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

[mfb]

Why would you aim at the axis of rotation for a peripheral docking?

I don't say you aim for it. The direction of approach is basically given by the launch site and the orientation of the space station. If you are unlucky, then that's the direction of your approach.

But I see that the axis of rotation of the station would need to be parallel with the orbital axis.

That orientation is probably unstable, but it depends on the size, mass and so on of the space station, and I guess a detailed analysis is beyond the scope of this thread.

 

[sophiecentaur]

I don't say you aim for it.

This comment b seems to be based on that assumption.

Space is three-dimensional. If your orbital maneuver to approach the space station leads to an approach aligned with the rotation axis, you have to slow down, and then start flying tangentially. Extra delta-v. If you want to leave in the same direction, you have to slow down again and speed up in a different direction. Even more delta-v.

All that presupposes that you would be using the conventional approach and then change for a tangential approach. Is that really a likely strategy? If your system would involve slowing down and then speeding up in a different direction then it is clearly not optimal. You seem to be making a lot of assumptions about using what would be old technology in such a system. 30m/s is a small fraction of the overall orbital speed so

I didn't write "rip things apart" anywhere in this thread. A.T. wrote something like that. I can also imagine components of a docking mechanism to break if docking doesn't work properly.

Sorry - wrong guy but the level of damage in the case of a failure is inherently less than is being implied here. The fail condition will only involve the ship leaving the contact point. I would have thought that a 'weak link' would be included, to limit the stress on ship or station. Perhaps AT could contribute here.

 

[mfb]

All that presupposes that you would be using the conventional approach and then change for a tangential approach.

No it is not.

The direction of your approach to the space station is determined by orbital mechanics: your launch site and the orientation of the space station. You might be lucky and have an approach where you can use the approach velocity to dock at the rim. In that case, you save a bit of fuel. But if you are not lucky, you need extra fuel to make that work out.
In general, your approach velocity will have some component along the spin axis. You have to remove that in both docking approaches. But only in one you also have to speed up in the other plane.

30m/s is a small fraction of the overall orbital speed so

It is a part that has to be delivered in a much more controlled way. Big rocket stages can deliver 9 km/s, but they won't do that with 1m/s precision.

 

[sophiecentaur]

No it is not.

The direction of your approach to the space station is determined by orbital mechanics: your launch site and the orientation of the space station. You might be lucky and have an approach where you can use the approach velocity to dock at the rim. In that case, you save a bit of fuel. But if you are not lucky, you need extra fuel to make that work out.
In general, your approach velocity will have some component along the spin axis. You have to remove that in both docking approaches. But only in one you also have to speed up in the other plane.It is a part that has to be delivered in a much more controlled way. Big rocket stages can deliver 9 km/s, but they won't do that with 1m/s precision.

I am now more aware of the importance of getting lined up with the plane of the station. I still don't see where 'speeding up in the other plane' has to come into it. Your planned rendezvous can be made on the basis of 30m/s speed difference from the start as easily as planning for a suitable speed of approach before using retros, prior to conventional docking. Yes, the calculations are harder but that is hardly a problem these days.

 

[A.T.]

I can also imagine components of a docking mechanism to break if docking doesn't work properly.

Yes, that's what I meant.

 

[sophiecentaur]

Yes, that's what I meant.

Hyperbole is not really helpful in Engineering discussions. The term "ripping" was over the top. You could also have said that the mountings of landing gear, thrusters and other engines should be strong enough to avoid them being "ripped out". Isn't it obvious that any system would be designed to accommodate stresses? The cost of such design would be included in a detailed study.

It is a part that has to be delivered in a much more controlled way.

Oh yes. But you are presupposing that it would inherently involve more fuel for a peripheral docking. From what I understand from your useful previous points about orbital considerations, you aim to be 'near enough' (energy-wise as well as positional) with your main engines and then fine tune with thrusters. What would make a different approach speed involve using more fuel?

That orientation is probably unstable,

Is that based on some figures, or is it just an assumption? Does the ISS orbit involve keeping the hull pointing tangentially to its course? That would require some fuel to keep it 'rotating' correctly as the internal masses are re arranged. A wheel, orbiting in the way I suggest, would need no such correction energy.
Please don't just argue against this as a matter of principle and give it some unbiased thought.

 

[Al_]

I'm not sure why your concerned with a reaction wheel.

Consider the arrival of new people on the station. Imagine it's a small station, at 1g, with a risk of nausea due to rotation. To quote the pdf in the link:
"The results of this study may be summarized:
1. 1.71 rpm: very mild symptoms.
2. 2.2 rpm: one subject threw up (he had a history of seasickness) but otherwise similar to 1.71 rpm.
3. 3.82 rpm: mild symptoms and subjects adapted within a day; adaptation was longer for the less resistant subject.
4. 5.44 rpm: highly stressful (except for the deaf subject) but most adapted in a day or so. Subjects with prior rotation experience did better than those without.
5. 10 rpm: highly stressful (except for the deaf subject); subjects could not complete all tasks. There was some adaptation over the two day run."

You slow the station down for the new arrivals, then build the speed back up slowly over several days. Everyone stays at peak performance.

 

[sophiecentaur]

Consider the arrival of new people on the station. Imagine it's a small station, at 1g, with a risk of nausea due to rotation. To quote the pdf in the link:
"The results of this study may be summarized:
1. 1.71 rpm: very mild symptoms.
2. 2.2 rpm: one subject threw up (he had a history of seasickness) but otherwise similar to 1.71 rpm.
3. 3.82 rpm: mild symptoms and subjects adapted within a day; adaptation was longer for the less resistant subject.
4. 5.44 rpm: highly stressful (except for the deaf subject) but most adapted in a day or so. Subjects with prior rotation experience did better than those without.
5. 10 rpm: highly stressful (except for the deaf subject); subjects could not complete all tasks. There was some adaptation over the two day run."

You slow the station down for the new arrivals, then build the speed back up slowly over several days. Everyone stays at peak performance.

That suggests that all the crew would need to go through this procedure - unless there were two wheels; one for newcomers and one for the existing crew. The outer wheel could have zero rotation for docking, making my idea irrelevant except that the docking ship would not need to rotate itself for docking.

 

[mfb]

Oh yes. But you are presupposing that it would inherently involve more fuel for a peripheral docking.

Yes, but that statement was purely about fuel for an aborted docking maneuver. Keeping that reserve will inherently need more fuel.
For the first approach, I said that it can need more fuel, depending on the orbits. It does not have to.

Is that based on some figures, or is it just an assumption?

It is based on some basic orbital stability considerations.

The ISS keeps a fixed orientation with respect to its flight direction, which means it rotates once per orbit. It has its largest principal axis in vertical direction. That configuration is stable. Any object where this axis is not in vertical direction all the time will feel a torque. A torque on a rotating space station will (in general) make its axis of rotation change.

I guess a space station could stay stable in any orientation if it has some mass at the end of a long tether. Otherwise things get more complicated.

 

[sophiecentaur]

Yes, but that statement was purely about fuel for an aborted docking maneuver.

OH, right. That makes sense. I would have to ask what percentage of a normal 'contingency' fuel load, that this could represent.

It has its largest principal axis in vertical direction

So would a ring, orientated with its plane in the same plane as its orbit.

 

[mfb]

OH, right. That makes sense. I would have to ask what percentage of a normal 'contingency' fuel load, that this could represent.

I posted the numbers a few pages ago. It is significant.

So would a ring, orientated with its plane in the same plane as its orbit.

It would have two axes with nearly the same moment of inertia, constantly changing their orientation. Adding some mass suspended from the hub that rotates once per orbit would stabilize it.

 

[A.T.]

Isn't it obvious that any system would be designed to accommodate stresses?

It is obvious that your system in general (the station, the ships, the connection) would have to accommodate higher stresses. This leads to heavier structures, which is bad for space travel. And for the connection in particular it's also the combination of very limited time to engage properly with immediate full loading that tries to pull it apart.

 

[sophiecentaur]

is obvious that your system in general (the station, the ships, the connection) would have to accommodate higher stresses.

Why would they be greater than those acting on the outer parts of the station anyway? We are already discussing a rotating structure with all the implied stresses it would entail. I can't show that the forces on a coupling mechanism would be greater than you'd expect from 'local g'. The station structure would need a lot of built in redundancy to deal with internally generated accidents.

immediate full loading that tries to pull it apart.

Not 'immediate' in any way. The peripheral and tangential paths would part at a very leisurely rate; actually, the separation would increase only at the rate of a body falling under local g. That is hardly immediate. Can you actually support your hyperbolic description with numbers? You are ignoring the great feature of this fail safe manoeuvre. If you miss docking you move clear. If you don't quite latch on, you still move away, relatively. What sort of mass are you envisaging for a suitable coupling? A length of kevlar webbing that can tow a large yacht behind a Severn class lifeboat (under storm conditions) is only a bit thicker than a standard car seat belt. Are you being misled by the dramatic clunking sounds that we always hear on blockbuster Space movies? Perhaps you could take a look at pictures of cable car suspensions that are also designed for storm conditions and full g. The sort of strap that would probably suffice could be only a bit more massive than Tim Peak's guitar.

 

[sophiecentaur]

It would have two axes with nearly the same moment of inertia, constantly changing their orientation. Adding some mass suspended from the hub that rotates once per orbit would stabilize it.

That doesn't tie in with what I thought I understood about gyroscopes (??) The ring is constantly rotating around its axis.

 

[mfb]

Not 'immediate' in any way.

The forces acting on the connecting structure go from 0 to ( spacecraft mass)*(local acceleration) within fractions of a second. That is a very high force gradient. Plug in numbers as suitable.

If you miss docking you move clear.

Or collide with the space station. It will be a soft collision if you don't miss it by a large margin, but that applies to a hub docking maneuver as well.

It would have two axes with nearly the same moment of inertia, constantly changing their orientation. Adding some mass suspended from the hub that rotates once per orbit would stabilize it.

That doesn't tie in with what I thought I understood about gyroscopes (??) The ring is constantly rotating around its axis.

Which part is unclear?

 

[sophiecentaur]

The forces acting on the connecting structure go from 0 to ( spacecraft mass)*(local acceleration) within fractions of a second. That is a very high force gradient. Plug in numbers as suitable.

That is not right. It is only the equivalent of going from a straight road into a 100m radius bend at 70mph. (Still using the extreme 30m/s for 1g local) The sort of stress is less than for a skater dropping down onto a half pipe and in that situation. the maximum g at the bottom will be 3 or 4 g, depending on the initial drop height. I would be grateful to see your sums that produce more than local g on the ship. It would be quite possible to have some resilience in the coupling (or perhaps in the crew seats) to take care of the transitional forces - not unlike going over a humped back bridge in a motor car or even applying the brakes (1g is a realistic value for good brakes on a good road). What sort of g forces do they experience at takeoff from Earth? Where is the 'extreme' situation that you are painting?

Which part is unclear?

If the station is a symmetrical ring and it is spinning in the same plane as its orbit then is that not a stable situation? The station's axis would always point to the same star. Did you include some other factor in your model?

 

[mfb]

It is only the equivalent of going from a straight road into a 100m radius bend at 70mph.

Within fractions of a second. A roller coaster might do that along the vertical axis (and only there), with a car you certainly wouldn't drive like that.

I said force gradient, not force. The rate of change of the force is very high. See jerk.

If the station is a symmetrical ring and it is spinning in the same plane as its orbit then is that not a stable situation?

See earlier posts: Not necessarily if the wheel is not perfectly symmetric.

 

[sophiecentaur]

with a car you certainly wouldn't drive like that.

Maybe not every day but everyone is (I hope) prepared to apply the brakes when necessary. Are you really claiming that the forces (step change included) would upset anyone who's fit for a trip to that space station? What happened to The Right Stuff? In a practical space station, the local g would be much lower than 10m/s², in any case. Can we just put that "ripping" word to bed please? A bit of resilience in the suspension would reduce the trauma to a very acceptable level - We can hit a bump in the car and the suspension takes care of some really significant impulses.
Someone mentioned a Straw Man and this looks like another one.

Not necessarily if the wheel is not perfectly symmetric

Right - but that would apply to any toroidal space station and you have not given a reason that would make my idea for orientation any worse than any other. I have assumed all along that the arriving ship's mass would be comparatively low. Moreover, its arrival would be expected and could be compensated for - just as the movement of large objects within the station.

 

[mfb]

Maybe not every day but everyone is (I hope) prepared to apply the brakes when necessary.

Car brakes don't lead to jerk as fast as the docking mechanism needs.

Are you really claiming that the forces (step change included) would upset anyone who's fit for a trip to that space station?

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.

Right - but that would apply to any toroidal space station and you have not given a reason that would make my idea for orientation any worse than any other.

Every orientation would need calculations for stability. My point is that you cannot simply choose the orientation to be convenient for docking. The stability is the main consideration.

 

[sophiecentaur]

You are assuming that full docking and initial coupling have to be at the same time. That's just not the case. You never seem to take Earthbound examples as valid but the hooking on procedure can be done on a resilient link. (Arrester wire) After a convenient time, during which the ship could move itself along the rail it can dock at a port.

The station would need to be stable or a wobble would be felt by everyone, like a rolling ship. A tolerance for contact by the visiting craft wouldn't be far different from what the station staff would need.

 

[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'.