For the Falcon 9R, and the Dragon spacecraft, the use thrusters for attitude adjustments on both the stages returning and to control the rotation of the descending Dragon.

You can see the return stage thrusters at 0:40, for example.

I'm wondering what would make reaction wheels impractical here - for the desired torque, would the weight of the reaction wheel and its associated motor and hardware outweigh the weight of the fuel/gas they are using?

Or perhaps the thrusters are more reliable?

It's logical that at a certain quantity of thrust ability, a reaction wheel could weigh less than the associated fuel and thrusters and piping. Is it just that the quantity of thrust they need is relatively low?

By 'reaction wheel' I do mean any currently available non-fuel using attitude controls - reaction wheels, momentum wheels, control moment gyroscopes, (anything else?).


2 Answers 2


Torque and/or mass.

You can go with small reaction wheels which can only barely affect the attitude, or with massive ones, which... still barely affect the attitude.

They are good for satellites that stay in the orbit for years, where you can afford hours per maneuver, but need years of operation. Meanwhile, RCS will run out of fuel quite fast, but it can provide a good torque rapidly. With a tight schedule for reentry at the right trajectory, Falcon 9 needs that quick reaction.

Any docking vessels require RCS for translation maneuvers for docking - it certainly does have some reaction wheels besides that, but RCS is still a must.


Let's have a look at the article linked from the Wikipedia page that talks about "a few hundred watts and 100kg of mass... thousands of newton meters of torque"

Let's ballpark the F9 1st stage at the moment of separation - before the most torque-intensive maneuver of turning 180 degrees - as 100 ton.

Let's approximate it as a rod of length 41.2 meters. The moment of inertia will be around 14,000,000 $kg m^2$.

I don't know how many thousands of newton meters of torque the sentence in Wikipedia meant, but let's generously assume a hundred thousands.

The angular acceleration will then be 0.007rad/s^2.

$ \phi = 0.5 \alpha t^2$. We need to accelerate for 1/4 turn, then decelerate for the remaining 1/4 turn to complete a 180 degree turn. Therefore half of the time taken will occur after $0.5 \pi$ radians of the turn. $t = \sqrt{ 2\phi \over \alpha} = \sqrt{\pi/0.007} s = 21s$. Double that for the braking half of the turn, for 42s for the whole maneuver.

42 seconds is still doable within time budget of F9 launch.

OTOH, the paper shows the CMG with highest available torque commercially available to be of order of 100Nm, not 100,000Nm. A CMG of 1000Nm (ten times stronger than the top one) will then take 420s to turn the stage around.

Meanwhile, on arm of, say, 20m (41.2m/2 minus margin not to place them not on the very tips) 100,000Nm (the same 42s from above) can be achieved with two 2500N rocket engines, which is a very modest value.

  • 6
    $\begingroup$ Are KSP's SAS modules massively overpowered compared to real-world CMGs, then? I'm able to maneuver quite well without RCS in KSP - on the order of seconds, not years. Is the torque capabilities and/or RCS system weight of the Falcon 9R specified anywhere for comparison to "a few hundred watts and 100kg of mass... thousands of newton meters of torque" from the wikipedia article? $\endgroup$
    – Ehryk
    Commented Feb 24, 2016 at 9:55
  • 7
    $\begingroup$ @Ehryk The KSP reaction wheels are simply working by magic. Remember; in the real world angular momentum is a conserved quantity. $\endgroup$
    – Taemyr
    Commented Feb 24, 2016 at 10:49
  • 5
    $\begingroup$ @Ehryk: In KSP the reaction wheels are the second greatest liberty the authors have taken against realism (the first being the planet densities). They are by far more unrealistic than unlimited restarts of rocket engines, no need for food or air supply, or the jetpacks with over 500m/s of delta-V. $\endgroup$
    – SF.
    Commented Feb 24, 2016 at 14:22
  • 1
    $\begingroup$ Is there a metric of how many orders of magnitude stronger they are than today's CMGs? 10x? 100x? $\endgroup$
    – Ehryk
    Commented Feb 24, 2016 at 15:17
  • 3
    $\begingroup$ KSP's reaction wheels are mainly to make up for the fact the the game engine can't do detorquing correctly. Try an RCS-only ship with a realistic amount of RCS, tumble it, turn on SAS, and watch it deplete its RCS to zero in a few minutes. I've been able to "fix" the reaction wheels for some time but I would need the engine fix first. $\endgroup$
    – Joshua
    Commented Feb 24, 2016 at 18:46

Somebody check my math. I have checked the math, shown edits. To turn the first stage around in one minute using a 2 meter diameter steel (or lead or tungsten or depleted uranium) wheel with a mass equal to a typical payload (~5000kg), you'd need to discharge a Tesla model-S Powerwall battery (about 5 kWh) in 10 seconds to spin it up to 1800 180 RPM, then dump the energy back into the battery again a half-minute later to stop it. Assume the first stage has 10% propellants remaining.

We can take a minute to pause and enjoy that image, but the thrusters have one thing going for them.

The nice thing about thrusters is that you can put them at the ends, where you can take advantage of the length of the body (r^2). That "little" 5000kg wheel in the middle is at a distinct disadvantage.

Structural engineers will shudder at the thought of applying the torque to the middle while most of the mass is at the ends - another advantage of having the thrusters at places where the mass is concentrated.

enter image description here

  • $\begingroup$ I didn't see the edit to @SF.'s answer while I was posting this. Much better math than mine! $\endgroup$
    – uhoh
    Commented Feb 24, 2016 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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