VERY HARD math problem, winner gets a special prize: TBD

My guess: 1348 ping pong balls.

My method is below: (my calculations are quick, so there are likely errors)

1. Diameter Sphere=96/pi=30.577

Diameter PP Ball=4.9634/pi=1.579

2. Earlier poster stating the ideal packing fraction of spheres is correct. You can see an excellent explanation in a college level Materials Science textbook. Look up the face centered cubic crystal structure. I calculated the largest cube that would fit in the sphere (may have been wrong doing this) as a cube with side length of 17.64. Using the 0.74048 packing fraction for this cube, I found 586 ping pong balls would fit in the cube.

3. You are left with 6 end caps that are portions of the sphere. I calculated layering the ping pong balls at ideal area packing fraction (I calculated 0.7854 ideal area packing fraction using the ideal fit of spheres). These layers will mesh perfectly with the cube, as the ideal volume packing of the cube is single layers of ideally packed spheres. Fit the single layers together and you have the ideal volume packing.

I found you can fit 5 more layers until you exceed the radius of the sphere and another layer of one ball will not fit. 5 layers instead of 4, because the layers of spheres fit together, so each layer adds the diameter of the ping pong ball*cos45 rather than the diameter of the ping pong ball.

I calculated the circular area of each layer and the number of balls that would fit in it based on the ideal packing. All five layers gave me an additional 127 balls in each cap. Multiply this by 6 for each cap and you have 762 total ping pong balls.

Cube + six end caps = 762+586 = 1348 total ping pong balls.

This is the best answer so far. Always break problems like this up.

guesser wrote:

2. Earlier poster stating the ideal packing fraction of spheres is correct. You can see an excellent explanation in a college level Materials Science textbook. Look up the face centered cubic crystal structure. I calculated the largest cube that would fit in the sphere (may have been wrong doing this) as a cube with side length of 17.64. Using the 0.74048 packing fraction for this cube, I found 586 ping pong balls would fit in the cube.

You did something wrong here. I will give you that the cube inside the sphere is 17.64. I get 1.59 diameter for the ping pong ball, so I'll assume rounding differences. With those two numbers, I should be able to put an 11 x 11 grid along the bottom of the cube for 121 balls. Ignoring efficient packing, I could do the same thing 11 times to fill up the cube for a total of 1331 balls.

mcgato wrote:

You did something wrong here. I will give you that the cube inside the sphere is 17.64. I get 1.59 diameter for the ping pong ball, so I'll assume rounding differences. With those two numbers, I should be able to put an 11 x 11 grid along the bottom of the cube for 121 balls. Ignoring efficient packing, I could do the same thing 11 times to fill up the cube for a total of 1331 balls.

There are likely arithmetic errors, I redid sections of the calculation due to mixing up the diameter and radius of the sphere early on.

That said, the efficient most packing of sphere isn't stacking layer upon layer directly inline on top of each other. That is simple cubic packing in Materials Science lingo and quite a bit less efficient than face centered cubic packing. I.e., the bottom layer may be 11x11, but the next one up will not fit 11 x 11 as it is offset from the first so the spheres on the second layer partially slide down into the gaps between spheres on the first layer. Some of the spheres at the edge of the second layer won't fit. Sorry if that is confusing, it's hard to describe without an image. The third layer will be 11x11, although it may or may not sit directly above the first layer.

Good reference here:

guesser wrote:

mcgato wrote:

I should be able to put an 11 x 11 grid along the bottom of the cube for 121 balls. Ignoring efficient packing, I could do the same thing 11 times to fill up the cube for a total of 1331 balls.

I.e., the bottom layer may be 11x11, but the next one up will not fit 11 x 11 as it is offset from the first so the spheres on the second layer partially slide down into the gaps between spheres on the first layer. Some of the spheres at the edge of the second layer won't fit. Sorry if that is confusing, it's hard to describe without an image. The third layer will be 11x11, although it may or may not sit directly above the first layer.

Clarification, the second layer might fit 11x11, it is unknown until we calculate the extra space at the edge of the 11x11 cube and account for the offset of the second layer to the first.

Also, if you have 11x11 on the bottom layer, you can stack more than 11 layers vertically due because the layer to layer spacing is less than the diameter of the packing spheres.

How do we interpret "~96in." and "~4.9634in."?

I'm having trouble with the "~", and why you can't specify the circumferences with more precision. After all 100in. is ~96in. and 5in. is ~4.9634in. Do you want an "~" answer?

Is "~96in." the inner circumference of the sphere? Is the sphere hollow? Is "~4.9634in." the outer circumference of the ping pong ball?

-rekrunnerthejerk

This should give you an estimation (assuming all the data is accurate):

http://www.randomwalk.de/sphere/insphr/spisbest.txt

Plot all of the points in Excel (or another program) and delete the first 13. Grab a Trend Line for all the data points and extrapolate to 4.9634/96 ~ 0.0517. I did this for a power series and got 5590 spheres. I think it had an R^2 value of 0.9974 or something like that.

I doubt the answer is exactly that (since the plots definitely don't follow a power series), but maybe it's in the ballpark of that figure.

FWIW, the maximum number of spheres that can fit in the larger one is simply the larger volume divided by the smaller volume --> 7235. This is obviously impossible, but that's the upper limit (not sure if that was mentioned in the thread already).

Just went for a run and gave this more thought. I'm recalculating my previous answer due to aritmetic to see if I can agree with calculations below that match an earlier answer.

The volume of the sphere is 4*3*pi*(15.278)^3 = 14940.

14940*0.74078=11067.5 cubic inches can be occupied by ping pong balls.

The volume of a ping pong ball is 4*3*pi*(0.7899)^3 = 2.0648.

11067.5/2.0648 ~ 5360 ping pong balls.

Sphere Density coefficient from wiki:

In geometry, close-packing of spheres is a dense arrangement of equal spheres in an infinite, regular arrangement (or lattice). Carl Friedrich Gauss proved that the highest average density – that is, the greatest fraction of space occupied by spheres – that can be achieved by a regular lattice arrangement = 3.14/3(sq rt 2) = ~.74048

So…

GIVEN:

C(L.sphere) = 96 in

C(S sphere) = 4.9634 in

1) using C formula C=2(3.14)r

a) r (L sphere) = 15.2866 in

b) r (S sphere) = .7904 in

2) using these r enter into the volume formula V=4/3(3.14)r^3

a) V (L sphere) = 14918.1164 cu in

b) V (S sphere) = 2.0621 cu in

3) Apply Gauss factorial from above…

V (L sphere) = 14918.1164*.74 = 11039.4061 cu in

4) Adjusted volume divided by volume of S sphere should provide our answer

11039.4061/2.0621 = 5353.47

5) therefore I would be comfortable guessing 5353 ping pong balls in a 96 inch sphere

i was wrong with my guess... made a sloppy error...

these guys who said 5350 something are about right...

are there any messageboards for fun math games like this... keep em coming OP

is it sad that this is the most interesting thread i have ever seen on letsrun?

I'm not going to venture an answer, but all you guys doing the closest packing thing are going to over estimate for two reasons.

1) The objects are being contained in a sphere. A sphere is not conducive to regular array packing.

2) You are all neglecting the edge effects, as well. The container is an impenetrable smooth surface, so the density of ping-pong balls drops to zero at that point.

I'll let you guys with too much time on your hands figure out the added corrections.

wendell wrote:

Sphere Density coefficient from wiki:

In geometry, close-packing of spheres is a dense arrangement of equal spheres in an infinite, regular arrangement (or lattice). Carl Friedrich Gauss proved that the highest average density – that is, the greatest fraction of space occupied by spheres – that can be achieved by a regular lattice arrangement = 3.14/3(sq rt 2) = ~.74048

So…

GIVEN:

C(L.sphere) = 96 in

C(S sphere) = 4.9634 in

1) using C formula C=2(3.14)r

a) r (L sphere) = 15.2866 in

b) r (S sphere) = .7904 in

2) using these r enter into the volume formula V=4/3(3.14)r^3

a) V (L sphere) = 14918.1164 cu in

b) V (S sphere) = 2.0621 cu in

3) Apply Gauss factorial from above…

V (L sphere) = 14918.1164*.74 = 11039.4061 cu in

4) Adjusted volume divided by volume of S sphere should provide our answer

11039.4061/2.0621 = 5353.47

5) therefore I would be comfortable guessing 5353 ping pong balls in a 96 inch sphere

I redid your calculations without any estimations. Pi =/= 3.14, V (L sphere) =/= 14918.1164, etc.

Here's a correct formula to use (capital variables for large sphere, lower case for smaller sphere):

# of ping pong balls = {[(4/3)*PI*(C/2/PI)^3]*[PI/3/sqrt(2)]}/{[(4/3)*PI*(c/2/PI)^3]}

You can type that into your calculator, or you can simplify to this (C=96, c=4.9634):

# = (PI*C^3)/(3*sqrt(2)*c^3)

# = 5357.836....

# = 5357

See how many balls you lost due to approximation (5353 vs. 5357)? Always use only what you start with, never approximate anything.

What is 23 alex.

lucKY2b wrote:

I'm not going to venture an answer, but all you guys doing the closest packing thing are going to over estimate for two reasons.

1) The objects are being contained in a sphere. A sphere is not conducive to regular array packing.

2) You are all neglecting the edge effects, as well. The container is an impenetrable smooth surface, so the density of ping-pong balls drops to zero at that point.

I'll let you guys with too much time on your hands figure out the added corrections.

1) I figured as much.

2) The OP never stated this was a real world application. If he/she had, information on the length of the boundary of the larger sphere would have been given. Therefore we have to assume there is no boundary. I.e. two spheres with half the diameter of the larger sphere would be contained in the larger sphere.

you are definitely correct, but i did not have a calculator with PIE handy...

but i would point out that 5357.836 SHOULD be rounded to 5358.

you are definitely correct, but i did not have a calculator with PIE handy...

but i would point out that 5357.836 SHOULD be rounded to 5358.[/quote]

Wouldn't you have to round down because the large sphere couldn't fit that extra fraction of a ping pong ball?

mcguirkthejerk wrote:

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.

There are no openings in a sphere so none will fit in artard!

More than enough.And,then some.