Space station artificial gravity - how to spin up to speed?

[mfb]

You are assuming that full docking and initial coupling have to be at the same time.

Replace "docking" by "initial coupling" if you prefer that phrase, that doesn't change my point.

 

[A.T.]

In the right orientation they would be okay, but that was not my point. They lead to high stress and design challenges for the docking mechanism.

It affects the design of the entire ship, to allow it be caught like this. Not just structural stability, but for example also its center of mass vs. position of the coupling mechanism.

 

[sophiecentaur]

It is pretty obvious that ships would be designed with docking in mind. The coupling arrangement would be in an appropriate place. Balance and 'swinging' would be factors but surely not difficult to deal with.
As for the 'shock' from a step function in acceleration, it's important to use some numbers. In 1s the station will rotate by 30m and the displacement of the docking point from the tangent course will be about 1m. Spreading the 'jerk' over 1s would be far more than needed but that's what 1m of stretch could achieve. And this is all based on 1g local gravity. How is this an issue?

 

[jbriggs444]

In 1s the station will rotate by 30m and the displacement of the docking point from the tangent course will be about 1m.

So you are assuming 1/5 gee at the docking radius, a station with a 30 m/s tangential speed and a docking radius of 450 m?

But you say that this is based on 1g local gravity. Please justify that remark.

 

[sophiecentaur]

So you are assuming 1/5 gee at the docking radius, a station with a 30 m/s tangential speed and a docking radius of 450 m?

But you say that this is based on 1g local gravity. Please justify that remark.

I was basing the 30m/s on a 100m (ish) radius and a required g of 10m/s², which may be unreasonable but I think the answer is right. I can see all sorts of reasons not to use those figures - mainly that Earth g is, apparently, not necessary.
This has been a long thread and the Kubrick dimensions crept in there at some stage. I stuck with a pessimistic 1g requirement because the stresses on any coupling mechanism would, in practice, be significantly less.

 

[jbriggs444]

I was basing the 30m/s on a 100m (ish) radius and a required g of 10m/s², which may be unreasonable but I think the answer is right.

The answer is wrong. Work it out.

Hint: SUVAT. 1 g over 1 second does not give one meter displacement.

 

[sophiecentaur]

The answer is wrong. Work it out.

Hint: SUVAT. 1 g over 1 second does not give one meter displacement.

I don't believe it's suvat because there is no actual gravitational acceleration on a 'falling' body. It's geometry at work as there's no field. A point on the periphery is about 1m away from the tangent after 1s. It approximates to two similar triangles with the angle of rotation being the same as the angle between the initial and final tangents.

 

[jbriggs444]

I don't believe it's suvat because there is no actual gravitational acceleration on a 'falling' body.

You are incorrect in this assessment.

Gravity is locally indistinguishable from a uniformly accelerating frame. A rotating frame is locally indistinguishable from a uniformly accelerating frame. SUVAT works for them all.

[That's why we have the station spin in the first place -- to set up something that is locally indistinguishable from gravity]

We can attack the problem in the other direction. You have a station with 450m radius and 30 m/s tangential velocity. The centripetal acceleration is given by $\frac{v^2}{r}$. Does that work out to one gee?

 

[A.T.]

A point on the periphery is about 1m away from the tangent after 1s. It approximates to two similar triangles with the angle of rotation being the same as the angle between the initial and final tangents.

Show your work.

 

[sophiecentaur]

I get the principle of that and I did consider it but what is wrong with the geometry? Also, until the ship is actually attached to the ring, it's not rotating at the same rate.
The illusion of gravity is limited. If you release an object from the periphery it will not follow a path that appears to be vertically 'downwards' That would have to the be radial but it's tangential and with no acceleration. Suvat fails for any appreciable distance of 'fall'.

 

[jbriggs444]

The illusion of gravity is limited. If you release an object from the periphery it will not follow a path that appears to be vertically 'downwards'

On the contrary. It will follow a path that appears to be vertically downwards -- at least until it attains a velocity significant enough for Coriolis to rear its head.

If you are standing in front of the counter on a space station trying to make an omelette and you drop an egg, it will land on the floor at your feet -- to a first approximation.

 

[sophiecentaur]

Show your work.

I can't draw diagrams on my iPhone, I'm afraid but my description of the diagram is reasonably accurate.
Must reconsider, I think.

 

[A.T.]

I can't draw diagrams on my iPhone

What radius and tangential velocity did you assume to get that 1m separation after 1s?

 

[jbriggs444]

30 meters/sec on a 100 meter radius is 0.3 radians after 1 second. Cos of 0.3 radians is 0.955. That's 4.5 meters separation. As expected for 0.9 gee by SUVAT.

Edit: $\frac{v^2}{r}$ is 9.0 meters/s² for that setup.

 

[sophiecentaur]

That figure I came up with was clearly rubbish (back of envelope let me down - sorry chaps) and the distance would be something like 4.5m. But that approach was totally unsuitable. What needs to be considered is how to deal with the effect of a step function of 0 to 1 g. This is the equivalent of what the suspension has to deal with if you suspend a car with its wheels not quite touching the ground and then let go. That is a trivial requirement and all cars will cope easily with that. It's essential for protecting the road surface as much as anything. The advantage for the docking mechanism is that it can afford to move much more than a car suspension. It can be a lot softer and would involve a much lighter weight mechanism. And, of course, from what's been written here, we are talking in terms of g/5, rather than g.

 

[David Pass]

If a space station has artificial gravity created by spinning, how can it best be spun up to speed? Little attitude rockets could do it, but they would use up fuel, and limit your ability to change the spin rate in future. What if you had an external wheel that you spin up very fast in the opposite direction? If you did this, would it cancel out the gyroscope effect because the total angular momentum of the whole would be zero, and make it easier to orient the station with respect to the Sun, spacecraft , etc.?

Could Feraday's Law of Induction be employed to generate sufficient energy to sustain gyroscopic motion? Would this effectively produce a perpetual motion device? I picture a system of tubes within the structure of the gyroscope. I'm thinking, though, that the structure would be to large to be practical.

 

[mfb]

Could Feraday's Law of Induction be employed to generate sufficient energy to sustain gyroscopic motion?

Using induced voltages would slow the rotation. That is the opposite of what we want.

You could use electric power (from solar cells) to induce a current in a tether, and use that to generate torque. JAXA tried that recently, although not for rotation.

Would this effectively produce a perpetual motion device?

There is no such thing.

 

[Sue Rich]

Has anyone considered doing penetrating ground radar? Is it possible the moon or Mars has fossil fuel like we do? If so couldn't that be used to run the artificial gravity mechanism?

 

[sophiecentaur]

Fossil fuels come from long dead life forms. There is no evidence (or reason to believe) that the Moon had life at any time.

 

[mfb]

And even if there would be hydrocarbons, there is no free oxygen to burn it.

Artificial gravity is interesting in space, but not so much on Moon/Mars.