Sunday, December 27, 2009

Balls Big and Small

If 4" diameter steel balls are packed into a cubic yard container, and 2" diameter steel balls are packed into another cubic yard container, which full container weighs more?

7 Comments:

Blogger Chris said...

If they are packed in a cubic arrangement, then they'll weigh the same.

December 27, 2009 5:19 PM  
Blogger Chris said...

I've not checked every detail (like can you fill the cubes), but I think they'll weigh the same as long as you pack them the same way.

December 27, 2009 5:22 PM  
Anonymous Anonymous said...

they both weigh the same.

December 27, 2009 9:09 PM  
Anonymous Wizard of Oz said...

Both containers would weigh the same if the balls were stacked directly on top of the ones below, so that each layer contains 18*18 = 324 balls of 2" diameter or 9*9 = 81 balls of 4" diameter.
But what if the balls were packed in between the ones in the layer below? I suspect (without doing the calculations) that you could get an extra layer of the smaller balls in its container, but not for the larger balls.
Then, for the smaller balls you would have 10 layers of 18*18 balls plus 9 layers of 17*17 balls.
This would then give you 5841 balls, which is 9 more than the 5832 that you would get with the conventional packing arrangement. This container would then weigh more than the other one.
Someone else will have to help me with the maths to determine whether an extra layer can be fitted in this way.

December 27, 2009 11:42 PM  
Anonymous Anonymous said...

given a container L^3
with balls diameter A and B

Da=1/2*Db

NA=integer(L/Da)
NB=integer(L/Db)


4/3*Pi*R^3/8=V/ball
density*V=mass/ball

if NA=2*NB then

Ma=NA^3*4/3*Pi*(Da)^3/8*density
Mb=NB^3*4/3*Pi*(Db)^3/8*density

Mb/Ma=(NB/NA)^3*(DB/DA)^3
MB/MA=(1/2)^3*(2)^3
MB/MA=1


Answer: Both containers weigh the same

Solution applies as long as
Number of B balls /Number of A balls= 1/8

This does not have to always be the case, as one could select from a wide variety of packing structures.

Cam

December 28, 2009 2:14 AM  
Blogger Chris said...

The sphere problem is intimately related to what is/was known as Kepler's conjecture, which proposes that the obvious densest packing is achieved when the layers of sphere's are packed in a hexagonal arrangement, and the alternate layers fit into the "gaps" of the previous layer.

It was shown to be "probably true" in 1988, but required computer assistance. Work is still ongoing to eliminate any doubt of the result.

Amazingly, the conjecture was proven to be true for something like 8 and 24 dimensional spaces first (with 8 and 24 dimensional spheres).

December 28, 2009 4:14 AM  
Blogger Chris said...

This post has been removed by the author.

December 28, 2009 4:21 AM  

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