Space station artifical gravity

[DaveC426913]

I've got a [strike]great[/strike] good head for intuiting physics but no post-2ndary math.

How fast must a cylindrical space station rotate to produce a given g-equivalent?

Say we standarize the gravity at a reasonable 0.5g and the station at 100 feet diameter. And maybe another station at 1000 feet in diameter for comparison.


I've been toying with some of the fun effects you could have on such a station.

Playing a traditional ball game, or any kind of traditional sport for that matter would be extremely awkward. Throwing up-rotation would cause the ball to tank; throwing down-rotation could very well cause the ball to float around the entire station and hit you in the back of the head. Running would be worse; you'd either trip and fall on your face or you'd float and lose traction.

Also, kids would be a menace in the halls, always launching themselves into orbit.

 

[mgb_phys]

Force is just = r w^2 ( w in rad/s )
so 100ft diameter = 16m radius, 0.5g would be 0.5rad/s = 3.1 seconds / rotation

 

[DaveC426913]

Hm. So at any given time, the occupants are moving at a speed of 100 ft/s. (300ft perimeter / 3.1 s). That's just over 60mph.

So, in fact, you could not launch yourself into "orbit". And only a good thrower could put a baseball into "orbit".


Actually, I don't understand your calculation. What is w? And how do you get from 16m / 0.5g to 3.1s?
(Also, I have edited line one of my OP.)

Does it get harder or easer in a larger station?

 

[mgb_phys]

w is omega - rotation velocity in rad/s (need to sort a problem with latex in opera)

F = m a = m r w^2, or w = sqrt( a/r )
With a = 0.5 g = 4.9m/s^2 and r = 100ft/2 = 15.24m
w = 0.56 rad/s, a full rotation is 2pi rad, so it takes 11s for a rotation.
(sorry must have got the last step the wrong way up )

So linear speed is 2 pi 15.2m in 11s = 8.5m/s = 19mph

ps. Was watching moonraker last night - they spin the space station to get artificial gravity (the commentary track says they copied it from 2001) but the don't walk on the EDGE of the spinning surface, gravity is along the axis of the station ie down = toward Earth - I imagine they have more artists and fewer scientists on staff than 2001.

 

[DaveC426913]

So linear speed is 2 pi 15.2m in 11s = 8.5m/s = 19mph

Huh! So you could launch yourself into "orbit"...


You know, now that I think about it, a lot of stories have got gravity and trajectories in rotating space stations all wrong.

For example, John Varley's Titan has it wrong. When Gabby fell from the core, she wouldn't be accelerating. She'd hit with the same radial speed she left the core with. Though it would be her horizontal speed that would get her.

Maybe I'd better read it again. I do remember that she had a very significant horizontal v. I don't recall if he indicated she was actually accelerating.

 

[Shooting Star]

Artificial gravity can be supplied by rotation, as long as you are still wrt the space station. As soon as you move, the Coriolis force comes into play, changing your direction.

In fact, if you are just at rest at a point in space, and the space station rotates about you, that can be viewed as the trajectory wrt the space station due to Coriolis force, if you do your calculations in the rotating frame.

 

[TVP45]

Artificial gravity can be supplied by rotation, as long as you are still wrt the space station. As soon as you move, the Coriolis force comes into play, changing your direction.

In fact, if you are just at rest at a point in space, and the space station rotates about you, that can be viewed as the trajectory wrt the space station due to Coriolis force, if you do your calculations in the rotating frame.

And, at 100 feet (or anything small) the Coriolis forces are going to do some bigtime Hallpikes on you. Oy and double oy!

 

[Shooting Star]

I searched and searched for the term "Hallpikes", but couldn't find the meaning of the word, though it sounds impressive the way you've written it...

 

[TVP45]

Oops. Sorry about that. I think that's called "stream of unconciousness" writing. A Hallpike (or sometimes called a Dix-Hallpike) is a movement where you quickly rotate your head to the side as you quickly lean back or forward. It sometimes produces projectile vomiting. If you want to check it out, look under otolaryngology terms on most medical sites.

 

[Shooting Star]

Yes, whenever I searched, it gave me medical related links, which I didn't explore further.

I am trying to understand why you made that statement. For maintaining Earth g as the centrifugal accn w^2*r, w must be high if r is Small.

But since Coriolis accn = 2 w X v', where v' is the velo wrt the moving frame, it would be greatest when v' is perp to w. If w is high because r is small, then it would do a great hallpike on the object.

Hallpike, hallpike...I must get the hang of how to use it. Some places it's given as hall pike.

 

[DaveC426913]

I am trying to understand why you made that statement...If w is high because r is small, then it would do a great hallpike on the object.

I think he's saying that people would experience some serious disorientation in a rotating frame. Coriolis forces and gyroscopic forces and inner ear balance and all that.

 

[Shooting Star]

Important point to note about pseudo froces

I forgot to mention one crucial point about the, uhh...hallpikes. Since these forces are inertial forces, they'll act on all points of a small body equally, like gravity. So, you won't really feel the hall-pikes. You'll be in free fall.

 

[TVP45]

If you simply kept your head in the same orientation, you would feel nothing. But, if you knelt and then stood up, if you nodded yes to someone's question, if you turned suddenly to look out the window (porthole?), then you would generally feel the effect. The folks at Houston who do centrifuge work regularly see this.

Keep in mind you've got 3 angular accelerometers in each inner ear and, when they don't agree with either the visual input or proprioception, or (in some cases) somasensation, one of the responses of the body (this is thought by some to be a remnant of early flight response) is to empty the stomach.

Large r gives only a very tiny difference within the small range of motion of the human body.

 

[Shooting Star]

I think he's saying that people would experience some serious disorientation in a rotating frame. Coriolis forces and gyroscopic forces and inner ear balance and all that.

That's why I immediately made the second post. If you are not touching anything, then you are simply weightless and no hall-pikes.

If you try to run along a radial path, oh yes, big time hall pikes or hall pukes. (I think -- haven't really calculated the differential force between, say, the head and the leg.

EDIT: This was posted before I saw TVP45's last post.

 

[Shooting Star]

If you simply kept your head in the same orientation, you would feel nothing. But, if you knelt and then stood up, if you nodded yes to someone's question, if you turned suddenly to look out the window (porthole?), then you would generally feel the effect. The folks at Houston who do centrifuge work regularly see this.

Keep in mind you've got 3 angular accelerometers in each inner ear and, when they don't agree with either the visual input or proprioception, or (in some cases) somasensation, one of the responses of the body (this is thought by some to be a remnant of early flight response) is to empty the stomach.

Large r gives only a very tiny difference within the small range of motion of the human body.

Quite agree with you but afraid to nod...

Well, the severity of the symptoms would depend on the angle between w and v'. You know more of the details, it seems.

I can't get hallpike out of my mind.

 

[TVP45]

Quite agree with you but afraid to nod...

Well, the severity of the symptoms would depend on the angle between w and v'. You know more of the details, it seems.

I can't get hallpike out of my mind.

Sorry to rambled off on a tangent. And, I didn't even explain it well; once I realized I had gone off, I didn't mention the otoliths which are really the culprit.

I have been following microgravity and artificial gravity for some years and just made a comment, without thinking, that others would not be expected to know. If you have any interest, you might Google Scott Wood, Owen Black, or Bernie Cohen and artificial gravity nausea. I think Scott Wood is probably the best guy; he's down at the centrifuge facility. I can never remember which sites permit linking and which don't, so I won't provide links. And, again, I probably answered far more than you were interested in.

 

[Shooting Star]

>And, again, I probably answered far more than you were interested in.

I live partly for that. Thank you for bringing the details to my attention. I do understand the overall picture, but these small details are helping to make up a complete description.

In Clarke's "Rendezvous with Rama", a lot of funny stuff about the effects of Coriolis forces are there, but I never thought about these small details.

 

[wysard]

Hm. So at any given time, the occupants are moving at a speed of 100 ft/s. (300ft perimeter / 3.1 s). That's just over 60mph.

So, in fact, you could not launch yourself into "orbit". And only a good thrower could put a baseball into "orbit".


Actually, I don't understand your calculation. What is w? And how do you get from 16m / 0.5g to 3.1s?
(Also, I have edited line one of my OP.)

Does it get harder or easer in a larger station?

Given that the Earth is rotating at about 1,000 mph and we don't "accidentally" launch into orbit I would have to say no. It's not about the speed, but the relative masses involved and the distance between their respective center of gravity.

Little station, spin fast. Big station spin slow. BUT.. (Always there is a but...) Because a station is bigger, there is more strain on the trusses holding the circumfrence to the centre. It's a trade off of size versus materials. Where material=mass this becomes an issue. Not only for accelerating or decelerating the spin, but also just to get it into orbit to assemble in the first place.

As for putting Coriolis force? Unless the station is spinning very fast it is unlikely to lead to odd mental effects, just ones where physics seems to take a trip into right-angle land. While Sci-Fi occasionally blunders in this respect with the effects of one side going (arbitrarily) clockwise and the other the reverse leading to odd disorientation effects the human inner ear acts as a counterweight to the problem in humans. Otherwise every time an aircraft pilot did a barrel roll they would be rolling the dice.

 

[Shooting Star]

Why is launching into orbit from Earth being compared to launching into orbit from a rotating space station? What relationship is there between the two situations except the word "orbit"? What is meant by orbit around a rotating space station? The last question may please be answered by the people who seem to know a lot about it. Otherwise, I'm hallpiking, and more seriously, I'm curious.

 

[Dale]

I think that by "orbit" in the space station they are referring to the situation where, in the inertial frame where the center point of the station is at rest, the object is stationary and the space station is rotating around it. In such a situation, neglecting air resistance, the object would have a nice periodic circular path in the station's rest frame. Essentially an "indoors orbit".

 

[DaveC426913]

Yep. Dalespam is correct. It's not an orbit at all of course. Sorry if I caused any confusion.


I'm not really sure what wysard is on about.

 

[Shooting Star]

I think that by "orbit" in the space station they are referring to the situation where, in the inertial frame where the center point of the station is at rest, the object is stationary and the space station is rotating around it. In such a situation, neglecting air resistance, the object would have a nice periodic circular path in the station's rest frame. Essentially an "indoors orbit".

I have my doubts. I'm sure that the concept of gravity has crept in somehow.

But the orbits about the space station are easy to find out if there is feeble or no gravity. If there is no gravity, the object has a constant vector v wrt to an IFR. You can kinematically transform to the rotating frame, without going into any force calculations.

 

[Shooting Star]

I just got the message from DaveC426913 as soon I had posted mine.

 

[Dale]

If there is no gravity, the object has a constant vector v wrt to an IFR. You can kinematically transform to the rotating frame, without going into any force calculations.

Yes, that is how I would approach it also. However, in the rotating frame such a path would be an accelerating path, so you would have to use ficticious forces (centrifugal and coriolis) to explain it in the rotating frame.

 

[Shooting Star]

Yes, that is how I would approach it also. However, in the rotating frame such a path would be an accelerating path, so you would have to use ficticious forces (centrifugal and coriolis) to explain it in the rotating frame.

That's right. It's quite a bit of fun doing it for the first time for this sort of simple problems. Then you are asked to find the level difference between the two banks of a river using Coriolis force, and then you get the real hall pikes. (I have already said that I can't get the word out of my mind!)

 

[Dale]

Then you are asked to find the level difference between the two banks of a river using Coriolis force, and then you get the real hall pikes. (I have already said that I can't get the word out of my mind!)

Wow! I didn't even think of that. Civil engineering projects inside one of these stations would be really painful. I wonder if you could design the path of an elevator or a roadway to eliminate hallpikes. Speed limits strictly enforced, no police or tickets needed!

 

[Shooting Star]

Wow! I wonder if you could design the path of an elevator or a roadway to eliminate hallpikes.

Again, Arthur C. Clarke's "Rendezvous with Rama". The elevator is treated in more or less detail.

 

[lilrex]

Going to the original post, the easy way to calculate G's is as fallows.

G = (v^2 / r)/32.2

Where G is the acceleration Earth's gravity, v is the velocity in feet per second, and r is the radius of the arc in feet.

Example: 180 fps on a 1000 foot arc will give you about 1g

And since the circle is 2000 feet in diameter that will give you 6283 feet of circumference divide 180 fps by that and you will get .02866 rotation per second or 1.72 rpm

Of course if I am wrong about this someone will correct me.

cheers

 

[Shooting Star]

Huh! So you could launch yourself into "orbit"...
…
For example, John Varley's Titan has it wrong. When Gabby fell from the core, she wouldn't be accelerating. She'd hit with the same radial speed she left the core with. Though it would be her horizontal speed that would get her.

Maybe I'd better read it again. I do remember that she had a very significant horizontal v. I don't recall if he indicated she was actually accelerating.

I too have forgotten the exact scenario in the book.

Actually, if we do the calculations in the rotating frame, then an object projected radially towards the circumference from the axis would tend to accelerate radially outwards due to the centrifugal force. But once it starts to move, the Coriolis force would change the direction. The Coriolis force cannot do any work, only change directions. But immediately the centrifugal force changes, because that is dependent on the distance of the object from the axis. The whole result is that the object spirals outward and when it hits the outer wall, it has a tangential component of velocity as well as a radial component. As you have said, the radial component remains the same, but the tangential component has increased. This can be interpreted as the centrifugal force doing the work and the Coriolis force changing the direction.

Looked at from a non-rotating IFR, the object traveled with constant velocity towards the outer wall until impact. Since the outer wall has only a tangential velocity, the mutual interaction between the body and the wall will have a tangential “bump”, a well as the bump due to the original velocity of the body directed radially outward.

Hope I haven’t bored you by repeating things you know better than me.

 

[KontaktMan]

Formula for centrafugal gravity

I have tried this formula but am getting strange results.
G = (v^2 / r)/32.2

First of all, I am not a physics guy at all, not even really good at math, so bear with me.
I followed the example that lilrex gives and if I'm not mistaken it goes like this:

G = (v^2 / r)/32.2
G = (180 fps ^2 / 1000 ft.)/32.2
G = (32,400 / 1000)/32.2
G = (32.4)/32.2
G = 1.0062111

All good for a 2000 ft. diameter space wheel.

OK, but I would like to do the calculations for a 2 mile diameter space colony. (Sc-Fi idea) So:

G = (v^2 / r)/32.2
G = (412 feet per second^2 / 5280 ft)/32.2
G = (169,744 / 5280 ft) / 32.2
G = 32.148484 / 32.2
G = 0.998400

Now, how can this be right? I would think the larger the diamater of the wheel is, the slower it needs to spin to equal 1G.

 

[D H]

Now, how can this be right? I would think the larger the diamater of the wheel is, the slower it needs to spin to equal 1G.

The larger wheel is spinning slower: From the numbers in your post, the 2 mile diameter wheel makes one revolution every 80.5 seconds while the smaller one makes one revolution every 34.9 seconds.

 

[KontaktMan]

The larger wheel is spinning slower: From the numbers in your post, the 2 mile diameter wheel makes one revolution every 80.5 seconds while the smaller one makes one revolution every 34.9 seconds.

Yes, I do see how the revolution is slower. I guess I was also expecting the speed to be lower as well. However, standing inside a 2000 ft. wheel you'd be going 180 fps and inside a 10,560 ft wheel you'd be traveling 412 fps. The concept is a little difficult for me to wrap my head around.

Do you think that would be felt more or less by a human being?

 

[kdv]

Yes, I do see how the revolution is slower. I guess I was also expecting the speed to be lower as well. However, standing inside a 2000 ft. wheel you'd be going 180 fps and inside a 10,560 ft wheel you'd be traveling 412 fps. The concept is a little difficult for me to wrap my head around.

Do you think that would be felt more or less by a human being?

You want the radial acceleration to be the same for any radius so you must keep $$\frac{v^2}{R}$$ constant. This shows clearly that if R is increased, the tangential speed v must be increased as well by a factor equal to the square root of the increase in radius. So multiplying by 4 the radius will increase by a factor of 2 the speed.

On the other hand the period is [itex] T = \frac{2 \pi R}{v} [/tex]. Since the speed does not increase as fast as the radius, the period decreases with an increase of radius.

In the case of the planets around the Sun, the acceleration decreases with distance and in that case an increase of distance actually corersponds to a decrease of speed.

 

[KontaktMan]

OK thanks. here's another related idea I've been struggling to have answered.

So far I've been speaking about the physics of a wheel (or a cylinder) to achieve 1G, but how about a sphere? Sure you could rotate it on it's axis to get your 1G at the equator, but could it also be twisted in another direction as it rotated so that no matter where you were on the inner surface you'd still be traveling at 412 fps, thus achieving 1G?

Impossible? Crazy?

 

[kdv]

OK thanks. here's another related idea I've been struggling to have answered.

So far I've been speaking about the physics of a wheel (or a cylinder) to achieve 1G, but how about a sphere? Sure you could rotate it on it's axis to get your 1G at the equator, but could it also be twisted in another direction as it rotated so that no matter where you were on the inner surface you'd still be traveling at 412 fps, thus achieving 1G?

Impossible? Crazy?

Impossible for a rigid sphere. For something non-rigid, it would onlybe possible for a very very short time until everything would rip apart. So, yes, it's crazy :-)