4
$\begingroup$

The purpose of a magnetic field is to shield against solar winds, which are relativistic ions produced by our Sun. This is a typical O'Neill cylinder with 4 islands. I'm looking for a cost effective and reliable way to simulate Earth's strength magnetic field for an O'Neill cylinder, which does not interfere with any instruments.

sketch of O'Neill cylinder

Image is taken from bagtagger@deviantart.

$\endgroup$
  • $\begingroup$ What instruments are you talking about? Having a magnetic field will always interfere with any readings. Instruments on earth are built to compensate for our magnetic field if necessary - your cylindrists (witty me) will have to do the same - but that would be obvious, so what's the actual question? $\endgroup$ – dot_Sp0T Sep 26 '17 at 6:46
  • $\begingroup$ I agree with @dot_Sp0T: what is the issue? Compensating for the magnetic field of a ship is ancient technology. If you know what kind of field you are creating, then you simply subtract that from your readings. There is no issue here. $\endgroup$ – MichaelK Sep 26 '17 at 6:49
  • $\begingroup$ @dot_Sp0T: compared to Earth the O'Neill cylinder is much smaller and to erect and maintain an Earth strength magnetic field at a certain height will be difficult. $\endgroup$ – user6760 Sep 26 '17 at 6:55
  • $\begingroup$ @user6760 so you're actually looking for a way to generate a stable magnetic field :D $\endgroup$ – dot_Sp0T Sep 26 '17 at 6:57
  • $\begingroup$ @dot_Sp0T: yes, most probably artificial one but should be at least as useful as Earth's. $\endgroup$ – user6760 Sep 26 '17 at 6:58
3
$\begingroup$

You have better to place your shield far from the cylinder (sunward, of course) and have the cylinder itself in the magnetotail, something similar to the NASA proposed gimmick to protect Mars terraforming.

This way you will:

  • need a weaker field
  • have it far from habitat and thus:
    • have a far weaker local magnetic field
    • have a far more uniform local magnetic field (way easier to compensate)
$\endgroup$
  • 1
    $\begingroup$ One thing the OP seems to forget is the O'Neil colony designs relied on the mass of the construct itself to provide protection from radiation. A magnetic shield would reduce the mass required for the colony, but add complexity in terms of extra equipment and station keeping of the magnetic shield device. $\endgroup$ – Thucydides Sep 26 '17 at 9:39
0
$\begingroup$

It will always interfere with instruments, but so long as it's either constant or predictable, then it can be filtered out from the instruments readings.

This is how a lot of earth based astronomy is performed, as earth has a magnetic field, it's atmosphere gets in the way of our images (and spectroscopy etc). But readings take from Earth are still useful because we can predict the impact on the readings, and subtract them from the result.


So how to generate it? A nuclear reactor and an electromagnet is one way - but that'd be power hungry. However you would have to constantly feed power into whatever magnetisation system you're using. Each particle it turns away from the ship will weaken it slightly. [work = force * distance. Force != 0 (it turns away), distance != 0 (the particle is moving while force is applied) therefore work != 0. If work != 0, then energy input is required].

Probably more efficient than a nuclear reactor and electromagnet running all the time is to use the reactor/electromagnet to magnetise a bar of iron. Then you can turn the reactor off until the iron is demagnetised to no longer be useful. You remagnetise it, and off you go again.

$\endgroup$
0
$\begingroup$

If you make the hull in a ferromagnetic alloy then the spin you put on it to generate internal pseudo-gravity will generate a magnetic field for the habitat as well. As to instrumental stability issues I think you've got to pick the lesser of two evils, is solar bombardment shorting out circuits more of an issues or magnetic distortions from the shielding? Any instruments that work reliably on Earth is probably going to be okay, really sensitive instruments are going to need a Faraday Cage but they probably do anyway.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.