"Bruce White" wrote in message
...
I tend to look ahead at projects, and when I get an idea, I think about it
a
while. I am wanting to make a piece that will require an icosahedron at
it's core to turn into a ball. My triangles will be 20 identical
equilateral triangles with 5-1/4" "legs" at 60 degrees, (I may make this
smaller, I haven't decided yet). These will all be glued together and
should form a icosahedron approximately 7-3/4" in diameter, which isn't
critical, I can live with just about any size over 6". This part I
understand. Now here's my dilema...

Your icosahedron can be thought of as two 5 sided pyramids, one pointing up
and the other pointing down, and these 2 pyramids joined together by a set
of 10 triangles facing up and down alternatively. Once you know that, it is
easier to visualize what the angle between 2 faces of the pyramid and
therefore the icosahedron, is. There are several ways of finding that
angle, some time ago I made an Excel spreadsheet to find that angle for
pyramids of any number of sides. The result I am getting matches the result
D J Delorie gets = 20.905. The actual angle between the face of the side
and the bevel is 69.095 degrees, but the complementary angle 20.905 degrees
is what you would set your blade to make the cut.

If anyone is interested in the Excel spreadsheet, it can be downloaded he
http://members.rogers.com/penate/dihedral.xls
The input needed are the width of the base of the pyramid (cell B4), the
number of sides of the pyramid (cell B6) and the height of the pyramid (cell
B5). In this case, that value has to be calculated separately by simple
trigonometry and is equal to:
height = 5.25 x ( SIN [ ArcCOS (0.5 / [ SIN(36) ] ) ] )
BTW, I used vector algebra in the spreadsheet, which was an overkill for
what I intended to do when I made the spreadsheet.

Guillermo