[PLUG-TALK] Geometry Question

[Denis Heidtmann]

Tony Rick wrote:
> 
> 
> On Thu, Dec 4, 2008 at 12:17 PM, Denis Heidtmann 
> <denish at dslnorthwest.net <mailto:denish at dslnorthwest.net>> wrote:
> 
>     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
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> 
> 
> I can't speak to the equator crossing path, but this article
> http://www.absoluteastronomy.com/topics/Truncated_icosahedron
> appears to confirm the assumption about being embedded in a sphere.
> See heading 'Canonical Coordinates'.
> 
> - tony
> 
> 
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Well, that link says I am wrong about the 0.46017 value for the ratio of the 
side to the sphere radius.  Their number comes out 0.4035...  I think I know 
where I went wrong.  So my angles are also in error.  I will recalculate the 
angles based upon their ratio.

-Denis