[PLUG-TALK] Geometry Question

[Tony Rick]

On Thu, Dec 4, 2008 at 12:17 PM, Denis Heidtmann <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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