# How does the torque transfer from the reaction wheel to the satellite body occur?

Let's imagine a cubic satellite orbiting the Earth. The reaction wheels are located along the $$x,y,z$$ axes. Let's assume that the sum of the external forces is zero. Then the equation of motion along the x axis, for example, can be written as follows (let's assume that there is no motion along the other axes):

$$J_x \dot{\omega}_x+J_{rw1} \dot{\omega}_{rw1}=0$$ (if the formula is incorrect, please correct me)

where $$J_x,J_{rw1}$$ - moment of inertia on x-axis and reaction wheel, $$\dot{\omega}_x,\dot{\omega}_{rw1}$$ - accelerations.

From the formula, we can calculate the acceleration at which the satellite will rotate and the direction of this acceleration, opposite to the direction of rotation of the reaction wheel:

$$\dot{\omega}_{x} = -\frac{J_{rw1}}{J_{x}}\dot{\omega}_{rw1}$$

My question is this: how does the torque transfer from the wheel to the satellite body occur in such a system? In a classical mechanical system (for example, a gear transmission), one gear presses/pushes another, and the latter, being fixed to the main body, rotates it in turn. Is it possible to describe the torque transfer from the wheel to the satellite body by some equivalent mechanical transmission or mechanical coupling?

• "In a classical mechanical system (for example, a gear transmission), one gear presses/pushes another, and the latter, being fixed to the main body, rotates it in turn. Is it possible to describe the torque transfer from the wheel to the satellite body by some equivalent mechanical transmission or mechanical coupling?" These parts of the question really confuse me, since a reaction wheel is a classical mechanical system. Commented Jul 30 at 23:03
• @ErinAnne What was meant was that in conditional gearboxes, for example, one gear pushes another, and it is clearly visible how the force/torque is transmitted. In a reaction wheel, this transmission is not like that.
– ayr
Commented Jul 31 at 3:34

Reaction wheels are no magic, they work pretty much the same as wheels on a car: a motor spins the wheel, the reaction force/torque acts on the motor, which is bolted to the chassis, so the chassis moves. In other words, assuming an electric motor, torque is created by Lorentz force between its rotor and stator, rotor exerts torque on the (fly)wheel, stator exerts opposite torque on the spacecraft body.

• yeah, so it turns out that a moment from the reaction wheel is applied to the satellite's body, but only in the opposite direction, right?
– ayr
Commented Jul 30 at 17:56
• @ayr Opposite to the direction of the moment applied to the flywheel itself, yes. That's the "reaction" part of the name. Commented Jul 30 at 19:43
• This is the desired torque transfer, the whee's raison d'être. In reality are there two more (presumably weaker) torques that must be considered that are perpendicular to this one, like when the spacecraft is constantly rotating to maintain nadir-pointing, or even faster maneuvers of say some kind of robotic activity? It's not the point of the question, but the motor and its mount must be able to easily absorb those and transfer them to the spacecraft.
– uhoh
Commented Jul 30 at 21:27

The torque doesn't transfer from the wheel, but from the wheel housing and gearbox. Newton's Equal Action/Reaction Law says that the torque on wheel and housing are equal and opposite. Torque on the wheel spins up the wheel. Torque on the housing, bolted to the satellite, spins up the satellite.

• Yes, the force of action is equal to the force of reaction. In space, the force of action is the moment of the wheel, and the force that reacts to it is the force that freely turns the satellite in the opposite direction.
– ayr
Commented Jul 31 at 17:03

This answer is also a response to and affirmation of @ErinAnne's comment:

"In a classical mechanical system (for example, a gear transmission), one gear presses/pushes another, and the latter, being fixed to the main body, rotates it in turn. Is it possible to describe the torque transfer from the wheel to the satellite body by some equivalent mechanical transmission or mechanical coupling?" These parts of the question really confuse me, since a reaction wheel is a classical mechanical system.

I think that the OP's "classical" really means "conventional" or "traditional" way of transmitting torque, like for example a "Power Take-Off" or PTO.

source click for larger

Power take-off (PTO) at the rear end of a John Deere tractor (year of manufacture 2001)

It turns out that yes, if we put a bright green1 tractor in space and attach a flywheel to its PTO, we'd have a nice model for a reaction wheel. If the tractor wants to spin up or spin down one way, it applies a torque (either through power or braking) to the wheel in the opposite way.

The only difference is that for a spacecraft's reaction wheel, the engine and gearbox would probably be replaced with something like a DC torque motor, the same way electric vehicles replace engines and transmissions with a direct-drive electric motor.

• You are right. The gearbox has high-speed shaft with a small gear, and low-speed shaft with a large gear. A motor is installed on the high-speed shaft. When it rotates, the torque from the high-speed shaft is transmitted to the low-speed shaft. In this case, in a toothed engagement, the rotation of the gears occurs in different directions automatically. And if we fix the large gear to the body of the spacecraft, we can control its position. Here, even such a concept as "reactive torque" for is absent. And in fact, this entire structure is really very similar to a reaction wheel, as you noted.
– ayr
Commented Jul 31 at 3:43
• @ayr what do you mean "reactive torque" is absent? Reactive torque is just a restatement of Newton's Third Law, it isn't absent in any mechanical system that obeys the laws of physics. Commented Jul 31 at 3:47
• @ErinAnne I mean, this concept is almost never used when working with gearboxes. Even when a gearbox is designed, all forces are simply called torques. And the physical meaning of this force is not a reaction to the application of another force.
– ayr
Commented Jul 31 at 3:53