[PLUG-TALK] Geometry Question

[Denis Heidtmann]

Richard C. Steffens wrote:
> My niece is very much into soccer. For Christmas, I want to make her a 
> wooden soccer ball. I am not into soccer, but having Googled for it, I 
> have found that a soccer ball is (I think) a truncated icosahedron, 
> having 32 faces, 20 of which are hexagons, and 12 of which are 
> pentagons. See:
> 
> http://en.wikipedia.org/wiki/Soccer_ball
> 
> I can figure out how to cut the hexagons and pentagons out of wood. What 
> I don't know is how to determine the angle of the edges based on the 
> thickness of the wood. I think I'll use 1/2" thick lumber. If I were 
> working with paper I'd just fold them and the angles would be whatever 
> they felt like. Wood, at that size, doesn't bend so well. So, I'm trying 
> to glue together these tiles, and it would be nice if the faces that get 
> the glue were actually in the same plane.
> 
> Any geometry wizards out there who know the formula, or can point me to 
> a reference?
> 
> As I'm thinking about this I realize that I need to decide on the 
> "diameter" of the ball. (I know, it's not a sphere, so it isn't a 
> diameter, but you get the idea.) So, I need to determine the size of the 
> edges of the tiles for a given "diameter" and then I'll have the three 
> sides of a triangle. Now, where did I put that book of trig tables...?
> 
> Does that sound right, or was high school too long ago?
> 
> TIA.
> 
I have made some calculations based upon some assumptions.  I assume that all 
vertices lie on the surface of a sphere.  One equator, selected for convenience, 
passes through two edges, 3 hexagons, and 4 pentagons. The equatorial path is 
along the edges, crosses opposite faces of the hexagons, and goes from a vertex 
to the opposite side of the pentagons.  I numerically determined that the ratio 
of the length of the edges of the polygons to the radius of the sphere must be 
0.46017 to have the polygons fit exactly.  The angles of the edges come out to 
be 65.83 degrees for the hexagons and 71.00 degrees for the pentagons.

If this approach is in error, it may be that the original assumption is not 
correct, and the angles of the edges would then not all be the same for the same 
polygon type.

If you want to go over my analysis, let me know and we can figure out a better 
way than text to communicate.

-Denis