2
$\begingroup$

Are spaceships using spin-gravity to simulate normal earth gravity on their inner surface practical considering materials available today (i.e., steel, carbon fibre, etc.)? Imagine the spaceship as described in Rendezvous with Rama, by Arthur C. Clarke.

A mathematical answer would be great, as I would like to model this in a spreadsheet.

$\endgroup$
  • $\begingroup$ Another way to look at it: consider a suspension bridge, and how much weight a steel cable can hold. To a first approximation, there isn't much difference between the center stretch of cable of a suspension bridge and a stretch of cable as part of a hoop. $\endgroup$ – TLW Sep 13 '19 at 3:47
  • $\begingroup$ When you consider the weight of the cable itself, there might not be enough strength left to hold a spinning ship together. Actually my question was with regards to more of a very large beer can spinning on its axis. The Rama example I gave was many kilometers in diameter. $\endgroup$ – Tom Cumming Sep 13 '19 at 20:22
1
$\begingroup$

I will show the mathematics for a feasibility calculation for such a rotating space station. The station you mentioned is a rotating cylinder station, rather than a concentrated mass rotating station. The equations for both are similar, differing by factors of 2. In general, if a cylinder can take the stress, then a point mass system with the same rotational speed can also take it.

To determine whether a spaceship like this is possible, we need to use the hoop stress equation for a thin-walled cylinder. So long as the thickness of the supporting material is less than 1% of the radius, this equation is extremely accurate. $$\sigma_\text{hoop} = \frac{Pr}{t}$$ where $P$ is the pressure, $r$ the radius of the cylinder, $t$ the thickness of the supporting material. But what is the pressure? We want to transform this equation into something more useful from a station-design perspective. Well pressure can be expressed as a function of acceleration $a$, mass loading pressure $p_\text{load}$ (units [$\text{kg}/\text{m}^2$]) and atmospheric pressure inside: $$P = ap_\text{load} + P_\text{atm}$$

So the hoop stress equation becomes $$\sigma_\text{hoop} = \frac{(ap_\text{load}+P_\text{atm})r}{t}$$ This formula should fit nicely into a spreadsheet.

Now for some actual numbers. We want $1g=9.8 m/s^2$ of acceleration, a station radius of $100m$ (that looks to be smaller than some of the images I browsed. But remember that's 200 meters to the other side!), and an average mass loading pressure (including the structure itself) of $2000 kg/m^2$, and the atmospheric pressure to $1\text{atm}=10^5\text{Pa}$. Lets set the thickness of the wall that holds everything in to be $10cm$. This would give you a useful loading pressure of ~$1100kg/m^2$ using steel for structure. Then the hoop stress is calculated to be $$\sigma_\text{hoop} \approx 120\text{MPa} \approx \frac{1}{4}\sigma_\text{yield,steel}$$ I quickly searched the yield strength of structural steel to be 400MPa. This is safely inside the structural limits of steel.

Note that this is the loading of an unsupported cylinder. Any radial supports, when carefully designed to manage localized stresses, will reduce this total stress.

So yes, using common materials today, such a station could be built.

What is stopping us? Mass. Lots and lots of mass. Launching things into orbit is hard. We could launch it in pieces, but even then there are numerous other difficulties not related directly to structural mechanics. This is one of the reasons it is useful to have manufacturing facilities on the moon using asteroids for material -- it removes the requirement to launch all the mass.

As a final thought, think about the scales involved. This thing is huge.

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Just the answer I was looking for! I'll spread-sheet it for sure. To be practical, you wouldn't ship it up, you'd, "live off the land", i.e., harvest what you have in space already. $\endgroup$ – Tom Cumming Sep 13 '19 at 2:23
  • $\begingroup$ As you say, this thing is, huge, hence my wanting to model it in a spreadsheet. I'd like to know where the limits are. The Rama example I gave is a "beer can" like ship many kilometers in diameter. $\endgroup$ – Tom Cumming Sep 13 '19 at 20:27
  • $\begingroup$ That is definitely going to stretch the limits of feasibility for sure. The actual issue you will face is the atmospheric pressure. You are essentially pressurizing said mega-beer-can to 1 atm, and will need the wall thickness to compensate. I think the limit will come when the outer edge speed gets fast enough to tear the disk apart. If it were just steel and no pressure, this would probably happen around a radius of 150km, but with atmosphere the radius will be far lower. $\endgroup$ – Quietghost Sep 13 '19 at 20:35
0
$\begingroup$

There is a difference in materials properties needed because the structure is mainly in tension instead of compression. Here’s how you can calculate some of the forces involved:

I will approximate the artificial gravity spacecraft to have its mass very concentrated around its outer extents, or where the gravity will be used.

Tension force on a tether, the easiest structure to calculate loads with, should be roughly equal to (mass of entire spacecraft)*(fraction of Earth gravity needed)*9.8m/s^2. This is Newton’s first law, F = ma. Acceleration is constant and radial-in for circular motion.

More than one tether, or for example an arrangement like spokes on a wheel, reduces the stress for each tether with an inverse relationship (2 tethers - half stress each). With a ring instead of two bodies and a tether, the ring experiences forces of its own from tension along the surface of the ring. Tethers can be seen as a reinforcement stiffening the ring, with tension set arbitrarily per design.

This can be made more accurate by taking an integral along the axis of the tether, integrating (differential mass)*acceleration with respect to radius from the Center of Mass of the whole spacecraft (tether and counterweight/other spacecraft included).

| improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ So is your answer "yes" or "no"? $\endgroup$ – uhoh Sep 12 '19 at 8:36
  • $\begingroup$ You mention a ring, but can you clarify what the tension would be on the ring if there was no tether? $\endgroup$ – Tom Cumming Sep 13 '19 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.