# Why is the helicopter in Rama (Arthur C. Clarke) subject to gravity?

In "Rendez-vous with Rama" by Arthur C. Clarke, an artificial gravity force is maintained in the O'Neil cylinder (thanks to its spin). However, according to my understanding of the centrifugal force, only objects bound to the cylinder and thus having a forced rotational velocity will be subject to this virtual gravity.

So, why does the helicopter used by the team to explore Rama undergoes gravity while "flying"?

[EDIT, after my discussion with @Hypnosifl]

A more precise question is: What about the 'gravity' on the helicopter if they move parallel to the central axis, with a velocity constant and parallel to the central axis, at any distance of the surface?

"constant velocity" = as seen by an inertial observer who wasn't rotating with the cylinder (meaning that it would seem to travel on a helix-shaped path by people rotating with the inner cylinder). (as per @Hypnosifls words)

• Helicopter? I remember a man-powered airplane....and it was a privately owned machine. – Organic Marble May 12 '17 at 3:50
• Yes, a "helicopter" since I don't a better name in English (I've read a translation of it, and did not find the name used in the original version). "man-powered" or not, I think, does not have incidence on the question or the answer. And the same for "privately owned" AFAIK. – Evariste May 12 '17 at 8:05

In order to hover (approximately) in place over the surface of the rotating habitat, the helicopter does need to exert a force "upwards", i.e. towards the axis of rotation.

This is because the habitat is rotating, which means that its surface is constantly moving around the axis. But the surface is also constantly being pulled towards the axis by the habitat structure supporting it — if it weren't, the habitat would simply fly apart. This centripetal force exerted by the habitat structure keeps the habitat together and the surface moving in a circle.

In order to stay above a particular spot on the surface, the helicopter must also move in the direction of the rotation at the same speed as the surface (or, actually, slightly more slowly, since it will be closer to the axis). But since the helicopter is not supported by the habitat structure, it must provide its own centripetal force (e.g. by thrusting against the surrounding air). If it did not, the helicopter would keep moving straight on a free-fall trajectory, which would quickly lead to a collision with the habitat wall: OK, you ask, but what if the helicopter did not try to stay above a particular spot on the surface, but simply floated in space at a constant distance and direction from the axis, letting the surface spin beneath it?

Well, if the interior of the habitat was in vacuum, this would actually work. (Of course, helicopters are kind of useless in vacuum, so in that case you'd better take a spacecraft instead.) However, if the habitat cylinder is filled with air (which it needs to be, to allow helicopter flight — not to mention letting the inhabitants breathe), there's a problem: wind.

You see, the air in a rotating habitat will also be rotating at (approximately) the same speed as the surface. How fast is that?

Well, according to Wikipedia, Rama has an interior radius of about 8 km (making its circumference a little over 25 km) and a rotation period of 4 minutes. Thus, the interior surface is moving around the axis at about 25 km / 4 minutes, or 375 km/h (= 233 mph). Near the surface, the air will also be moving at about this speed, so a low-flying helicopter trying not to follow the surface, but to remain stationary with respect to the central axis, would have to fight a 375 km/h headwind.

The current world record for fastest helicopter flight is a little over 400 km/h (249 mph), so this actually does seem to be (barely) possible, even using current technology. Besides, going higher up towards the axis would not only reduce the required speed, but would presumably also lower the air density, and thus drag. Yet, even so, the power needed to achieve such extreme airspeeds would certainly be much greater than that needed for simply hovering over the surface.

(That said, if you did have a big rotating space station and a fast helicopter, and managed to pull off this trick, you'd observe something interesting: when flying anti-spinward at exactly the right speed to cancel out the rotation of the surrounding air, the helicopter would be able to remain aloft even with the propeller tilted sideways, at a 90° angle to the surface. Also, anyone inside the helicopter would effectively be in free-fall.)

• I suspect that a "speed" of 375 km/h would be easier to achieve for a helicopter which only has wind speed to worry about, and not the effects of gravity. – Ajedi32 Dec 1 '14 at 15:03

This is really more of a physics issue, see my answer on the physics stack exchange here--objects away from the ground in a rotating cylinder will still appear to be subject to gravity, their inertial paths will naturally cause them to come crashing to the inner surface of the cylinder (similarly, see this page for a short explanation of why someone jumping off a cliff on Rama would appear to fall to the ground like on Earth). And if the helicopter wanted to hover above a fixed point on the inner surface, it would itself need to rotate in a circle, causing those inside to experience a centrifugal force. Rama is supposed to have a inner diameter of 16 km according to the wikipedia page, so the center would be 8 km above the surface--assuming the helicopter flies at a height much less than this, the difference in centrifugal force due to it traveling on a smaller circle will be minimal.

If the helicopter has some significant velocity relative to the inner surface, then it will experience departures from the behavior of gravity on Earth, in the form of the Coriolis effect. In physics terms, if you want to use a rotating reference frame to analyze motion (a coordinate system in which points on the inner surface of the cylinder have position coordinates that don't change with time), the only differences with an inertial reference frame would be the need to assume two fictitious forces, the "centrifugal force" (the radial force that simulates gravity) and the "coriolis force". The centrifugal force is given by the equation mv^2/r, so the acceleration is v^2/r; as mentioned above Rama has an inner radius of 8 km = 8000 m, and the "Plot summary" section of the Rendezvous with Rama wiki page mentions the rotation period is 4 minutes, so the velocity of a point on the inner surface must be 2*pi*(8000 m)/(240 s) = 209.44 m/s, so the acceleration at the inner surface is (209.44 m/s)^2 / (8000 m) = (43865 m^2/s^2)/(8000 m) = 5.483 m/s^2 (a little over half Earth's gravitational acceleration of 9.8 m/s^2). Likewise, for a helicopter hovering at a height D meters above a point on the surface the velocity would be v = 2*pi*(8000 - D m)/(240 s) and the acceleration would be v^2/r = v^2/(8000 - D m) = 4*pi^2*(8000 - D m)/(240 s)^2 = 0.0006854*(8000 - D m)/s^2. In both cases, the centrifugal force acceleration is a vector pointing straight down in the rotating frame, which needs to be added to the vector for the acceleration due to the Coriolis force to find the actual acceleration of the helicopter in the rotating frame (see vector addition).

The acceleration due to the Coriolis force is a little more complicated to figure out, but if you're interested, the equation is given by -2*(Omega x v). Here Omega is a vector whose magnitude is equal to the angular rotation rate of Rama, which is 2*pi radians/(4 minutes) = 0.02618 radians/second, and its direction is defined to be along the axis of rotation of Rama (the central axis of the cylinder). And (Omega x v) is a vector formed by taking the cross product of Omega with the velocity vector of whatever moving object you're considering, like the helicopter. The magnitude of the cross product just be the product of the magnitudes of Omega and v times the sine of the angle between them, so if the helicopter's velocity is parallel to the central axis of the cylinder then the Coriolis force is 0 because sine(0) = 0, whereas if the helicopter's velocity is at a 90 degree angle relative to the central axis of the cylinder (it's traveling straight up or straight down or along the inner circumference of the cylinder), then since sine(90)=1, the magnitude of the Coriolis acceleration can just be found by multiplying the magnitude of Omega (0.02618 radians/second) and the magnitude of the velocity, and then multiplying by 2.

Just to pick an example, if the helicopter is traveling at 45 meters/second (162 km/hr or about 101 mph) in a direction along the circumference of the inner surface, the acceleration due to the Coriolis force would have magnitude 2 * (0.02618/second) * (45 meters/second) = 2.356 m/s^2, and the direction would in this case be either straight "up" or straight "down", depending on the details of whether the station was rotating clockwise or counterclockwise and whether the helicopter's direction of travel on the inner circumference was clockwise or counterclockwise. So when this vector was added to the centrifugal force vector, the people in the helicopter would feel either slightly heavier or slightly lighter than if the helicopter was just hovering over a fixed point on the inner surface with zero velocity in the rotating frame--if they were flying close to the ground where the centrifugal acceleration is 5.483 m/s^2, they might feel as light as 5.483 m/s^2 - 2.356 m/s^2 = 3.127 m/s^2, or as heavy as 5.483 m/s^2 + 2.356 m/s^2 = 7.839 m/s^2. But they would have to fly considerably faster than modern helicopters typically fly in order for the Coriolis force to completely cancel the centrifugal force and make them feel completely weightless.

Note that adding the centrifugal and Coriolis vectors only gives you the acceleration in the rotating frame of Rama's surface, which may not be the same as the acceleration measured on board the helicopter. As an analogy, for a falling helicopter on Earth the acceleration in the Earth frame is 9.8 m/s^2, but people on board the helicopter will feel weightless (zero proper acceleration as measured by an accelerometer on board) as long as the helicopter is in free-fall. But as long as the helicopter maintains a velocity in the rotating frame that's close to constant, the acceleration in the rotating frame should match what's felt by those on board (and the helicopter's velocity should naturally tend to be fairly constant as long as its rotors are spinning since that will give it a fairly constant velocity relative to the air, which means a relatively constant velocity relative to the ground due to drag between the air and the ground, so Motti's answer is correct in that sense).

• That's a lot more than I expected @Hypnosifl! However, I understand you assume the helicopter travels with a radial velocity. What if (as in the novel), the helicopter "takes off" from the axis of the cylinder. It will spin at the same rotation speed than the cylinder, but will have no velocity in a direction along the circumference... If it comes near the surface (thanks to the pilot's efforts), what would it undergo more and more gravity? – Evariste Nov 30 '14 at 15:26
• Can you elaborate on what you mean by "takes off from the axis of the cylinder"? When I talked about the cylinder's central axis I was talking about the axis of rotation through the center of the cylinder, 10 km away from the "ground"--do you mean it's taking off from there, or taking off from the ground but moving parallel to that axis, or are you thinking of a radial axis from the ground to the center rather than the rotation axis? – Hypnosifl Nov 30 '14 at 15:36
• Yes, they take off from the cylinder's central axis , 10km from the "ground". Later in the book, they take off from the ground itself. – Evariste Nov 30 '14 at 15:40
• So the question is: "What about the 'gravity' on the helicopter if they move parallel to that central axis, anywhere in the cylinder, with no initial radial velocity?". – Evariste Nov 30 '14 at 15:46
• At that point the centrifugal force should be zero, so the only force in the rotating frame would be the Coriolis force, and if they keep traveling along the central axis that won't change. But when you say "no initial radial velocity", are you also asking what they'll experience if the pilot later creates some radial velocity so they move towards the ground? – Hypnosifl Nov 30 '14 at 15:47

If I'm not mistaken the cause is that the atmosphere in the cylinder will be rotating together with the inner surface of the cylinder (due to drag). Therefore even an air-born object will still have rotational velocity since it's dragged along with the air surrounding it.

• Even in a vacuum, if you analyze things in a rotating reference frame the total apparent force felt on the helicopter will be a combination of the centrifugal force and the coriolis force, see my answer for details. – Hypnosifl Nov 30 '14 at 15:00
• I started answering this but it appears that @Hypnosifl gave a much better answer (he put some actual science to my intuition) – Motti Nov 30 '14 at 15:01
• I also added one last paragraph to my answer after I realized why the drag between the cylinder and the air is still quite important to understanding why people on board the helicopter tend to feel an "artificial gravity" force--in a vacuum an object away from the surface would tend to be accelerating significantly in the rotating frame, which would mean the apparent force felt by the object would differ significantly from the fictitious force in the rotating frame of Rama's surface. – Hypnosifl Nov 30 '14 at 15:21
• I remember an ultra-light human-powered airplane, but no helicopter. One of the later books, maybe? – Joe L. Nov 30 '14 at 15:29
• Sorry, I read the book in French and was not sure about the original word used by Clarke. In French, it's "libellule" (that we can translate as "dragon-fly" in English), but I assumed "dragon-fly" was not the word in the English version. So I used "helicopter" to simplify as a first approximation. – Evariste Nov 30 '14 at 15:35