marbles packed in sphere of minimum size

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galactus

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Here is a cute problem I ran across in an old textbook I found out in the shed.

I have posted a half-a**ed Pant diagram. The best I could get it.

"A set of 4 marbles of 3/4 inch diameter each are to be packed inside a sphere. What is the *exact diameter of the smallest container possible?".

* exact means no decimal approximation
 

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This is a old problem from crystallography. The closest packing possible would be a hexagonal packing.

The marbles will form a triangle in the base and and form a tetrahedra.

Then we would locate the CG of the tetrahedra and that would be the center of the sphere.

A little bit of "tedious" (well a lot bit of - in my opinion) algebra and we would have it. I had done this in my undergraduate days (where the heck did I put that notebook...).
 
This problem is in the cones, cylinders, and spheres section of an old college geometry book I have. I don't think it is meant to involve anything too fancy-schmancy. The book is "College Geometry, a Discovery Approach" by David Kay, second edition

The solution given is \(\displaystyle \frac{3\sqrt{6}}{8}+\frac{3}{4}\)

There is another neat one involving the height of stacked up cannonballs.

Suppose there are 10 cannonballs of diameter 9 inches stacked up in a pyramidal stack. How high is it?.
 
galactus said:
This problem is in the cones, cylinders, and spheres section of an old college geometry book I have. I don't think it is meant to involve anything too fancy-schmancy. The book is "College Geometry, a Discovery Approach" by David Kay, second edition

The solution given is \(\displaystyle \frac{3\sqrt{6}}{8}+\frac{3}{4}\)

There is another neat one involving the height of stacked up cannonballs.

Suppose there are 10 cannonballs of diameter 9 inches stacked up in a pyramidal stack. How high is it?.
Click to expand...

I said it was tedious - not fancy calculation (certainly not schmancy).

These calculations have a very practical application - in metallurgy - this number is needed to calculate the energy required to flatten "sheets" (rolling foils).

10 cannonbslls again will be stacked as tetrahedra (stacks of triangles 6-3-1). Again "tedious" geometry and algebra - nothing too fancy-schmancy.
 
I tried the "four marbles in a sphere" problem, but I keep getting \(\displaystyle \frac{3\sqrt{2}}{4}+\frac{3}{4}\)

This is a little larger than what the book gives. I was just wondering if anyone came up with the book solution or got the same solution as I did.

I checked out those sphere packing sites, previously.
 
If the marbles have radius 'r' then the distance of the CG of the tetrahedra to one of the vertices = ?(6)/4 * (2r)

Then the diameter of the circumscribing sphere would be D = 2 * [?(6)/4 * (2r) + r] = ?6 * r + 2r

r = 3/8

D = 3*(?6)/8 + 3/4

Galactus - if you already have not done so - can check out

http://www-personal.umich.edu/~hoaglund/tetrahed.html
 
Thanks Subster :D

I see my dumb mistake. I even tried the tetrahedron method and made a mistake which lead me down the primrose path.

As a matter of fact, I have a primrose blooming right now. How appropriate :D

Thanks for the site. I did not see that one.

The centre of a tetrahedron is the intersection of two space heights (1,2,3). It is centre of gravity, centre of the sphere through the four corners, and centre of the largest sphere, which still fits inside the tetrahedron (4). ... You get two formulas for r and R with the help of the Pythagorean theorem and H=R+r:
There is \(\displaystyle r=\frac{\sqrt{6}}{12}\cdot a \;\ R=\frac{\sqrt{6}}{4}\cdot a\).
 

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"A set of 4 marbles of 3/4 inch diameter each are to be packed inside a sphere. What is the *exact diameter of the smallest container possible?".

1--Draw 3 circles of equal diameter, 3/4 inch, at centers A, B and C, each circle touching the other 2,AC horizontal with C above.
2--The centers form an equilateral triangle.
3--The centroid of ABC is at D at the intersection of the 3 medians.
4--DC = 2/3 of median AE (E at mid point of AB.
5--With D as center, draw a 4th circle of 3/4 inch diameter representing 4th marble atop the other 3.
6--To lower left of figure, draw a line, mn, perpendicular to projected AB and parallel to EC.
7--Representing a vertical slice through the 4 marbles through EC, project E and C onto mn.
8--On the projection line of point D, locate point F on mn.
9--On the same projection line for D, locate point D (3sqrt6)/16 above mn (to be verified later).
10--Center of enclosing sphere lies at geometric centroid of tetrahedron ABCD on line FD at 1/4FD above F at G.
11--CE = (3/4)(sqrt3)/2.
12--FC = (2/3)(3/4)(sqrt3)/2 = (sqrt3)/4.
13--FD = (3/4)^2 - (sqrt3)/4 = (sqrt6)/4.
14--FG = (sqrt6)/16.
15--Radius of enclosing spher R = FD - FG + 3/8 = (sqrt6)/4 - (sqrt6)/16 + 3/8 = 3(sqrt6)/16 + 3/8.
16--Sphere diameter D = 2R = 3(sqrt6)/8 + 3/4.
 
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