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I know that the "gravity" in an O'Neill Cylinder is from Centrifugal effect. Calculated by:
F=mω^2r
F is centrifugal Force
m is the Mass
ω is the Angular velocity
r is the distance from origin

However, in this equation, would a roof of a building be considered as "the origin" for the purposes of the equation, or is that always the centeral axis of the cylinder?
Would having a completely enclosed (airtight) room make a difference? What if the room had a higher or lower ambiant air pressure?

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  • $\begingroup$ r is the distance from the axis of rotation. The other stuff you mentioned makes no (appreciable) difference. (Strictly speaking, you will get some real gravity from having a roof above you, but the amount thereof is so negligible as to be irrelevant to anything but incredibly sensitive detecting devices. Unless the "roof" is neutronium, in which case you have other issues 🙃.) $\endgroup$
    – Matthew
    Nov 30, 2021 at 20:05

2 Answers 2

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r, in this case, is the distance from the center of the cylinder. The radius of the cylinder, if you will.

It doesn't matter whether a room in an O'Neill has a roof or not, or if it's hermetically sealed, or if it has a higher or lower than normal air pressure; the gravity in it will be the same. Just like on Earth.

What will matter are the terms in the centrifugal force equation you cited: the mass of whatever's being weighed (i.e. a person, or a vehicle, or whatever else you may be interested in that's sitting inside the cylinder- this term does not refer to the mass of the cylinder itself), the angular velocity that the cylinder is spinning at (in revolutions per minute, or radians per second, or whatever units you find convenient), and the radius of the cylinder.

There are two ways in which an object (with a fixed mass) may be able to experience different amounts of gravity in different parts of an O'Neill cylinder. These both effectively boil down to changing the angular velocity or radius terms in the equation.

First, if there's a train or roadway or other such vehicle route encircling the inside surface of the cylinder, the velocity of the vehicle will add to the angular velocity of the cylinder. If the vehicle is traveling in the same direction that the cylinder is rotating, then any passengers and cargo will experience increased gravity for the duration of the trip, proportional to the speed of the vehicle. If the vehicle travels in the opposite direction of the cylinder's rotation, the passengers and cargo will feel less gravity.

In the special case that the vehicle is moving at exactly the same speed that the cylinder is rotating, but in the opposite direction, its passengers will feel no gravity at all. An observer outside the cylinder will see the cylinder rotate around the vehicle, while the vehicle itself stays perfectly still.

Second, if there are any tall buildings inside the cylinder, anything on the upper floors will experience less gravity than on the lower floors. The radius term refers to the distance between the center of rotation (i.e. the central axis of the cylinder) and whatever is being measured. If you're standing on the inside surface of the outer hull, then your r is practically the same as the radius of the cylinder. If you're standing on the top floor of a tall building, significantly closer to the cylinder's central axis, then your r will be smaller, and you will experience proportionally less gravity.

There are also some tricks you can play involving Coriolis forces, but those only kick in when you're moving at some velocity that isn't stationary in the cylinder's reference frame. Elevators may seem to tilt to the side if they ascend or descend fast enough; aircraft flying inside the cylinder may tend to follow some very counterintuitive trajectories; and long-range sharpshooting across the middle of the cylinder will be almost impossible without computer targeting assistance.

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  • $\begingroup$ A really refreshing approach to the characteristics of centripetal force in a rotating cylinder. Missing is that centripetal force does NOT act perpendicular to the cylinder, but at a tangent to it. On a frictionless surface, the object will travel 'sideways' along the surface, opposite the direction of rotation. A ball bearing, for instance, will rotate around the inside surface of the cylinder. The air mass rotating with the surface will tend to counter this effect, if there are walls 'pushing' the air and applying momentum to it. $\endgroup$ Nov 28, 2021 at 4:11
  • $\begingroup$ "centripetal force does NOT act perpendicular to the cylinder, but at a tangent to it" ... uh, no? "Centripetal" means "center-seeking". When an object moves in a circle, whatever force is causing it to do that is called "the centripetal force", and is always directed toward the center of the circle, perpendicular to the object's path. For a ball on a string, it's the tension force in the string; for person standing in an O'Neill cylinder, it's the normal force between their feet and the floor. $\endgroup$ Nov 28, 2021 at 4:21
  • $\begingroup$ Inertia says the object will continue to move in a straight line unless acted upon by an outside force. The object is traveling at right angles to the perpendicular of the cylinder, around an orbit. The applied force accelerates around the orbit instead of a straight line in a tangent perpendicular to the cylinder. This accelerating force is applied by the floor of the cylinder. Remove the cylinder, and the object will NOT travel perpendicular to the cylinder, but at a tangent to the orbit. The person is NOT 'pressed' against the floor, but accelerated around the orbit. $\endgroup$ Nov 28, 2021 at 4:31
  • $\begingroup$ The cylinder keeps moving in front of the person's direction of travel, and thus the person keeps colliding with the interior surface of the cylinder. $\endgroup$ Nov 28, 2021 at 4:31
  • $\begingroup$ If the person were holding on to a rope attached to the axis of the cylinder, the person would be stretched, not compressed, and the rope would be in tension. The feet would want to keep moving away from the hands holding the string. In a cylinder, because the accelerating force is applied by the floor, and not a rope, the person is compressed or pushed against the floor and the floor is in compression with the person. The tension is in the cylinder walls preventing the cylinder wall from breaking up. $\endgroup$ Nov 28, 2021 at 4:41
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An enclosed room would have no effect on the pseydo-gravitational pull, and the origin will, indeed, always be the rotational axis of the cylinder. Air pressure will likewise have no effect on the forces involved, though it may have an effect on the individuals present in the enclosed room.

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    $\begingroup$ I believe the air pressure would have an effect on the mass of the air pushing you. And therefore play a part in the equation. $\endgroup$ Nov 28, 2021 at 3:54
  • $\begingroup$ The point about the mass of the air pushing down on the person is an interesting concept. It would not affect gravity, but it WOULD affect the pressures on the person. However, the question is specifically about gravity. $\endgroup$ Nov 28, 2021 at 4:02
  • $\begingroup$ The question is about Pseudo-gravity, specifically one caused by the Centrifugal effect, which has mass as one of the variables of the equation. If that variable is set to zero (ie: a vaccum) then there is no Pseudo-gravity. If the mass of the air doesn't factor into the calculations, then what is the "mass" variable of the equation? $\endgroup$ Nov 28, 2021 at 4:32
  • $\begingroup$ The m term in the equation is the mass of whatever you're weighing. If you plug the mass of your body, the radius of the deck you're standing on (i.e. the distance between you and the cylinder's central axis), and the angular velocity of the cylinder into that equation, and then step on a bathroom scale, you should get the same number on the scale and out of the equation. $\endgroup$ Nov 28, 2021 at 4:46
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    $\begingroup$ As Someone Else said, m is the mass of the object under consideration — the one you’re calculating F for. If it’s a person, it’s the mass of the person. Etc. $\endgroup$
    – Daniel B
    Nov 28, 2021 at 5:08

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