calculating container needs

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calculating container needs

Postby jc » Sat Jul 26, 2008 5:29 am

I would like to place 8million 1 millimeter beads in a baby bottle. How large would the bottle have to be to contianthe beads?
jc
 

Re: calculating container needs

Postby Guest » Sat Jul 26, 2008 7:19 pm

jc wrote:I would like to place 8million 1 millimeter beads in a baby bottle. How large would the bottle have to be to contianthe beads?


It would have to be a big, strong baby to lift the bottle. I get about 5.66 L.

You may wish to read the Wikipedia article on "close packing" (Google for it)

The volume of one sphere is (4*pi/3)*r³ or (pi/6)*d³, using the diameter.

In an optimium lattice structure, close-packed spheres will occupy pi/sqrt(18) of the volume. This is only true for the lattice shape and any practical package will have minor errors at the walls (as long as minimum dimensions are several times the sphere dimension).

Thus (pi/sqrt(18))*Vc = (pi/6)*n*d³
and Vc = (0.5*sqrt(2))*n*d³

This is 4*sqrt(2)*10^(-3) m³ or 4*sqrt(2) liters. While this evaluates to 5.6585. . . liters, I would recommend rounding up at least to 5.66 L to allow for wall effects.
Guest
 

Postby Guest » Sat Jul 26, 2008 10:49 pm

My answer above assumes the beads are carefully packed in optimum arrange. There is another topic "random close packing" when spheres are not ordered, but "dumped" in and then tapped or vibrated to settle them as much as possible.

Apparently in that case, the packing factor is only 64% and that is the limit with adequate vibration (it can be less). Using 64%, I get 6.55 L
Guest
 


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