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 out: http://www.ugcs.caltech.edu/~peterw/portfolio/polyhedra/ which contains a few photographs, and http://www.ugcs.caltech.edu/~peterw/studies/platonic/ 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)/8]. We can immediately find the degree measure of one such arc, which is ArcSin[1/(2R)]. 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: (0,0) /\ (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! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum