0:06

Okay now we were looking at an old nominal big omega.zero.

Â The wheel's not accelerating.

Â We don't launch that way.

Â We launch with everything locked down and then we release it.

Â And in fact, the craft will have some spin after it gets kicked off.

Â And then we engaged its wheel and momentum happens and

Â how does it settle in the end as you're spinning up?

Â There are some classic answers we can go through.

Â 0:30

So To begin with, what's commonly done with these dual spinners,

Â they're released where it's spinning about an axis that's not aligned with the wheel.

Â Wheel was aligned with b1, so the spacecraft, let's assume it's spinning

Â about b2 or b3 or some combination thereof, something orthogonal to b1.

Â So we want it to be doing this in the end, spinning skinny, but

Â right now it's just doing this kind of a tumble.

Â 0:55

And then the reaction wheel is spun up.

Â If this is spinning, and it has inertia of one, and

Â a spin rate of one, then inertia times spin rate gives momentum of one.

Â Just non dimensional units.

Â I'm going to spin up my wheel until its momentum magnitude matches that

Â of the space craft after launch which is one in this case, right?

Â Now, in this kind of a dynamical system is Is the angular momentum vector conserved?

Â 1:37

>> The wheel and the spacecraft.

Â >> Okay so for the system I'm talking about wheel and spacecraft system.

Â >> Okay. >> Is that an internal torque or

Â an external torque?

Â >> It would be an internal.

Â >> Exactly. So there is still no external torque.

Â So for the combined hub and wheel system, the total momentum vector which

Â is what we [LAUGH] used to derive the equations of motion like h dot equal to L.

Â That h still has to be fixed as seen by inertial observer.

Â So we can use that.

Â 2:09

So then we want to really study what happens to the attitude as we do this

Â spin up, right?

Â And we're going to go through this.

Â So the initial spin rate, we said, we're tumbling here, right?

Â About b2, just without loss of generality, I'm tumbling at a rate 1,

Â inertia 1, gives me momentum 1, that's what I have here.

Â And spinning up the momentum of the wheel relative to the body,

Â that was the wheel inertia times big omega.

Â Big omega was omega of wheel relative to b, right?

Â 2:34

I'm simply having omega dots, that tells me how little h is changing with time.

Â And we're just going to constantly increase it right now.

Â We're not doing any input shaping or something like that,

Â just going to ramp it up linearly until you hit the desired speed.

Â And the desired speed will make little h be equal to big H, that's what we said.

Â So the momentum of the wheel relative to the craft is same as the total momentum

Â of the system.

Â 2:59

So the manuever time you can figure it out.

Â Well at the end I have to have this much.

Â That's my constant rate.

Â You can quickly compute.

Â This is going to take a hundred seconds, a thousand seconds depending on

Â how quickly you're ramping up your wheel speed.

Â Right? This is how long it will take to

Â reach this total amount of momentum.

Â So we can compute that.

Â Nominally we had launching like this, we're spinning right and

Â then this wheel is going to be spinning up until its momentum magnitude matches

Â this spacecraft momentum at the time T0, and this is going to line up.

Â 5:03

>> Well no, we are spinning up >> Okay, so you are assuming

Â >> It's locked in, and we're there, right.

Â But it is dependent on the direction of these vectors.

Â Just because little h is equal to the magnitude of the initial big H

Â doesn't mean the two directions have to be the same.

Â And that's essentially, this is our initial h that we have, this system.

Â This is how we launched, wheels were locked down, then we're speeding it up.

Â 5:28

And, at the ends the wheel has a vector

Â magnitude that is the same as the initial magnitude of the system.

Â Little h was equal to big H, those are the scalars.

Â But you could have two vectors that have the same magnitude but

Â not be in the same direction.

Â And then this is actually what the spacecraft body must be doing.

Â So this momentum vector plus this one add up to be the same as the original.

Â We argue we cannot change initial momentum.

Â 5:57

So, by doing this maneuver, we're not guaranteed, we, typically,

Â when spacecraft launched this way, they want it to line up like this.

Â We talked about the Boeing 501s.

Â They have geostats that always point at the earth, skinny ones like this.

Â You want them to be nice and even.

Â They're not doing this big wobble.

Â They call this coning motion.

Â This is your coning angle that you can see right here.

Â So we'd like it to be zero, but it's not typically zero when it happens.

Â And there's different ways to explain what is happening, but basically this momentum

Â matching guarantee doesn't enforce that those two things actually line up.

Â 6:33

Now the analysis of this gets rather complicated.

Â Typically we have to,

Â now I will show you some numerical results to kind of talk about this.

Â We don't have nice analytic answers to guarantee these behaviors, but

Â we can study this stuff.

Â I mean there's lots of theories behind this but I'm giving it a spin.

Â Initially with Tom Blake you're not omega 2 but omega 3 for 30 degrees per second.

Â Have wheel inertia, spacecraft stuff, everything's locked down.

Â This gives me an initial big H, momentum.

Â And now I'm going to go ahead and start spinning up that wheel at a constant rate.

Â So I just took the equations of motion you saw earlier, last class, and

Â I'm integrating them with these initial conditions and also tracking the attitude.

Â I'm not just tracking omega but I used Euler angles and, I forget what I used,

Â something, maybe MRPs.

Â And at two hundred seconds, I have finished that maneuver.

Â And you saw then at two hundred seconds, we were close to being lined up like this.

Â But there was still a little bit of a wobble left, that happens there.

Â And so

Â that amount of wobble, that's what's happening with these amplitudes here.

Â We can predict that now and see what's happening.

Â So we get close but not quite there.

Â Here I'm taking a thousand seconds, because I'm taking a lot more time.

Â So it'll take about a minute to run and

Â you can see all these simulations running and it keeps wobbling and spinning.

Â But it's a much more gradual process.

Â So what's going to happen now in this process is It'll take,

Â obviously takes more time to roll up, but if you look at the final omega two's

Â before it was like plus minus seven degrees and even these axis were bigger,

Â we will have less of a coning angle at the end of this maneuver.

Â So when It gets close there we'll want to take a look at that.

Â And that's the general trends we have in these systems.

Â With a dual spinner we're just spinning it up, we don't want to be too aggressive

Â because you end up with this huge momentum vector off to the side.

Â You really have, then you have to do other stuff.

Â You want to have a coning angle as small as possible so

Â a little patience goes a long way.

Â Take a little bit longer to spin this thing up and

Â at that point, you have a smaller coning angle.

Â So it will get there.

Â We're about halfway there.

Â The way we get rid of this coning angle, you could use thrusters,

Â in the end if you want to change the attitude.

Â You could apply h dot equal to l, apply some thrusters, in the right control.

Â And change that momentum vector of the body, and puff it back into place.

Â But now we're using fuel, we hate using fuel in space.

Â There's an elegant method that people have devised which called the mutation damper.

Â Anybody heard of a mutation damper before?

Â 8:59

No?

Â What it does is you put basically a ball in a tube of slush,

Â oil, something highly viscous.

Â So it creates friction.

Â So we're losing energy.

Â All right?

Â And what we want to have is spin about omega one in the end and so

Â you make that damper, you line up that tube of slush along either b2 or 3.

Â So if you're still wobbling with the conic motion,

Â you're always actually exciting that slosh, and you're losing energy.

Â And then people have shown this stuff,

Â at the end it would actually get rid of the coning motion completely.

Â And that goes there.

Â We've used this as a 50 turn project in the past where people have to integrate

Â this and study the spin up and go through that.

Â It's a fun project actually.

Â So take away notes is spin up,

Â just because you match momentum magnitudes doesn't mean you've matched the vectors.

Â Teaching those vectors and patience goes a long way.

Â To get rid of the coning angle, you want to put a mutation damper orthogonal.

Â That'll get rid of the energies of these modes here over time, and

Â it'll settle in place.

Â And there's all kinds of, you can look up mutation dampers.

Â But here's the results.

Â So with two hundred seconds you see where you are.

Â A thousand seconds, you know, five times longer,

Â we've improved it by a factor of two.

Â There's really, it's an asintotic thing.

Â So if you want to get even closer, you might take a long time.

Â That's where, how patient are you, you know.

Â A little bit of mutation damping can go a long way, and

Â get rid of the coning motion.

Â 10:23

Other things are, we've seen plots.

Â I'm just going to go through this, reasonably high-level.

Â But just to show you, plots have also been applied not just for rigid bodies, but

Â also systems of rigid bodies like these wheels and dual spinners.

Â The energy here, this is really the energy of the spacecraft only,

Â here, not the real energy relative to the spacecraft.

Â So not the part with i w over two, big omega squared.

Â So that's called E star.

Â The total momentum energy of the system is as before, the momentum sphere just

Â has to be constant, even with internal torques acting because they're internal.

Â They're not changing the total momentum of the system.

Â Which is kind of nice, that's a constraint.

Â So we can rewrite the energy ellipsoid of the spacecraft

Â in terms of these coordinates.

Â And with a little math this is how you, this is what you come up with.

Â So because the energy that you have, the momentum about the one axis is due to

Â the wheel and due to the spacecraft, so you have to pull out the wheel momentum.

Â And then divide by the right stuff to come up with the energy terms

Â of the spacecraft about the one axis.

Â The two and three axis look like before.

Â But this is a sphere, this is still an ellipsoid, but it's a shifted ellipsoid.

Â If your h one two threes are essentially your X, Y, Z, independent coordinates.

Â If you go look up the classic ellipsoidal equation.

Â If you didn't have minus h this would be a centered ellipsoid that's there, and

Â with the minus h, it actually shifts that ellipsoid along the one access.

Â So as we spin up the wheel that energy ellipsoid is going to move.

Â And that's what you see here.

Â Initially we launched the spacecraft spin is here.

Â We were spinning purely about b3 in that numerical problem I showed you.

Â Then as we spin up the wheel

Â you can see these are the momentum coordinates that we have to satisfy.

Â That's the total momentum coordinates of the spacecraft.

Â It intersects energy in here.

Â And the energy ellipsoid starts to shrink inside and of the spacecraft cause as

Â we spin up the wheel the spacecraft tends to de-spin and change it's orientation.

Â Although momentum we'd like it to go into the wheels.

Â So there's different critical points.

Â At one point this ellypsoid intercepts in this nice figure eight motion.

Â This kind of a separate trick stuff that we've seen with rigid bodies.

Â And then when the momentum magnitude matches, this is what you end up with.

Â Just because you match momentum magnitudes doesn't guarantee that you've despun

Â the primary body completely.

Â 12:51

And this is where a mutation damper would help get rid of that remaining energy and

Â make that actually attractive and collapse onto a point.

Â And that's what would be happening there.

Â So I'm just kind of showing this more as a pictorial thing.

Â So you've seen this.

Â It's not something that'll be on the exam.

Â But these plots, these graphical visualizations of momentum and

Â energy are quite powerful.

Â And if you read a lot of classical attitude control,

Â you will see these kind of tools being used in different areas.

Â