Calling all math wizards

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...

These triangles will be made of 3/4" thick lumber, and I need to figure out what the angle to cut the inside miters on the triangles so they all glue together nice and snug into a roundish type of structure.

Any help??

Bruce

According to my math, and

a sphere of outer radius R you need triangles with edge length 1.323*Rand bevel angle 20.905 degrees.The minimum thickness of the wood must be 0.26 * R, any extra thicknessbecomes the thickness of the resulting sphere. Thus, for a sphere of radius 3 7/8, you want triangles with edges

5.127" long and wood (1+T)" thick (T = thickness of sphere wall).

guess my math is lousey

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 here:

input needed are the width of the base of the pyramid (cell B4), thenumber of sides of the pyramid (cell B6) and the height of the pyramid (cellB5). In this case, that value has to be calculated separately by simpletrigonometry 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 forwhat I intended to do when I made the spreadsheet.Guillermo

Pardon Me but what is a icosahedron?

Its a hedron with 20 polygons for sides :-)

OK.. Have you seen a soccer ball? it is basically a polyhedron having 32 sides or polygons, 12 pentagons and 20 hexagons. Icosahedron is just a solid having 20 sides. Etymologically, "ICOSA" means 20 in Greek (I know, it's all Greek to you :-)

Guillermo