## June 4, 2012

### Geodesic Vacuum Spheres

Since a standard thin walled pressure vessel is not adequate to the construction of a vacuum airship it will be necessary to try an alternative system.  Much preferred to thin-walled structures for compression are geodesic structures.  Although geodesics are found commonly in nature and there are many examples of early engineers using similar principles (the Roman arch and dome being examples) they were first described analytically and thoroughly explored by Buckminster Fuller in the 20th century.
Geodesic domes, or braced domes, consist of a large number of members arranged into a regular polyhedron.  Domes typically use a truncated, or cut off, shape rather than the full polyhedron.  For our purpose here the full shape will be used.  Although any regular polyhedron is acceptable, the most common is Fuller's original icosohedron proposal.  An icosohedron is a 3 dimensional shape with 20 equal faces, gamers will immediately identify the basic shape as a 20 sided die.

The length, l, of any side of a regular icosohedron of radius R is given by:
$l=1.5015R$
For a large sphere this will still result in member buckling.  To resolve this each triangular face of the shape will be subdivided into a number of other triangles.  Each side will be formed by f number of d length members, where f is called the frequency of the shape and d is the chord length.  Using the buckling criterion d can be found:
$d = \frac{pi}{k} \sqrt{\frac{EI}{P_{cr}}}$
If each member is composed of a thin hollow tube the area moment of inertia can be estimated as:
$I = \pi r^3 t$
Now we'll consider the compressive failure of each member, if the member is designed to fail in compression at the same point if fails by buckling the relation between member radius and wall thickness is found:
$A = \frac{P{cr}}{\sigma_U} = 2 \pi r t$
Substituting this into the equation for chord and dividing that by the side length to find frequency.  Rounding up:
$f = 1.5015 \pi k \frac{R}{r} \sqrt{2 \frac{\sigma_U}{E}}$
This is a rough estimate as not every member of the structure will be the same size.
Using this frequency though will let us calculate the node positions on the first face.  These can be described in spherical coordinates:
$\phi = arctan(\frac{U_2}{U_1 - U_2})$
$\theta = arctan(\frac{\sqrt{U_1^2 - 2 U_1 U_2}}{f-U_1})$
where the factors U1 and U2 are varied from 0 to f.  These represent movement from one vertex of the triangular face (0,0) along two sides to the remaining vertices.  This will allow us to calculate the location of each node on the first face, we can then find the length of each member of that face.  The members of the face will appear on the final structure 20 times, once for each face.
Another post will contain an analysis and the code for it.

Reference:
Ramaswamy, Eekhout, and Sureesh.  Analysis, Design, and Construction of Steel Structures. "Chapter 8: Braced Domes." http://www.scribd.com/doc/36287836/Analysis-Design-and-Construction-of-Steel-Space-Frames#download