0:05

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

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

if you just have these all 0.

Gimbal rates at zero, gimbal acceleration is zero.

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

these earlier equations simplify to this form.

And that's nice.

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

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

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

omega dot, back substitute it so

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

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

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

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

that's what you come up with.

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

1:00

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

there's not gamma tau term in there.

And that's good.

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

Js times big omega dot.

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

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

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

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

gs, gs transposed omega dot.

Here we have the full inertia tensor times this.

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

wheel frame inertia.

1:52

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

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

Those two actually cancel.

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

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

This will be the full answer.

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

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

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

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

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

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

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

So we command a twist rate and

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

But that's bunch of lectures in the future.

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

of the spacecraft with a single reaction wheels.

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

two and three like we did with the dual spinner.

2:58

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

Now that's a single device.

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

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

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

we don't back substitute here.

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

The big omega times the small gimbal rate,

which is still a very big torque acting on it.

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

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

gyroscopics.

But this is the full torque level solution.

3:37

Now we've done single VSCMG devices.

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

with reaction wheels you may have three, four.

I've seen some with six and more.

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

also maybe for torque.

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

your maneuvers require one and a half Newton meter.

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

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

And in essence everything we've done still holds.

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

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

The first gs axis, the second gs axis.

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

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

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

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

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

So we just carry out that kind of a calculation.

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

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

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

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

So it allows you to write in a little

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

If you did it like this,

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

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

notations to simplify.

So if you can do one wheel,

really doing multiple wheels is just a bookkeeping problem.

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

5:36

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

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

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

So if you just go back and

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

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

that's the kinetic energy.

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

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

lots of algebra later.

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

We've solved this power equation and just for

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

If an external torque is acting on it.

So the way you want to look at this though,

this is kind of called the Work-Energy principle.

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

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

so to speak with that.

Omega is body relative to inertia.

6:37

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

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

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

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

This patterns holds, and

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

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

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

Saves you lots of days of algebra.

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

the really complicated multi VSCMG equation.

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

7:21

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

a single reaction wheel.

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

instead of we only have to account for the transverse and

gimbal axis directions of the wheel inertia in this IRW.

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

So it's just a summation again, right?

We have this h vector.

We saw this h vector with a dual spinner.

At some point I defined the momentum of the second,

the dual part, relative to the body like that.

And that's also appearing here.

This is kind of a common notation.

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

You can write now your equations of motions and

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

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

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

l mean once l know what the motor torques are,

l essentially decoupled these two differential equations.

l can get omega dot directly.

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

then you know now what big omega dot is.

You substitute it in as a known and

you get your big omega dots, the wheel acceleration.

So this gives you a way,

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

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

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

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

In that case, this projection matrix is nothing but

100010001 which is an identity matrix, right?

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

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

you're off by a degree or two?

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

alignment assumption.

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

differential equation, and

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