Thread: Easy questions View Single Post 2005-09-07, 22:46   #12
wblipp

"William"
May 2003
New Haven

22·32·5·13 Posts Quote:
 Originally Posted by THILLIAR This does not fly. Show us all your caculations to come up with that distance.
But I already did show all the calculations. And Ken already provided links to MathWorld that get the same result.
What part do you think is wrong?

I'll say it again with more words.

We start with three balls on the bottom layer that are nestled together in an equilateral triangle, and a fourth ball that is nestled into the hole.

The upper ball is centered over the centroid of the equilateral triangle formed by the centers of the lower three balls.

The centroid is 2/3 of the way from any vertex to the opposite side's midpoint.

The length of that segment is 5*sqrt(3) = 8.66 cm

The distance from any vertex to the centriod is 2/3 of this distance
10* sqrt(3) / 3 = 5.77 dm

We can form a right triangle from the center of any ball on the base to the center of the upper ball by going from the lower ball's center to the centroid then straight up to the upper ball's center.

The hypotenuese of this triangle is two radii, 10 cm.

The leg from the lower ball to the centroid is
10 * sqrt(3) / 3 = 5.77 cm.

The other leg is the interlayer distance that is, the vertical distance from centers on one layer to centers on the next layer. From Pythagorus, it must be 10 * sqrt(6) / 3 = 8.16497 cm

That's a complete derivation, twice, using nothing more than basic geometry, and supported with a link to Mathworld's derivation of the same value. If it's wrong, I deserve more than "that doesn't fly." You can point out errors in my derivation, or you can provide an alternate derivation, but unsubstantiated disagreement is not an acceptable response at this point.   