I'm trying to run a Matlab simulation for desaturation of a reaction wheel using magnetorquers.

But there is still something that I'm not entirely sure of :

I know that when the reaction wheel reaches its max speed ( in terms of rpm), it becomes saturated and will have 0 acceleration. Assuming that you stop the reaction wheel motor after saturation, will the angular velocity generated by the reaction wheel still cause it to turn? I think this is true as it would then justify using magnetorquers to externally counteract this angular velocity, but I am not entirely sure.

Also, this is my current plan to desaturate the reaction wheel:

When the reaction wheel is saturated, a control signal will be sent to active a P controller with the control law : m=-Kp(href-hw) where m is the magnetic moment of the magnetorquer, hw is the momentum of the reaction wheel and href is a reference momentum. This P controller would then be connected to a magnetorquer ( Let's call the this torque tmag).

tmag would then be added into the dynamic equations for the satellite.

Would this work?


  • $\begingroup$ "Assuming that you stop the reaction wheel motor after saturation, will the angular velocity generated by the reaction wheel still cause it to turn?" Why wouldn't it keep turning? Shouldn't it keep turning if you stop prior to saturation as well? How is this situation different from plain inertia? $\endgroup$ – called2voyage Sep 21 '16 at 16:41
  • $\begingroup$ For the record, I did upvote. I love having a Matlab question. I'm just a little unclear on your scenario. $\endgroup$ – called2voyage Sep 21 '16 at 16:41
  • $\begingroup$ I'm a little confused because on earth frictional forces would cause the wheel to stop rotating after awhile, but that would not happen in space since there is no friction, or gravity. $\endgroup$ – John Sep 21 '16 at 16:43
  • $\begingroup$ That's not quite true. There can still be friction in space where surfaces are in contact for reasons other than gravity. The question is what are the possible sources of friction on the reaction wheel. My guess is that it is not enough to keep from spinning for a while. $\endgroup$ – called2voyage Sep 21 '16 at 16:45
  • $\begingroup$ Ah, so it would mean that the reaction wheel wouldn't stop spinning even after the motor is stopped. Which is why one would need to desaturate it, especially when it reaches saturation. In that case, would my proposed Matlab plan for the desaturation controller work? $\endgroup$ – John Sep 21 '16 at 16:48

If a reaction wheel is spinning at maximum speed and you turn off the motor it will slowly decelerate because of friction. Potentially, depending on how it's designed, it might regeneratively brake as well, which will convert the stored momentum back into electricity. Because of the conservation of angular momentum, as the wheel slows down, the satellite will experience angular acceleration in the opposite direction.

The purpose of torque rods is not really to de-spin wheels that have spun up as the result of a maneuver. Torque rods are used to desaturate the momentum that builds up in a spacecraft over time due to external disturbance torques caused by effects such as atmospheric drag, gravity gradients, and solar pressure. These torques are small, but integrate up and if left unchecked will eventually require wheels to be spun fully up, which makes the wheels useless for maneuvering.

There are lots of good papers on the canonical control laws for torque rod control. A common algorithm goes by the name B-dot, because the commanded dipole is proportional to the time-derivative change in the local Earth-generated magnetic field. Here's one example.

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    $\begingroup$ Nice discussion, so yes or no - would it work? It's (roughly) the same amount of angular momentum - a) bringing a saturated wheel down to zero and b) constantly torquing the spacecraft to keep the wheel near zero for a time long enough that it would have saturated if you hadn't. However, operationally they are different, but are they different enough that the proposed scheme would not work? $\endgroup$ – uhoh Sep 27 '16 at 5:19
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    $\begingroup$ The typical approach is not to wait until a wheel is saturated to activate torque rods but to have them active essentially all the time (as long as the momentum is above some hysteresis band). If you let your wheels saturate then you've lost control authority. Torque rods take time to work, so best to plan ahead. Personally I'd stick to one of the common control laws because they're known to work $\endgroup$ – Adam Wuerl Sep 29 '16 at 5:31
  • $\begingroup$ Sounds very good and I agree. The question is not "what is the best way" or "how would you do this" though. It's great to give guidance and I hope it's given the OP's attention. Can you add a yes or no somehow - it would work (poorly and not reliably) or it would not work? $\endgroup$ – uhoh Sep 29 '16 at 10:14
  • $\begingroup$ @AdamWuerl I apologize for the long absence, I was having my mid term tests. Thank you for the clarification, I do have a clearer idea now. So the purpose of the torque rods is not really to reduce the momentum of the reaction wheels, but to directly reduce the overall momentum of the satellite. But from your reply, it seems like torque rods only serve to counteract disturbances, but aren't they used to prevent reaction wheel saturation as well? Also, I have heard of B-dot , but that's used for de-tumbling and not de-saturation. $\endgroup$ – John Oct 2 '16 at 12:11
  • $\begingroup$ Decreasing wheel vs. satellite momentum are similar: satellite momentum is the vector sum of the wheels and everything else. Closing a control loop around either will likely work. Wheels will spin up due to disturbance torques, so desaturating the body momentum desaturates wheels too. You're right that B-dot doesn't work here, so people typically use M × B control. That is shorthand for observing that torque is the cross-product of dipoles from rods and Earth, then using that to pick M so the resultant torque is as desired per torque rod math. $\endgroup$ – Adam Wuerl Oct 3 '16 at 4:13

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