How would I calculate the size and angle of this space station ring?

Question

This is an idea for a space station on the surface of Eris (but could work on other celestial bodies), using rotating rings to simulate gravity. The idea is that these rings would sit horizontally on the surface of the dwarf planet much like you would expect in space. Instead of the floor of the rings being the inner flat edge, the rings would be slanted forming a sort of downward cone shape. You could imagine it much like how a freeway or racetrack is banked around turns.

Ideally the rotation would produce centrifugal force pulling the occupants towards the edge to simulate gravity, however, the slant of the rings would counteract the existing low gravity of Eris which is roughly 1/12 of Earth's. The closest real world comparison I could think of would be the Gravitron amusement park ride, but on a massive scale.

I imagine that first I will have to determine both the size and speed of the ring(s), but how would I calculate the slant angle? Is there a formula that I could use to plug in the measurements of the station and then calculate that value?

Please let me know if there are other values I might need for this as well. My primary goal is for station occupants to feel a force of 1/2 G pressing downward with none or minimal force pulling them towards either edge of the ring. Is this even feasible? Assume structural integrity and necessary energy are sufficient and that the ring is well balanced (or at least that some system maintains balance).

Answer 1

Let Eris's gravity be $g$, and the angular speed of the ring be $\omega$ and its radius be $R$.

The below assumes the ring is small compared to the size of Eris, i.e. real gravity is essentially constant direction and magnitude across the ring.

To have an effective gravity of $G$, the angular speed needs to be $\omega = \frac{(G^2 - g^2)^{1/4}}{\sqrt{R}}$ (for the given values this turns out to be $\sqrt{\frac{2.2m}{R}}$ radians per second, or $\sqrt{\frac{200.5 m}{R}}$ rpm), with an angle (between the "floor" of the ring and horizontal) of $\theta = arctan\left(\frac{\sqrt{G^2 - g^2}}{g}\right)$. For the given values this comes out to an angle of 80 degrees (or 10 degrees from vertical).

Also note that if the ring has multiple levels at significantly different radii $r$, the angle needs to be adjusted to be $arctan\left(\frac{\sqrt{G^2 - g^2}}{g}\frac{r}{R}\right)$.

In general, if $G >> g$, $\sqrt{G^2 - g^2}$ is about $G >> g$. Then (small-angle and pi ~ 3 approximations), the angle will be about $\frac{g}{G}*60$ degrees from vertical.