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I'm currently reading Arthur Clarke's Rendezvous with Rama. Rama is a 50 km long, 16 km diameter cylindrical alien artifact that is rotating fast enough to provide a 0.6G artificial gravity on its long inner surface. The characters in the book enter Rama through an air lock at the hub at one end of the cylinder's central axis, where they are weightless; they then descend a series of ladders and stairs, getting heavier as they go, until they reach full weight at the surface.

One character has a flying contraption with him. He sets off from the hub, weightless, with the intent to ride along the axis the entire length of Rama. He is warned not to drop down towards the surface, as lowering at all will increase the weight (or centrifugal force) he feels. But I'm not sure if this is true.

Centrifugal force is not like gravity, which reaches out to pull you down. If there is nothing pushing him along the rotation, he shouldn't feel any centrifugal force at all, right? Of course, the air itself is going to be moving along with Rama's rotation, but would the air near the hub be moving fast enough to exert a serious sideways force on him, causing him to take part in Rama's rotation? Theoretically, if there was no air, he could lower himself almost all the way to the surface (which from his vantage point would be moving by very fast), and still remain weightless the entire time.

So, am I thinking about this entirely wrong? Or is Clarke?

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    $\begingroup$ The air inside is rotating along with the cylinder. The "flying contraption" is embedded in the air. You asked, "but would the air near the hub be moving fast enough to exert a serious sideways force...?" No. It wouldn't. That is the reason the character was told to stay near the center line. If he flew too close to the "ground", then the air in that region would be moving fast enough... $\endgroup$
    – Solomon Slow
    May 25, 2016 at 17:23
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    $\begingroup$ FWIW, to achieve 0.6g acceleration, the cylinder needs to do one revolution in approx 231 seconds, so the speed at the surface is approx 217 m/s = 781 km/h. $\endgroup$
    – PM 2Ring
    Dec 3, 2018 at 2:12
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    $\begingroup$ Having just read this bit I too am unsure about the physics displayed. TBF the only other example I have of this sort of setup in this way is Babylon 5's S2 finalé. In B5 there is a train that runs down the centre of the cylinder in the zero-gravity area and someone blows it up leaving Sheridan floating out towards the ground. The issue is that he is effectively weightless and moving towards a ground that is spinning at high speed so that even though he's moving relatively slowly, when he impacts he'll definitely die. No thought was given to the air movement by JMS. Reading Rama the use [...] $\endgroup$
    – TheoGB
    Oct 20, 2022 at 10:00
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    $\begingroup$ [...] of the term gravity definitely feels wrong since really he is presumably meaning the rotational force from the air. The whole thing is presented as if the ground is moving relative to the flier but I feel like that couldn't possibly be the case since fluid dynamics will affect the air. The whole section read wrongly to me and the answers here haven't really provided the maths to convince me that Clarke's correct. $\endgroup$
    – SuperCiocia ♦
    Oct 21, 2022 at 1:00

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It's been a while, since I read the book, but I believe that his flying contraption runs at low air speed. And of course the atmosphere rotates with the can (give or take a little).

That being the case as he moves out he picks up roughly the same speed that he would have standing on a surface at the same "height" and consequently experiences roughly the same pseudo-gravity. He can alter that a little be flying spinward for extra pseudo-gravity or anti-spinward for less. But low air-speed makes this a relatively small effect.

Next you should ask yourself how a device designed to fly in a gravity field will maneuver in the absence of one. Or more importantly how a pilot used to flying with gravity would handle the situation. I'm betting that he'd intentionally move down to where he had enough pseudo-gravity to orient the craft. That shouldn't take a lot and he can adjust his control surfaces for less lift but at least the craft will fly like he expects it to.

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    $\begingroup$ My recollection from the novel is that the pilot does have to "descend" a bit at one point because the gravity at his "altitude" is too weak and it's not handling the way he expects it to. $\endgroup$
    – Michael Seifert
    Oct 20, 2022 at 11:54
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I just found your question because I was thinking the same, as I read the passage about 'Dragonfly.' As Dmckee said, there must be aerodynamic effects from the rotation (the air-bike wouldn't move otherwise), but it doesn't make sense that he'd feel .3g or whatever, since he's not in a gravity well. It would be more like scuba diving in a pool with a current, or a slow-moving river. You only feel "centrifugal forces" when something is changing the direction of your linear acceleration, which, in this case, nothing is.

When the crew is ascending the staircases, they WOULD indeed feel a pseudo-gravitational force vector from that effect, that would lessen as they approached the axis of rotation.

They could also float all the way back down to the external surface from the 'pole,' but they'd be in for some road-rash when they made contact.

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With no air, you are correct.

The air makes this entirely different. I don't recall mention of there being heavy winds near the surface, meaning that the air is moving with the cylinder. The air is accelerating outward ("downward"), and the dragonfly would be accelerated with it. Along the hub, the air would not be moving much, and there would be a gradient of steadily increasing wind-speed as you move away from the hub.

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  • $\begingroup$ Why do you say that the air is accelerating outward? $\endgroup$
    – PM 2Ring
    Dec 3, 2018 at 2:27
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In addition,

If on the ground, running one direction would be very hard, and the other direction might cause you to come off the floor.

So,

Flying with rotation, very slowly, would completely zero out the friction of air rotation flow, if that happens at all. Literally at all? Air flow from wind would like be farcin excess of rotational flow.

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    $\begingroup$ Note that there is an "edit" button so that you don't have to post a second answer to the same question. $\endgroup$
    – Kyle Kanos
    Dec 2, 2018 at 18:23
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You are the one thinking about it incorrectly. Let's start off with your mistake. Why would the wind NOT be moving at EXACTLY that same speed that the cylinder rotates? Are there ANY forces at all that would slow it down? No, there are not. The only friction in the system is from the walls of the cylinder. So the only forces involved will KEEP the air spinning in sync with the cylinder, and nothing is acting to slow the air down. This means that the air will create a constant gradient with the radius of the cylinder as if you were touching the solid ground at any point in the process. So Arthur got the physics exactly correct, and your problem was framing Rama as if it were part of a larger system, and not the entirety of the system as it actually is.

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