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We have wrapped up for pre-motion, right?

Â So that's a spacecraft tumbling.

Â There's different ways to write it.

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

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

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

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

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

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

Â 'cause that doesn't always work out.

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

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

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

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

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

Â Now we're going to look at different systems.

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

Â It's still being built for some of them.

Â I think Cassini is actually a dual spinner.

Â It has this huge platform that spun at a different

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

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

Â Then we're gonna look at also gravity gradients.

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

Â differential gravity to stabilize the spacecraft.

Â So, but those are all passive weight.

Â We don't have any active feedback sensing.

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

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

Â nonlinear control for the feedback control developments.

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

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

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

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

Â specifically for the attitude control problem.

Â All right?

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

Â So let's look at this.

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

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

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

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

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

Â where my inertia tensor just becomes a diagonal.

Â All right?

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

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

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

Â All right.

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

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

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

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

Â You know, the brakes failed and they clamped up

Â and if the spacecraft tumbles the wheel tumbles, right?

Â But if you have a non-zero Omega,

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

Â We're not looking at flexing at this point

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

Â So, let's start looking at these things.

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

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

Â And the first thing we need is to get momentum.

Â Different ways to write this.

Â IS is my inertia tensor of the spacecraft.

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

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

Â and you can see this flywheel is symmetric

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

Â along the spin axis.

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

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

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

Â and for this wheel because of the symmetry assumption.

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

Â I is actually the combined spacecraft plus wheel inertia tensor,

Â and it's also gonna be diagonal.

Â A diagonal plus a diagonal gives you a diagonal.

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

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

Â For convenience.

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

Â Different ways you can add up the contributions.

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

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

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

Â It doesn't have to be.

Â If it's not coincident,

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

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

Â That's what you typically do.

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

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

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

Â which makes sense that Omega is B relative to N.

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

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

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

Â So we have I is equal to IS plus IW.

Â And if I'm saying now,

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

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

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

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

Â That Omega was defined relative to the body frame.

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

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

Â this is how it comes up conveniently.

Â So, that makes it easy.

Â So now we've got this,

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

Â this part, alright?

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

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

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

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

Â If you use B frame components.

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

Â this inertia which is IW times Omega,

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

Â So, out of the end result,

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

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

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

Â This is nothing but the inertia tensor times this.

Â So, those things added up.

Â Different ways you can get to it, alright?

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

Â where I kind of doing this implicitly.

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

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

Â Different ways you can approach it.

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

Â and these are simple derivations again.

Â This is prime stuff for a final exam.

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

Â So, now we have H written out,

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

Â Anyway, it just, it all adds up.

Â So, we need H dot equal to L.

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

Â Why?

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

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

Â Same thing is what we did with a single rigidbody,

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

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

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

Â So, you'd have the body frame derivative

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

Â Cause Omega BN, the crossed products vanish.

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

Â that gets you Omega till the I Omega.

Â So, this part is just like a single rigidbody,

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

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

Â Essentially.

Â That's one way to look at it.

Â This second part.

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

Â a little h, little - a lowercase h.

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

Â So, IW a - this is a little h.

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

Â it can spin down.

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

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

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

Â Omega two B two, Omega three B three,

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

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

Â and then everything is moved over to the other side.

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

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

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

Â to those are the wheel gyroscopics that come in.

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

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

Â and your wheel is spinning and you're tumbling,

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

Â Alight?

Â But that's it.

Â I have now derived the equations of a dual-spinner

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

Â So, let's look at this more.

Â This is written in a complete

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

Â This still works even without that assumption.

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

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

Â it simplifies like with a single rigidbody.

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

Â and three direction, we can rewrite into this.

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

Â That's how it impacts a spacecraft.

Â