Saturday, July 9, 2011

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

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