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For the sake of fixing some image in your mind, imagine you want to practice some sport in a rotating cylinder world: whether it be launching a javelin, strike a tee at the golf club or scoring a 3 points shot on the ball field, some sort of ballistic trajectory will be involved in most of the cases.

On Earth we know that, if we neglect the interaction of the object with the air and we are below escape velocity, the trajectory will be an elliptic arc. In a rotating cylinder world I think the apparent gravitation field would be different than on Earth, I even doubt it could be even called a "field".

How would that work on a rotating cylinder world?

What are the ballistic trajectories in a ring world?

For the sake of helping the calculation, if needed, assume

  • a cylindrical world, with 1 km radius, rotating at 0.95 rotation/minute.
  • neglect drag and aerodynamic effects (Coanda effect, lift, etc.), thus assume the launch is happening in a vacuum
  • arbitrary direction and velocity of launch
    • neglect the real gravity due to the cylinder mass

Alongside with the mathematical relationships, I would also appreciate a graphical comparison with respect to the Earth case.

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    $\begingroup$ Sadly I don’t have time for a full answer, but I suspect the keyword here will be Coriolis. $\endgroup$ – Joe Bloggs Jul 5 at 8:05
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    $\begingroup$ @JoeBloggs, take your time for an answer. I am not in a hurry. $\endgroup$ – L.Dutch - Reinstate Monica Jul 5 at 8:06
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    $\begingroup$ @Cyn, I don't think neither of the two applies here. $\endgroup$ – L.Dutch - Reinstate Monica Jul 6 at 3:39
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    $\begingroup$ "ring world" confused me with Ringworld vibes—i.e., a world with radius on the order of AUs. "O'Neill Cylinder" is probably the best name for these much-smaller habitats. $\endgroup$ – imallett Jul 7 at 5:17
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    $\begingroup$ You switch between "ring world" and "cylinder" in your question, which makes answering difficult. The inner surface of a ring world can have no perceptible curvature, for example. $\endgroup$ – rek Jul 7 at 12:25
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Whenever you're wondering how particles would move in some sort of accelerated reference frame, you have two options: analyze it from an inertial frame or analyze it from the accelerated one. They each have their own strengths and weaknesses. Analysis from an inertial frame requires nothing more than good ol' Newton's laws, but requires you to keep track of some oftentimes confusing coordinate transformations, while using an accelerated frame lets you ditch the transformations but requires you to add new 'fictitious' forces-- in the case of a uniformly rotating frame like this one, these would be the Coriolis and centrifugal forces. The latter approach can be good for getting a rough intuitive idea of what will happen and can be useful in calculations for a trajectory that has no analytical solution. Fortunately, with the constraints you've imposed, there actually exists a relatively simple analytical solution, so we'll analyze this from an inertial perspective.

First things first, we need to set up our inertial and accelerated coordinates so that we can relate them and transform our understanding in an inertial frame to one in our accelerated frame. Let's set up our inertial coordinates so that the origin is at the bottom of the cylinder with the z axis pointing radially inward, the y axis tangential to the cylinder wall, and the x axis outward along the axis of the cylinder. The origin and orientation of the inertial coordinates do not change over time. Meanwhile, the accelerated coordinates are denoted by primes, and point in the same direction at $t=0$ but unlike the inertial coordinates, rotate along with the cylinder. Below is a diagram that hopefully makes this clearer:

enter image description here

Here, $\omega$ (technically w in the picture because I couldn't figure out how to get greek letters in my drawings) is the angular velocity of the cylinder and is pointed in the x direction along the cylinder axis. Let $R$ denote the cylinder radius. Now, the conversion between our accelerated and inertial frames is the composition of a translation and a rotation. It's difficult to wrap your head around without fiddling around with it for a bit, so I'll just leave the result:

$$\mathbf{x}=A\mathbf{x'+b}$$

Where $\mathbf{x}$ are the inertial coordinates, $\mathbf{x'}$ are the accelerated ones,

$$A= \begin{bmatrix} 1 & 0 & 0 \\ 0 & cos(\omega t) & -sin(\omega t) \\ 0 & sin(\omega t) & cos(\omega t) \end{bmatrix}$$

$$\mathbf{b}= \begin{bmatrix} 0 \\ Rsin(\omega t) \\ R(1-cos(\omega t)) \end{bmatrix}$$

Now, the accelerated coordinates are the ones we care about, since they describe how stuff looks from the viewpoint of someone on the cylinder. So, we invert the prior equation to obtain

$$\mathbf{x'}=A^{-1}\mathbf{(x-b)}$$

Now the hard part's over. All that's left is to note that a particle will travel in a straight line in the inertial frame, since we are ignoring any gravitational effects. If the initial velocity of the projectile is $(v_x,v_y,v_z)$ as measured in the accelerated frame, it is straightforward to show that

$$\mathbf{x}= \begin{bmatrix} v_{x}t \\ (v_{y}+R\omega)t \\ v_{z}t \end{bmatrix}$$

By imposing the constraint that at the time of impact, $y^2 + (z-R)^2 = R^2$, we can determine that

$$t_{impact} = \frac{2Rv_z}{v_z^2+(v_y+R\omega)^2}$$ Note that if $v_y=0$, then in the limit of small $v_z$ we have $t_{impact}=(2v_z)/(R \omega^2)$, which agrees with the prediction of a parabolic trajectory under uniform gravity with an acceleration of $R\omega^2$.

When you finally do all of the matrix multiplication, you end up with the parametric equation:

$$\mathbf{x'}= \begin{bmatrix} v_{x}t \\ [(v_{y}+R\omega)t-Rsin(\omega t)]cos(\omega t) + [v_{z}t+R(cos(\omega t) -1)]sin(\omega t) \\ -[(v_{y}+R\omega)t-Rsin(\omega t)]sin(\omega t) + [v_{z}t+R(cos(\omega t) -1)]cos(\omega t) \\ \end{bmatrix}$$

When I have more time later, I'll add some graphs in here that demonstrate what this looks like for various trajectories, but the general gist is: velocity components that angle along the axis of the cylinder don't have much of an effect on trajectory, while those that are angled along the -y direction can have an immense impact at speeds on the order of 100 m/s given the radius and angular velocity specified in your question.

EDIT: Trajectory Graphs

First things first, it's important to note that what trajectories look like for this problem depend heavily on the initial velocity of the projectile, so I recommend playing around with some graphs yourself to get a more complete understanding-- desmos is a good tool for this. 3D plots of the trajectory are a little busy, so instead I've plotted the projections of the trajectory into the x'z' and y'z' planes. For all plots, the solid blue line is the particle trajectory as seen from the rotating frame, while the dashed red line is the parabolic trajectory expected for a uniform gravitational field of magnitude $R\omega^2$. On the y'z' plane plots, the dotted black line gives the location of the cylinder wall.

First things first, what happens if you lob a ball directly upward at about $10 m/s$? Well, in a normal gravitational field, that's easy-- it should go straight up and down. However, in our cylinder world, the Coriolis force will push it in the y' direction. It still goes straight up and down in the x'z' plane, so we will omit that graph because it's boring. You can see that the ball moves about a meter in the y' direction, which probably wouldn't be very noticeable since it's hard to throw something directly upwards anyway.

So, now let's try throwing it a bit harder. It wouldn't be too crazy to throw a ball at $20 m/s$, so let's do that at an angle of 30 degrees with respect to the horizontal in the positive y' direction. We again ignore the x'z' trajectory because it is trivial. We can see that in this scenario, our trajectory still looks parabolic but is noticeably shorter that what we would expect from a uniform gravitational field.

enter image description here

If we throw the ball with the same speed and angle but in the -y' direction, the ball instead travels much further than we would expect.

enter image description here

If we throw the ball at the same speed but in the x' direction, the projection of the trajectory in the y'z' plane looks exactly the same as it did for when we threw the ball straight up, since the Coriolis force does not act on x' components of velocity. Meanwhile, the x'z' projection looks very similar to a normal ballistic trajectory:

enter image description here

Let's ratchet things up a notch and get a professional to throw a $40 m/s$ fastball again at 30 degrees from the horizontal in the y' direction. By now, you'd definitely notice something was up-- the ball is only going about half the distance you'd expect:

enter image description here

If you throw it with the same speed but in the -y direction, not only does it go over twice the distance you'd expect, but its trajectory now noticeably deviates from a parabola:

enter image description here

At this point, we might as well go for broke and get Ryan Winther to hit us a $100 m/s$ golf drive at 6 degrees from the horizontal. If he decides to hit the ball in the -y direction, he's in for an interesting hole:

enter image description here

This is about the peak for strange deviations to trajectories. If you continue shoot a projectile with greater and greater speeds, it will hit the cylinder so quickly that the accelerated frame will not have time to deviate significantly from the inertial one and so the trajectory will approach a straight line. For instance, if we shoot a railgun directly upwards at $3 km/s$, its trajectory looks like:

enter image description here

To summarize: things thrown solely in the x' direction will have fairly normal trajectories, as will things thrown in the y' direction at either very small or very large speeds. However, at intermediate speeds on the order of $100 m/s$, projectiles shot in the y' direction will land short an projectiles shot in the -y' direction will go much further than expected.

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This is actually easy to calculate. There is no "gravity", so you just sum your starting velocity with your launch velocity. This makes your launch vector a straight line.

Here's what makes it confusing: While someone from space would see a straight line, an observer on an adequately large cylinder world would see an apparently normal ballistics curve. This is because they will move with the projectile starting at the exact speed that was displaced at launch, but as the ring world turns them back into the shot, they will accelerate up towards the projectile as though it were falling toward the ground under the influence of gravity.

This phenomenon is comparable to Einstein's Elevator Thought Experiment whereby the acceleration of gravity and the acceleration of a moving frame of reference are equivalent.

What happens when firing parallel to the rotation of the cylinder.

enter image description here

What happens when you fire perpendicular to the cylinder:

enter image description here

At least this is how it would generally work on a mega structure like a Ring World or even a Halo Array. Since you are describing more of a Stanford Torus, things get a bit weird because the "world" is so small, and the rotation so slow.

Here is a JavaScript to calculate exact trajectories: https://jsfiddle.net/nosajimiki/k98z2h1a/201/

Some interesting conclusions based on the calculator are:

1 - Firing with the rotation has an effect similar to increasing gravity, the projectile will fall faster because you are adding to its angular velocity. A ring with a higher velocity will be less affected by this phenomenon.

2 - Firing away from the rotation causes the object to fall slower up to the point you match the speed of rotation and you lose all "gravity". If you exceed the speed of rotation, your projectile will be able to "fall" again. A ring with a higher velocity will be less affected by this phenomenon.

3 - There is a slight drift in the direction of rotation when firing perpendicular to the ring. This is proportionate to how high your shot goes compared to the radius of your ring. As your shot goes "up" it is also crossing space parallel to the ring, its speed stays the same, but it crosses a larger angle of the ring for its speed making it drift in the direction of the ring's rotation. A shot fired downward from an aircraft drifts in the opposite direction. A ring with a larger radius will be less affected by this phenomenon.

4 - Every sport is affected a little differently . Sports that involve high velocity low angle projectiles such as baseball will be less affected by side-drift; so, they will probably be played perpendicular to the station. Sports with higher arching low velocity balls such as basketball would be less affected by gravity distortion and more affected by side-drift, but probably still be aligned in the perpendicular to prevent one team from having to contend with higher gravity than the other. Less adversarial sports with high angled, fast projectiles like javelin and golf will probably be played in the parallel to the ring with all contestants throwing/driving in the direction of rotation to prevent dangerous low-G and side-drift conditions, and course will be smaller than on Earth since the harder you launch your projectile, the more it must contend with "gravity". Planet sized rings and larger will probably not need special regulations on course and field alignment.

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    $\begingroup$ They are not going to see a normal ballistic arc at all. The Einstein Elevator example doesn't apply because it involves only linear acceleration. For the rotating space station, imagine throwing a ball perpendicular to station's rotation. The ball is going to to appear to deflect to the right or the left (depending on which direction you threw it). Snipers making long-range shots have to deal with this issue as over very long ranges they have to take the rotation of the Earth into account as it will appear that it deflects the bullet's path. $\endgroup$ – Keith Morrison Jul 5 at 16:15
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    $\begingroup$ The effect you are referring to is negligible for all but the most precise of calculations. It might be more precise to say that over any scale that the ring's curvature is "apparently flat', you will see an apparently normal ballistic curve. It is true that the direction you need to compensate for in a ring is the inverse as on a planet, but the only reason it would be more noticeable on a torus is if the torus is smaller than the Earth which I addressed. I'm just waiting on clarification from the OP to put in the exact way to calculate that difference. $\endgroup$ – Nosajimiki - Reinstate Monica Jul 5 at 17:43
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    $\begingroup$ @KeithMorrison: DING! I knew the Coriolis effect was going to make an appearance on this question. I just knew it. $\endgroup$ – Joe Bloggs Jul 5 at 22:05
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As others have noted, outside of the torus the throw is just a straight line.

A 1 km world rotating at 0.95 rotations/minute has a velocity of 1.9 pi km / minute, or 100 m/s, or 360 km/h.

So you have to add a 360 km/h velocity vector to the vector you throw the Javelin at.

In the rotating frame of reference, there are some interesting cases.

If you throw it directly counter-rotating at 360 km/h, parallel to the ground, it will appear to hover and fly at a fixed speed above the "ground" of the ring.

If you throw it direction with the ring at 360 km/h, parallel to the ground, it will fly off at 360 km/h and fall twice as fast as it "should", roughly. This works because when you rotate something 2x as fast, the "gravity" generated also goes up by 2x, and when thrown at 360 km/h + 360 km/h it is emulating the ring spinning twice as fast.

If you throw parallel to the ground perpendicular to the direction of rotation, it will fall just as if you dropped it, but also move horizontally as it falls.

Throwing up is interesting. For a simple case, imagine you throw it both up and in such a way that in the non-rotating frame of reference, the Javelin is still moving at 360 km/h. This corresponds to the Javelin's arc, instead of being inset by ~1.5 m and then shooting out at 360 km/h perpendicular to the ring (the "drop" case), it instead flies off at an angle from that location.

Another fun one to consider is throwing it "backwards" at 360 km/h, then adding some of "hubward" velocity. Outside of the ring, this corresponds to the Javelin moving directly towards the hub.

For someone outside the hub, the Javelin floats towards the hub, crosses it, then comes down on the other side facing "backwards".

To someone on the ring, the Javelin flies off at 360 km/h. It floats upwards, not falling, but as it gains height it also slows down; it always takes 2 seconds to fly "around" the world, regardless of how high it is, but higher up that "around" is a shorter distance.

It reaches the middle. At the middle, it appears to be spinning around its center every 2 seconds.

When it passes the middle it appears to be flying backwards at increasing speed. Eventually it floats down to human height, where someone could grab the "blunt" end as it flies past at 360 km/h, or let it plow into the ground.

The vertical component can be as large as you want here.

More generally, horizontal velocity of throws varies with height, and objects rotate strangely with regards to creatures standing on the ring, warping throws away from what you'd expect to see.

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    $\begingroup$ Nice answer, many interesting cases. Here's another one: Fire an arrow in the direction of the hub (rotating frame of reference of the shooter). As the arrow is fired straight up, the horizontal speed component (non-rotating reference frame) does not change at launch, so it won't hit the hub. From the shooters point of view, it will be deflected spin-wards, fly forward in a large arc, and then come down as precisely vertical as it was shot. $\endgroup$ – cmaster Jul 5 at 20:22
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Heh heh. You make me remember my high school geometry classes and polar coordinates.

The thing to keep in mind is there is no gravity in this case. The path a projectile takes will just be a straight line while the ring rotates under it.

Diagram showing initial velocity, rotation, and apparent path

So in the coordinates of a person watching from outside the ring the path is the red arrow, and its equations are just this.

$$ x(t) = x_0 + v t $$ $$ y(t) = y_0 $$

That's just linear motion straight across. But in the rotating coordinates of the ring, the object will move according to the blue line. And that gives you the following.

$$ x^\prime = x(t) \cos ( \omega t + d) + y(t) \sin ( \omega t + d) $$ $$ y^\prime = - x(t) \sin ( \omega t + d) + y(t) \cos ( \omega t + d) $$

$d$ is an angle that gets you $y^\prime$ to be zero at $t=0$. That way, in the coordinates of the ring the object starts at the observer standing on the inside of the ring. That's pretty easy to get.

$$ \tan d = \frac{y_0}{ x_0}$$

$\omega$ is the angular velocity which is just such as to give 0.95 rotation per minute in your example. So that means $\omega$ in "per minute" is the following.

$$ \omega = 2 \pi \cdot 0.95 = 5.969 $$

So, you find $x_0$ and $y_0$ by symmetry. Rotate things around so the path is horizontal. Then substitute in to get $x^\prime$ and $y^\prime$. And you get the direction of the toss relative to the rotation by changing the sign of $\omega$.

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    $\begingroup$ Have you allowed for the initial tangential velocity at launch? 99m/s is non-negligible. $\endgroup$ – Separatrix Jul 5 at 14:30
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    $\begingroup$ @Pelinore - the question says "neglect the real gravity due to the ring mass". Also, on the inner surface of a ringworld the I believe the gravitational vectors from all the bits of mass that make it up should cancel out, for the same reason they cancel out for an observer inside a hollow spherical shell. $\endgroup$ – Hypnosifl Jul 5 at 15:54
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    $\begingroup$ the question says "neglect the real gravity due to the ring mass" but fair enough, I missed that bit :) $\endgroup$ – Pelinore Jul 5 at 16:15
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    $\begingroup$ Initial tangential velocity is vital here. If you discount it then all your calculations will be thrown off by the rotation of the ring. IIRC Throwing the ball spinwards or anti-spinwards shouldn’t change the distance travelled from the initial starting point (in the rotational reference frame) $\endgroup$ – Joe Bloggs Jul 5 at 16:30
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    $\begingroup$ "by definition a ring world has it's sun in the centre at the same distance as an earth type planet" — Nope. That would be true for a "real", Niven-style Ringworld, but the OP's "example" diameter (a measly 1km) and RPM suggest something closer to Clarke's Rama (which had "sun" in channels along the ring surface). There may be a light source, but it is unlikely said source is massive enough to have significant gravity. $\endgroup$ – Matthew Jul 5 at 18:12
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Okay, first thing's first. Linear speed.

You are rotating at 0.95 rotations per minute, with a 1 km radius. 2πr gives us a circumference of 2π km or 6.283185307 km. If you travel 0.95 of that a minute, then your velocity on the ring world is 5.969026042 km per minute, or 358.1415625 km/h. I'll convert that to m/s for all remaining calculation, so that is 99.48376736 m/s (My numbers are long because there has so far been no need for significant figures)

Next, let's have a look at the centrifugal acceleration. Now, before you all get up me in the comments, I'll just say I KNOW that there's no such thing as centrifugal force. But technically, the force holding objects down to the ring cannot be called gravity, and to avoid confusion later on, I don't want to call the force holding things down inertia either. So, for the purposes of this question, I'll call it centrifugal force, and if you have a problem about it, you can go cry to Newton.

Anyway, the formula for centrifugal acceleration is F = v^2 / r. This means that the force is (99.4837673 m/s)^2 / 1000 m, and the centrifugal acceleration(our planets equivalent to gravity) is about 9.89 m/s^2, or just 0.08 m/s^2 greater than Earths. So if you were blindfolded on this world, you most likely wouldn't notice that you weren't on Earth (unless you fell off).

Those are the effects the viewer would notice. So far, I have not seen an answer that addresses the initial speed of the javelin, so I will address that. Once the javelin leaves the hand of the thrower, it is no longer being acted on by the rotation of the ring world. You also said to discount any forces that include the air resistance. This means that once the javelin leaves their hand, the velocity will not change. By definition, the path taken by the javelin is a straight line. However, any object thrown will experience two phenomena. First of all, it has a velocity of 99.483 m/s added to it tangential to the the circle at the point it was thrown. Second of all, while it is in the air, the ring world will be rotating. This leads to some... interesting phenomena. For example, say there is a tennis ball cannon set up to shoot at precisely the same speed as the ring world is rotating in the opposite direction. Due to there not being any air resistance, and no gravity, the tennis ball will seem to leave the cannon, and then it will remain floating until it hits something. Also, something funny. OP mentioned javelins. An Olympic javelin thrower can launch a javelin at a speed approaching 100 km/h. If the thrower somehow throws their javelin straight up, it will take approximately 5.207 seconds to hit the ground. In those same 5.207 seconds, the javelin thrower will travel approximately 518 meters. The javelin hits the ground 550 meters away from the spot where it was thrown, which means if the javelin thrower runs at 6.39 m/s, slightly slower than the average human 100 meter sprint, then there's a chance. Just a chance. That the javelin thrower could be hit by his own javelin.

OP, you asked for an examination of the projectile trajectories. The answer is, technically there are none. Any object thrown will travel in a straight line until it hits something on this world. However, the velocity will always have 99.483 m/s added to it in the direction of rotation. And you asked for graphs:

An arc with 3 vectors going from one point.

This is the javelin thrower. The vertical arrow represents the speed of his throw, the horizontal arrow represents his speed from the rotation, and the diagonal arrow represents the total velocity

An arc with a line bisecting.

This is the path taken by the javelin.

Thanks for reading!

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I want to first apologize to any physicists reading this. Feel free to edit misuse of terms such as speed and velocity, I might use them wrong.

Also feel free to skip to the "real fun part". It will give you a sense of how weird throwing stuff may get.

Introduction

The most important counter-intuitive thing you need to acknowledge is that the force you feel and mistake for gravity on the ring world is in fact just you moving really fast around the circumference. Think Homer Simpson in his Ball of Death. You stop moving, no more force, just you floating in the space.

If you jump, there are no forces acting on you, so you move in a straight line, as all the answers here have correctly pointed out. You also have some initial velocity you got by standing on the rotating ring world and that is why while you jump up or even counter-rotation, you still move rotationward for an observer with your trajectory being a straight line. This is obviously not what you wanted to know because who cares about some high and mighty observers, right?

Balls don't fall back or make spirals

When you are on the Earth, jumping direction has almost no significance (Coriolis are negligible). However in your scenario, the rotation has a lot of significance. As H Franklin calculated in his answer, firing a tennis ball at about 100 m/s counter-rotationward would counter-act the initial force and the ball would hang in space for an outside observer and would rotate around the world with the same speed as the world of 0.95 rotations per minute from your point of view.

Now if we gave it the same counter-rotation speed but also a slight vertical speed, things would become much more interesting. For your observer, the ball is slowly rising from where you threw it until it hit the world on the opposite side. Now for you, it still files around the world with the 0.95 rotations per minute speed, but also gains altitude at the same time. The trajectory would therefore be a spiral until it hit the center and then spiral again until it hit the ground again. Thing to note here is that it has the same tangential and vertical speed, therefore from your point of view, its horizontal speed (horizontal meaning moving around the circumference, what is horizon anyways?) would decrease as it gained altitude.

With this we can already see that elliptical/parabolical approximations of the trajectories don't hold very well but let's continue.

Quick maths

Let's make new coordinate system. X will mean the distance you would have to travel on diameter to be directly under the object of interest. Y will mean the distance of the object from the closest point on the circle (i.e. the distance from you if you are standing right underneath it).

First we will compute these coordinates for an observer with a given origin, from where the ball was thrown. In our picture that is the point B. The ball always gets the velocity of the rotation (direction of the spin, tangent where we stand, in our case vector BF) and the velocity we give it with our throw. Together it will produce the straight line we discussed for the outside observer with constant speed. We know the speed so we know the standard coordinates at any moment. So the problem is now to translate those coordinates to our knew system.

Let's take the point D. I will denote $\phi$ the angle FBD, $r$ the radius AB and $s$ the distance traveled BD. We are interested in $x$ being the length of the arc BE and $y$ as the distance ED.

We know that AE has also length $r$, so $y = r - t$ where $t$ is the length of DA. We can calculate $t$ with the law of cosines, i.e. $t^2 = s^2 + r^2 - 2rs \cdot \cos (90^{\circ} - \phi)$

The new x coordinate can be calculated from the angle of BAE denoted $\alpha$ and computed with the law of sines, $\sin \alpha = \frac{\sin ( 90^{\circ} - \phi)\cdot t}{s}$

The last step to have your own experience is to account for your own rotation from the observers point of view. That means adding the distance you travelled due to spin to the x coordinate.

Real fun part

Throwing balls up

Now let's imagine would happen if you were to throw a ball up giving it the same vertical speed as is the speed of spin and for the sake of having nice numbers we'll say that is $1$ and we'll also assume the radius to be $1$. In terms of the last paragraph, $\phi$ would be 45 degrees, the speed of the object would be $\sqrt{2}$. The distance it would have to travel from B to C would also be $\sqrt{2}$, meaning it would fall back down in exactly $1$ time unit. For an outside observer, it would fall exactly one quarter of the circle in the direction of spin, but where would we be in that moment? The spin speed is $1$, which would mean we need $2\pi$ time units to finish one whole rotation and about $1,57$ to finish a quarter. But in the same time the ball we threw vertically landed a quarter of the circle from the point it was thrown, making it seem to us that it fell a bit in front of us.

Playing catch alone

Now for few last points of observation, the ball is traveling in a straight line with constant speed. That means that the higher it gets, the faster it will change the x coordinate, because the closer it is to the center, the faster the tangential speed will become with the same velocity. Now this would allow us to do some neat tricks. Imagine we want to throw a ball and catch it again. We found out already that if you throw it just up, it falls in front of you. So when we combine these two points, you need to throw it up and back a little and as it will rise, it will seem to you that it changed direction and started moving back to you and a bit in front of you in the direction of the spin and after it starts falling, it will change direction once more and fall back to you. If you draw that trajectory, that sure as hell is no ellipse.

Thrown ball

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I can tell you how best to calculate this, but the formulae are more complicated than I can figure out off hand:

You should look at the flight of the (say) javelin in a non-rotating reference system, in which it will not be subjected to any force once it leaves the hand of the thrower (ignoring aerodynamic effects, as you say we should), which means that it travels in a straight line in this reference system.

Then you simply have to look at the speed with which the javelin leaves the hand of the thrower (including the rotation speed) and the three-dimensional direction it is thrown, which can be figured (though not easily) from the direction of rotation, the direction of the throw relative to the direction of rotation, and the angle the javelin is thrown 'upwards' from the 'floor' of the cylinder.

Once you have the direction, you can calculate where the linear course of the javelin intersects with the cylinder, and the time it takes can be calculated as the ratio between distance and speed. This time, in turn, tells how much the cylinder has rotated since the javelin was thrown.

This is fairly uncomplex, but rather complicated, and I am afraid that my spatial geometry is a bit rusty.

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    $\begingroup$ The entire point of the question is to find out the equations in the rotating frame of reference of the habitat. That's why is says "hard science". The querent knows full well how rotating habitats work. $\endgroup$ – AlexP Jul 5 at 9:12
  • $\begingroup$ @AlexP: And I provided guidelines for how best to find those equations, which may not be a whole answer, but certainly a step along the way. I would have put the answer in a comment if it wasn't much too long for that. $\endgroup$ – Klaus Æ. Mogensen Jul 5 at 10:07
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    $\begingroup$ Seems like a hard science answer to me, this is the mathematically simplest way to determine what point on the surface the thrown object will impact and its height above the surface as a function of time, and the question doesn't say anything about having to calculate the answer in a rotating reference frame. $\endgroup$ – Hypnosifl Jul 5 at 13:29
  • $\begingroup$ "All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc." $\endgroup$ – rek Jul 7 at 12:30

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