How many softballs will fit into a given space?

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Request for Question Clarification by mathtalk-ga on 18 Feb 2004 15:27 PST

Hi, jimhart-ga:

Are you asking for the largest number of softballs that will fit into
an asymptotically unbounded volume, expressed as a ratio of the
collective percentage of the volume taken up by the softballs
themselves?

regards, mathtalk-ga

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Clarification of Question by jimhart-ga on 18 Feb 2004 15:43 PST

we have the size of the room. They told us that we're using a 12"
softball. It would be easy if we could stack them as if in a small
box. We'd only have to use the ball as if a cube. However, they are to
be randomly stacked. So, I'm trying to figure out how to estimate the
unusable space between the balls. you see?

Thanks
Jimhart.ga

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Clarification of Question by jimhart-ga on 18 Feb 2004 15:45 PST

oh, also: the 12" ball has a diameter of 3.82" if that helps at
all....is there a standard formula for this?

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Request for Question Clarification by mathtalk-ga on 18 Feb 2004 20:47 PST

Hi, jimhart-ga:

There are figures for the volume percentage of the optimal packing,
for random packings, and for specific ones like the cubic packing that
you mention.

If the room is big enough, relative to the diameter of the softball,
then using such percentages should give you an "engineering" answer to
the question.

regards, mathtalk-ga

It depends on how they are packed, the shape of the space, whether
you're allowed to squeeze on the sides of the space, and if so, how
much pressure you can apply and the constitutive relations for the
softballs!

Apparently softballs shouldn't pack as densely as M&M's - from Princeton:
http://www.princeton.edu/pr/news/04/q1/0212-candy.htm

My guess is very few extra balls are possible over the vertical and
horizontal stacking = LWH/diameter cubed, provided the LWH are each
exact multiples of the soft ball diameter, even if the room smallest
dimention is 100 times the ball diameter. For not exact multiples;
LWH/.99 diameter cubed is a good estmate, but will yield more balls
than will fit if the total is less than one million, for some
dimention combinations. If 1% denting of the balls is allowable
LWH/.98 diameter cubed may be typical.  Neil

These past answers might provide useful information:
http://answers.google.com/answers/threadview?id=45574
http://answers.google.com/answers/threadview?id=58891
http://answers.google.com/answers/threadview?id=119494