Imagined the 10 balls are stacked in 3 levels: 6 on the bottom, 3 in the middle, and 1 on top, making a triangular pyramid.
Started off writing complex formulae for spheres, and trying to find tangent points and intersections! :-/
Then realized an easier way:
The height of the pyramid is a sum of:
height from ground to centre of level 1 (=d/2)
height from centre of level 1 to centre of level 2 (=h?)
height from centre of level 2 to centre of level 3 (=h?)
height from centre of level 3 to top (=d/2)
=> height of stack = d+2h
We find h by realizing that the centre of level 1 and 2 (and consequently levels 2 and 3) form a triangular pyramid with all sides of length d. The height of such a pyramid, h = d x sqrt(2/3).
=> height of stack = [1 + 2 x sqrt(2/3)] x d = 2.633 x d
Edited on August 4, 2004, 2:02 am
for Rixar - the tetrahedron must be a perfect one, of side length d, as when the speheres are forced together, touching, there is a uniform separation between the centres of length d/2 + d/2. Maybe start imaging just 4 spheres, then repeated on two levels to get the picture. Does that help?
Edited on August 4, 2004, 2:08 am
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Posted by David
on 2004-08-04 01:50:07 |
No Subject
