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

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

Â we're just going to use linear stability analysis.

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

Â linearize and make small, and small departure approximations.

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

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

Â But it falls very much the same steps,

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

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

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

Â That's where you do that.

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

Â These are just equilibrius, right?

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

Â But this equilibria is stable cause,

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

Â This part weighs more than this one,

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

Â or at least towards it.

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

Â But this is also an equilibria.

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

Â that equilibria towards the other one.

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

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

Â and make this completely stable if we wish?

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

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

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

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

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

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

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

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

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

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

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

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

Â So, how do we describe that here?

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

Â And you've seen this differential equations before.

Â This is how I can relate my Omega BO.

Â But these angles now are B body relative to orbit.

Â They're not body relative to inertial.

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

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

Â You can map into the DCMs.

Â That's one approach, right?

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

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

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

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

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

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

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

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

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

Â into the body frame, you multiply times here.

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

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

Â That's what you end up with.

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

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

Â You can write this motion using whatever angles,

Â you guys are very good now with kinematics.

Â So, good.

Â Now we have, we'll make a B...

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

Â So, we can add them up.

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

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

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

Â and the big Omega accounts for that the frame itself

Â is in a constant rotational state, right?

Â So, all these terms add up there.

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

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

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

Â So, now if we want approximate small departures,

Â we can actually linearize this expression for small motions.

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

Â and the associated yaw pitch roll rates.

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

Â What happens there?

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

Â And then that's the result you get.

Â Let's look at the first line.

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

Â Here we have sign of pitch times yaw rate.

Â Well sign of pitch linearizes to pitch right?

Â Sign of X becomes approximately X for small departures.

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

Â We drop it, right?

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

Â times sine of small, linearizes to the small.

Â So, that's the first order term.

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

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

Â That's our first order approximation.

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

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

Â And in the homework, you do it with MRPs.

Â So, good.

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

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

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

Â Right?

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

Â this term is going to vanish,

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

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

Â Right?

Â So, this is still rigorous.

Â No approximations.

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

Â so there's no bigger make a dot.

Â Everything else receives an extra dot over it.

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

Â first order approximation for Omega, and we can start.

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

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

Â There's an extra step we have to do.

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

Â We have a general expression,

Â we need to now get an approximation

Â of the gravity gradient torque around one of these equilibrius.

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

Â So, let's look at that.

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

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

Â times zero, zero something.

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

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

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

Â the position vector is written this way.

Â And there's no approximation here yet.

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

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

Â So, this is still...

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

Â This is still perfectly valid.

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

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

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

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

Â but we're not doing that here.

Â So, we're looking at small departure motions.

Â One thing that stands out is the same one.

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

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

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

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

Â And in this system we call that axis three,

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

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

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

Â which is kind of nice.

Â So, this is a general expression.

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

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

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

Â So that comes in there and another cosine to one.

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

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

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

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

Â Right?

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

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

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

Â Right?

Â That's what we need there.

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

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

Â So, good.

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

Â we've got a linearized version

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

Â That's how we define all this stuff.

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

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

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

Â So, we've got this linear,

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

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

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

Â Plugging in these Omegas that's what you get.

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

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

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

Â this is still Omega squared.

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

Â at this point, drop second order terms.

Â So, we can look at that.

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

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

Â and in between things cross coupled with yaw and roll.

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

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

Â It's going to be second order.

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

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

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

Â That's what I've written here.

Â That's the pitch equation, right?

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

Â two axis, three, positive pitch.

Â That's basically, that's this motion.

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

Â So now, let's look at that result.

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

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

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

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

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

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

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

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

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

Â Three is positive, a real number squared guaranteed positive.

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

Â I haven't seen anybody build such a structure.

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

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

Â for the pitch motion to be stable.

Â Now, again, what I1 and 3 mean?

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

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

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

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

Â this would be gravity gradient stable for pitch motion.

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

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

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

Â So, the pitch motions actually would be stable.

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

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

Â This would deviate.

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

Â There's other shapes.

Â I1 had to be bigger so you could fly.

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

Â And let's see.

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

Â This is I2, if I do it.

Â How do I have to do this then?

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

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

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

Â right?

Â So, it's different.

Â That's what comes out of it.

Â If you want to pitch motion to be stable,

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

Â is bigger than the 03 axis inertia.

Â Now, that just stabilizes one of the angles.

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

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

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

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

Â they're all small.

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

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

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

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

Â and the pitch are perfectly decoupled from the other two.

Â So, pitch this is what we need.

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

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

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

Â and papers that deal with this topic.

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

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

Â That's what they've done.

Â And there's probably historical reasons for that.

Â But this is how they rewrite it.

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

Â And we have to look at the stability of this.

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

Â We look at the characteristic equation, right?

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

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

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

Â - would give me a four dimensional state.

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

Â So, you end up with a fourth order polynomial.

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

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

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

Â Imaginary tipping implies oscillatory motions,

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

Â And we just need real parts of it, right?

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

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

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

Â Now there's different criteria.

Â Some of you have taken classes on this,

Â you might have recognized this Routh Hurwitz criteria

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

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

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

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

Â you are not hand-deriving these conditions.

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

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

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

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

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

Â which is the quadratic form.

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

Â That's how you find all four possible routes cause

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

Â So, that's what we're going to use.

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

Â These are the conditions in terms of those betas.

Â This is how those terms are defined,

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

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

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

Â That's the equivalent kind of condition.

Â This one I added.

Â This came from the pitch conditions,

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

Â So, if you think of that condition,

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

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

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

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

Â Right?

Â Then the product is going to be positive.

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

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

Â But that restricts it to two quadrants.

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

Â looking at how they are defined.

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

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

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

Â And these four conditions you're seeing here.

Â If we plug those visually,

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

Â This is what you end up with.

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

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

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

Â or this slither here.

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

Â which gets a little bit more complicated to visualize,

Â but that's what happens mathematically.

Â This is what guarantees that we have stability.

Â So, what does this mean?

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

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

Â - in this region here, just little white region,

Â you have a condition where I2 is the least inertia.

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

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

Â It won't just start to tumble about other axes

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

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

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

Â So, those are the two possible regions.

Â We typically fly this kind of a scenario.

Â That means I2 is the largest inertia.

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

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

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

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

Â With the wings going out,

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

Â And then the axis along from wingtip to wingtip.

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

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

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

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

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

Â Wings are sticking out, noses up.

Â You want that to line up with your orbit normal.

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

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

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

Â That's what I have to do.

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

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

Â which is my local vertical.

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

Â Well, it used to fly.

Â That was its gravity gradient stabilized orientation.

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

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

Â But that's where these inertia ratios come from.

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

Â This is a first order analysis.

Â Higher order terms are ignored.

Â In particular flexing has been ignored.

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

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

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

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

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

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

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

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

Â does that give you gyroscopic stack and, you know,

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

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

Â