E-4 Mensuration Question

1. May 15, 2005

extreme_machinations

What Is The Maximum Number Of One Inch Diameter Spheres That Can Be Packed Into A Box 10 Inches Square And Five Inches Deep?

2. May 15, 2005

DeathKnight

Well along the width of the base 10 spheres can fit same is the with the lenght of the base. Along the height of the box 5 sphere can fit. So the total no. of spheres is 10x10x5=500 spheres. That's what I think.

Last edited: May 15, 2005
3. May 15, 2005

HallsofIvy

That looks to me to be much harder than a "E-4" problem! (I interpreted "E-4" as "fourth grade, elementary school!) Suppose you put the second layer of spheres not directly on top of the first layer, but "between"- that is each sphere in the depression between 4 adjacent spheres. "Sphere packing" in general is a very hard problem.

4. May 15, 2005

DeathKnight

I worked on it and found out that maximum no. of spheres in only possible if they are placed directly on to each other. For this particular problem if the spheres are placed as you have said than there are two possibilities. Either the no. of spheres that can be packed is (81+100+81+100+81)=443 or it is (100+81+100+81+100)=462. They both are less than 500. I can be wrong though.

5. May 15, 2005

saltydog

I would like to see some proof of this. Would analyzing the 2-D analog be any help? You know, circles in a square. I mean, just on top of one another, well, that's just doesn't seem to be a problem of any stature.

One concern I have DeathKnight is that you implicitly assume you can get only five levels. However, when you place the balls between each other, might you not be able to fit a sixth level?

Edit: Perhaps I shouldn't say implicitly. Have you in fact verified that you can't fit a sixth level?

Last edited: May 15, 2005
6. May 15, 2005

DeathKnight

Damn I was wrong. You can definitely add the sixth level which makes the no. of max spheres 543. I was making a simple error.

Last edited: May 15, 2005
7. May 15, 2005

saltydog

Well, I noticed something interesting when you consider the 2-D analog. Note the attached plot. I can't get 6 levels in 2-D but that's apparently because in 3-D, the height of the levels drop slightly due to the 3'rd dimension. This is enough it appears, to get the top one in. The next question then is what ratio of the dimension of the container to the dimension of the balls is needed to assure that the last level just fits?

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8. May 17, 2005

extreme_machinations

hexagonal close packing[placing spheres in the voids of the layer beneath ] and cubic close packing assures maximum usage of the available space in cubes which is = 74% . whereas in square close packing[placing spheres on top of each other and side by side as suggested by deathnight]
the maximum occupied volume is only 52.2%

im just not sure if this can b extended to a cuboid .
if the maximum occupied space in a cuboid is same as in a cube ,then the problem gets simplified .

available volume of cuboid = 74% of 500 ................eqn 1
volume of one sphere =4/3 {pie}{1/2}^3...............eqn 2

now if we divide eqn 1 and 2 we get the number of spheres as[706].

which is greater than 543 , but im not sure about this cos the packing efficiency fo cuboid could be different.

9. May 18, 2005

extreme_machinations

Can Somebody Please Confirm This ???