Cross-section shape of water distribution in a spinning torus

May 2014
Hi. I'd like to know how to roughly calculate the distribution of water inside a torus that is spinning. More specifically, for narrow channel inside it (so the cross-section/profile would be a slightly curved floor with vertical walls).

I know the water level is not going to be flat, as the higher volume of the water at the centre is going to force some of the water out to the side. But it's not going to match the curvature of the floor, either. I'm guessing it will be somewhere in between?

Is there an relatively straight forward way to calculate this if I know the basic dimensions, such as the force, the minor radius of the torus, and the width of the walls inside it?

I'm asking because I'm creating a mock-up of a rotating space station in AutoCAD and want find a way to draw the correct profile of the water level for a large central river inside it. So it doesn't need to be accurate, just close enough to appear real, and constructed with basic geometry.


PHF Helper
Jun 2010
Morristown, NJ USA
Since you mentioned that this is a model of a space station, presumably it is in orbit. Hence gravity can be ignored, and the spinning of the space station causes "artificial gravity" in the direction radiating from the center of spin. So the water will move as far to the outside as possible, filling the inside edge of the outer rim.
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Apr 2008
Bedford, England
Restating ChipBs answer in AutoCAD terms.

The surface of any liquid in a rotating space-station will be a section of a cylinder, where the axis of the cylinder is the axis of rotation of the space station.