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Date: 12/15/96 at 23:39:47
From: Sam Matz
Subject: Geodesic domes

I'll keep this short. For a little over a year and a half I have been
obsessed with geodesic domes, spheres, and Bucky Fuller. I have 
searched on the web for different things but recently I have been 
looking for ways of calculating all of the lengths of the segments, 
the angles, etc. I was hoping you could shed some light on the 
subject. I know I haven't been real specific, but anything you can 
tell me about all of the math involved in geodesics would be very much 
appreciated. Thanks. 

From a Freshman at T.H.S.     

Date: 12/16/96 at 16:15:50
From: Doctor Pete
Subject: Re: Geodesic domes

The topic has been an interest of mine for several years now, and I 
have gone so far as to build paper models of geodesic spheres.  Check 

which contains a few photographs, and   

which contains information regarding the Platonic solids (in 
particular the regular icosahedron).

I do plan on including the mathematics of how the dimensions of the 
sphere were calculated, but I've been rather busy as of late.  I'll 
provide a few bits and pieces here, though.

Given a regular icosahedron with edge length 1, consider the central 
projection of its edges and vertices on its circumscribed sphere.  
We get something of an "inflated" icosahedron, where the edges have 
become arcs of great circles on the sphere.  The radius of the 
circumscribed sphere is calculated to be R = Sqrt[(5+Sqrt[5])/8].  
We can immediately find the degree measure of one such arc, which is 

Now, divide each edge of the icosahedron into v equal parts.  
v is the *frequency* of the desired geodesation.  On each face, 
draw a series of parallel lines so that each face is dissected 
into v squared smaller triangles.  For v = 3, we have this:

               (1,0) /__\ (0,1)
                    /\  /\
             (2,0) /__\/__\ (0,2)
                  /\  /\  /\
           (3,0) /__\/__\/__\(0,3)
                 (2,1) (1,2) 
so we have dissected a representative face into 3^2 = 9 parts. Then 
consider the central projection of these lines onto the surface of the 
sphere. The result is a v-frequency geodesation of the sphere. Note 
that in the above diagram, I have labeled the sub-vertices of the 
dissected face (the center should be labeled (1,1)). Clearly, the 
distance between any two adjacent sub-vertices is 1/v, if the 
icosahedral edge length is 1. Since (0,0), (3,0), and (0,3) are 
actually points on the sphere, the distance from the center of the 
sphere to these points is simply R. On the other hand, the other 
points do not touch the sphere, so the distance from the center to 
these points is strictly less than R, and is a function of the 
frequency v and the label. With some careful consideration and 
computation, one finds that the distance from the center to some point 
(m,n) on the face is given by:

     d(m,n) = Sqrt[(R^2+1)v^2 + (m+n)^2 - (m+v)(n+v)]/v 

This, in turn, gives the angles subtended by the various sub-edges of 
length 1/v.  In particular, the degree measure of the angles near the 
center of the face will be larger than those nearer to the edges and 
one can see by symmetry that one need only find the "horizontal" 
angles, e.g., the angle between (1,0) and (0,1), or (1,2) and (0,3).  
By the Law of Cosines, this is:

              d(m+1,n)^2 + d(m,n+1)^2 - 1/v^2
  Cos A(m,n) = ------------------------------- 
                    2 d(m+1,n) d(m,n+1)

This is a fairly straightfoward calculation for a computer.  More 
complicated formulae exist for non-icosahedral geodesation schemes.

I hope that was interesting -- if you'd like to know more about where 
I got these equations from, don't hesitate to e-mail me.  

-Doctor Pete,  The Math Forum
 Check out our web site!   
Associated Topics:
College Polyhedra
College Probability
College Triangles and Other Polygons
High School Polyhedra
High School Probability
High School Triangles and Other Polygons

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