Originally Posted by wblipp
This is easily calculated by starting with the equilateral triangle from 3 balls
on the bottom, and observing that you can get from the center of any of
these balls to the center of the upper ball by going to the centroid then up.
Going straight up forms a right triangle, so we have a right triangle with
the hypoteneus of 10 cm (2 radii) and one of the legs 10*sqrt(3)/3.
Pythagorus tells us the third leg, which is the vertical distance from
the center of layer 1 to the center of layer 2, is 10*sqrt(6)/3.
about 8.165 cm.
Packing in 12 layers will use up 11 interlayer distances
plus 5 cm on each end, totaling about 99.815 cm.
This is tight. We could shrink this box to 100 x 99.49 x 99.82.
I'd be surprised is anyone can wiggle enough to fit another ball.
