# Producing artificial gravity

1. Jan 4, 2004

### TSBasilisk

I'm a Junior, and I am participating in a competition. A team I am on annually joins in the ISSD competition, or International Space Settlement Design competition, which is sponsored by NASA engineers. Having participated since my Freshman year, I am currently working on the structural engineering of our proposed station.

However, I am stuck in a rut. My section deals with the production of gravity on the station. Last year, I managed to utilize an equation, which this year seems to work as effectively as a wet noodle for a weapon.

So, I was wondering if anyone can assist me in finding a viable equation for calculating artificial gravity production, or in translating my older equation, which I sadly forgot to fully detail.

Here is the closest translation of the equation from last year:
N= 1/2pi * square root of a/s

Where N=rotational speed(rotations per second/minute)
a=gravity produced(m/s/s)
s=distance from center of rotation(m)

The arrangement of the figures is what confuses me, along with units to utilize, although I'm fairly sure the posted units are correct. I can't find any source for this equation online. I remember that last year this equation gave me a lot of grief before I got it working. If there is a better equation, I'm more than willing to try it.

2. Jan 4, 2004

### Staff: Mentor

Your equation is equivalent to that for centripetal acceleration for a rotating system:

$$a = \frac{v^2}{r}$$

Where a is the acceleration, v is the speed, r is the distance from the center. You can relate speed (v) to rotational frequency (f) by: v=2&pi; r f.

Hope this helps a bit.

3. Jan 4, 2004

### TSBasilisk

Thank you. I'll have to slog my way through the rest of these problems, and hope I don't come out with ridiculously large numbers.