Hard sci-fi question: rotational artificial gravity space station

Quickleaf · Apr 17, 2016

Taking a Bernal sphere for example, what would conditions on a space station be like in regards to gravity?

From NASA and Wikipedia sites, I've read that some areas of the station would spin (have artifical G) and others would be stationary (zero-G). For example, the "residential sphere" would spin, and centrifugal force would simulate gravity so buildings would be constructed on the inner walls of the sphere.

Would centrifugal force artificial G effectively operate like the gravity we experience? Or would there be some sort of weird curve that falling objects would experience?

Within the spinning sections, there'd 1/3rd to Earth-normal G (depending on specifics) along the "equator" of the section, with diminishing G the closer one gets to the axes. A lot of the Bernal sphere designs mention Zero-G recreation areas near the axes, though it probably wouldn't be true zero-G but more like low-G, right?

And my big question: What would it be like to be in the center of the spinning sphere? Imagine taking the sphere and dropping a tube from one axis to the other. Along that tube, what kind of "gravity" would exist? Since centrifugal force is strongest away from the center of the spinning object, would it essentially be zero-G?

Morrus · Apr 17, 2016

Assuming it's the right rotation rate, on the inner surface it'll be 1G and indistinguishable from Earth. Depending on the size of the sphere, gravity might decrease rapidly so that the second floor of a building might be noticeably lower G. Climbing ladders could be odd! If it's small enough, your head might be in perceivable lower G. With a larger sphere (like the one illustrated above), that effect would not be noticeable unless you climbed several floors at least.

The centre of the sphere would indeed be zero G.

Actually, we have a couple of pretty scientific types on the boards here, so @Umbran or @freyar might have more to add. I'm sure there's an equation or something you can use to determine the gravity at a given distance from the 'wall' for sizes of sphere/speeds of rotation.

RangerWickett · Apr 17, 2016

My understanding is that if you entered the sphere at its axis, you'd just float, and the world would spin around you.

If you went to where the equatorial plane crossed the axis, then pushed yourself toward the 'ground,' you still wouldn't feel any gravity. You'd just travel at whatever speed you accelerated to (and would slow down due to air resistance). However, the ground beneath you would be spinning, possibly quite fast. If you drifted to the ground, at some point you'd probably move into the path of something like a tree or building, and it would hit you and then drag you along the path of rotation. Only then would you start to experience gravity, as the rotational movement tries to move you away from the center, and the ground keeps you from flying away.

Check out this calculator: http://www.artificial-gravity.com/sw/SpinCalc/SpinCalc.htm

As an example, a 38 meter radius ring (239 meter circumference) rotating at 3 revolutions per minute produces an effective 1/3 G at 'ground level.' From a stationary observer it's moving at 43 kph, or 27 miles per hour. So even if there were no obstacles, if you drifted into it'd be like bouncing into a wall while traveling at 27 mph. Probably not fatal, but pretty unpleasant.

If you have the same ring and spin it almost 5 times per minute, you get a velocity of 44 mph. You probably don't want to fly into a house at that speed. So a well designed station would have ladders or elevators down from the central spur that would slowly increase your tangential velocity.

Morrus · Apr 17, 2016

That calculator is fantastic! Really useful tool for this sort of thing.

MarkB · Apr 17, 2016

It would feel effectively like normal gravity at ground level, but thrown objects might not follow exactly the path you expect, due to the rotating frame of reference. It's a tricky thing to visualise.

Quickleaf · Apr 17, 2016

Awesome feedback

Thanks!

My example is a 1 km diameter (500 m radius) "squashed" sphere.

[MENTION=1]Morrus[/MENTION] That's exactly one of the questions I'm trying to determine. How high up do you have to climb a building for there to be noticeable change in gravity. I think it would affect how high-rise type buildings were constructed, since the shearing forces (might be using the wrong term) between regular G and lower-G would require stronger building materials. Plus it might suggest activities happening at the upper levels of high-rise buildings would be substantially different...for example moving construction activities to the lower-G zones for increased efficiency.

[MENTION=63]RangerWickett[/MENTION] Really helpful on how to visualize entering at the zero-G "fixed" axis and seeing the entire station spin around you. I suspected some kind of shuttle or elevator would be necessary, but hadn't conceived of exactly why...

I plugged a 500 m radius in and got a Tangential Velocity (or "rim speed") of 156 mph, which would be "splat your dead" for anyone moving or falling from the zero-G axis to the ground...in scientific terms

[MENTION=40176]MarkB[/MENTION] That's another one of my questions. I mean, nothing we throw on Earth actually travels straight, technically. But in the rotational artificial G environment I'm wondering if it would be more obvious...or would it basically be a case of "throwing a baseball while in a moving car"? In other words, if everything/everyone is rotating at the same rate in relation to each other, there doesn't appear to be any change from Earth-standard gravity (assuming 1 g centripetal acceleration).

But what happens if I punt a football down a field or fire a railgun at the elevator/shuttle tube along the central axis when the station is rotating at 1.3 rpms and the rim is spinning at 156 mph?

Umbran · Apr 18, 2016

Quickleaf said:
My example is a 1 km diameter (500 m radius) "squashed" sphere.

How do you want it "squashed"? Here's a note: the sphere is in many ways a bad idea. A cylinder is a better bet. The centrifugal force felt depends upon the distance from the axis of rotation. So, if you spin the thing so you feel a full G at the equator, anywhere "north" or "south) of the equator has lower force. If you want more surface to be at 1 G, use a cylinder.

[MENTION=1]Morrus[/MENTION] That's exactly one of the questions I'm trying to determine. How high up do you have to climb a building for there to be noticeable change in gravity.

It depends upon the size of the sphere. If your change in elevation is negligible compared to the radius, you won't feel much difference. If you are talking about a 100 meter tall building in a 1 KM radius sphere, you'll notice it.

I think it would affect how high-rise type buildings were constructed, since the shearing forces (might be using the wrong term) between regular G and lower-G would require stronger building materials.

If you can build the sphere in the first place, building materials are probably not your biggest problem. Remember that we *are* talking about cfreating a very large structure *in space* already.

[MENTION=40176]MarkB[/MENTION] That's another one of my questions. I mean, nothing we throw on Earth actually travels straight, technically.

Well, be careful there. When we are talking about deflection due to spin, it is actually that the object is moving straight, and the Earth is spinning under it!

But in the rotational artificial G environment I'm wondering if it would be more obvious...or would it basically be a case of "throwing a baseball while in a moving car"? In other words, if everything/everyone is rotating at the same rate in relation to each other, there doesn't appear to be any change from Earth-standard gravity (assuming 1 g centripetal acceleration).

Again, the difference will depend in general on the size of the sphere - the larger the sphere, the less noticeable the effect will be.

In general, if you are throwing directly along or against the spin (so "east" or "west" - or what you'd probably want to call "spinward" and "anti-spinward"), what you'll see is your throw fall short or go farther than you'd expect. If you throw north or south, you'll see it go wide to the left or right some.

But what happens if I punt a football down a field or fire a railgun at the elevator/shuttle tube along the central axis when the station is rotating at 1.3 rpms and the rim is spinning at 156 mph?

Not really in a position to work out the exact numbers at the moment, I'm afraid. Perhaps later.

freyar · Apr 18, 2016

There's a lot of good stuff here already, so I'm not sure exactly what to add. As people have already said, if you're not already rotating with the station, you won't feel any "gravity." Also, the Coriolis effect will be a lot more pronounced. So, if you throw a ball "straight up," it will come back down but will not come back to you. From your perspective (standing at one spot on the inner surface of the station), the ball will curve off as it flies up and down.

Quickleaf · Apr 18, 2016

Umbran said:
How do you want it "squashed"? Here's a note: the sphere is in many ways a bad idea. A cylinder is a better bet. The centrifugal force felt depends upon the distance from the axis of rotation. So, if you spin the thing so you feel a full G at the equator, anywhere "north" or "south) of the equator has lower force. If you want more surface to be at 1 G, use a cylinder.

Thanks for the feedback [MENTION=177]Umbran[/MENTION]! I'd looked at the O'Neill cylinder early on. My understanding, however, was that it might be advantageous (in terms of medical health, crystal production, maybe other things I'm unaware of) to have areas of low-G or micro-G in addition to 1 G. Is that wrong?

In general, if you are throwing directly along or against the spin (so "east" or "west" - or what you'd probably want to call "spinward" and "anti-spinward"), what you'll see is your throw fall short or go farther than you'd expect. If you throw north or south, you'll see it go wide to the left or right some.

Is there an easy way (or perhaps a site) I could approximate how severe the range increase/decrease or divergence would be? My hunch is that we're talking something relatively slight, maybe 10% difference at 100 m throw/shot, and not a severe change?

Umbran · Apr 18, 2016

Quickleaf said:
My understanding, however, was that it might be advantageous (in terms of medical health, crystal production, maybe other things I'm unaware of) to have areas of low-G or micro-G in addition to 1 G. Is that wrong?

Medical health? Well, older folks might like *slightly* lower G, but not really low-G or micro-G. We have to spend a lot of effort mitigating and repairing the effects of extended low-G, rather than wanting to use it much.

There are certainly industrial uses, but all you need to do then is put those facilities *on the axis*, rather than down on the very, very valuable habitation real estate.

Is there an easy way (or perhaps a site) I could approximate how severe the range increase/decrease or divergence would be? My hunch is that we're talking something relatively slight, maybe 10% difference at 100 m throw/shot, and not a severe change?

There are a whole bunch of variables - size of the sphere, direction you throw, speed of the projectile, and so on. Even the geometry may matter if you expect the projectile goes far enough that you can't consider the ground to be flat under it. I don't know of any calculator made for it. It is all soluble stuff, but there's lots of details.

I'd just say... I'n not want to play baseball in it.