0:05

A review from last time, we talked about momentum controlled devices,

Â let's do a quick thing.

Â What was the big benefit, Kevin,

Â of momentum controlled devices versus thrusters to do attitude control?

Â >> Can have a more precise control of the attitude.

Â >> Okay, I mean, we can give just a really small torques on it.

Â Though that's slowly changing because people are coming up with very

Â interesting micro thrusters that can also do nano Newton meters of force.

Â And so historically that's been precisely true, 10, 20 years from now, we'll see.

Â [LAUGH] It's evolving, but that's definitely so

Â we can make some more forces.

Â What else, Mariel?

Â >> You don't have to use fuel.

Â >> No fuel, that is the big benefit.

Â So regardless of, well, those little micro devices are probably very efficient,

Â they may not take much fuel.

Â But it's still a finite resource.

Â Versus these wheels, it's purely electrical.

Â Part of that is a motor, an electric motor, that we drive.

Â You give it a voltage, it spins up, all right?

Â So with reaction wheel, you put 1 Newton meter in on that torque,

Â you get 1 Newton meter back.

Â It's a 1 to 1 relationship, which is nice.

Â But we have to also develop the motor torque equations, which we'll do today.

Â What we did last time was developed equations of motion of a spacecraft with

Â the single variables b, c, and g, or at least, I outlined how you develop them.

Â There's lots of little steps in between that you're doing.

Â But so today too, we need extra equations of motion that we'll talk about

Â today when it comes together.

Â What is the benefit of a CMG device then compared to a reaction wheel?

Â Mandar, what do you think?

Â >> CMGs use a high spin rate, which stabilize,

Â as opposed to reaction wheels, which need to spin.

Â >> So it's the highest spin, what gives you a stability?

Â 1:52

>> No, sorry, they use high spin rate to generate the torque.

Â >> Generate the torque, exactly.

Â There's a huge gyroscopic coupling.

Â If your wheel is spinning quickly, and my arm is my torque vector,

Â if you're twisting that arm, you're changing that momentum vector.

Â And h dot equal to l, a change in momentum, must be a torque.

Â That's that huge gyroscopic torque that can be taken advantage of it.

Â Great, so we get huge torques.

Â If you have to maneuver very quickly or maneuver very large objects,

Â like space stations and really big satellites, CMGs are definitely preferred.

Â Otherwise, you would need huge wheels.

Â What's the challenge with reaction wheels though?

Â 2:27

They're simple, they're cheaper, what's the big issue?

Â So less torque, right, but what else?

Â >> So there can be static dynamic imbalance of the wheel.

Â >> I didn't think about that, but that's correct.

Â You could have an imbalance.

Â That's always the case, but that's also true for a CMG.

Â No mechanical device is ever perfectly balanced, all right?

Â So and a CMG actually moving at a very high frequency is going to give you this

Â very high frequency jitter that you just have to make sure your system handles,

Â and it's not an issue.

Â Every structure kind of absorbs some of that jitter, acts like a low-pass filter.

Â So, people have to design that.

Â Yeah, so imbalances, if you're doing a reaction wheel,

Â you go from slow speed to high speed.

Â And you go through all the frequencies, so

Â you may want to model if you have structural residences.

Â If the sense of platform has a 10 hertz first mode stiffness,

Â you may not want to hang out at 10 hertz a long time with a wheel.

Â because you would be exciting that frequency.

Â So there might be things like that to consider.

Â But what's the fundamental challenge?

Â Why can't we just use reaction wheels and apply a torque indefinitely?

Â Andrew.

Â >> Desaturate?

Â >> Desaturate, right, at some point, you physically just can't go faster.

Â The bearings break down, there might be heat dissipation issues.

Â There's all kinds of stuff, an issue.

Â So with CMGs, we're going at a fixed speed nominally.

Â So I don't have any speed limits because I'm just running at a fixed speed.

Â I can always keep twisting.

Â What is the problem we have with CMGs then?

Â 3:52

Casey. >> We have gimbal lock.

Â >> Gimbal lock, right?

Â So with the reaction wheels,

Â the axis about which we produce a torque is fixed as seen by the body.

Â because its wheel is bolted down, right?

Â And we had some pictures on this, let me just go down a few.

Â I like it because it looks beefy, it's really bolted down,

Â this spin X, the motor axis doesn't change as seen by the body.

Â So it's always, you put on this 1 Newton meter on this wheel,

Â it always gives you a torque about that axis, no matter what the body is doing.

Â Versus if you go look at CMGs, which we had here, now, this is our spin axis.

Â But then there's a gimbal axis, and that varies with time.

Â And the gyroscopics we were hearing about earlier,

Â that's always about that transverse axis.

Â If this is my spin axis and that's my gimbal axis,

Â then the change in momentum is about the third axis, that's the GT axis we called.

Â And that axis varies as you gimbal.

Â So I could only get that big gyroscopic torque about a single axis.

Â And what can happen with gimbal lock is with multiple devices,

Â those axes can all become coplanar.

Â In fact, that's what would happen if you would just go there,

Â take the spacecraft, and an astronaut just goes there and keeps pressing against it,

Â just to be an irritant, that spacecraft will hold an attitude.

Â But eventually, all the gimbals will line up,

Â such that all the GT axes are orthogonal to that torque being applied.

Â And then you have gimbal arc, and everything starts tumbling.

Â So any momentum exchange devices has some momentum limitations in terms of speeds or

Â orientations or how you cannot absorb infinite amount of momentum onto a system.

Â That's just not physically possible.

Â There's always a finite limit that we're seeing.

Â So good, this is what we went through.

Â Now deriving equations of motion,

Â what was the one overriding equation that drives all this stuff?

Â >> [INAUDIBLE] >> H dot=L, so Casey,

Â let's go through this quickly.

Â H was what for this system?

Â >> The total angular momentum.

Â >> Of the dynamical system, right?

Â And here we have the spacecraft, we have a gimbal structure, and

Â then we have the wheel within that gimbal structure.

Â So we actually have three components.

Â What was the dot, Kevin?

Â >> The initial derivative.

Â >> Right, the derivative as seen by the inertial frames.

Â Immediately transport theorem, right, we're going to have to use that a bunch.

Â And then the L?

Â >> External torque.

Â >> The net external torque acting on it, so motor torques don't go in that L.

Â Motor torques are internal systems,

Â where the spacecraft is being pushed off relative to the gimbal frame.

Â And the gimbal frame pushes off relative to the wheel frame,

Â those are internal torques.

Â They don't show up in that L that we had, good.

Â So, we went through this H dot=L, how hard could that be, right?

Â And it's a lot of details, a lot of book keeping.

Â So, some rigor is helpful here.

Â Just notationally, to review, because we see the same,

Â [COUGH] excuse me, today, gs is our spin axis of the wheel,

Â gt is the transverse axis then, and gg, that's the gimbal axis.

Â So for gimbaling gamma dot, wheel speed, big omega, all right,

Â that was kind of the rough breakdown.

Â We had the gimbal frame that we actually used a lot in the derivation.

Â There's also body frame with a typical b1, 2, and 3, but you see in this system,

Â it tends to be easier to write everything in the gimbal frame.

Â 7:39

>> Does the wheel inertia.

Â >> Because of the symmetry, exactly.

Â If you did the DCM, this is a rotation between a wheel fixed frame and

Â the gimbal frame.

Â The only difference is the rotation about the one axis, the GS, and

Â if you put in a DCM, pretty imposed with the one axis rotation.

Â Because of the symmetry here, you end up stuff that cancels,

Â and that doesn't matter in the end, so you get the same thing.

Â And I show that in Mathematica quickly without doing it by hand.

Â But that's a trick we use.

Â So that's why in the end, we define the wheel frame, but

Â we never actually use it afterwards once we get past this step.

Â And say well, the wheel tensor, we can express it easily in a diagonal form also

Â with a G up here, instead of a W.

Â We went here, we took our DCM, so if we have these three axes,

Â we did this kind of when we worked with direction cosine matrices, right?

Â The BN matrix, you could write as B1 transpose, B2, B3 as rows or N1,

Â 2, and 3 as columns.

Â So instead of BN, we have BG, so

Â the G frames become the three columns that you have.

Â We can express that.

Â If you then do the matrix math here, which we did last time, you can write it as

Â the principle inertias times the outer products of the base vectors.

Â It's a very convenient form, because we know a lot of algebra on this.

Â Any questions on these things too?

Â If there's any questions on last material, yes, Casey.

Â >> The gimbal frame is literally the frame holding the wheel, right?

Â 9:04

>> Yes, gimbal frame is this one here GG, so this axis lines up always with that.

Â So, on a general CMG device, this is the only one that's fixed as seen by the body.

Â That's where you bolt down the CMG.

Â But the other two axes as you gimbal, so if this is my gimbal axis,

Â the GS axis of the wheel and the GT, those are varying, they rotate about GG.

Â That's what's happening there, yep.

Â So GS, GG, and GT, that's where they come from.

Â Good, so we went through that.

Â So now to get H, as you were saying earlier,

Â H has to be the total angular momentum of this dynamical system.

Â And here we have three rigid body components.

Â We account for all of them.

Â Spacecraft to hub itself is easy, that's just inertia tensor times omega BN,

Â we've done that before, a single rigid body.

Â The gimbal frame has the same kind of formulation,

Â the gimbal inertia times the angular velocity of gimbal relative to inertial.

Â And we write that one as a sum of angular velocity of gimbal relative to body

Â plus our typical body angular velocity.

Â And this one is nothing, but gamma.tensor GG, and there was ways.

Â I showed you vector ways to do this math or matrix ways, I actually think later,

Â here, we did matrix ways.

Â So this is kind of the core stuff when you write these momentums,

Â just remember it has to be relative to the inertial again.

Â You don't just put in the angular momentum of the wheel relative to the body.

Â You need the full momentum of the system, H dot = L,

Â inertial angular momentum that went there.

Â Good, run through this, then there was a variety of derivatives we had to take,

Â which hopefully, you could do with the transport theorem,

Â that was pretty straightforward.

Â The more interesting ones is what were omega S, omega T, and

Â omega G, what were those omegas?

Â 10:44

Usually, we have omega 1, 2,

Â and 3, which is the B1, 2, and 3 frame component, right?

Â Kevin, do you remember?

Â >> The components of the omega vector in G frame.

Â >> Yes, so the omega S is the GS component of omega, and omega T is the GT,

Â so instead of using omega 1, 2, we've taken the exact same omega vector.

Â Just instead of using the body frame, we're using this gimbal frame.

Â We're breaking it up into three different components, right?

Â And just convenient because those transpose,

Â those mappings happened everywhere.

Â So now this is a scalar.

Â So Tebo, I think last time, you answered this one.

Â Taking a derivative of the scalar,

Â with respect to what frame are we taking this time derivative?

Â >> You don't take it with respect to a frame, it's a scalar.

Â >> A scalar, but the scalar is a product of a bunch of vectors,

Â a dot product essentially.

Â So over here, what frame do you get to pick?

Â >> Whichever one you'd like.

Â >> But? >> It has to be the same.

Â 11:40

>> So if you have A times B, you can't say,

Â well, I will take the Q frame derivative of A.

Â And then later on,

Â if chain ruled the N frame derivative of B, once you pick a frame across a product,

Â across a scalar part of that product, you're pretty much committed, all right?

Â And this is if you do it inertial, you can plug it in, and

Â I think I also showed last time that if you did it in the body frame, for example,

Â or the gimbal frame, that's another one you could pick.

Â You'll always get the same answer as you would expect, all right?

Â So this was a good exercise of taking derivatives of scalars since scalars

Â are a function of vectorial quantities, to put that together.

Â So just definition for today, you see this again.

Â J is nothing, but

Â the summation of the inertia of the frame plus the wheel within the frame.

Â because these sums appear everywhere, so we define J as that.

Â 13:16

One particle replaces the next particle.

Â So, anyway, so keep that in mind, when you do these simulations,

Â in this case, you're not going to simulate this in this class, but 6010, you would.

Â These are the things you would have to update every time step, 6010 is fun.

Â There's some really interesting code you're writing there.

Â You guys think estimation is bad, you have no idea.

Â This is fun stuff.

Â So now we have our equations of motion,

Â this is kind of where we ended up with last time.

Â This is a rigid spacecraft plus a single VSCMG device, and

Â we still have the external torques that act on it, right?

Â That's not motor torques, that's gravity gradient torques or

Â radiation pressure, all this good stuff.

Â And as before, if it's a reaction wheel, what all vanishes?

Â We'll see that again today.

Â What all goes to 0 if it's a reaction wheel, instead of a VSCMG?

Â Endor?

Â >> Gamma dot terms?

Â >> Gamma dot, yeah, anything else?

Â 14:13

Gamma double dot also goes to 0, so this vanishes, this vanishes.

Â All this other stuff vanishes, and in fact,

Â you get only this term, you get this term and this term.

Â If you go back and look at your dual spin equations of motion,

Â let me just put this up.

Â With the dual spin stuff, you had exactly these same terms.

Â The one difference was, we didn't have GS GT GG, there,

Â the dual spinner, we lined up our wheel with B1.

Â But in essence, B1 then is GS, all right, that is that axis.

Â So just make that replacement, it's a 1 to 1 thing.

Â This is a slightly more general version,

Â where your wheel doesn't have to line up with the B1 axis.

Â That's what we found.

Â If it is a CMG, Kevin, how much simplifies?

Â