problem
You will have to compute a three dimensional integral
What I don't know:
What the integral is, otherwise I could solve it. 
~96in.
~4.9634in.
96/4.9634= 20.7164437 ping pong balls.
so 20 full ping pong balls.
youre welcome 
doubt it can be solved with simple math, it requires advanced algorithm, probably written in specific program (otherwise it would take ages to calculate all diameters and points of contact). Yes, I can write one, but I'm lazy bum.

^^ Your an idiot, thats just the circumference, you need the volume.

Close packing of spheres can ideally achieve 74% density. (pi/(3sqrt(2))
Big sphere is 14940 cubic inches
Each little sphere is 2.065 in
14940/2.065*(pi/3sqrt(2)) = ~5357 spheres
a decent estimate 
are we allowed to smash the ping pong balls? or the gianr sphere?

Ping pong balls deform easily. Are we supposed to assume that the balls will remain perfect spheres, or are we allowing for them to flatten to some extent as we force more balls into the larger container?

I'm monitoring the responses. With that said, if your question does not receive a response, assume it's a stupid question. Pleb, thanks for the taking a stab, I'm positive it's closer than most other guesses. Tomtom i wish you weren't so lazy.

No time to dig into the math right now, but some quick calculations make it around 3788 ping pong balls. If I think about it more the number with probably come down slightly, but I have to go run.

i worked off the above poster's assumption that sphere density would be about 74%
that said... i ean the volumes for the LARGE SPHERE and SMALL SPHERE using V=4(3.14)r^2 and got
L SPHERE = 2935.02 in^3
S SPHERE = 7.845 in^3
if we figure in that .74 factor on the large spehere, that leaves us with about 2171.9 od sphere volume consumed by small spheres.
Divide 2171.8 by small sphere volume of 7.845 and we get approximately 276.8 ping pong balls in the larger sphere
Let me know if this works out for you.
wedlock 
mcguirkthejerk wrote:
I'm monitoring the responses. With that said, if your question does not receive a response, assume it's a stupid question.
There's really no need to be a dick about it. I'm not sure why you believe that my question was stupid anyway. Whether you are assuming that the balls remain rigid or you are allowed to compress them will make a difference in the number you can fit. 
plebs wrote:
Close packing of spheres can ideally achieve 74% density. (pi/(3sqrt(2))
Ideally. Yet when you try to put 5 inch diameter spheres into one 9 inch diameter. Only one will fit, and you'll have one sphere of volume 65,45 into one of volume 381,7, density 17,15%. 
Mr. Physics, really?
No. you can not compress the ping pong balls. 
wiki says that in reality it would achieve about 63.4% packing (it even has a reference).
http://en.wikipedia.org/wiki/Sphere_packing
Thus it is probably closer to 0.634 * 14940.4 / 2.111 = 4487.
plebs wrote:
Close packing of spheres can ideally achieve 74% density. (pi/(3sqrt(2))
Big sphere is 14940 cubic inches
Each little sphere is 2.065 in
14940/2.065*(pi/3sqrt(2)) = ~5357 spheres
a decent estimate 
It was a stupid question.

tomtom wrote:
Ideally. Yet when you try to put 5 inch diameter spheres into one 9 inch diameter. Only one will fit, and you'll have one sphere of volume 65,45 into one of volume 381,7, density 17,15%.
Well, duh, but with such a large sphere and tiny smaller ones, it is likely very close to ideal. Notice I said "a good estimate" 
I think wedlock is close. However, it is likely less than 276.8 because:
1) I assume you can't have fractions of a pingpong ball and
2) If you superimpose a 96 inch circumference sphere on a pile of most densely stacked [face centered cubic packing] pingpong balls of circumference 4.9634 inches, some of the pingpong balls are going to be cut through by the surface of the sphere.
Example: The diameter of the 96 inch sphere is about 30.5577, and the diameter of the ping pong ball is about 1.58, so about 19.34 ping pong balls fit along the axis of the big sphere. Except you can't have a fraction of a ping pong ball, so you can only have 19 balls.
I don't know how to figure out how many fractions of ping pong balls get cut off. But since 19/19.34 = .9824, I'm guessing that since 276.8 x .9824 = 271 and change, the correct answer is closer to 271. 
Sorry, forgot to cube to get volume. My new answer is 5353.486 x .9824 = about 5259 balls. So plebs was much closer.

come on... wrote:
^^ Your an idiot, ...