Artificial Gravity - No Gravitics

[CarlP]

Howdy!

The thread on near-future, Solar System-based campaigns really spoke to me. I'm working on developing one of them right now. I'm gathering together all of the vertically-oriented spaceship designs, since I don't want artificial gravity in my universe (well, okay, in my solar system.)

One of the things I'd like is rotating space stations, for inducing gravity. The question is, does anybody know of a formula for calculating this? I've got the old Traveller research ship deckplans, which is a classic wheel-shaped. I'd like to do other plans, of different sizes. I'd like to know what sort of rotational speeds (approximately) would be necessary for 1.0 G, and maybe even for lesser G levels, too.

I've got a bunch of sites I've found, but reading them makes my head hurt. So I'm asking here, in case anybody has encountered something like this. Thanks.

Carl

 

[Psion]

Originally posted by CarlP:
One of the things I'd like is rotating space stations, for inducing gravity. The question is, does anybody know of a formula for calculating this? I've got the old Traveller research ship deckplans, which is a classic wheel-shaped. I'd like to do other plans, of different sizes. I'd like to know what sort of rotational speeds (approximately) would be necessary for 1.0 G, and maybe even for lesser G levels, too.

It depends on the radius.

What you want to do is set the centripetal force* to the force due to gravity (weight). Since your mass doesn't change, you have to set the centripetal acceleration equal to the acceleration on the earths surface due to gravity. Long story short, I think (someone check my numbers) the numbers crunch out like this:

T = 2*pi*(r/g)^.5

T = period (number of seconds for 1 revolution)
r = radius
g = simulated gravitational acceleration, 9.8 m/s^2 = 1g

in units of g's and meters:

T = 2.006 * (r/g)^.5

where r = radius in meters, g = "earth g's", and T = seconds.

Quick test: 1 g, one mile space station. Should be about 80 seconds. (1 mi = 1609 meters)

T = 2.006*(1609/1)^.5 = 80.47 seconds. Yup. Try that.

* - note for phsyics head onlookers -- yes I know it's not a real force. Thx.

 

[robject]

Hey, how much volume does a one mile radius space station displace? Assume the wheel's cross-section is 3m 'tall' (the floor is the outer wall, you know) and, ah, how about 15m wide?

That would be the outer cylinder minus the inner cylinder, right? So...

Let's just say the inner radius is 1600m and the outer one is 1603m, ok? Rough estimates are no problem.

h = 15m
pi x h = 47m
r1 = 1600m
r1^2 = 2,560,000sqm
r2 = 1603m
r2^2 = 2,569,609sqm


Ring Volume = pi x h x ( r1^2 - r2^2 )
= 47m x 9609m^2
= 451623m^3
= more than 32,000 tons!

Nice starport...

 

[CarlP]

Sweet! Thanks so much for that information. It looks exactly like what I was looking for.

Although, after posting, I found out that the formulae that I was seeking are in Heinlein's "Space Cadet", which is sitting on the shelf in my library. I need to read the classics more.

Again, I appreciate the response.

carl

 

[Evo Plurion]

CarlP - Don't forget about the coriolis effect. On spinning space stations, it's the apparent force caused by the fact that your head is moving slower than your feet because it's closer to the station's centre. The rate of rotation can only go so high before the coriolis effect begins to induce dizziness and nausea. I recommend that you have a look at Winchell Chung's website, Atomic Rockets, where he's assembled lots of really useful information on realistic spacecraft design, artificial gravity, rocket engines, etc.

Good luck!

 

[Psion]

FYI, the smaller the space station, the worse the coriolis effect. It also has the effect of making it harder to run in the direction of rotation.

For more info, see here and here.

 

[robject]

OK, the big question, then, is: what size station would reasonably minimize the coriolis effect? Anyone have biometrics on this?

Ten miles' radius? More?

Anything that large will probably have a counterweight on one side, and only a percentage of the 'hull' as a living area. Hey -- that's perfect, because that makes lots of room for fuel storage. Or how about a large hydroponics and/or park area with a curved lake?

Sounds like the optimal rotating station is also its own ecosystem.

Let's see... 10 miles, appx 16,000m. In this case, might as well make the cieling 30m 'tall', and the 'floor' 100m 'wide'... gonna be truly gargantuan...

h = 100m
pi x h = 314m
r1 = 16,000m
r1^2 = 256,000,000sqm
r2 = 16,030m
r2^2 = 256,960,900sqm

Ring Volume = pi x h x ( r1^2 - r2^2 )
= 314 m x 960,900 m^2
= 301,722,600 m^3
= more than 22 million tons!!

Egads...

Maybe another solution would be to spin the station at a slower speed, perhaps to simulate 0.7 G's for instance, or even 0.5 G's. That would allow smaller stations to be built.

 

[Psion]

Originally posted by robject:
Maybe another solution would be to spin the station at a slower speed, perhaps to simulate 0.7 G's for instance, or even 0.5 G's. That would allow smaller stations to be built.

Probably fairly common. Not only for that reason, but also limits in material strength.

 

[Evo Plurion]

And on visitors and travellers embarking on low-g spacecraft or planets/moons.

 

[Telsor]

I think if you're looking at a radius like that, you have to be realistic and say you're not going to have a 'ring'...more a bar-bell type with 2 living areas at the ends of a struct..the strut needs only be big enough to hold the 2 ends together..

You might also have a low/zero-g section in the middle for docking/science, and some kind of communications/transport to link these up ( probably set with counterweights ).

So the 'ship' would look like this.

(-------O-------)

Exact size depends on materials available, g desired ( required? ) and how much room you want in each section. Just remember, the ends need to match, although one could be deadweight ( an asteroid? ).

 

[TJP]

Originally posted by robject:
OK, the big question, then, is: what size station would reasonably minimize the coriolis effect?

I remember reading from somewhere (helpful, eh?), that in order to prevent disorientation and nausea (caused by too fast rotation speed), the maximum rotation speed would be 1 RPM (for general populace).

Knowing that, you can determine the maximum felt gravity on your station when minimizing the coriolis effect, i.e. the T in the formula above is 60, plug in your station's radius (r) and you just resolve g.

 

[GBoyett]

Here's what Transhuman Space says about torus stations and spin gravity

Torus Hulls
(Note these units are in feet with 1dton = 500 cubic feet, opposed to 14m^3)

Spaces(dton)=R * r^2 * 0.04
R - Torus Radius r - Cross Sectional Radius

To convert to other units
Spaces(dton) * 500 = Volume(cf)

Volume(cf) * 0.0283 = Volume(m^3)

Spin Gravity

Maximum Spin Gs = R/300
R- Radius of spin path

For larger stations (absolutely no motion sickness)
Max Gs= R/1000