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I am currently in the process of sizing my flywheel for a reaction wheel that I intend to build as part of a project. I was wondering how to calculate the maximum acceleration and torque deliverable by my motor after I attach the flywheel to it. The motor I am currently using is this Faulhaber 2610T006B, shown below:

https://fmcc.faulhaber.com/details/overview/PGR_4563_13825/PGR_13825_13814/en/SG/

I am guessing that the mass of my flywheel plays a part, but I am unsure how to do the calculations.

I am aware of the equation \$\tau\$ = \$I\$/\$\alpha\$ , I current have the \$\tau\$ that I require, so I would need \$\alpha\$ to find the moment of inertia required by my flywheel, so that I can size it to meet that requirement.

Thanks.

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  • \$\begingroup\$ Inertia, Momentum , mass, radius and rate of change in speed determine torque required dummies.com/education/science/physics/… \$\endgroup\$ – Sunnyskyguy EE75 Oct 14 '16 at 17:21
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    \$\begingroup\$ The mass of the flywheel has no effect on the maximum obtainable reaction wheel torque, it only influences how long a given amount of torque can be generated. How great the torque is at any given moment is directly proportional to the motor current, and motor current alone. \$\endgroup\$ – jms Oct 14 '16 at 17:22
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The problem - turn a cubesat using a reaction wheel.

In physics, conservation laws are very useful, as it enables you to cut through lots of error-prone and confusing calculation, and find the important thing at once.

Consider the following abstract diagram of a CubeSat and its reaction wheel (motor between them not shown)(yes, I have thought about the rotation sign convention for the diagram)(lengths of vectors not to scale)

enter image description here

Assuming no external torques (secular or magnetorquing), then at all times the sum of their angular momenta will be zero by the Conservation of Angular Momentum $$ L_{cubesat} + L_{reaction} = 0 $$

As the angular momentum = angular speed x Moment of Inertia, we can also write $$ w_{cubesat}J_{cubesat} + w_{reaction}J_{reaction} = 0 $$

If we want to spin the reaction wheel, keeping its speed less than some limiting speed (mechanical and policy), in order to spin the cubesat at at least some speed, then we can re-write this as $$ \frac {J_{reaction}}{J_{cubesat}} >= \frac {w_{cubesat}}{w_{reaction}} $$

For the numbers used in the other post, we want to turn the cubsat through 90 degrees in 30 seconds.

There are several ways to do this, a few exemplars are illustrated below

enter image description here

At one extreme, the blue curve is the conceptually simplest, to accelerate the cubesat as quickly as possible with an impulse of torque, and then allow it to coast at constant speed for 30 seconds, before braking it hard at the end. The area under the speed curve is the angle turned. Half a minute at 0.5 rpm is one quarter of a turn, or 90 degrees.

Other trajectories are possible that have the same width (30 seconds) and same area under the curve (quarter turn).

At the other extreme, the red curve shows the lowest acceleration trajectory possible. In order to get the same area under the curve, we need twice the peak angular velocity.

The yellow curve uses less torque (ie motor current) than the blue trajectory, and a lower peak speed than the red one.

The green curve is what we would see if the motor was actually unpowered during the 'coast' phase, bearing friction would slow it slightly, which would produce a small braking torque on the cubesat rotation.

Which of these trajectories is preferable depends on system mass, energy consumption, and possibly a few personal preferences. As a rule of thumb, the optimum is not to be found at either extreme.

We can calculate the required flywheel inertia and the motor torque independently.

The calculation is somewhat pre-empted by the motor having already been chosen. It has a max speed of 6k rpm, a max torque of 3.5m, and a MOI of 800n.

Flywheel Moment of Inertia

The published Jcubesat is 30m kgm2.

As an assumption, the maximum motor speed permitted for this manouevre is 3k rpm, to allow a little in hand. Other max motor speed choices can be accommodated by simply changing the MOI of the flywheel.

If we want to support a peak cubesat speed of 0.5rpm, Jreaction must be at least 1/6000th of Jcubesat, at 5u kgm2, so Jflywheel needs to be at least 4.2m kgm2.

If we want to support the slow acceleration manouever with a peak cubesat speed of 1rpm, then Jreaction must be at least 1/3000th of Jcubesat, at 10u kgm2, so a Jflywheel greater than 9.2u kgm2.

Motor Torque

We can calculate the motor torque either with respect to the cubesat or the flywheel, we don't need to do both, as we know they each have the same magnitude of angular momentum. Let's do it with respect to the cubesat.

First calculate how close we can get to the impulses trajectory

With the rated motor torque of 3.5m Nm operating into 30m Jcubesat, the acceleration is 3.5m/30m = 117m rad/s2. The required final cubesat speed is 0.5rpm = 52m rad/s. This gives an acceleration time of 52m/117m = 444mS. This time is so much less than 30 seconds, that it's not unreasonable to call this an 'impulse'.

Now calculate torque required for the lowest torque trajectory.

The final cubesat speed is 1 rpm = 104m rad/s, obtained in 15 seconds. That's an acceleration of 104m/15 = 6.9m rad/s2. The torque required to accelerate the J of 30m kgm2 at an acceleration of 6.9m is 30m x 6.9m = 207u Nm, a fraction of the motor's rated torque.

Less than twice this torque is required to get the yellow trajectory, with its corresponding saving in peak speed and so flywheel mass.

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  • \$\begingroup\$ I just noticed that if we were to use the red curve ( slow ramp), the satellite would be turning at twice the required slew rate. So if I wanted a max slew rate of 3 degrees/second, I'd have to set my slew rate in the calculations as 1.5 degrees/second, so that the highest speed is then 3 degrees/second? \$\endgroup\$ – John Oct 17 '16 at 17:16
  • \$\begingroup\$ It depends what the requirement is. If the requirement is to move through 90 degrees in 30 seconds with the low torque red triangular profile, then the max slew must be 6 degrees/s, to make the average slew rate for the 30 seconds 3 degrees/s. If, however, the requirement is for a maximum slew rate of 3 deg/s, then 90 degrees in 30 second can only be acheived with the blue impulses trajectory. Which is it? \$\endgroup\$ – Neil_UK Oct 17 '16 at 17:48
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$$M_{mot} = M_{load} + J\alpha +F\Omega$$
Where \$M_{mot}\$ is torque delivered by the motor, \$M_{load}\$ is the load torqe, \$J\$ is the system moment of inertia: motor rotor + load, \$\alpha\$ is the angular acceleration, \$F\$ is the friction constant, \$\Omega\$ is the angular velocity.

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  • \$\begingroup\$ Is Mmot the torque delivered by the motor under no load conditions? Also, how do you calculate the load torque? \$\endgroup\$ – John Oct 14 '16 at 17:53
  • \$\begingroup\$ that's the wheels and mass moving load or some other if any load, otherwise I would assume F can be used, also back EMF reduces available current with speed, no load full speed current determines F and RPM/V gives max speed no load \$\endgroup\$ – Sunnyskyguy EE75 Oct 14 '16 at 17:55
  • \$\begingroup\$ @John This is the equilibrium equation: motor vs. load, inertia, friction. It's a balance equation. If we ommit the friction and accelration, then motor torque equals to the load torque, makes sense or not? You can't calculate the load toqrque, you can measure it with torqductor. But instead calculating inertia from the torque, which you can't know, you can make a simple penduluum to measure the moment of inertia. \$\endgroup\$ – Marko Buršič Oct 14 '16 at 18:23
  • \$\begingroup\$ I understand the equation. But I can't really see how it helps me design my flywheel, I've edited my post to include more details of what I'm trying to do. \$\endgroup\$ – John Oct 14 '16 at 18:27
  • \$\begingroup\$ In the end I only know the torque that I require, I still need the angular acceleration to find my required moment of inertia \$\endgroup\$ – John Oct 14 '16 at 18:28

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