I'm looking at a ship that is pretty big. About 720 feet from nose to tail, with two stonking big rockets on the sides, these will flip for deceleration. At roughly the same rotational pivot point, in the middle of the ship there is, for want of a better word, a "Slot", inside which sits a bloody great big hamster wheel. This will be "roughly" 240' diameter, or about 750' circumference. (About 60' across its edge.) The habitation section is on the very outer edge of the wheel, and is about 10-12' "high" with the floor on the outer ring and the ceiling closer to the centre.

The rotation is along the shape of the ship, rather than at 90 degrees. So it appears like a huge wheel in the middle of the vehicle, (a bit like those old one wheeled gyro toy cars they made years ago...)

Q1 Given the dimensions, how fast would that ring need to spin for the passengers on the outer "habitation" section to experience as close to Earth type gravity as possible?

Q2 follow up... Also, with the option of either Short powerful acceleration + equally heavy deceleration and coasting in the middle, vs. slow accelerate to halfway point and slow decelerate back to zero, would either model dramatically impact the passengers in the hamster wheel? Basically, would the ship need to hit a "cruising speed" before deploying the rotation on the habitation ring?

  • $\begingroup$ SpinCalc $\endgroup$
    – Alexander
    Jul 18, 2019 at 18:53
  • 1
    $\begingroup$ One question per post please, and possibly not something that a quick google search will show you immediately. $\endgroup$
    – L.Dutch
    Jul 18, 2019 at 19:02

2 Answers 2


A1: 9.81 m/s² = v²/R = omega² * R = 4 * pi² *frequency ² * R

so frequency = 1/(2pi) * SQRT(9.81 / R) = 1/(2pi) *SQRT( 9.81 m/s² / (240 feet*0.3 meter/feet))

= 0.0587 hertz = 3.52 RPM

EDIT: Extra information, the outside of the hamster wheel will be moving at 26.5 meters per second (velocity = omega * R = 2*pi*frequency * R). This means that realistically, the entry port to your hamster wheel will have to be in the middle.

A2: Any acceleration by the ship in the forward direction will obviously impact anyone inside the ring by pulling them towards the place the ship is leaving. If you suddenly accelerate at say, 10 meters / second² (or about 1 g) everyone in the hamster wheel would fall over as the one currently moving towards the front of the ship would fly gently upwards only to come crashing down brutally 10 seconds later as the hamster wheel as revolved a half turn and they are suddenly experiencing about 2g's or 20 meters/second². In addition to this, having a flywheel in the middle of your ship will cause your ship to move involuntarily sideways as you accelerate.

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    $\begingroup$ That's great. Thank you, I won't have anyone try exercising by running the centre to "keep up with" the rotation. (That was one of the reasons I actually wanted clarity on this.) $\endgroup$
    – Tommy
    Jul 18, 2019 at 19:01

The formula for centripetal acceleration is the square of the rotation rate (in radian/sec -- 2*pi radians is a full revolution) times the radius of rotation. Grab a calculator.

Now, the easiest way on the passengers is to accelerate at around .01g, turn over, and decelerate at the same rate. This is low enough you can just build a little slope into the floors and no one aboard will notice. It's also a bit less "that's impossible!" than a higher acceleration for a habitat ship, and can easily be faster than a big boost (which would have to be done with the ring stopped, everything restowed after boost and before deceleration) with a coast phase.


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