In the previous post, Gravity in space, I've talked about a way to “create” gravity inside a ship. The ship must be rotating someway, that's it. The fake gravity is the conseguence of the centripetal force exerted on you by the inner surface of the cylinder or torus — when an object is moving in circle, you know there must be a force which “disturbs” its inertia and makes the object steering.
An interesting and important property is that the more you go near the rotation axis, the weaker the “gravity” becomes. On the axis the effect disappear. It's like being in space, but you escape gravity by going towards the rotation axis. In the case of the torus, you can't escape, because the rotation axis can't be reached from inside the ship.
Yes, there are other ways to artificial gravity. If there are no masses, the illusion of gravity must be given by something else. One option is the exploitation of dynamics effects, as seen. Say thank you to acceleration. Acceleration is what makes your velocity vector change: it can become “shorter” (you're slowing down) or “longer” (you're going faster), or it can stay the same length, but change direction, or it can do both: changing in length and direction. The solid curved ground of the ship does this to us: it changes the direction of our velocity vector (only the direction). It's what the centripetal force does, if you remember Newton — F = ma, the force exerted on an object is equal to the (inertial) mass of the object multiplied by the acceleration (the “rate” at which its velocity vector change). In the general case F and a are vectors. We are rotating altogether with the cylinder; that is, our velocity vector (seen from a place which isn't rotating altogether with the cylinder), tangent to the surface which determines our circular trajectory, is rotating. (Instead from your point of view, you are at rest, unless you're running.)
Now, is that the only way we can apply a force? Of course not. Take again a look at the Newton's second law of motion, F = ma. Where's the rotation? Not hardwired! It only says that there's an acceleration. So the secret is this: acceleration. Here it is our second option: we can keep a constant acceleration in a specific direction, and the bodies inside the ship would believe there's a gravity force “pulling” them in the opposite direction — provided that they stay on a floor perpendicular to the direction of the acceleration. It's like being into an elevator in the space. (On earth, a free falling elevator gives the emotion of no-gravity, just before dying in the impact when it hits the soil.)
The acceleration should be near to 10 m/s2, since gravity on earth is a little bit less than that (but lower values could be ok to make humans comfortable), and it's not a constant on all over the globe, although it doesn't change too much.
The problem is that your ship must be always accelerating. In a specific direction. It means its speed increases. Always. After 10 seconds, you are travelling at 100 m/s if you began from 0. If you want to stay around a stellar system, you must “steer”, i.e. accelerating so to change the direction of the velocity vector. You want to do it with very small “lateral” acceleration, in order to avoid to shake the passengers. But the longer the ship is traveling, the faster it's going, and so small “swerves” will result in larger and larger curved trajectory.
This way to achieve artificial gravity is ok only to give (artificial) gravity in long journeys, for instance aiming towards Trappist system. If you need orbiting around a planet, you will dislike this method very much.
There's another limit to this method. You'll be going faster and faster, and you'll start to deal with relativistic effects. The most important thing here is that the energy you need to keep that constant acceleration (and so the artificial gravity inside the ship) isn't itself constant, because of the relativistic effect on the mass of the ship. Anyway, before going relativistic, I rule this method out because it doesn't suit the idea of space station orbiting around a planet.
I've briefly read about another method which uses magnetic fields. I don't know anything about it, but I rule it out, too: before designing my spaceship that uses that technology, I would like to read a lot of literature on the effects of exposition to strong magnetic fields for a very long period of time.
So, the easier and more obvious way to obtain artificial gravity in space is the one I've already explained empirically, and its ground is well known non-relativistic mechanics.
Rama of Randezvous with Rama is a simple giant rotating cylinder, that's why inside it resembles a world where gravity exists. It is 54 km long — but this isn't important — and has a diameter of 20 km. This is important, altogether with the speed of rotation, which is, according to Wikipedia, 0.25 rpm.
The “gravity” on the ground — that is, if you are on the inner surface of a cylinder with its internal diameter of 20 km and rotating at 0.25 rpm — is a little bit less than 7 m/s2. The gravity inside Rama is less than the gravity on earth, but not too much smaller.
Of course the walls which separate you from the external space have their thickness, so the cylinder seen by outside must have a diameter bigger than 20 km, or viceversa, the internal cylinder diameter is smaller than 20 km. I've supposed the former is true.
I'm not going to give a full explanation but I will drop some useful formulæ.
The circular motion is the key, and in particular we like very much the uniform circular motion.
Let's begin with the angular frequency.
The speed says us the amount of space a body travels per unit of time. When you say 1 m/s, what you mean is that in a whole second you advance of a single meter. Angular frequency is the same concept, but applied to angle instead of “space”. The International System of unit for angles is the radian, not the degree. Hence the angular speed is measured in radians per second, rad/s.
You can convert between radians and degrees easily, consider that π rad = 180°. If you have α rad and you want to know it in degree:
If you reason in degree but need the value in radians (as trigonometric functions in many programming languages require),
If I say that the angular speed is 3.14 rad/s, I'm saying that in a single second our object swiped an angle of 3.14 radians (use the formula above to discover that it is almost 180°). A complete revolution is 2π — this is the value, in radians, of a round angle. Therefore an angular frequency of 2π rad/s means that the object complete a whole revolution in a single second. It's 1 revolution per second. Since there are 60 seconds in a minute, 2π rad/s is 60 rpm. This is the formula:
Rama rotates at 0.25 rpm. It's 1/4 of a whole revolution in a minute, i.e. 2π/4 radians per minute, that is ω = π/120 rad/s. (In the formula I've put a conversion of unit — minutes in stead of seconds —, nothing thrilling, but it can be useful.)
Let's take a point on the surface of the cylinder. We can “reduce” it to the motion of a point on a circle, a flat thing. Because the motion we're interested in happens all in a plane; so we can stick to a 2D world.
The perimeter of a circle is 2πr, with r being its radius. It means that when a point complete a revolution in a second, i.e. it goes at the speed of 2π rad/s, the point travelled for the whole circle, that is its perimeter, which is 2πr. And the speed is the space divided by the time, so
This makes a relationship between the speed of our point v, the radius r of the circle and the angular frequency ω. What's that exactly? It's the modulo (the “length”) of the speed vector of our point.
We need the acceleration that makes that speed vector rotate of the right amount. This acceleration vector is orthogonal to the speed vector. This is what we have each time an acceleration causes only a rotation of a vector, without changing its length (and our length is kept constant since our speed is constant). Such an acceleration is our centripetal (it points towards the center of the curve) acceleration. Our gravity.
It's formula is simple this:
Let's use these equations plugging the data of Rama. We know that it rotates at 0.25 rpm, revolutions per minute. We want this as rad/s. I've given the formula above, so: 2π/60 × 0.25 = 0.026 rad/s. This is ω. In the SI we use meters, not kilometers, so Rama has a diameter of 20000 meters, i.e. a radius of 10000 meters. Let's plug these numebers:
(I've used more digits, therefore a higher precision, hence the “...” for the other digits)
This is the “gravity” of Rama when you stay 10 km far from the rotation axis.
In order to give a nice gravity experience, and in the same time a good illusion to stay on a flat ground, we must choose radius and revolution period (the time needed to complete a revolution) with care.
