alphaQuad napsal(a):
> "Mike the Elder" <nomail@nomail> wrote:
> 
>> ..... I suppose I should do one now ;-)
>>
>> Best Regards,
>> Mike C.
> 
> Yes you should!
> 
> Academic question:
> 
> While your on it, ponder if all points can connect 6 edges, as opposed to what
> segment3.png shows; points connecting 5, 6, and 7 edges.
> 
> Wondering if mathematical magic can do it, or if my design depends on limited
> num stacks/slices, or if it is even possible. Seems a structural weakness and
> design flaw where 7 edges connect with adjacent 5-edge connections.
> 
> aQ
> 
no. You need vertices (points) of order 5. Assuming you only want to use 
vertices of order 5 and 6, you need 12 5-order vertices as in the 
icosahedron, or 6 4-order vertices, or 4 3-order vertices.
It follows from the Euler's characteristic:
	V+F=E+2
V-number of vertices
F-number of faces
E-number of edges
2-euler's characteristic of the sphere
if we assume no double edges(diangular faces) we can write E >= 3/2 F.
establishing F <= 2/3 E into the previous formula we get:
V + 2/3 E >= E + 2
V - 2 >= 1/3 E
6V - 12 >= 2 E
If you definitely want no 5-order vertices you can still use toroidal 
domes (V+F=E+0) instead of spherical ones. or you can use partial spheres.
-- 
You know you've been raytracing too long when...
you ever saw a beautiful scenerey and regretted not to take your 6" 
reflective ball and a digital camera, thinking "this would have been a 
perfect light probe"
-Johnny D

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