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So now we found equilibrius, the next step is stability about this equilibrius.

And like with the dual-spinner, at this stage,

we're just going to use linear stability analysis.

So, we're going to come up with these equations,

linearize and make small, and small departure approximations.

And we're doing it in one way and in your homework you actually duplicate this,

but do it with MRPs instead of all our angles basically.

But it falls very much the same steps,

so that's why I'm going through this part a little bit quicker.

This is much more about the high level thing, make sure the concepts make sense.

The mathematical details, you will do it in your homework anyway.

That's where you do that.

So, now we're looking at how can we use gravity gradients.

These are just equilibrius, right?

And we know that this is an equilibria, and this is an equilibria.

But this equilibria is stable cause,

if you have a small departure, it's visually very easy to see.

This part weighs more than this one,

so it can have a restoring force that gets you back to the equilibrium point,

or at least towards it.

It won't be asymptotic, it'll just oscillate.

But this is also an equilibria.

And if you perturb it, you're going to be driven away from

that equilibria towards the other one.

And so, that local motion is not going to stay linearly stable.

So, how can we identify this for all three axes,

and make this completely stable if we wish?

So, we used the same O frame again, RC, everything is there.

This is from different notes for some- instead of having N, I have a big Omega.

That's in the orbit rate but it's the same thing, just different label.

So, that's your orbital speed that you would have,

it's basically nu over your rate, circular orbit radius cubed.

That gives you, in radians per second, your orbital speed.

And we're going to now look at, we're assuming the principle chorded frame.

That's a key assumption, so inertia tensor in the body frame is there.

And the motion, now we do allow non-zero angular motion, right?

If this is my equilibrium, I want to bump it and see what happens with small motions.

So now I have to account for some Omega BO's in this development, right?

That's how you get your equations of motion about this state.

So, how do we describe that here?

I'm going to use all the angles three, two, one, yard pitch roll sequence.

And you've seen this differential equations before.

This is how I can relate my Omega BO.

But these angles now are B body relative to orbit.

They're not body relative to inertial.

If we needed body relative to inertial, you know how to do that, right?

We know how to add, three, two, one, all of our angles.

You can map into the DCMs.

That's one approach, right?

Then multiply amount and pull them out again, if you have to.

So, this is what's going to go into here.

This is the constant orbit rate, but it's about 02, not about a body axis.

So, I need the DCM to go from the orbit to the body frame,

and I'm using the same three, two, one angles.

So, that's this classic one you've derived,

this is the one we've seen in the first part of the class.

So, if I need to map my orbit motion, which is basically zero, big Omega, zero,

that's the orbit rate in the orbit frame into a vector,

into the body frame, you multiply times here.

So, this is going to pull out the second column at zero, big Omega,

zero times a three by three gives you that big Omega, times the second column.

That's what you end up with.

If you want to choose different attitude parameterizations you can do it.

It's just going to have MRPs here, or CERP, or quaternions, or whatever you want to use.

You can write this motion using whatever angles,

you guys are very good now with kinematics.

So, good.

Now we have, we'll make a B...

Both Omegas are now written in the B frame, which is nice.

So, we can add them up.

I add up both matrix representations of it, and so this is the math that you get.

So you can see the yaw pitch and roll rates, again,

are the motions of the body relative to the orbit frame,

and the big Omega accounts for that the frame itself

is in a constant rotational state, right?

So, all these terms add up there.

But that's now my Omega BN parameterized this way.

This is good for anything, there's no approximation in these kinematics.

We did assume that the spacecraft is on a circular orbit but that was about it.

So, now if we want approximate small departures,

we can actually linearize this expression for small motions.

Big Omega's not a small term, but we can assume now my yaw pitch, roll angles,

and the associated yaw pitch roll rates.

These are all going to be small oscillations, slow oscillations.

What happens there?

So, they're all treated as small variables, and we're linearizing them all about zero.

And then that's the result you get.

Let's look at the first line.

That's already linear, that's just your roll rate.

Here we have sign of pitch times yaw rate.

Well sign of pitch linearizes to pitch right?

Sign of X becomes approximately X for small departures.

So, that's small times small, it's small quadratic.

We drop it, right?

Here we have a cosine of small, linearizes to one,

times sine of small, linearizes to the small.

So, that's the first order term.

And that's what you see in the first line here.

So, that's what we end up with only these two terms.

That's our first order approximation.

You can do the same steps for the other two axes.

And in fact, I encourage you to do it on your own, I don't need to do this for you.

And in the homework, you do it with MRPs.

So, good.

Now, this isn't the body frame, we need, in the end, Omega dot.

But we can use our handy dandy identity that Omega is Omega BN,

so the N framed derivative, and the B framed derivatives are the same.

Right?

So, I don't have to do any cross-product stuff all I have to do is,

this term is going to vanish,

and all I have to do is take a derivative of these three matrix components,

and that is the inertial derivative of this particular Omega BN vector.

Right?

So, this is still rigorous.

No approximations.

It just means here, big Omega is constant - I'm on a circular orbit -

so there's no bigger make a dot.

Everything else receives an extra dot over it.

Now, I have a first order approximation for Omega dot,

first order approximation for Omega, and we can start.

Well, we're almost ready to start applying it to here.

We can linearize this term with that, this term will still be make it times Omega.

There's an extra step we have to do.

What we're still missing is the gravity gradient torque.

We have a general expression,

we need to now get an approximation

of the gravity gradient torque around one of these equilibrius.

How is that torque vary if I pitch, roll, yaw slightly, you know, around it.

So, let's look at that.

We did this math already earlier, this is how we get to this stuff.

Now, we have a very particular BO matrix in terms of yaw, pitch and roll,

times zero, zero something.

So, we're going to pull out the third column of this Bo matrix,

which is nothing but these sines and cosines, times RC.

So, this is now my RC one, two, and three, the B frame vector components of

the position vector is written this way.

And there's no approximation here yet.

And you can substitute this expression into this earlier expression we derived,

where we have products of RC ones, twos, and threes, and this gives you this answer now.

So, this is still...

We haven't made any small angle approximations yet in this expression.

This is still perfectly valid.

You can now describe your gravity gradient torques as a function of yaw, pitch,

and roll angles of the body relative to this orbit frame.

That's what this yaw, pitch, and roll means.

And, again, you can make inertial tensors specific to make it go to zero,

but we're not doing that here.

So, we're looking at small departure motions.

One thing that stands out is the same one.

If you look at this, because it's the third column with the DCM there,

there is no yaw angle in here, which actually makes somewhat physical sense.

Right?' We line things up, we know that if this is lined up, it's a zero gravity gradient.

That's true, that answer is true regardless of how I rotate about that axis.

And in this system we call that axis three,

and for a yaw, pitch, roll, that's a three to one rotation.

So, any three rotation doesn't really affect that gravity gradient torque.

That's what you find duplicated here by having used these all the angles in this sequence,

which is kind of nice.

So, this is a general expression.

Now we want to linearize this cause we're looking at linear results.

What if we have not just general big motions, but I'm just having a little wiggles, right?

These angles are going to be small cosines become one, sine of two axis, just two X again.

So that comes in there and another cosine to one.

Here you have a two- Theta becomes, sign of two Theta becomes two Theta.

But here you have a sine times a sine, so that small times small,

that's going to be higher order and drops out.

So, generally, while yaw doesn't appear, you do get a torque about the third body axis.

Right?

Nominally, the lining up B one, two, and three with 0, one, two, and three.

You do get a torque but it's actually really small.

That's the one that kind of helps you, though, restore yawing motion if needed.

Right?

That's what we need there.

But if you linearize, it's small enough to where you can drop it,

and it doesn't appear there, so that becomes a second order term that you would have.

So, good.

Now, we have R. With a little bit of math,

we've got a linearized version

of gravity gradient in terms of three, two, one, order angles relative to the orbit frame.

That's how we define all this stuff.

So, at this stage we're ready to start to put this together.

This is still our fundamental equations of motion of a rigid body,

s single rigid body, not a dual-spinner or something but close enough, right?

So, we've got this linear,

and we're plugging in those conditions including that the inertia tensor is diagonal.

This one, we've seen before, it gives you these terms, these differences,

with Omega twos and threes, Omega ones and threes and all the products of Omega's.

Plugging in these Omegas that's what you get.

And then the gravity gradient torque, it only had first and second components,

the third one dropped out completely in the linearization that we have.

So, the last step you have, because this is Omega till the I Omega,

this is still Omega squared.

Linearizing Omega wasn't quite enough, you still have to,

at this point, drop second order terms.

So, we can look at that.

And the pitch equation, the faded one is actually a nice example.

You've got pitch acceleration here, you've got pitch angles here,

and in between things cross coupled with yaw and roll.

But if you look at these terms, it's yaw rates times roll rate.

That small time small, drops out, yaw rates times yaw angle, again, small times small.

It's going to be second order.

Roll times roll rate, second order, and roll times yaw angle, also again small time small.

So, the second equation, once you linearize and drop the second order gyroscopic terms,

all you're going to be left with is this term, and this term.

That's what I've written here.

That's the pitch equation, right?

So, if we had- I'm flying along this way this is my one axis,

two axis, three, positive pitch.

That's basically, that's this motion.

If you're flying along like this, how does this motion, how is it going to be stable?

So now, let's look at that result.

When you linearize, these things always look at spring mass damper systems, right?

And there's no damping in this system, it's gravity, it's a conservative force.

It's like, it just adding a springiness to the system, that's all you get out of it.

So, there is never asymptotic convergence with gravity gradient stabilization,

it's just a marginal stability like a spring and a mass.

And if it's deflected, it's going to oscillate but it shouldn't get bigger or smaller,

at least not within the linear stuff that you're doing.

So, this is our stiffness that we have to look at,

and to make this pitch motion stable, we need this to be positive.

Three is positive, a real number squared guaranteed positive.

I2, principle inertias, they're all positive - you never have negative principle inertias.

I haven't seen anybody build such a structure.

So, the only way to make this positive negative is through the relative size of I1 and I3.

So, the way this is written I1 actually has to be bigger than I3

for the pitch motion to be stable.

Now, again, what I1 and 3 mean?

These are the principle inertias we've linearized about the O frame.

So I1 is the inertia of the craft nominally aligned with the 01 axis,

and the I3 is the inertia of the craft nominally about the 03 axis to orbit radial axis.

So, if you look at the spacecraft, pretty distinct inertias,

this would be gravity gradient stable for pitch motion.

I'm flying this way, so the inertia about this axis, this inertia is I1.

It's definitely much bigger than the inertia about the gravity axis three, right?

This is much skinnier than this, and that would be stable.

So, the pitch motions actually would be stable.

If you reverse that, well, now all of the sudden,

you have a negative stiffness, and that's how it appears mathematically.

This would deviate.

But there's other ways you can do this as well.

There's other shapes.

I1 had to be bigger so you could fly.

Then this, so this way you could fly it as well.

And let's see.

No, that wouldn't work, I3 is now this.

This is I2, if I do it.

How do I have to do this then?

So, if I want this to be larger, I fly it there, and this to not to be the skinniest one

- but I could fly, I could fly this way basically.

And then the pitch motion, which kind of makes geometric sense, this would also be stable,

right?

So, it's different.

That's what comes out of it.

If you want to pitch motion to be stable,

you have to line up your spacecraft such that O1 axis inertia

is bigger than the 03 axis inertia.

Now, that just stabilizes one of the angles.

It doesn't guarantee that yaw and roll are going to be stable, in this case.

So we have to look at the other two differential equations.

Now, the fate of one actually- let me look at that one again -

here, there were some faded dots that appeared but then multiply times here and here,

they're all small.

The only one that's first order is going to be this product.

And the same thing down here, there's a [inaudible] times this.

Everything else is going to be dropping out a second order.

So, the theta dots drop out of the roll and pitch at the roll and yaw equations,

and the pitch are perfectly decoupled from the other two.

So, pitch this is what we need.

Now we're focusing on roll and yaw motion, which to first order decouple.

This form of writing it, takes a little bit of algebra - quickly a few steps.

This is just a classic form you find in a lot of textbooks

and papers that deal with this topic.

People come up with these KR and Ky ratios which are differences,

inertia, two minus the other, two minus the other, divided by the third basically.

That's what they've done.

And there's probably historical reasons for that.

But this is how they rewrite it.

So, now we ended up with second order coupled differential equations.

And we have to look at the stability of this.

The way you do this in linear controls, a linear analysis, dynamic analysis class.

We look at the characteristic equation, right?

You write this stuff up into first order form, get the determinant of the,

you know, one minus, that matrix sends s, get the determinant.

This in the first order form - because the second or differential equation

- would give me a four dimensional state.

Roll, pitch, sorry, yaw, roll, yaw-rate, roll-rate.

So, you end up with a fourth order polynomial.

What has to be true of the roots of this characteristic equation?

Well, they can't have any positive real parts.

These roots can be real, they can be imaginary, that's fine too.

Imaginary tipping implies oscillatory motions,

real roots implies asymptotic convergence or divergence depending on where they are.

And we just need real parts of it, right?

So, all four roots we have to make sure they're either on,

basically, not on the positive plain but on the negative,

or the imaginary axis, to guarantee some form of stability.

Now there's different criteria.

Some of you have taken classes on this,

you might have recognized this Routh Hurwitz criteria

that you can use, and I'll apply that here in a moment.

I'm not going into the details, it's really not important.

If you haven't seen this, there's whole courses you can take on this.

This is more about the results for the level of this class,

you are not hand-deriving these conditions.

But, in essence, we've got a quartic term,

we've got a quadratic term and then a zeroth order term.

And that's something we can take heavy advantage of.

It's easier to get roots because you could read,

instead of solving for Lambda, you basically solved it for Lambda squared,

which is the quadratic form.

And then the square root of that gives you the actual roots.

That's how you find all four possible routes cause

it's always plus minuses and how does that work.

So, that's what we're going to use.

Now, skipping a bunch of derivations, this is what you find on these kinds of topics.

These are the conditions in terms of those betas.

This is how those terms are defined,

and this is the inertia ratios, how you have to highlight this.

This is what would guarantee that all the roots are not positive.

In the KR-KY space, which has these ratios, how do you size stuff?

That's the equivalent kind of condition.

This one I added.

This came from the pitch conditions,

and the pitch conditions written in KR-KY just mean KY has to be bigger than KR.

So, if you think of that condition,

it gives you one half of the whole KR-KY space

that drops out because you have to be on the plus side.

This one, too, actually, KR, X times Y has to be positive.

Well, that means that your X and Y are either positive or X and Y are both negative.

Right?

Then the product is going to be positive.

So that means your answers have to be either in the quadrant.

So if you're looking at it- Now, we'll see the quatrains in a moment.

But that restricts it to two quadrants.

Just instead of using X, KR-KY written in terms of inertia,

looking at how they are defined.

This really boils down to this second inertia has to either be the largest -

so that's the inertia about the 02, that's the inertia about the orbit normal axis,

it's all equivalent - it has to either be the largest inertia or the least inertia.

And these four conditions you're seeing here.

If we plug those visually,

and you can get discovered Matlab, mathematica, and plot these things out.

This is what you end up with.

So, we have these conditions that we had to be, one had to be bigger than the other.

The product had to be positive which put this in this quadrant or this quadrant.

And the other ones that inequality gave us this region that were stable,

or this slither here.

The slither is because of this other constraint that we had from this equation,

which gets a little bit more complicated to visualize,

but that's what happens mathematically.

This is what guarantees that we have stability.

So, what does this mean?

This means, in these regions, if I have my -

and this one has basically, it boils down to I2 so the largest inertia

- in this region here, just little white region,

you have a condition where I2 is the least inertia.

In both of them, if I have small departures, not just in pitch but in yaw,

or in roll as well, that these motions would stay stable.

It won't just start to tumble about other axes

cause you could be stable this way but then unstable about this axis,

and that wouldn't be good for the space station;

they don't like to flip upside down, right?

So, those are the two possible regions.

We typically fly this kind of a scenario.

That means I2 is the largest inertia.

I'm sure many of you have seen the images of the space shuttle,

it's flying, its happy, its nose is typically pointing up.

For the space shuttle the nose is actually the axis of least inertia.

So, it's kind of like the skinny side, right?

With the wings going out,

and everything, that's the axis of maximum inertia like this flat plate.

And then the axis along from wingtip to wingtip.

If you look at a side profile, it's more than looking at the nose,

but it's not as much as looking at the broad side.

So that's the axis of the two axes, it's the axis of intermediate inertia.

What this says is, you have to pick your maximum inertia,

which is basically looking straight out from the cargo bay from the shuttle, you know.

Wings are sticking out, noses up.

You want that to line up with your orbit normal.

So if you're flying along this way - this is my local gravity direction -

my axis of flight direction, and then my normal axis is this way.

I have to line up my cargo bay with the orbit normal, essentially.

That's what I have to do.

Then I still need, I want bigger than I3, so the inertia about my long track axis,

which is this axis, has to be bigger than the inertia about my orbit radius axis,

which is my local vertical.

That's why the space shuttle tends to fly this sidestepping motion.

Well, it used to fly.

That was its gravity gradient stabilized orientation.

It didn't take any fuel, any thrust, little astronauts wake up bump the wall.

That was enough to just kind of stay close, and it didn't go crazy, you know.

But that's where these inertia ratios come from.

If you're going to fly one here, it's trickier.

This is a first order analysis.

Higher order terms are ignored.

In particular flexing has been ignored.

So, if we have it's not rigid but things could move,

this quickly can become unstable as well, which you have to be really careful, ok?

So, historically, People like this one but, again, first order analysis is nice.

Life isn't first order, life isn't linear, it's non-linear and messy,

and these higher order terms might have, actually, a strong impact,

especially the flexing part, you know, the small attitude.

If it were rigid, this is all rigorous and right.

But if you have flexing happening all of a sudden with the motions,

does that give you gyroscopic stack and, you know,

the stability you have is this big and the gyroscopic is this big,

the gyroscopic is going to overrule and take you all over the place.