Counter Rotational Gee's

[Knife]

OK, as sci fi guy's we should all be aquainted with the idea of counter rotational or even rotational gravity. Notable examples are 2002, B5, Mission to Mars, just about ever space station shown in the 60's and 70's sci fi.

My question is this: How viable is this really?


Some where, and I can't remeber where, I remember something along the line of if you have a "area" 1000meters long rotating at one revolution per minute it simulates 1 G. Is this correct, if so is it scalable as in 500meters rotating a 1rpm makes 1/2 G?

If this is not correct, then what is the formula. I know I could look it up in other places but I consider this place a resource too. I'm especially interested in some of you engineers opinions out there.......

[Phil]

Isn't it actually due to centaped(t?)al force?

The inertia of the object or person in the rotating device "wants" to push the object in a straight line, but because the device is revolving it ends up as a force along the radius of the rotation.

[Wicked Pilot]

The equation you are seeking is a=(v^2)/r

a is the radial component of linear acceleration, or the g that you want.

v is the linear velocity of what ever is rotating

r is of course the radius

[Wicked Pilot]

Isn't it actually due to centaped(t?)al force?

You are very wise indeed. It is actually centrepetal force that exerts a force on you, pushing you towards the center of rotation. The force you feel pulling you down is called centrifugal, but is in fact only inertia trying to hold you on the tangent path. Centrifugal force is a "phony" force.

[Phil]

Towards the centre of rotation?

Wouldn't it be away from the centre?

[Knife]

Thanks for the formula. Being so far from a math guy that is possible, what is the ^ mean? Been a while since algerbra and trig.

[Knife]

OK, there is the math. Now on too opinions, would it work as one would intend or would it work but make everybody dizzy and sick, or would it not work at all?

[Andrew J.]

Thanks for the formula. Being so far from a math guy that is possible, what is the ^ mean? Been a while since algerbra and trig.

I think it's just "to the nth power."

10^2 is ten squared, 10 ^3 is ten cubed, etc.

[The Dark]

It will definitely create gravity, but too fast of a rotation causes all sorts of balance and nausea problems, due to fluids in the inner ear telling the body it's moving while the eye says you're standing still. Most people can handle 1 RPM, but 3 RPM is enough to make almost anyone lose it. A good site for astro calculations is http://www.bumply.com/2300/astro.htm.

[Knife]

So let me get my tired old brain around this in lamen terms.

A(or G's) is equal to the rate of rotation squared, divided by the radius of the object rotating.


Correct or am I a dumb fuck?

[Durandal]

The biggest problem with sci-fi writers is that they don't get under what circumstances rotational gravity will actually work. You can't just set something spinning and magically get gravity, like in Babylon 5.

Here's how a to generate gravity with a space station and rotational motion. First, you must build the station in a circular shape. Now, let's say we build it to look like Babylon 5 (a giant turd in space), but let's assume that it's more uniform. Just a giant cylinder. Now, let's set it spinning about it's long axis at sufficient velocity to produce 1g of acceleration.

Will the entire station experience 1g of acceleration due to the spin?

NO

Only the people on the inside surface will experience the pull. If you are on the central axis, you will experience no pull at all. Why? Because, as the station spins, you can observe any point on the surface to constantly be undergoing a change in linear direction (acceleration). Thus, you're not actually being pulled outside , but your body wants to keep going in a tangent line to the circle. The net result of this is the "feeling" that you're being pulled outside. This is the same effect you feel when rounding a corner in a car at higher-than-you-should speeds. If you were oriented horizontally in the car, you'd experience a pull toward the outside of the curve.

So, the bridge on Babylon 5 is grossly misplaced, and the crew should be floating.

[Wicked Pilot]

A(or G's)

Standard derived unit for acceleration is m/s^2. 9.8m/s^2 is equal to the gravitational pull on earth at sea level.

is equal to the rate of rotation squared,

No, it's the linear velocity squared. You can find linear velocity by taking the circumference and multiplying it with the rate of rotation.

divided by the radius of the object rotating.

The distance from the center of mass to whatever perticular point you are trying to calculate.

Correct or am I a dumb fuck?

You are not correct, however you are certainly not a dumb fuck. A dumb fuck is not someone who ask questions, but someone who assumes they are always right regardless.


Here's an example. Suppose you have a Babyon 5 style space station. The rotation section has a diameter of 500 meters, a radius of 250 meters. This station rotates twice every minute. Suppose you want to know the how many Gs are experienced by someone standing on the lowest floor.

First you need the linear velocity at that point. The lowest floor is farthest from the center of mass, or 250 meters. We can then calculate the circumference, it is the product of the diameter and pi. (500m)(3.14)=1570m. Take that and divide it by the rotational velocity of twice per minute, or 1/30 rotations per second. (1570m)(1/30s)=52.3m/s

Now, we take our origional formula a=v^2/r, and plug n' chug. a=(52.3m/s)^2/(250m)=11.0m/s^2. Since one g equals 9.8m/s^2, that person standing on the first floor is experiencing 1.1g.

[Knife]

A(or G's)

Standard derived unit for acceleration is m/s^2. 9.8m/s^2 is equal to the gravitational pull on earth at sea level.

is equal to the rate of rotation squared,

No, it's the linear velocity squared. You can find linear velocity by taking the circumference and multiplying it with the rate of rotation.

divided by the radius of the object rotating.

The distance from the center of mass to whatever perticular point you are trying to calculate.

Correct or am I a dumb fuck?

You are not correct, however you are certainly not a dumb fuck. A dumb fuck is not someone who ask questions, but someone who assumes they are always right regardless.


Here's an example. Suppose you have a Babyon 5 style space station. The rotation section has a diameter of 500 meters, a radius of 250 meters. This station rotates twice every minute. Suppose you want to know the how many Gs are experienced by someone standing on the lowest floor.

First you need the linear velocity at that point. The lowest floor is farthest from the center of mass, or 250 meters. We can then calculate the circumference, it is the product of the diameter and pi. (500m)(3.14)=1570m. Take that and divide it by the rotational velocity of twice per minute, or 1/30 rotations per second. (1570m)(1/30s)=52.3m/s

Now, we take our origional formula a=v^2/r, and plug n' chug. a=(52.3m/s)^2/(250m)=11.0m/s^2. Since one g equals 9.8m/s^2, that person standing on the first floor is experiencing 1.1g.

Thanks man, I hope I got it now.

[NecronLord]

The biggest problem with sci-fi writers is that they don't get under what circumstances rotational gravity will actually work. You can't just set something spinning and magically get gravity, like in Babylon 5.

Here's how a to generate gravity with a space station and rotational motion. First, you must build the station in a circular shape. Now, let's say we build it to look like Babylon 5 (a giant turd in space), but let's assume that it's more uniform. Just a giant cylinder. Now, let's set it spinning about it's long axis at sufficient velocity to produce 1g of acceleration.

Will the entire station experience 1g of acceleration due to the spin?

NO

Only the people on the inside surface will experience the pull. If you are on the central axis, you will experience no pull at all. Why? Because, as the station spins, you can observe any point on the surface to constantly be undergoing a change in linear direction (acceleration). Thus, you're not actually being pulled outside , but your body wants to keep going in a tangent line to the circle. The net result of this is the "feeling" that you're being pulled outside. This is the same effect you feel when rounding a corner in a car at higher-than-you-should speeds. If you were oriented horizontally in the car, you'd experience a pull toward the outside of the curve.

So, the bridge on Babylon 5 is grossly misplaced, and the crew should be floating.

Rama is much better, [Don't force me to read the book to check the formula fits]

Edit;

B5

People standing on the exterior will experience the spin, or rather will have the appropriate KE transvered inot them. And pay a visit to deep space, eventually

People standing on the exterior with magenetic boots, will get a lot of pain!

[Durandal]

People standing on the exterior will experience the spin, or rather will have the appropriate KE transvered inot them. And pay a visit to deep space, eventually

People standing on the surface of the station will be thrown off instantly in the absence of any kind of anchoring force.

People standing on the exterior with magenetic boots, will get a lot of pain!

Depending on how strong the magnets are, their legs may be ripped off, leaving two stumps clinging to the station .

[NecronLord]

People standing on the exterior will experience the spin, or rather will have the appropriate KE transvered inot them. And pay a visit to deep space, eventually

People standing on the surface of the station will be thrown off instantly in the absence of any kind of anchoring force.

as I said, visit deep space... in time...

You could stand on the axis, sort of.

People standing on the exterior with magenetic boots, will get a lot of pain!

Depending on how strong the magnets are, their legs may be ripped off, leaving two stumps clinging to the station .

hence the pain

[XPViking]

Be prepared for a stupid question. Wicked Pilot mentioned 9.8m/s^2 is equal to the gravitational pull on earth at sea level. Does it change? That is, would I feel more Gs if I was standing at the top of a mountain or is the effect negliable? Is this the same forumla used to calculate how many Gs a person would feel on different planets?

XPViking

[Colonel Olrik]

Be prepared for a stupid question. Wicked Pilot mentioned 9.8m/s^2 is equal to the gravitational pull on earth at sea level. Does it change? That is, would I feel more Gs if I was standing at the top of a mountain or is the effect negliable? Is this the same forumla used to calculate how many Gs a person would feel on different planets?

XPViking

First, you must know where the 9.8 comes from.

The attraction force between two masses m1 and m2 is given by

F = G*m1*m/d^2. G is constant. d is the distance between the two mass centres.

to calculate the gravitional acceleration you (mass m1) suffer on a planet (mass m) you just have to equal that force to m2*a, giving

a = G*m/d^2.

For Earth's mass and radius, you'll have a = 9.81 m/s^2.

Now, G and m are constants. But d is not. You can say that on top of a mountain d is bigger than at surface level, thus a is lower. But when you compare Earth's radius to the highest mountain you see that the difference is negligible. So, a = 9.81 can be said to be constant anywhere on Earth

[FreeWare]

So, a = 9.81 can be said to be constant anywhere on Earth

Except the Earth is a bit ellipsiodal.

So, if such a rotating device can only simulate gravity by centrepetal force wouldn't you begin to float if you jumped off the surface inside the device? Would gaining the same speed as the inner surface make you fall back on it?

[The Dark]

So, a = 9.81 can be said to be constant anywhere on Earth

Except the Earth is a bit ellipsiodal.

So, if such a rotating device can only simulate gravity by centrepetal force wouldn't you begin to float if you jumped off the surface inside the device? Would gaining the same speed as the inner surface make you fall back on it?

It would depend on the size of the habitat. If it is small enough that a human's jumping ability would cause one to (for lack of a better term) "travel to a lower gravitational region," then it is possible that they would return much slower, since gravity would be weaker. They would not float unless they jumped perfectly to the axis of spin, since there is gravity everywhere else, merely in varying amounts. A practical 1G space station would be too large for such antics, however, as a station small enough to jump-float in would have to rotate very quickly to maintain gravity, and as I mentioned before, 3 RPM is roughly the human limit.

[NecronLord]

3RPM?

you sure you don't mean 3 rps?

[Enlightenment]

3RPM?
you sure you don't mean 3 rps?

No, he's right. Three revolutions per minute is the maximum rotational rate humans can cope with. Anything higher will result in nausea.

[The Dark]

3RPM?

you sure you don't mean 3 rps?

The truly weird fact is that most people have enough trouble with a sustained 1 RPM. Think about carnival rides for a moment. Most of the spinning ones travel at probably (rough estimate, been a while since I visited a theme park/carnival) 20-30 RPM. After getting off one of those, I know I'm certainly not up to working, and no way could I do anything while on the ride. The human body is very sensitive to motion, it's just that we are so used to the normal effects that we don't notice it any more.

[Patrick Degan]

So, the bridge on Babylon 5 is grossly misplaced, and the crew should be floating.

Um, not quite. The command dome is not located at the station centre. It still experienced rotational forces so there would be some gravity there. But it would be less than 1g to be certain. Passengers on the core shuttle monorail would be at near 0g. They sort of dealt with this in the pilot movie (remember when Sinclair and Garibaldi were riding one of the cars and were in harnesses) but never again —though they did depict near-0g at the core in the second season closing episode when Sheridan had to dive out of the monorail car when two Centauri left a bomb on board.

You are correct in your general observation that the series goofed in showing a consistent 1g on all levels of the station. Some should have been <1g, other zones should have had >1g. But I mark this as a concession to artistic license, and probably one more for production convenience, since it would have increased the budget and complicated filming (or post-production editing) to show varying gravity on the station.

[Enlightenment]

You are correct in your general observation that the series goofed in showing a consistent 1g on all levels of the station. Some should have been <1g, other zones should have had >1g. But I mark this as a concession to artistic license, and probably one more for production convenience, since it would have increased the budget and complicated filming (or post-production editing) to show varying gravity on the station.

The irony of all this, of course, is that it would have been cheaper--and perfectly accurate--if they'd gone for a Rama-like smooth cylinder or the real-life O'Nieil Island Three concept rather than an ugly variable-radius monstrosity. Sometimes, respecting physics can actually save the TV people money...