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I was recently reading a very good sci-fi story recounting, in loving detail and with a good deal of accuracy, the first space program of an alien race, with strong (intentional) similarities to the American and Soviet endeavors. One thing caught my attention, though: a capsule in distress vented extra $O_2$ in order to spin-stabilize itself in preparation for extremely-manual retro-rocket activation (which was expected to be much less precise than normal, as the retro-rockets had to be activated with a jury-rigged flare gun). But as far as I could tell, the $O_2$ system had not been designed for the purpose, so presumably had only a single vent (rather than two deliberately placed to create spinning, which would be very strange, or two placed to cancel thrust, which would not have been useful). Would a continuous vent over some time actually create a proper spin and create a stable attitude, or would it instead just spiral outward and become uncontrollable?

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  • $\begingroup$ @SF's answer points out that if the location doesn't happen to be just the right place, the venting will add angular momentum about one axis while removing it about another. (I think I should say "along" the axis but it sounds weird - angular momentum vectors are drawn perpendicular to the plane of rotation) $\endgroup$ – uhoh Mar 12 '16 at 4:41
  • $\begingroup$ What is the title/author of that book out of curiosity? $\endgroup$ – Steve Apr 12 '16 at 20:15
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The short answer is "maybe, if they were spectacularly lucky, but it's a story, so that's OK."

Longer answer:

There is quite a lot buried in the question in terms of assumptions, but here's a go.

  • Just for a vent, if the thrust direction is radially out from the centre of mass then there will be no torque and no spin up. It would need to be off radial. Details: it would be most efficient if tangential.
  • I presume what you mean by "rather than two deliberately placed…" is that if you have two vents you could arrange them so that you get a "pure" torque with no vehicle translation. If there is just one vent then there will be some translation regardless of whether there is some torque as per the first point. If there is nothing to crash into nearby then it's probably not a problem.
  • The point of spinning for stability during a rocket motor firing is so that the thrust vector is in a known, deliberate direction and doesn't drift because of minor thrust misalignments or an initial attitude drift rate from another cause such as asymmetric drag. The significance being:
    • You have to know (i.e. measure) which direction you want to fire the retro-rocket in the first place.
    • You have to be able to point the spin axis along the direction of the intended retro-rocket firing to better than the plausible thrust misalignment. That could mean a reaction wheel (which could give you your stabilisation spin anyway), another thruster, or another vent in the right place and a means of confirming (measuring) the achieved direction of the spin axis.
    • The degree of spin-up has to be meaningful for the size of retro-rocket, potential misalignment and the amount of drift that could otherwise take place, plus drift from initial rates. This could be quite uncomfortable if crewed (think 60rpm).
  • Objects don't automatically become stable when they spin. A solid flat spinning-top would, but anything with a different shape (e.g. pencil like) and some sort of energy damping from appendages, liquid slosh etc could actually become less stable, or rather transfer its spin axis away from the desired one.

I have a dim recollection from college (a long time ago) of being taught that an early US unmanned mission actually suffered from precisely the last problem because it had whip-aerials and it was not foreseen that this would cause the pencil-shaped upper stage to become unstable.

EDIT:

See SF and Uhoh's answers for one extra constraint I missed, viz. relationship of the created spin to the axis along which the motor is mounted. My points above about measuring angles etc addressed the also vital but distinct concept of the orientation of the spin to the desired thrust vector in inertial space.

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  • $\begingroup$ Your presumption is correct, thanks. I'll wait a bit longer to make sure no one has any salient points to add, but I suspect you'll get the checkmark. $\endgroup$ – Nathan Tuggy Mar 11 '16 at 4:35
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Off-axis yes, but in-plane.

The thrust vector must be in the plane perpendicular to the spin axis and passing through the center of gravity of the rocket.

And of course oriented opposite to the spin direction.

In this picture, the spin axis goes lengthwise through the rocket, so a thruster located at the height of the center of mass of the rocket will work. enter image description here

It doesn't need to be tangent to the rocket surface (though it will be most efficient if it is), but if the thrust vector is either pointing off the plane surface, or not attached to the plane surface, the rocket, while losing its original spin, will begin entering a new spin, with its plane of rotation passing through the center of mass and containing the thrust vector.

Yes, it will introduce some translational motion too, but it will be quite minimal comparing to what you normally face with orbital motion - nothing to worry about. That would be an issue when docking, but with one thruster you're definitely not docking on your own anyway.

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Luckily their capsule was not space-station-shaped!

venting into space - uhoh!

@SF.'s answer points out that the vent needs to be on the plane containing the center of mass to avoid adding spin about a different axis. If the vent is ON the station, it better not be narrow at that point!

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  • $\begingroup$ Good point, well made, I had a nagging suspicion I had missed something out! $\endgroup$ – Puffin Mar 13 '16 at 0:56

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