A review from last time, we talked about momentum controlled devices, let's do a quick thing. What was the big benefit, Kevin, of momentum controlled devices versus thrusters to do attitude control? >> Can have a more precise control of the attitude. >> Okay, I mean, we can give just a really small torques on it. Though that's slowly changing because people are coming up with very interesting micro thrusters that can also do nano Newton meters of force. And so historically that's been precisely true, 10, 20 years from now, we'll see. [LAUGH] It's evolving, but that's definitely so we can make some more forces. What else, Mariel? >> You don't have to use fuel. >> No fuel, that is the big benefit. So regardless of, well, those little micro devices are probably very efficient, they may not take much fuel. But it's still a finite resource. Versus these wheels, it's purely electrical. Part of that is a motor, an electric motor, that we drive. You give it a voltage, it spins up, all right? So with reaction wheel, you put 1 Newton meter in on that torque, you get 1 Newton meter back. It's a 1 to 1 relationship, which is nice. But we have to also develop the motor torque equations, which we'll do today. What we did last time was developed equations of motion of a spacecraft with the single variables b, c, and g, or at least, I outlined how you develop them. There's lots of little steps in between that you're doing. But so today too, we need extra equations of motion that we'll talk about today when it comes together. What is the benefit of a CMG device then compared to a reaction wheel? Mandar, what do you think? >> CMGs use a high spin rate, which stabilize, as opposed to reaction wheels, which need to spin. >> So it's the highest spin, what gives you a stability? >> No, sorry, they use high spin rate to generate the torque. >> Generate the torque, exactly. There's a huge gyroscopic coupling. If your wheel is spinning quickly, and my arm is my torque vector, if you're twisting that arm, you're changing that momentum vector. And h dot equal to l, a change in momentum, must be a torque. That's that huge gyroscopic torque that can be taken advantage of it. Great, so we get huge torques. If you have to maneuver very quickly or maneuver very large objects, like space stations and really big satellites, CMGs are definitely preferred. Otherwise, you would need huge wheels. What's the challenge with reaction wheels though? They're simple, they're cheaper, what's the big issue? So less torque, right, but what else? >> So there can be static dynamic imbalance of the wheel. >> I didn't think about that, but that's correct. You could have an imbalance. That's always the case, but that's also true for a CMG. No mechanical device is ever perfectly balanced, all right? So and a CMG actually moving at a very high frequency is going to give you this very high frequency jitter that you just have to make sure your system handles, and it's not an issue. Every structure kind of absorbs some of that jitter, acts like a low-pass filter. So, people have to design that. Yeah, so imbalances, if you're doing a reaction wheel, you go from slow speed to high speed. And you go through all the frequencies, so you may want to model if you have structural residences. If the sense of platform has a 10 hertz first mode stiffness, you may not want to hang out at 10 hertz a long time with a wheel. because you would be exciting that frequency. So there might be things like that to consider. But what's the fundamental challenge? Why can't we just use reaction wheels and apply a torque indefinitely? Andrew. >> Desaturate? >> Desaturate, right, at some point, you physically just can't go faster. The bearings break down, there might be heat dissipation issues. There's all kinds of stuff, an issue. So with CMGs, we're going at a fixed speed nominally. So I don't have any speed limits because I'm just running at a fixed speed. I can always keep twisting. What is the problem we have with CMGs then? Casey. >> We have gimbal lock. >> Gimbal lock, right? So with the reaction wheels, the axis about which we produce a torque is fixed as seen by the body. because its wheel is bolted down, right? And we had some pictures on this, let me just go down a few. I like it because it looks beefy, it's really bolted down, this spin X, the motor axis doesn't change as seen by the body. So it's always, you put on this 1 Newton meter on this wheel, it always gives you a torque about that axis, no matter what the body is doing. Versus if you go look at CMGs, which we had here, now, this is our spin axis. But then there's a gimbal axis, and that varies with time. And the gyroscopics we were hearing about earlier, that's always about that transverse axis. If this is my spin axis and that's my gimbal axis, then the change in momentum is about the third axis, that's the GT axis we called. And that axis varies as you gimbal. So I could only get that big gyroscopic torque about a single axis. And what can happen with gimbal lock is with multiple devices, those axes can all become coplanar. In fact, that's what would happen if you would just go there, take the spacecraft, and an astronaut just goes there and keeps pressing against it, just to be an irritant, that spacecraft will hold an attitude. But eventually, all the gimbals will line up, such that all the GT axes are orthogonal to that torque being applied. And then you have gimbal arc, and everything starts tumbling. So any momentum exchange devices has some momentum limitations in terms of speeds or orientations or how you cannot absorb infinite amount of momentum onto a system. That's just not physically possible. There's always a finite limit that we're seeing. So good, this is what we went through. Now deriving equations of motion, what was the one overriding equation that drives all this stuff? >> [INAUDIBLE] >> H dot=L, so Casey, let's go through this quickly. H was what for this system? >> The total angular momentum. >> Of the dynamical system, right? And here we have the spacecraft, we have a gimbal structure, and then we have the wheel within that gimbal structure. So we actually have three components. What was the dot, Kevin? >> The initial derivative. >> Right, the derivative as seen by the inertial frames. Immediately transport theorem, right, we're going to have to use that a bunch. And then the L? >> External torque. >> The net external torque acting on it, so motor torques don't go in that L. Motor torques are internal systems, where the spacecraft is being pushed off relative to the gimbal frame. And the gimbal frame pushes off relative to the wheel frame, those are internal torques. They don't show up in that L that we had, good. So, we went through this H dot=L, how hard could that be, right? And it's a lot of details, a lot of book keeping. So, some rigor is helpful here. Just notationally, to review, because we see the same, [COUGH] excuse me, today, gs is our spin axis of the wheel, gt is the transverse axis then, and gg, that's the gimbal axis. So for gimbaling gamma dot, wheel speed, big omega, all right, that was kind of the rough breakdown. We had the gimbal frame that we actually used a lot in the derivation. There's also body frame with a typical b1, 2, and 3, but you see in this system, it tends to be easier to write everything in the gimbal frame. We have the angular velocities of gimbal relative to the body, wheel relative to gimbal, so you can add, subtract them as needed. And why was it in the end, if we assume this gimbal frame is principle and the wheel frame, we've picked one that lines up with symmetry, it's principle, why is it that the wheel inertia is the same in the wheel frame as it is in the gimbal frame? What was the reason why that was case because that's not generally true, right, if you have two different frames? And that's only true for IW, not for IG. IG, if you look at this, it has three distinct inertias. IW has a spin axis inertia, and then the two other inertias are the same. >> Does the wheel inertia. >> Because of the symmetry, exactly. If you did the DCM, this is a rotation between a wheel fixed frame and the gimbal frame. The only difference is the rotation about the one axis, the GS, and if you put in a DCM, pretty imposed with the one axis rotation. Because of the symmetry here, you end up stuff that cancels, and that doesn't matter in the end, so you get the same thing. And I show that in Mathematica quickly without doing it by hand. But that's a trick we use. So that's why in the end, we define the wheel frame, but we never actually use it afterwards once we get past this step. And say well, the wheel tensor, we can express it easily in a diagonal form also with a G up here, instead of a W. We went here, we took our DCM, so if we have these three axes, we did this kind of when we worked with direction cosine matrices, right? The BN matrix, you could write as B1 transpose, B2, B3 as rows or N1, 2, and 3 as columns. So instead of BN, we have BG, so the G frames become the three columns that you have. We can express that. If you then do the matrix math here, which we did last time, you can write it as the principle inertias times the outer products of the base vectors. It's a very convenient form, because we know a lot of algebra on this. Any questions on these things too? If there's any questions on last material, yes, Casey. >> The gimbal frame is literally the frame holding the wheel, right? >> Yes, gimbal frame is this one here GG, so this axis lines up always with that. So, on a general CMG device, this is the only one that's fixed as seen by the body. That's where you bolt down the CMG. But the other two axes as you gimbal, so if this is my gimbal axis, the GS axis of the wheel and the GT, those are varying, they rotate about GG. That's what's happening there, yep. So GS, GG, and GT, that's where they come from. Good, so we went through that. So now to get H, as you were saying earlier, H has to be the total angular momentum of this dynamical system. And here we have three rigid body components. We account for all of them. Spacecraft to hub itself is easy, that's just inertia tensor times omega BN, we've done that before, a single rigid body. The gimbal frame has the same kind of formulation, the gimbal inertia times the angular velocity of gimbal relative to inertial. And we write that one as a sum of angular velocity of gimbal relative to body plus our typical body angular velocity. And this one is nothing, but gamma.tensor GG, and there was ways. I showed you vector ways to do this math or matrix ways, I actually think later, here, we did matrix ways. So this is kind of the core stuff when you write these momentums, just remember it has to be relative to the inertial again. You don't just put in the angular momentum of the wheel relative to the body. You need the full momentum of the system, H dot = L, inertial angular momentum that went there. Good, run through this, then there was a variety of derivatives we had to take, which hopefully, you could do with the transport theorem, that was pretty straightforward. The more interesting ones is what were omega S, omega T, and omega G, what were those omegas? Usually, we have omega 1, 2, and 3, which is the B1, 2, and 3 frame component, right? Kevin, do you remember? >> The components of the omega vector in G frame. >> Yes, so the omega S is the GS component of omega, and omega T is the GT, so instead of using omega 1, 2, we've taken the exact same omega vector. Just instead of using the body frame, we're using this gimbal frame. We're breaking it up into three different components, right? And just convenient because those transpose, those mappings happened everywhere. So now this is a scalar. So Tebo, I think last time, you answered this one. Taking a derivative of the scalar, with respect to what frame are we taking this time derivative? >> You don't take it with respect to a frame, it's a scalar. >> A scalar, but the scalar is a product of a bunch of vectors, a dot product essentially. So over here, what frame do you get to pick? >> Whichever one you'd like. >> But? >> It has to be the same. >> So if you have A times B, you can't say, well, I will take the Q frame derivative of A. And then later on, if chain ruled the N frame derivative of B, once you pick a frame across a product, across a scalar part of that product, you're pretty much committed, all right? And this is if you do it inertial, you can plug it in, and I think I also showed last time that if you did it in the body frame, for example, or the gimbal frame, that's another one you could pick. You'll always get the same answer as you would expect, all right? So this was a good exercise of taking derivatives of scalars since scalars are a function of vectorial quantities, to put that together. So just definition for today, you see this again. J is nothing, but the summation of the inertia of the frame plus the wheel within the frame. because these sums appear everywhere, so we define J as that. And then, yeah, the fun, which you did in homework, there's some sub-results there. We do all this, and then the final definition is IS, which is the inertia of the hub itself, plus the inertia of the VSCMG combined, gives us the spacecraft inertia. So this one now, does this I vary as seen by the body frame? >> Yes. >> Yeah, this one is fixed. It's a rigid hub, but if this is a variable speed CMG as you're gimbaling, that wheel, the mass distribution does change. If you just had a reaction wheel, does J vary as seen by the body, if you only have a reaction wheel? All right, no, because that wheel, even though you're spinning it up and down, because it's perfectly symmetric balanced, there's no change in mass distribution. One particle replaces the next particle. So, anyway, so keep that in mind, when you do these simulations, in this case, you're not going to simulate this in this class, but 6010, you would. These are the things you would have to update every time step, 6010 is fun. There's some really interesting code you're writing there. You guys think estimation is bad, you have no idea. This is fun stuff. So now we have our equations of motion, this is kind of where we ended up with last time. This is a rigid spacecraft plus a single VSCMG device, and we still have the external torques that act on it, right? That's not motor torques, that's gravity gradient torques or radiation pressure, all this good stuff. And as before, if it's a reaction wheel, what all vanishes? We'll see that again today. What all goes to 0 if it's a reaction wheel, instead of a VSCMG? Endor? >> Gamma dot terms? >> Gamma dot, yeah, anything else? Gamma double dot also goes to 0, so this vanishes, this vanishes. All this other stuff vanishes, and in fact, you get only this term, you get this term and this term. If you go back and look at your dual spin equations of motion, let me just put this up. With the dual spin stuff, you had exactly these same terms. The one difference was, we didn't have GS GT GG, there, the dual spinner, we lined up our wheel with B1. But in essence, B1 then is GS, all right, that is that axis. So just make that replacement, it's a 1 to 1 thing. This is a slightly more general version, where your wheel doesn't have to line up with the B1 axis. That's what we found. If it is a CMG, Kevin, how much simplifies? >> You only get rid of the omega [INAUDIBLE] >> That's it, whoopdy do, we're still left with 6 million terms. So now, especially when you have lots of CMG, that's why you can see the mathematics of CMGs, way more complicated right away. In fact, if you have a dual spinner constant speed, even this time goes to 0, gets more simpler. But this is where we ended up last time.