4
$\begingroup$

I am in the process of designing a reaction wheel to be used in a cubesat for my final year project. I have already chosen my motor ( Faulhaber 2610T006B ) , and am currently in the process of sizing my flywheel. But I have gotten stuck halfway through. The following is my working:

$$Desired \space slew \space rate: 3° per \space sec \approx 0.0523599 \space rad \space per \space sec$$

$$\theta = \frac12 \frac\tau J t^2$$

$$For \space 90° rotation \space at \space 3° per \space sec: \space Time \space taken = \frac{90}{3} = 30s$$

$$\therefore \tau_{min} = \frac{2\theta J}{t^2} = \frac{2\times0.5 \times \pi \times 0.03}{30^2} = 1.047 \times 10^{-4}Nm,$$

$$where \space \tau_{min} = Minimum \space torque \space required \space to \space achieve \space 90° rotation \space within \space 30s$$

$$For \space reaction \space wheel, \space \tau = I\alpha,$$

$$where \space I = moment \space of \space inertia \space of \space flywheel,$$

$$\alpha_{max} = max. \space angular \space acceleration \space of \space the \space motor(\frac{rad}{s^2})$$

$$Set \space \tau_{min} = \tau , \therefore I = \frac{\tau_{min}}{\alpha_{max}}\rvert$$

$$$$

$$$$

$$Finding \space \alpha_{max},$$

$$F = ma, \space a = radius \times \alpha_{max},$$

$$where \space m = weight \space of \space flywheel = 50g \approx 0.05Kg, \space a = linear \space acceleration \space in \space \frac{m}{s^2}$$

$$\require{extpfeil}\Newextarrow{\xRightarrow}{5,5}{0x21D2}\tau = Fdsin(\theta) \xRightarrow[\theta = 90°]{} Fd, where \space d = length \space of \space the \space rotor = 0.006meters$$

$$\therefore \tau \space should \space be \space the \space torque \space required \space to \space move \space the \space load, what \space should \space \tau \space be?$$

As can be seen from the working above, I am trying to achieve a slew rate of 3 degrees per second. But I am having trouble finding alpha max, that is, the maximum angular acceleration when my flywheel is attached to my motor. I also forgot to mention that J is the principal moment of inertia about one axis and is 0.03kgm^2.

My flywheel will be a cylindrical shape, once I am able to find alpha max, I can find my required I(moment of inertia of flywheel) and then begin sizing it.

Am I approaching this the right way? And how should I find alpha max?

I would also like to ask if I were to factor in all three reaction wheels operating at once, how much would my requirements for momentum storage change by?

Thanks!

$\endgroup$
3
$\begingroup$

Think of reaction wheel torque and momentum as separate parameters that are separately sized.

Max wheel momentum is based on the moment of inertia (MOI) of the wheel about its rotation axis multiplied by the maximum rotation rate. The max rotation rate is dictated by the electrical properties of the motor, friction, and balance. The last two depend on the design and the manufacturing. That said, you can start with a rule-of-thumb number for sizing like 6000 rpm, which is roughly what these commonly used small RWs have for max top speed. (Disclosure: I have used this companies wheels.)

You want your satellite to have a max slew of:

$$0.0523599 \space rad \space per \space sec$$

So you can back out require wheel MOI directly from that:

$$ MOI_{wheel} = MOI_{spacecraft}\frac{\omega_{max_{spacecraft}}}{\omega_{max_{wheel}}}$$

That is, the larger your spacecraft MOI, the larger your wheel's MOI needs to be; the faster you want the spacecraft to rotate the larger the wheel; the faster the wheel can spin the smaller the wheel. Holding the other variables constant of course. Plug in 6000 rpm, 3°/sec and your spacecraft MOI and you're done (use the same units of course).


You were on the right track for solving for required torque. Think of the maneuver as a trapezoid. Wheel speed will increase linearly to max speed, then sustain there, then decrease linearly back to zero. Maneuver duration in angle and time are set and you know max spacecraft rotation rate, so you can solve for required slope of curve and use that to back out required torque.

There's lots of freedom to increase the torque of a wheel by increasing the size of the electric motor that drives it and the power consumption during acceleration. Sanity check your results against existing technology.

$\endgroup$

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.