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Artificial gravity by spinning? 
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#1
Feb1213, 12:38 PM

P: 13

Every time I see a SF pic which has some simulated gravity by spinning I can't imagine this actually working. The latest SF I watched (mission to mars) had a wide rotating cillinder somewhere in the middle of a much longer cillinder which formed the main ship.
Now imagine yourself moving from the front of the ship to the back of the ship, at some point you'd float through the spinning artificial gravity room. If you would float exactly through the center would you spin about your own axis? If you're floating a bit off center you would supposedly be swung to the outside of the cillinder? What force would be responsible for this effect? How can the rotational force of the room be transferred to you if you're not touching anything? I can imagine a simple earth based experiment by having a long hollow tube standing upright spinning around it's longtitudinal axis. If you'd drop a ball in there I don't think it would be swung out (sticking to the wall) before it actually hit the spinning bottom, which would finally transfer the rotational speed and swing it out? 


#2
Feb1213, 12:49 PM

P: 643

The key seems to be making sure that anyone in the craft rotates with the wheel. Then, if the centripetal acceleration is 9.8 m/s^2, one can just use the equivalence principle to point out that this is locally indistinguishable from gravity.
So you sort of have a point, just a rotating disc isn't enough, something to make people rotate with the rotating disc is also necessary. If they're on the "floor" (the outside of the disc,) friction's enough to be responsible for this in most cases. 


#3
Feb1213, 01:05 PM

P: 718




#4
Feb1213, 01:12 PM

P: 718

Artificial gravity by spinning?
Have you ever ridden one of those fairground rides that look like big spinning flying saucers? You start out strapped to the wall so that the harness can pull you around as the ride is tangentially accelerating from rest up to its operational speed. Once it reaches that speed, the harnesses are released and the only force you feel is the sensation of being pushed straight back against the wall. However, the operator sitting in his booth in the center of the ride is obviously unaffected by this. 


#5
Feb1213, 04:24 PM

P: 13

So once you're spinning with it, would you keep spinning with it (no drag)?
Would it be safe to grab hold of the outer edge from a relative standstill (assuming the outer edge is spinning with a speed that equals 1g)? What happens if you throw a ball up into the air to about the center of rotation? I'm guessing it would just come down normally into your hand again. Even if you threw it past the center of rotation? I'm guessing yes to the last one too, but it's a bit tricky somehow. Finally, do you feel this is a realistic aproach for future spacecraft? Thanks a bunch! 


#6
Feb1213, 04:54 PM

P: 778




#7
Feb1213, 06:07 PM

P: 13

Thanks for the great little anim, however it's a bit confusing too because I'm not sure how the ball got to the center at the start of the movie.
Was it thrown up by someone sitting at 90 degree from the orange dot? Or is it comming down aftersomebody at +90 degree from the orange dot threw it up? 


#8
Feb1313, 04:06 AM

P: 4,232

The ball is shot from the center towards the dot.
Similar animation: 


#9
Feb1313, 11:00 AM

P: 3,264



#10
Feb1313, 03:15 PM

Sci Advisor
PF Gold
P: 2,741

[QUOTE=A.T.;4268409]The ball is shot from the center towards the dot.
OK, then it didnt answer what k!rl asked which was if you were out on the edge ( on the floor ) and tossed the ball up would it come back down to where you were? I suspect yes, as it has the same momentum as you and it have when you release it for the same reasons as if you throw an object up in a moving velicle ... train, car, plane etc Dave 


#11
Feb1313, 06:35 PM

P: 211

No, it would fall ahead in the direction of rotation. For instance, if the ship rotates directly, it would fall to your righthand side.



#12
Feb1313, 07:12 PM

P: 705

Take the following example... Suppose you build a round rotating cylindrical space ship with an inside radius of 1000 m. Rotate it at a rate of 0.1 radian/sec in a counterclockwise direction. Center it in a coordinate system where (0,0) is at the center axis of rotation. I am standing on the inside surface. At the exact time when my coordinates are (1000,0) I throw the ball toward the center at a rate of 100 m/s. The ball must travel in a straight line at constant velocity until it lands (newtons 1'st law). The components of the balls velocity are 100 m/s in the X direction and 100 m/s in the Y direction. It's easy to see that after 10 seconds the ball will contact the inside of the cylinder at coordinates (0,1000), ∏/2 radians counterclockwise from the position it was thrown from . During that 10 seconds the cylinder will have rotated 1 radian, which is less then ∏/2. 


#13
Feb1413, 03:48 PM

P: 13

Thanks for clarifying everybody, seems sort of obvious in hindsight as always. Someday interstellar jue de buoles will be the favourite past time in the space resorts! =)
When the mass in such a gravity room is unevenly divided it may cause the spacecraft to wobble though. 


#14
Feb1413, 04:06 PM

P: 7

Here's an explanation that might assist in understanding. http://www.youtube.com/watch?v=Otmg0knGtE
A space station example starts at 26:22 in the video. 


#15
Feb1413, 04:07 PM

Mentor
P: 12,081

If the station is not extremely large, all those effects from a positiondependent centrifugal force and coriolis force would be notable. 


#16
Feb1413, 06:27 PM

P: 13

Thanks for the link telecomguy. I've seen some of Walter Lewins lectures before, I like his style!



#17
Feb1513, 06:41 PM

P: 1

After 50 years of building things and programming I am retired and writing a science fiction book. I want to get things right but am having problems visualizing some of the effects of artificial gravity by spinning and the math and small examples have not helped. I have spent a few hours working the math in examples and getting numbers that don't mean much to me (see examples at the end).
My 'ship/space station' is an O'Neill colony large tube but bottle shaped. It is 50km long and 30km in diameter at the big end and 10km at the small end and rotates around the smaller axis, of course. It is filled with air, landscaped on the interior and has two rivers that curve around the interior DNA style from one end to the other. My questions are: 1. What will be g at the small end if the ship rotates fast enough to create a 10 m/s equivalent centrifugal force at the big end? Will the g force be more or less at the small end? I think it will be less but I'm not sure and then, by how much less? 2. Will the Coriolis force be sufficient to drive the rivers flow? Which direction (small end to large end or the reverse) must they flow? (I will have them pumped in the opposite direction.) Or will the difference in g drive them from one end to the other? 3. Will the Coriolis force be sufficient to stir the interior atmosphere? Will the interior atmosphere float free of gravity and be stirred by the interior mountains and buildings to any degree? Would it build up momentum and start revolving with the structure? 4. If you jump from a large building in the interior would you float free until something turning around on a straight vector (to/from where) smacked you? Could you fly with wings in zero gravity up to the center where the sun tube burned off the glue and you hung there roasting slowly? 5. If people built buildings on the exterior of the tube the g force would be greater the farther the floor from the center of rotation; right? By how much, though? Would a 25 story building be crazy or would the g force just be a few tenths more and be barely felt? (itex)R=1500^2m(\itex) radius at the big end (itex)R=500^2m(\itex) radius at the small end (itex)ω^2 1500 = 0.082 m/s(\itex) force of gravity on earth (rounded) (itex)ω^2 500 = 0.14 m/s(\itex) (itex)ω^2 = 0.3 rad(\itex) how fast a rotating frame of reference must rotate to simulate the force of gravity on earth. (itex)T = 2∏/ω = 78.5 sec(\itex) (itex)T = 2∏/ω = 44.8 sec(\itex)  Nick "Age does not protect you from love. But love, to some extent, protects you from age."  Jeanne Moreau 


#18
Feb1613, 12:02 PM

Mentor
P: 12,081

For constant angular velocity ω, gforce is proportional to the radius r: ##a=\omega^2 r##
Your "height" difference would be like a steep mountain, and air would be very thin at the small end. The Coriolis force will give some deviation from a straight line  but you should reduce the "height" difference to get nicely flowing rivers. 


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