Thread: Easy questions
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Old 2005-09-07, 01:48   #8
wblipp's Avatar
May 2003
New Haven

3×787 Posts

I agree with Ken's solution at at least 1254 balls are possible.
The additional wiggle room in every direction is small, so I doubt that
repacking a few boundary layers would allow you to squeeze in an additional ball.

I'll try to clarify the packing. Start with 10 balls in a row along one wall on the bottom of the box.

Next snug a row of 9 balls next to the 10.
Alternate rows of 9 and 10 balls.
Looking down on the box, the centers of the balls lie on
equilateral triangles, so each row is 5*sqrt(3), about 8.66 cm further along.
11 rows takes up 10 interrow distances from center to center, plus
a radius of 5 at each end, about 96.6 cm total.

The next layer starts with a row of 9 stacked on
the first two rows of the bottom layer.
The center of this row lies above the centriod of the
equilateral triangle of the balls on the bottom row.
The centroid is 1/3 of the way from the base.
So the center of this first row of the second layer is
5+5*sqrt(3)/3, about 7.887 cm from the wall.

The next rows on this layer have the same inter-row spacing, 8.66 cm.
Adding 10 interrow spacings plus a 5 cm radius for the last row,
we can fit in 11 rows on this layer, too - reaching to 99.49 cm.

The first, and all odd layers, have 6*10+5*9=105 balls.
The second, and all even layers, have 6*9+5*10=104 balls.

The only issue remaining is to verify that 12 layers are possible.

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 inter-layer 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.
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