0:05

Now, we're going to look at how this all simplifies.

Â If we have a single reaction wheel, we mentioned earlier,

Â if you just have these all 0.

Â Gimbal rates at zero, gimbal acceleration is zero.

Â That means this whole frame is locked down, relative to the body,

Â these earlier equations simplify to this form.

Â And that's nice.

Â So now, what we can do too is you could rewrite this.

Â We're going to use the motor torque equation to get rid of big omega dot.

Â So I'm taking that extra differential equation, I'm going to solve for

Â omega dot, back substitute it so

Â I get my spacecraft accelerations in terms of the motor torque.

Â One thing, I can show you this in identity, this term here,

Â with a little bit of algebra, you can show that it relates to omega cross gs.

Â They just, if you carry out the cross product terms,

Â that's what you come up with.

Â So that's what I used to replace this with this part.

Â 1:00

The next thing is this is the motor torque equation without gimbaling, so

Â there's not gamma tau term in there.

Â And that's good.

Â And so you can see in here, there is a Js times big omega dot.

Â Js times big omega dot.

Â I could then directly substitute this, I can take Js times this, bring it over

Â to the left hand side and substitute this back up in here, I can do that.

Â If you do that, you end up with a term here that's going to be Js.

Â This is Js times this, you're still going to have a Js over here,

Â gs, gs transposed omega dot.

Â Here we have the full inertia tensor times this.

Â And this inertia was the space craft inertia plus the gimbal and

Â wheel frame inertia.

Â 1:52

But since, when you substitute this back in there, you get this extra term.

Â It's just like this last term you have here.

Â Those two actually cancel.

Â So, for the reaction wheel form that I'm showing you, people often ignore this and

Â include that extra term, which doesn't make momentum work.

Â This will be the full answer.

Â You can then, so redefining that, I can rewrite these groups in terms a little

Â bit, but this is more the classic form that you would see for reaction wheels.

Â For reaction wheels we come up with control loss that I'll show one example of

Â then, where we come up with torque level control loss.

Â CMGs as I cover in 6010, we don't come up with motor torque level control loss.

Â We come up with gimbal rates as Mario is pointing out right?

Â We want to take advantage of that gyroscopic torque as we twist.

Â So we command a twist rate and

Â then you have a sub server system that implements that twist rate.

Â But that's bunch of lectures in the future.

Â So this is the classic equation of motion of a wheel,

Â of the spacecraft with a single reaction wheels.

Â But oriented in a very general way, gs, gT, don't have to line up with B one,

Â two and three like we did with the dual spinner.

Â 2:58

That's what you'd end up with in there, so good.

Â Now that's a single device.

Â If you do a single CMG, that's really not much, as you guys mentioned earlier,

Â this one term vanishes, but you're left with everything else.

Â This is still the form that we use because we don't solve for the motor torques,

Â we don't back substitute here.

Â In fact, this is the term that gives us the huge gyroscopic effect.

Â The big omega times the small gimbal rate,

Â which is still a very big torque acting on it.

Â So, in fact, to control the valves of this you find they come up with laws to come up

Â with these desired gimbal rates to give you all the commanded steering and

Â gyroscopics.

Â But this is the full torque level solution.

Â 3:37

Now we've done single VSCMG devices.

Â Typically you don't fly a single one, we fly clusters of four at least with VSCMGs,

Â with reaction wheels you may have three, four.

Â I've seen some with six and more.

Â Just depends on, you might use more than three for redundancy but

Â also maybe for torque.

Â Maybe a single wheel that you can get off the shelf gives you one Newton meter, but

Â your maneuvers require one and a half Newton meter.

Â So you may double up on the wheels just to get more bang there.

Â So we have to deal with clusters, systems of them.

Â And in essence everything we've done still holds.

Â It just comes down to book keeping problem if you have multiple wheels.

Â Then where we had one wheel, that is going to be a gsI, right?

Â The first gs axis, the second gs axis.

Â The same thing for the transverse and the gimbal, everything has 1 through n.

Â And there's some convenient projection matrices where I'm just making a three by

Â N matrix of the N number of CMG devices, all the projections calculated.

Â This inertia, it was before the HUB plus the single VSCMG,

Â now list a HUB plus a summation of all those, all right.

Â So we just carry out that kind of a calculation.

Â And the rest of the gyroscopic terms have grouped this into these tau vectors.

Â Those are the gyroscopic terms and within them they're n by ones you will see.

Â Each one of these torques acts about gs1 through gsN axes.

Â And then the gT1 through gN axis, and the gG one to the gGN axis.

Â So it allows you to write in a little

Â bit more compact way, this way whereas before, we had gs times all those terms.

Â If you did it like this,

Â it would take you probably two pages to write out every single term.

Â So it allows you to cheat a little bit, and just use some substitutions and

Â notations to simplify.

Â So if you can do one wheel,

Â really doing multiple wheels is just a bookkeeping problem.

Â All the hard part is done, now I just want to have a nice bookkeeping method.

Â 5:36

What we can look at for examples too is kinetic energy.

Â We have the kinetic energy that the hub, great, and then you got the wheel.

Â It has a spin rate relative to the body and the wheel also can gimbal.

Â So if you just go back and

Â look at omega W relative to N, it has both big omega and gamma down in there.

Â Its inertia, the combined inertia system, you can write this all out,

Â that's the kinetic energy.

Â The beauty and you guys aren't doing it in this class, 6010 gets to do this fun.

Â But you take the derivatives of this, plug in the equations of motion,

Â lots of algebra later.

Â You come up with the same looking power equation that we've had before.

Â We've solved this power equation and just for

Â the rotational part you proved that omega dotted with L was the power that you have.

Â If an external torque is acting on it.

Â So the way you want to look at this though,

Â this is kind of called the Work-Energy principle.

Â This torque is the torque that acts on the body relative to inertia.

Â You're pushing off with thrusters, you're really pushing off inertial space,

Â so to speak with that.

Â Omega is body relative to inertia.

Â 6:37

Big omega here, that's the wheel speed of the wheel relative to the gimbal frame.

Â Us is the motor torque acting between the wheel and the gimbal frame.

Â Little gamma dot is the angular velocity between the gimbal frame and the body.

Â And ug is the torque acting between the gimbal frame and the body.

Â This patterns holds, and

Â you'll see this in a lot of analytic dynamics text materials.

Â Once you've recognized it and know how it gets there,

Â instead of taking this derivative and getting there it's really powerful.

Â Saves you lots of days of algebra.

Â But anyway, so we have a nice power expression even for

Â the really complicated multi VSCMG equation.

Â We don't do much with it in this class, but other classes you might see them.

Â 7:21

So the next step is now let's we did the equations of motion for

Â a single reaction wheel.

Â If we apply the same principle, we have this IRW,

Â instead of we only have to account for the transverse and

Â gimbal axis directions of the wheel inertia in this IRW.

Â But instead of just one, we have n of them.

Â So it's just a summation again, right?

Â We have this h vector.

Â We saw this h vector with a dual spinner.

Â At some point I defined the momentum of the second,

Â the dual part, relative to the body like that.

Â And that's also appearing here.

Â This is kind of a common notation.

Â And you put this all together, with a little bit of algebra, it's not much.

Â You can write now your equations of motions and

Â nice matrix form where us is the stack, the n by 1 of all the motor torques.

Â And then you can get out of this your angular rate of acceleration.

Â The nice thing about this form is l don't have big omega dot showing up anymore.

Â l mean once l know what the motor torques are,

Â l essentially decoupled these two differential equations.

Â l can get omega dot directly.

Â And that same motor torque you applied to the motor torque equations, and

Â then you know now what big omega dot is.

Â You substitute it in as a known and

Â you get your big omega dots, the wheel acceleration.

Â So this gives you a way,

Â we've kind of analytically one way solved the sets of equations.

Â That's good. So that's if you have a system of them.

Â A popular configuration of gs, that's all the projection axis is you line up.

Â The first wheel with b1, the second wheel with b2 and the third wheel with b3.

Â In that case, this projection matrix is nothing but

Â 100010001 which is an identity matrix, right?

Â And then the gs just drop out and you end up with this form.

Â But this form is nice because if you even try to do it perfectly with b1 what if

Â you're off by a degree or two?

Â This form allows you to solve it in a much more general way without that perfect

Â alignment assumption.

Â But that's a differential equation where we do control we'll use this particular

Â differential equation, and

Â come up with a control development to do stabilizing that you'll see.

Â