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We have wrapped up for pre-motion, right?

So that's a spacecraft tumbling.

There's different ways to write it.

We can predict the angular velocities if it's axisymmetric analytically in a closed form.

If it's not axisymmetric general,there were these homogeneous

undamped duffing equations that we can write,which is kind of elegant,

that in your current homework you will actually apply this different ways.

What we're moving to next is different ways to start,

not just having a rigidbody by itself spinning and trying to be stable,

'cause that doesn't always work out.

A common design that moved us from torque-free motion,

where people looked at natural equilibrias of a -one single solid spacecraft,

that's the simplest thing you can build, there's no moving parts.

But it has limitations about which axis you have to pretty much spin about

axis of max inertia if you wanna turn stable with energy loss.

Now we're going to look at different systems.

The Dual-Spinner is a pop --- used to be a very popular design for GEO satellites.

It's still being built for some of them.

I think Cassini is actually a dual spinner.

It has this huge platform that spun at a different

range than the lower platform and it helps the stability as well.

And so that's a very classic result that still applies.

Then we're gonna look at also gravity gradients.

That's the way people build satellites with it ---exploit

differential gravity to stabilize the spacecraft.

So, but those are all passive weight.

We don't have any active feedback sensing.

Where am I pointing, where should I be going, and how do I close this loop.

That will be then in the third part of the class where we talk about

nonlinear control for the feedback control developments.

We're kind of putting --- we're gonna have a little bit of theory on control theory.

You get all the basics basic that we just need for this class.

I'm not giving a whole class on nonlinear control,

but this is the specifics you need, and this-and this is how we apply it

specifically for the attitude control problem.

All right?

So, now dual spin and gravity gradient, those are passive ways to stabilize a spacecraft.

So let's look at this.

What we can write is --- I'm just modeling my spacecraft like this.

This doesn't necessarily have to be an axisymmetric body.

I just drew a cylinder cause it was easy to draw.

What I do have is a coordinate frame B, that's a principle coordinate frame,

which we now know means, I've chosen an option

where my inertia tensor just becomes a diagonal.

All right?

So, you have that and I'm lining up my spin wheel with

where its spin axis is lined up with one of my principal axis,

and just without loss in generality, I just called it the first one.

All right.

So, that's how this thing is lined up.

So, this flywheel can have a speed, big Omega,

that's the speed it has relative to the body.

If you zero big Omega, that means this body now acts as a single rigidbody.

You know, the brakes failed and they clamped up

and if the spacecraft tumbles the wheel tumbles, right?

But if you have a non-zero Omega,

that means the wheel is moving so it's a way of having now, basically, two rigidbodies.

We're not looking at flexing at this point

but it's a multi-body system that we're actually solving with this.

So, let's start looking at these things.

We're going to use, as before, we use H dot equal to L for a single rigidbody.

We're going to do H dot equal to L for this two-body system.

And the first thing we need is to get momentum.

Different ways to write this.

IS is my inertia tensor of the spacecraft.

IW is the inertia tensor of this flywheel, and this is a full three by three, all right?

But, since I've chosen a B frame where IS is a diagonal,

and you can see this flywheel is symmetric

so it has a unique inertia - I'm just calling that inertia, IW -

along the spin axis.

And then there's two transverse inertias that are equal.

But in the B one, two, and three frame it's also gonna be an IS diagonal matrix.

The B frame is both the principal frame for the spacecraft,

and for this wheel because of the symmetry assumption.

So, from that, actually, you can have a combined sector.

I is actually the combined spacecraft plus wheel inertia tensor,

and it's also gonna be diagonal.

A diagonal plus a diagonal gives you a diagonal.

So, I'm just calling these I one, two, and three.

The wheel inertia about B one, its spin axis, is just going to be IW the scalar of it.

For convenience.

Now, that's the inertia tensor, now we want to write angular momentum.

Different ways you can add up the contributions.

One way is, you could say, 'OK, the spacecraft is spinning',

and you can see, this is actually lined up in the CG.

The CG of the wheel, the CGA of the spacecraft, they are coincidence here.

It doesn't have to be.

If it's not coincident,

that part of the mass is like having a bolt one meter removed,

you just use parallel axis theorem, and adjust the inertia tensor of the spacecraft.

That's what you typically do.

So, without loss of in generality we can typically write it this way.

But here, I'm showing you one way where I'm saying, the total inertia times Omega,

that's really gonna be the inertia of the spacecraft times Omega,

which makes sense that Omega is B relative to N.

That's the classic angular momentum you've seen.

Then, the wheel momentum is gonna be the wheel inertia times Omega W relative to...

And-and I'm gonna break that up actually into mathematics here.

So we have I is equal to IS plus IW.

And if I'm saying now,

IS times Omega BN, that's the momentum of the spacecraft.

Sorry, IS times Omega BN, and IW, I need the angular velocity of W relative to N.

But W relative to N is simply gonna be big Omega times B one plus Omega B relative to N.

This is your angular velocity of the wheel frame relative to the body, alright?

That Omega was defined relative to the body frame.

So, like in that exam problem where you had feet,

that second angled part one defined relative to the first link, you know,

this is how it comes up conveniently.

So, that makes it easy.

So now we've got this,

so this is really gonna be expanded as IW times Omega BN plus IW times

this part, alright?

This and this combines to IS plus IW times Omega BN,

plus this part is really a three by three that you have.

And this in the B frame, 'cause that's the principal frame of the wheel,

and the angular velocity is just gonna be Omega zero, zero, in the B frame.

If you use B frame components.

So, you can see out of this all you're gonna get is

this inertia which is IW times Omega,

and these other ones don't matter cause they're all multiplied times zero.

So, out of the end result,

this thing just becomes IW, big Omega, and that's in the 1 axis.

So, if you carry this out, you would get this zero, zero.

And that's the same thing as B one hat times that.

This is nothing but the inertia tensor times this.

So, those things added up.

Different ways you can get to it, alright?

So, here I'm doing a little bit more of a direct way,

where I kind of doing this implicitly.

OK, this is the angular momentum as if the wheel were locked,

and then I'm adding the angular momentum of the wheel relative to the body.

Different ways you can approach it.

But try different ways cause it should always give you the same,

and these are simple derivations again.

This is prime stuff for a final exam.

You know, I can- something you can do, you should be able to do.

So, now we have H written out,

and this tensor, I'm not specifying any particular frames at this point.

Anyway, it just, it all adds up.

So, we need H dot equal to L.

We're gonna choose to differentiate this vector as seen by the body frame.

Why?

Because we know it's seen by the body frame, the-the I primes will be zero.

But then we have to add Omega crossed, there is momentum vector again, alright?

Same thing is what we did with a single rigidbody,

we're applying it to this dual-spinner now.

This part looks just like the rigidbody, which is actually have the inertias of two

rigidbodies, if one were locked relative to the other, and it gives you the same results.

So, you'd have the body frame derivative

of Omega, which is the same thing as the inertial derivative of Omega, alright?

Cause Omega BN, the crossed products vanish.

So, this gets you exactly the same parts plus Omega cross this vector,

that gets you Omega till the I Omega.

So, this part is just like a single rigidbody,

just a t- if that t- If that's- if that's,

if that wheel locks in, it is a single rigidbody, alright?

Essentially.

That's one way to look at it.

This second part.

People typically call this momentum of the wheel relative to the body,

a little h, little - a lowercase h.

You see this often in the literature, so I'm gonna use the same notation.

So, IW a - this is a little h.

So, as seen by the body frame, this wheel can change speed, it can spin up,

it can spin down.

So, I'll have an h dot in the B one, and then Omega crossed hB1.

This vector, it has to be equal to the external torque.

So, if you carry this out, you can plug in Omega as being Omega one B one,

Omega two B two, Omega three B three,

and you cross with B one, you know how to do that, the B one B ones cancel,

the B two B ones give minus B three, and then the B three, B one, gives you minus B two,

and then everything is moved over to the other side.

So, we have, like before I'll make a dot equal to this, that's like a single rigidbody.

I have dropped L at this point saying, 'Let's look at a torque-free motion case',

and the rest of the gyroscopics are also brought over to the right-hand side

to those are the wheel gyroscopics that come in.

If you change the wheel speed, you get some effect along B one.

If you just are spinning and you're tumbling with the spacecraft

and your wheel is spinning and you're tumbling,

there's some cross coupling terms that appear in here.

Alight?

But that's it.

I have now derived the equations of a dual-spinner

where the spin rate doesn't even have to be constant.

So, let's look at this more.

This is written in a complete

tensor notation, so this doesn't have to be a principal frame actually.

This still works even without that assumption.

But if we assume it's a principle frame, we can write this as a diagonal.

This other part, Omega till the I Omega, if this I is a diagonal,

it simplifies like with a single rigidbody.

And the other terms, we had something in the B one, two,

and three direction, we can rewrite into this.

This is the gyroscopics of having a spinning or accelerating wheel there.

That's how it impacts a spacecraft.