## June 25, 2012

### Potential of a Geodesic Vacuum Sphere

As I said before, a conventional thin-walled pressure vessel would be incapable of bearing atmospheric pressure without weighing more than the air it displaced.  However using the methods described in my previous post on the subject it should be possible to describe a geodesic sphere that can.
Recall that a geodesic shape is an ordinary polyhedron composed of stiff members made to bear only compression forces in one direction.  These members can be kept short be subdividing the faces of the shape, thus preventing the members from buckling.  The final shape is then covered in a strong fabric or shell that bears tensional loads and holds the vacuum.

The equations outlined before allow us to describe a single face of the final polyhedron.  It is possible, quite easily using a pen and paper, to show that the number of internal (non-edge) members on each face is equal to:
$3 \sum_{i=1}^{f}(i-1)$
Since there are 30 edges on a regular icosahedron, this means that the total number of members, which we will call n, is a simple function of frequency:
$n = 30f + 60\sum_{i=1}^{f}(i-1)$
Leaving that for a moment, we can perform a stress analysis at a single vertex of the shape.  At each vertex five members meet at a point.  For that point to remain fixed in space those five members must exactly counteract the vertical component of ambient pressure pushing down on that point.  If we consider these five members alone, a structure called a pentacap, then the pressure could be approximated as the ambient pressure acting on a circular area equal to the two dimensional plan of the pentacap.  The angle one edge makes with the vertical can be found by the law of cosines, using this angle we can then calculate the radius of the plan:
$A_{plan} = 0.262 \pi R^2$
Using this angle will also tell us the relation between the vertical component of member force and the full member force borne.  Both the angle and the relation between the pentacap radius and polyhedral radius is fixed for a given shape, and so the relation between member force and pressure and spherical radius becomes:
$F=0.823P_{i}R^2$
We can calculate the necessary relation between member radius, r, and vessel radius, R, from the frequency as shown before.  This will also allow us, for any frequency, to find the maximum member length.  We then substitute the member force into the equations for member buckling and compressive failure to find the necessary member radius and thickness for a given frequency.
I'm afraid to admit to a mistake in my previous post, the equation for radius ratio in terms of frequency should be:
$\frac{r}{R} = \frac{1.5015}{\pi} \frac{k}{f} \sqrt{2 \frac{\sigma_C}{E}}$
Member length then becomes, in terms of frequency and area:
$d = \frac{1.5015}{f} \sqrt{\frac{A \sigma_C}{0.262 \pi P_i}}$
By multiplying this by the area we calculate material volume used in the member:
$V = Ad = \frac{1.5015}{f} \sqrt{\frac{\sigma}{0.262 \pi P_i}} A^{3/2} = \frac{1.24}{f} \frac{P_i}{\sigma_C} R^3$
Now, multiply this volume by the material density and the number of members previously found.  This total mass must be less than the volume of air displaced by the sphere.  If we simplify this as the circumscribing sphere volume:
$1.24 \frac{P_i}{\sigma_C} \frac{n}{f} \rho < \frac{4}{3} \pi \rho_i$
We'll just simplify this:
$\sum_{i=1}^{f}(\frac{i-1}{f}) < \frac{1}{2}(0.113 \frac{\sigma_C}{P_i} \frac{\rho_i}{\rho_{mat}} - 1)$
In a future post this will be used to find the optimum frequency for various potential materials.  This will then give the member sizes appropriate.

#### 1 comment:

1. Spaceship Earth and all other spaceship planets are spherical, so perhaps the spaceships we build should also be spherical.

A geodesic evacuated sphere need not withstand atmospheric pressure at sea level. Rather, it could be evacuated gradually as it increases in altitude. It could begin its acsent filled with helium, and then pump it out as it reaches its maximim altitude, which may be around 15 miles high. Dumping thousands of lbs of helium would then push the sphere even higher where we'd switch to propellers, then rockets to reach a stable low earth orbit.