Okay, so we've done a review of what these devices are. There are certain benefits, everything has drawbacks, now we need to get into the math of it. How do we get the equations of motion of these things? So we're going to be stepping through this carefully. We are going to do it for spacecraft with a single VSCMG. So that's a single variable-speed control-moment gyroscope. That's a single axis control-moment gyroscope. There have been a few that have flown this way. Yes, Jordan? >> Sorry, going back really quick to the CMGs, do you know if there are any that use like flux spinning to do like, so that could just levitate the bearing and rotate them? >> There are, yeah. because, especially, well, with the variable speed, the Air Force had a nice program, Fred Levy was running that, to look at energy storage, because as you spin up this wheels, as you de-spin them, you can actually pull and you could turn mechanical energy in the dimo effect. You can turn into electrical energy, so this could replace a battery. But you have to spin to pretty high speeds to make them efficient. So they're looking at 30, 40,000 RPM's, really fast and mechanically with worrying about friction and little imbalances is quite tricky. So there's some that are looking at magnetic bearings, levitation, and those kind of things. Not necessarily flux pinning, I haven't heard that one being used but, magnetic bearings, absolutely, you will find some of those devices looking at that. Yep, so there's a lot of mechanical challenges and how to do this well, robust, it survives launch, where you get shaken to death, almost hopefully, not slight. So anyway, so think of VSCMG as a Hybrid, we'll have the math for the CMG and for the reaction wheel all embedded in one. So we can talk to this once. I want you to understand this path, deal the parts of just specialties of it, that's going to be our goal. So goes with time, to wrap this up. So very convenient because we can run both but, this VSCMG's are devices being researched and done actually phoned some small sats via CMG and tested those as well already. So this is not just theoretical, this is starting to happen. But this is stuff I did way back for my PhD, it's one of my chapters in the PhD. So we're going to try and look at this. One of the niceties I'm going to highlight is we talked about CMGs having singularities. What we'll find is VSCMG actually avoids this by move, by changing the wheel speeds, I can actually produce torques about axis or to the tork direction. And that can be useful, it's highly redundant, each device if it's a variable speed you could use it as a reaction wheel, and as a CMG. So that's two control modes that are for devices that gives you eight control modes, and it's still a three degree problem. So you've got a lot of redundancy built in which is sometimes nice, and also combine power energy storage, something we just discussed. That's something people have been researching and looking at and would be nice, that we didn't need batteries as much. So how do we get equations of motion? It all basically boils down to this one little, simple equation, all right? How hard could that be? It's kind of like, hey you're doing your whole dissertation f=ma, that's it. Right, if you're doing orbits, trajectories, unstable manifolds. All you're doing, and Daniel's laughing, that's really that's all you're doing is f=ma, right? How hard is that, in attitude it's h.equal to l how hard can that be? And I'll show you here this is the hint of that. If you want some more fun go chat with Cody and John from my lab, they've been doing these imbalanced VSCMG equations that are quite interesting. So let's review this, what is h.equal to l? What is l here? Start in the right hand side, Andre? >> [INAUDIBLE] >> Be more specific. >> [INAUDIBLE] on the body. >> Which body? because we now have multiple. >> Rigid body. >> Only on the rigid bodies and not on the wheels? No, external torque. >> Exactly, right? If this is a dynamical system, this applied to blobs in space, jello flying through space. But it was the net external torque acting on this dynamical system. Now we don't have blobs but we will have systems of rotating, it's a multi-rigid body problem essentially that we're solving here. So that's just the net external torques. We don't need motor torques. Whatever that reactional spin torque is doesn't appear here, because that's an internal torque. The Gimbal access motor torque does not appear here it's also an internal torque. So just the externals, gravity gradients, solar radiation pressure drag, maybe you have thrusters to augment this. Because maybe for momentum dumping people still use thrusters sometimes, that could be good. Now what is the h on the left hand side? Robert. >> On the tip of my tongue, I'm just- >> The total angular momentum. >> Total angular momentum, right? All of this dynamic assistance. So if it's a blob, it's just the entire thing, it's not quite a blob here. For us it's a spacecraft, plus the wheels, plus the frame whatever holds these things together. Everything that has momentum, angular momentum, we have to account for. And then what's this Baud up here, what type of time derivative are we taking? And as you know we often for good reason do things in a body frame, but transport theorem is going to be all over this. We have to do that, right? We're also about which point have we taken torks in momentum here? Center of mass, actually. So this either has to be a center of mass point or taking moments and torques about an inertial point. Then h equal to l is true. If you're talking about a body fix point, that's anything else than center of mass, there was extra turnums that have to appear here. So again these inbalance things makes things immediately much more complicated. So h equal to l. We can assume this is all about the center of mass of this system. All the wheels are balanced so even when I'm trying gyrating I'm not really changing. When I'm putting the gimballing angles in I'm not changing the center of mass of that CMG configuration, I'm just changing where the axis is pointing. I'm changing the inertias but not center of mass, so that makes huge simplifications. So we need some frames to define this stuff, and I've got this gimbal frame. So you will have gs is my spin axis, so this is your wheel, if it's just the reaction wheel this is all you would have. A reaction wheel in a gimbling frame, this is how we can twist that spinning wheel, GG is the gimble axis, and that other. And the CMG devices solves a pancake with the motor knob sticking out left and right, it's actually the gg axis. And so gs is the spin axis, gimble axis, and then gt is at the transverse axis, it's the third axis that cannot complete a full frame. GS crossed to GT gives you plus GG, okay? So that's what these are, we've defined them. Big omega here is going to be the wheel speed and we're treating it here as time varying. Later on we can freeze it to make it a CMG device, not a variable speed CMG. Gimbel angles are defined through gamma, so the gimbel rate at which I am twisting these things is gamma dot, and that's what that's defined. So that's the rotation rate about this gg axis, so kinematics is going to come back in. And then so the full frame is gs, gt, gd, that's the g frame that we have. This is the one we'll use a lot. So let's talk about angle velocities, the angular velocity of the gimbal frame relative to the body. This is where we attach this device to a body, because the motor is taking it and it's twisting it left and right, right? And so this is the twist rate and the axis about which you're twisting is gg. So the angular velocity of the g frame relative to the b frame is just going to be gamma times gg. It's the classic kinematics we did in the first part of the class. There's a wheelframe, I'm going to introduce this quickly, and then within a few steps, we're going to ignore it again. We'll make some arguments why we can ignore it. So the wheelframe needs three axis. I'm picking principal axis, so the spin axis, ds, I can reuse it just to be lazy. That's my first one and then I have a wt and a wg which are two axis that are fixed relative to the wheel, so as the wheel rotates, these axis are going to rotate with the wheel, right? So if you look at the wheel frame, you can go ahead and get the angular velocity of the wheel system relative to the gimbal frame, and that's simply going to be the wheel's spin rate times the wheel spin axis, that's it. So if you sat on the wheel frame it's set all this gimbal frame, look at the wheel you would see the wheel's spinning about that axis at that rate and vice versa. So now we have our two. If you need omega of w, relative to b, you would just add up the two of course, and that's how, so we can get any combinations of omegas that we need. Let's talk about inertias. I'm assuming this thing is built in a way such that I have three principal inertias of this gimbal contraption and they'd line up with the gs, gt, gt frame. Recently good assumption, may not be perfect but it's a good simplifying in the analysis that we're doing right now. So I have IGS, IGG, GT and IGG. So IG is the inertia's of the gimbal frame along the ST and G axis. That's where the subscripts are coming from. For the wheel, we can write that in the wheel frame, you're going to have an inertia about here so that's going to be the wheel inertia about the s axis. And then I have Iwt, because the wheel is perfectly symmetric, right? No imbalances, no other warping, all these kind of stuff. It's going to be the same inertia about the any two orthogonal axis in the plane of the disc, of the wheel, right? So I just have Iwt, Iwt. Now, here I'm saying the wheel inertia in the gimbal frame, is actually going to be the same as wheel inertia expressed in the wheel frame. Why does that happen? I think this is really, that's an inertial tensor. This is a matrix representation of the inertial tensor, where I picked the w frame. Here I'm claiming if I picked the W frame or the G frame, they both give you the same answer. Why do you think that is? >> There's a degenerate Eigenvalue in the w inertia. >> You're getting mathematical. No, but you're right. There is a repeated Eigenvalue in here. That means that these axes are actually, the Eigenaxes that generates these inertias are not unique. If we go back and look at that. We talked about Eigenvalue of an axis symmetric body. Where we had IS and IT in some of the torque free motion of an axis symmetric body, all of that stuff. So you don't have to pick. I could really, to get this diagonal from I just have to pick this axis and I have to pick any two axis that are orthogonal to GS will do. If any two axis will do, I can always pick GT and GG if you wanted to, and they'll give you the same answer. So actually I wrote up a quick mathematical thing. So really I'm going to evaluate this now, I'm pulling in library. I'm going to define now here this as a diagonal matrix in this case the g frame, I've got this. Now, since GS is my first axis, I'm doing what we call m1 basically that was the one axis rotation that we needed with an angle. So, you'd have 1, 0, 0 or I can do metrics form of wg, right? So that's the classic note, that's the rotation because we're doing one axis rotation where the GS is the first axis. So if I translate here from the gimbal frame to the wheel frame, this is what I could do, I get this answer and then you can identify what does this cosine squared the same thing sine squared. Those things are just going to drop out to b1, and after all this math you get nothing but exactly the same result again even though we did this code of transformation. So if you don't believe the geometric arguments, you can quickly do the math and validate and go, yeah. Actually, this inertia tensor is diagonally both in the w frame and in the g frame, that's all we need, because that saves us a lot of transpose without then end up all you're doing is creating a lot of trigs that become one in the end and not needed. Good, so that's kind of helps explain that one last slide. Now I need to get some corner frames and attitudes. We have the body attitude, we know how to describe that already, MRP's, DCM, Cortanians whatever you want to wish. We need to get the attitude of the gimball frame relative to the body frame. And if you go back and look at how we defined DCM's, right? It was basically the mapping from N to B and I'm just going to write that one out. If your recall BN ended up being n1, n2, n3 in B frame components, that was one way to write it or each row, Was the b1, 2 and 3 in N frame components. So if we go back and look at those definitions, if you have the three axis, you can actually construct the DCN very quickly. And in fact here, I'm using this one, so instead of having, what do i have here? Let me go back. Here we have BG instead of BM. So instead of n1, 2, 3, I have the G first, second and third axis I put in. And then the B frame component which is how we define them and that's give you to DCM. So that works. So if we have BG. Now if I need to get the inertia tensor which was diagonal in the G frame which is great. But I typically will need it in the end in the body frame, we'll do a lot of body frame relative derivatives, I would have to reimposed multiply with the DCM, right? To map it from one frame into another frame and you can do that. The mathematics here become quite simple, IG is a diagonal and this one I can write out like this. So, let's just kind of do this by hand once. So, if I have BG is equal to my first vector, my second, my third, and then I need to compute BG times I in G frame components times B, now G transpose, right? That would give me IG and the body frame. Now, if you write it in the matrix form like this, it's going to be nothing but gs, gt, gg that's the BG times this diagonal which was Igs 0, 0, 0 Igt, 0, 0, 0 Igg times the same matrix transposed. So, this transpose becomes gs transposed, gt transposed gg transposed, right? And now you can just start to carry out all this matrix math here. If you look at the second part, this is going to be Igs times gs. You can think of this like a vectrix, matrix of vectors that we talked about earlier. So out of this, all you'll going to get is Igs times gs hat transpose Igt, the second row, 0 times this, 0 time this, drops out, gt transpose plus Igg, gg transpose. So that's that part times this, so, [LAUGH] GS's, let's see. Actually that's not right, hold on. This is not a single line. This is a three by three times a three by one. This is not plus, that's wrong. This would have to go here, and this goes here. This should be a three by one. It's a three by three times a three by one. That should be a three by one, one by three times three by one. Now all I have to do is gs times this term, gt times that term, gg times this term, and that's where you get the final answer. So from here, if I go backwards, IGs, Gs, Gs transpose plus these other terms. That's all we're doing. So it's a nice analytic way that we can do these projections from the inertia of one frame to another. And you will see this is a handy result to what we'll use several times as we go through that. So that's what was done here. This one also in the gimbal frame, I can just do the same thing, but instead of IGs, IGT, I have IWs, IWT and that's it. It's a little bookkeeping. So you get exactly the same result. So now we have the inertia's written in two different frames. Good, next step, Angular Momentum. You guys said H had to be the angular momentum of the total system. The system will be composed of a single VSCMG, so it's a spacecraft, that's part B, that's the body. Then I have G that denotes the inertia and momentum of the gimbal frame that has some mass and it rotates and does stuff. And then you have HW which is the momentum of just the wheel itself spinning about that. So we have to add up all these momentums. And each one of these momentums is of these bodies relative to the inertial frame. So for the body this is dead simple, you've done this. I'm using IS now, that is my inertia tensor of the spacecraft. You'll see a different I appearing later on. So this one right now is just of the spacecraft itself. It has the center of mass of the reaction wheel, CNG devices already accounted for in that. But what we're not doing is the momentum of the wheel-disk part spinning about the center of mass of those disks. So a lot of it is already lumped in there. And this part will be easy to go forth. The other ones get more complicated. So for the gimbal frame, it's the same math at the first step. We have IG times, now I need omega G relative to N and omega G relative to N is omega G relative to B plus B relative to N, right. Earlier we kept calling this just omega, here and being a little explicit because we've got lots of different frames and lots of different omegas. So it's a little bit easier to keep track of. So we need both. This one was simply gamma dot times Gg and that's put in here. And the other one is omega BN. The inertia one I've written with this projections into G frame components already. So, IG times omega BN gives me this answer, and IG times this one will give me this answer. Now, again, I'm jumping through a few steps directly. So let me just look at it once. This is a tricky one to do yourself in the homework and different ways to solve this. One thing you could say is, you know you have IG like this, and omega G relative to B is just gamma dot times Gg. Well Gg times Gg here transposed just gives you 1. That's good. Gg times Gt transposed, what's that going to happen? The dot product between Gt and Gg. Zero, they're orthogonal, right? So that drops out. And same thing with the Gg times Gs. They're also orthogonal, they drop out. So that's one way to do it. I think on the next slide I show you another way with the matrix method. That will give you the same answer hopefully. It's the same physics. Now these terms, gs transposed with omega BN will happen a lot. If you look at, if we were showing the definition, let us recall how we typically write omega. So omega BN, we typically write it as omega 1b1 + omega 2b2 + omega 3b3. So we have a reframe, the omega 1, 2, 3s are nothing but the 1, 2 and 3 access components. Omega 1 is really defined as b1 dotted with omega BN, or in matrix form this is equivalent to b1 transpose omega BN, all right? Now, here we don't have b1 dotted with omega, we have gs dotted with omega. So if you have omega BN, gs hat, this is really what is going to give you, if you think of this, if this is your omega vector. And here is your gs vector. It's giving you the gs component of that omega vector essentially. And we're just going to call that, instead of omega 1, 2, 3, we've already used 1, 2, 3 for the body frame stuff. So I can't reuse the same names, I'm just going to use omegas, omegat and omegag for the gs, gt, gg axis of that stuff. So a slight notational thing but it helps us. So that's what you see defined right here. Where we have Omega s is gs hat transpose with omega. Omega T is gT transpose with omega and gg is the projection of g onto omega. So the omega vector, this is again without any subscripts, we imply it's omega BN. At some point it just gets old to write all that stuff over and over again. So this is omega BN is equal to these three vector components of that along the g axis. So if you use that definition then you have here gs transpose times omega and that becomes nothing but this little omega s. So, the same momentum expression here can be rewritten a little bit more compactly in this form and that will help us in same label keeping that we do. So, now we have to angular momentum vector off the g frame relative to the inertial frame. What we need next is the angular momentum of the wheel. >> [INAUDIBLE] >> Itself. Then it's the same formula. You need the angular velocity of the wheel relative to inertial, and we know the angular velocity of the wheel relative to gimbal. That was big omega times gs. That was the spin axis side. Gimbal relative to body, that was the gamma dot times Gg, that's the gimbaling rate. And then of course, this is the classic spacecraft body rate that you'd have. If you plug that in you now have three of these terms to evaluate. And I will just to show you different ways to do it, I'll do it different ways here. This first term I've written out, and it's going to be this inertia tensor which we rewrote earlier as a result of the principal inertias times these outer products times omega. And in here quickly you can see again gs hat transposed omega. That's what we defined to be omega s. The projection of the gs axis onto omega was the omega s component. Same thing with omega t and omega g. So that's one way we can get that first term. That gives us this one. Now we want to get the other two. I'm showing you a matrix representation, for some people this might be a easier way to think. Both of them should all give you the same stuff. This term I can write in G frame components as this diagonal matrix. G relative to B, that was gamma dot times Gg. So, in g frame component it's 00 gamma dot, all right? And then you just do your typical matrix math with which will give you 00 IWt that along the third axis, as the vector representation of that. That's the third axis is Gg, so you always get back the same answer. You could do the same thing also for this one, just here omega w relative to g was big omega times gs, so it's big omega along the first axis of the wheel frame. This inertia tensor in the wheel frame. You carry out the matrix math, you get just something in the first axis, which is the gs representation. So, this ties back to the, you know, week one stuff that we did. Vector matrix representation. You should be able to go back and forth and whatever's most convenient It'll get you there, it should be the same answer. Now, we have these, we add them up to the other term, you combine them, this gives us now the angular momentum of the wheel that we need. Okay, great. A few more things, as we take derivatives, we'll want inertial derivatives and, of course, we'll use transport theorems because a lot of these inertia tensors are fixed by the body frame. And also some of these other axis have benefits. So, the gs, gt, gg axis you could write analytically if you know their orientation at launch. I locked them down and gs was pointing here and gt was pointing here, and there's no gimble angle, right? And after launch, you rotated 15 degrees, you can actually compute with your classic cosines and sines. What are the current orientations of those axes? gg stays always the same, hopefully, that means you bolted it down correctly, all right? The gimbal axis is supposed to stay body fixed, always. But for a VSCMG, the GSGT, we have to treat generally as time varying. So, you can write this and in code in fact, it would simulate this. If I know my gimbal angle I can't compute at any instant. If I know my initial orientations at launch what are my current orientations of this frame of the wheel? How far has it twisted? All right, so we will need derivatives of this stuff. If we take body frame derivatives, different ways you can do it. If you want to, you could just use this expression here and take derivatives. because these are actually fixed as seen by the body. And cosine becomes minus sine, which is right here. Sine becomes cosine, and you can see the derivative of this is going to be the gt direction and vice versa. And the derivative of this one is just common sense, is going to be zero. That's one way. Or you can also use a transport theorem and say, well wait a minute, instead of taking the b frame, what if I took the g frame derivative of this stuff? And if you have the g frame derivative of gs that is zero, I see by the frame itself these axes don't move. Plus you would need the omega g relative to b, which is gamma delta gg, crossed with this axis, you get a g, a third axis cross with gs, the first axis which gives you plus the second one. You can see the transport theorem, you get the same answer. Different ways you can approach it. I like the transport, it's very compact, you can get there very, very quickly without writing it out. So, we can get the body frame derivatives, the key is the third one, that's where it's locked down. That one doesn't vary, the other two are a function of the gimbal rate. If you need inertial derivatives, which we can do it here, we've chosen to do b-frame derivatives. We've already got these answers as a sub-result, plus the omega crossed these axes again. And you can do that, and carry out, if you write omega. The easy trick here is to get this answer in your homework I would say. Write omega node as omega 1 b1, omega 2 b2 because you're crossing the gs. Instead write omega as omega s gs, omega t gt, right? Use the g frame components of omega and those definitions and that's where these terms come in. And then, it's just a classic cross products between vectors in the same frame. So, 1 crossed 2 gives you 3, 3 crossed 1 give you minus 2, all the usual rules. And that's how you can get there then as well. So, that would work. So then, you should derive this yourself then. I'm just kind of showing you the highlights and how we go after it. Since we have omega s, omega t and omega g, the vector components of the body rate in the g frame components. We will also have derivatives of this. Omega s, omega t, omega g are scalars. So, what type of derivative am I taking on the left hand side? Tebow. >> Time derivative it's- >> As seen by what frame? >> Doesn't matter. >> Doesn't matter, right? Just a scalar, so that stuff comes back again. If it's just a scalar it's just a time derivative. Over here, Tebow, what type of derivative have I chosen to use here? >> An inertila derivative. >> Inertial derivative. But that was purely a choice, I didn't have to use an inertial derivative. I could have used any, as long as I'm consistent, you can't use these derivative chain rules into one part in a and one part in b. If you've taken a, make everything a, if you've taken b, make everything b. The math will work itself out. So, you can do that, so that term is just okay, the angular acceleration dotted in the the gs direction, great. This term here, gs hat transposed omega. If you go back and look at gs hat is here, so we have all this math times, what did we do, times omega, gs hat transposed omega. So, if you can see here, gs hat, this stuff, transposed omega, gt transposed with omega, what does that give me? >> Omega t. >> Omega t, exactly. That was the projection. Gg transposed omega gives you omega g. And you're going to have an omega g, omega t with that product and here omega t, omega g with the product. 1 plus, 1 minus, they both cancel and all you're left with is here times gt times that term, that second term, which was gt transposed omega which gives you omega t. Let's say we did body frame. How is that going to look different? Let me do that by hand. So, just so I can prove, so if we use this formula, I don't need to fire here, okay. If we use this formula as Tebow says, that's just a time derivative. Now, let's say, I'm going to use body frames. I'm going to use primes as a body frame instead of writing ddt and b's and all that stuff. You would have gs transposed prime, right, times omega plus gs hat omega BN prime. Well, one thing, omega BN prime, how does that relate to the initial derivative? It's the same thing, right? Omega, the B and the N frame is the same. So, that's kind of cool. We get right away gs hat transposed omega BN dot, the inertial derivative. So, that's good. This one, we need the body frame derivative of gs, and if you go back, we did that actually and I had showed you the results. Here we go. The body framed derivative of that, different ways you can derive it, but I'll use the transport theorem, was nothing but gamma dot gt. And if I plug that one in, gamma dot gt transposed omega BN, this whole thing is nothing but the definition of that and you get the same answer. So, as expected, hopefully, we took here, b frame derivatives, in the slides I'm taking n frame derivatives, I get the same answer both ways. So, that's always a good thing to practice then you know you're doing all your math correctly. So, you do this and you apply this now for all the three axes and now we have the inertial derivative that you will need as a sub result once we start to take H dot and start grinding. So, we have that. One other thing we'll use is you will see a lot in the math, we have ig plus iws appearing. And that's the inertia of the wheel plus the inertia of the gimbal frame that holds the wheel and twists. And so, we can sum them up. If you see js, it's nothing, but basically the inertia of the combined wheel gimbal frame system. So, just a notational thing, because there's lots of stuff happening. And now the fun, you need to do this yourself, I'm showing you the highlights. [LAUGH] But this is now where you apply your excellent skills on transport theorem and differentiation, we need H dot = L, so I would say since we've broken it up as H of the hub B H of the gimbal and H of the wheel, do each separately. It just helps manage all the algebra. Otherwise you're bound to get one letter crossed with another. Do one by at the time. So if I take H dot using the transport theorem. I take the derivative as seen by the gimbal frame, because I'm using a lot of gimbal frame components. But then I need to add plus omega of the gimbal frame, relative to the inertial, right? because dot is still the inertial derivative, cross with that H expression, carry all that algebra. This is what you get, I'll let you do that. You guys are very good at this now. You do the same thing for the gimbal stuff and this is what you get, the gimbal frame stuff. And then for the body itself, well that's pretty trivial, we've done this before for a single rigid body, you regain the same derivative expression, right? Now we pack it together H dot = L is a sum of all three. We're going to use this definition, J was the inertia of the gimbal frame and the wheel, the combined VSCMG device. This is the inertia of the rigid part of the spacecraft. So this one is actually constant, as seen by the body frame. This one isn't, because inertia, everything is spinning and twisting. And I now is the combined spacecraft inertia, that's the full thing. The rigid part plus all these spinning, twisting wheel parts, all combined. That means I in this notation actually becomes time varying. If you simulate these things, every time step, you have to recompute where are these axes, where are these inertias and sum up all these inertia contributions again. So it becomes even as seen by the body, I is time varying but IS is the rigid part. That's what you took advantage of when you did the earlier differentiation of HB, that it was fixed as seen by the body frame. Yes, Louis? >> How do you guys account for the center of mass of the gimbal system, in the body or no? >> So all of these things have been taken in account for. IS here actually already accounts for the parallel axis part. That if the CG of the wheels aren't mounted in the center of the spacecraft, center of mass of the spacecraft, but they're offset somehow. That is a fixed offset, and that's already accounted for actually here when we derive this stuff. So here these are just the wheel parts that that's the momentum and inertia is spinning about their own perspective center of masses. So center of mass is all subtlety. I'm glossing over somewhat, but this is rigorous, this works, it's just you have to account for depending on where you place them, okay. There's this much mass out there, that M distance squared term is already put into this part. because that part is fixed, relative, it's a balanced wheel. If it's inbalanced, your question would be very good, because then with an imbalance, as I'm rotating, then the center of mass of the wheel is actually orbiting inside the craft and then this wouldn't work. You'd have to be much more careful about which points you do it so. Being balanced allows us to, as complicated as it is, and I'm sure you'll agree in the homework, being balanced makes this much, much easier. So that's a key assumption, so now we throw this all together. H dot = L, and I combine terms, which you will do as well, you can come up with these kind of a terms. Now this is the I inertial tensor of the hub, the rigid plus all the spinning stuff. You can write it out into this form, which is our equation of motion of the spacecraft with a single variable speed CMG. L is still the net external torque we were talking about earlier, so that could be your gravity gradient torque acting on this or deviation pressure, if we have thrusters acting for generating pure torques. But it's all about the system center of mass. And then inside you have all these terms. So the first thing I always do is let's make sure these equations make sense if we simplify them. If we bolt down like on launch, we have no gimbling, we have no wheel acceleration, everything is locked down for launch, I should get back the equations of motion of a single rigid body, which was this part and this part and the L. Everything else should vanish. So if you look at that and you see, okay, if I have gamma dot, if I'm locked down, there's no gimbling going on. So gamma dot has to be 0. This vanishes, this vanishes, let's see, gimbal acceleration vanishes. That's 0 and this is 0, so no gimbling makes a lot of them vanish. If you take the remaining equations you've got one here, IWS omega dot, and you've got this is all vanished, that vanished. You got this term that's non-zero, and this term that's non-zero. Go back and look at your dual spinner equations of motion. There we lined it up with the real axis being along B1 or something. So if you make GSB1, you get exactly the same equations again, so that's a nice validation. If the wheel is locked down, then big omega is 0. That means this term vanishes as well, this term vanishes as well, and there's no wheel acceleration. So, as expected, all these terms vanish once you lock it down, which is always good. Now, so this is good for both devices, but you can see the gimbling adds a lot of terms. There's gamma, gamma, gamma, all the stuff that happens here. What if you have a reaction wheel mode, we talked about that. We basically get a double spin equations of motion, way simpler. What if we make it a classic CMG? In that case, the wheel's speed has to be constant, not variable speed. Let's see how much it simplifies then. One term, woop-de-doo. You got all this other stuff, so the CMGs really aren't that much easier than the variable speed CMGs, you just now accounted for all the other stuff spinning and gyrating. >> Why is it that they haven't flown VSCMGs? >> They have, and in fact every CMG is in fact a variable speed CMG because it launches spun down and they have to slowly spin it up. But what they do is a simple control that they, it's complexity of the stuff. Here, let me show you what generates the control that you have. We have body rates are pretty small. I mean, we're talking small. People freak out when this craft goes more than two degrees per second, that's nothing. The wheels will be going way, way faster. So CMGs, we're looking at 10,000 rpm. So the big omega is going to be the big honking elephant. So this is small, gimbal rates are small, this stuff is all small. What's going to give us the big torque? It's going to be this term, right here, that's the wheel speed, which is huge. IWs, which is your biggest inertia of this disk because we move the mass out about the GS axis, that give us that ring, the pancake shape. And gamma dot is directly proportional to that. So, if you hold the fixed wing speed, this is the control torque that we would use in a control application, and that's what they're after. You could make everything else variable, but that's really more in a later research, we've been finding in how we can avoid singularities. But the trick is, with the reaction wheel, what generates the torque is this one. I'll show you the motor torque equation in the next class. But as you accelerate a wheel you get, with a minus sign, an opposite torque back onto the craft. But you only get one Newton meter in, you get one Newton meter out. Here though, if I gimble, with a little effort, the faster the wheel is going, the bigger the torque I get. And that's what they've been focusing on. So, devices could do this. The issue is the motor torque about the spin axis typically are not that strong. If you have to muscle through a singularity acting like a reaction wheel, you're simply going to saturate that little motor very, very quickly. So the way they're designing it, they're doing the minimal effort required, which means it tracks it within a simple bound, but it's not very tight bound, and it doesn't have much. It may take, it just needs you to spin it up and keep it roughly at the same speed, that's all they need. So it would be a hardware change. So some have flown them like this and actually demonstrated that you can do it, it's just a little bit more complexity in the control and people, air force in particular have been looking at these form of energy storage devices and to account for them. There's extra complexity, they feel like I can do it. Or some CMG manufacturer who was chatting with me during my PhD too, and they're saying well, we could also take a reaction wheel off the shelf and a CMG off the shelf, and combine both devices to get the same mathematical benefits. Off the shelf is always cheaper than trying to custom design specifically for something. So until these become common, there's different ways you can do it. But for here, this is kind of, where are we going to go? So what we can see, my final comments on this is, with these equations of motion, you see now there that as the wheel speed grows, we can get a very large torque. But also there's limits in each reaction wheel, how fast you can spin it and that's that big omega dot. The gimbling is a nice mode, but you will also have singularities. Actually, I think I could show those as well. This will be something about the gt axis, that's one axis. If I have four devices, I'm going to have four gt axes. And depending on how you've lined it up, you can have all those axes become coplanar. And if your control is outside of that, which will happen commonly with external disturbances, that's a gimbal lock situation. Because these axis about which I get these nice generous torques vary with time, it just makes a lot of things more complicated. But that was the singularity that we have so, anyway so you can see, CMGs, definitely a lot of benefits, the amplification but also mechanical challenges, more expensive. But you get a lot of bang for your buck, way bigger torques than what you get with a reaction wheel, but it comes at a cost and the controls get more challenging, too.