Artificial gravity calculations

From the equation, we can deduce what number you need in order to bring this rpm
times pi up to 30, which will then yield a product of 1.

(3.1459 x rpm)=30

therefore, at RPM= 9.536

Now, do the same with R, which brings it to 1, and cancels out the denominator

Thus, in order to produce 1g, need R = 9.81m, spinning at 9.54 rpm
In that case, circumference = 61.7 m, spinning rate = 35.33 km hour

Or, you can vary R in meters, so as to calc. g

If you use 6 rpm, you approximate Mars gravity

since you obtain by plugging 6 into the rpm variable, you obtain
0.396 g with R = 9.81m