# Thread: Artificial Gravity

1. Newbie
Join Date
Apr 2004
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2
My question is fairly straightforward but I cannot seam to find an answer that does not require a degree in higher mathematics. What is the simplified formula to compute the artificial gravity on a spinning space station? If the station is 1000 meters across and rotating at 1 rpm what is the gravity? And does that scale up, would a 1500 meter station rotating at 1.5 rpms have the same gravity?

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Charles Sizemore
Morality; is doing the right thing and not caring if any one is watching.

2. Member
Join Date
Apr 2004
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27
This is a fairly straight-forward problem, once you have the main equation. The equation that you need is the equation for centripetal force -

F = m W^2 r

or, perhaps more usefully, in terms of acceleration,

a = W^2 r

where F is force, a is acceleration, W is angular velocity (in radians per second) and r is the radius of the rotation.

To create a centripetal force equal to gravity, the acceleration must be equal to 10 m/s/s. Using the numbers you give above, the &#39;gravity&#39; would be as follows:

1 rpm = 0.01666 rev per second = 0.105 radians per second

a = (0.105)^2 x 1000
a = 11.025 m/s/s

So, gravity would be a bit too strong in this space station. As you can see from the equations, the alternative station that you describe (1500m, 1.5 rpm) would have much stronger artificial gravity - in fact, it would be 37m/s/s - almost 4 times normal gravity&#33;
The larger the wheel is, the slower it needs to rotate to give the same effective gravity.

3. It has not much to do with higher math, only with understanding the basic principles of gravity.

- 1 g = 9,8 m/s*s, which is nothing more then accelleration.
- On a spinning wheel, the a always happens towards the center of the wheel.
- the formula for a is: a = v*v/r.
- the formula for v is: v = 2 * pi * r / Totaal time for one rotation.

Combining the formulas would give you:
a = 39,5 * (r / T^2)
.... (to calculate it in gees, just divide a by g.)

As you can see in the Formula, a is dependend of (r/T^2)
To achieve the same a with different radius, you only have to make sure that:
(r1/T1^2) = (r2/T2^2) or T2^2 = T1^2 * (r2 / r1)

As you can see from T2^2 = T1^2 * (r2 / r1), your assumption is not correct.

4. Newbie
Join Date
Apr 2004
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Originally posted by SpaceErnie@Apr 27 2004, 08:29 AM
This is a fairly straight-forward problem, once you have the main equation. The equation that you need is the equation for centripetal force -

F = m W^2 r

or, perhaps more usefully, in terms of acceleration,

a = W^2 r

where F is force, a is acceleration, W is angular velocity (in radians per second) and r is the radius of the rotation.

To create a centripetal force equal to gravity, the acceleration must be equal to 10 m/s/s. Using the numbers you give above, the &#39;gravity&#39; would be as follows:

1 rpm = 0.01666 rev per second = 0.105 radians per second

a = (0.105)^2 x 1000
a = 11.025 m/s/s

So, gravity would be a bit too strong in this space station. As you can see from the equations, the alternative station that you describe (1500m, 1.5 rpm) would have much stronger artificial gravity - in fact, it would be 37m/s/s - almost 4 times normal gravity&#33;
The larger the wheel is, the slower it needs to rotate to give the same effective gravity.
Man a realy regret not paying attintion in math class, I am about to ask some really dumb questions but that is the only way to correct my ignorance.

> This is a fairly straight-forward problem, once you have the main equation.
> The equation that you need is the equation for centripetal force -
>
> F = m W^2 r

? F is force, and W is angular velocity, a radian is 57.3 degrees, what are m and ^

> or, perhaps more usefully, in terms of acceleration,
>
> a = W^2 r
>
> where F is force, a is acceleration, W is angular velocity (in radians per
> second)and r is the radius of the rotation.
>
> To create a centripetal force equal to gravity, the acceleration must be
> equal to 10 m/s/s. Using the numbers you give above, the &#39;gravity&#39; would
> be as follows:

? m/s/s

> 1 rpm = 0.01666 rev per second = 0.105 radians per second

? how did you get the 0.01666 revs per second

> a = (0.105)^2 x 1000
> a = 11.025 m/s/s
>
> So, gravity would be a bit too strong in this space station. As you can see
> from the equations, the alternative station that you describe (1500m, 1.5
> rpm) would have much stronger artificial gravity - in fact, it would be
> 37m/s/s - almost 4 times normal gravity&#33;
>
> The larger the wheel is, the slower it needs to rotate to give the same
> effective gravity.

Sorry for any frusteration at my stupidity and thank you for your response.

5. Moderator
Join Date
Jul 2003
Posts
4,555
The ^2 means squared. For example, 10^2 = 10 x 10 = 100, or you may need to cube a number, 3^3 = 3 x 3 x3 =27.

m = mass.

In the equation, you would multiply the mass, times the angular velocity squared, times the radius (one half of the distance across the widest part of the wheel) of the wheel.

If I I remember my physics classes correctly:

m/s/s is meter per second per second, Earth&#39;s gravity is 9.8 m/s/s. If you drop something it will accelerate towards Earth at 9.8 meters per second each second

6. Established Member
Join Date
Jan 2004
Posts
300
Question then. To shoot a rocket into space, you need to take this negative effect of 9.8 m/s/s into account. Basically, you have to move at a spped greater than 9.8 m/s/s in order to get any lift at all?

7. that&#39;s something you can calculate yourself sphinx.

If you give a bullet an initial speed of X, then this speed will be reduced with ~10 m/s/s

Say X = 900 m/s or three times the speed of sound.
then it would take 90 seconds to reduce that speed to zero.

Only thing you have to do is sum every speed the bullit has passed, (whole speeds is sufficient enough) and average them.

So at first you have 900 m/s and at 90 seconds you have 0 m/s
The most simple formula to sum sequential things up is:
(first speed + last speed) multiplied by (half of total number of speeds or in this case seconds)

(Sum 900 to 0) = 45 * 900 = 40.500 meters.

So the bullet came only ~ 40,5 Km high in the Air.

if you throw something up with an initial speed of 0,5 meters a second, it just gets less high.

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