Thread: Asteroid Driver
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Old 01-03-2021, 09:49 AM
cfuruti cfuruti is offline
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Originally Posted by Damraska View Post
The spun section provides 15,700 square feet of space under artificial gravity, equivalent to about 8 moderately sized homes. I have not done calculations for the rate of spin but estimate this cylinder should provide constant gravity equivalent to one fourth Earth normal, enough to stave off the debilitating effects of prolonged life in micro-gravity.
(sorry, I'm used to international units) If
  • the inner radius of the spinning section is r meters
  • the desired pseudogravity is a m/s˛
  • the section spins at R rotations per minute,
The centripetal acceleration of constant circular motion is a = r omega˛, where omega is the angular velocity in rad/s.
rad/s = 2 pi RPM / 60,
therefore the acceleration at the "bottom" floor is a = r (2 pi RPM / 60)˛ = 4 pi˛ R˛ / 3600
  • r = 900 a / (pi˛ R˛)
  • R = (30/pi) sqrt(a/r)
For instance, if you want a = g/4, about 2.5 m/s˛,
  • if R = 5 RPM, r should be about 9.13 m
  • if r is 10 m, R should be about 4.78 RPM
Now you can estimate minimum RPM and radius for acceptable mechanical constraints and practical limits.
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