So what we want to do next is actually look at motor torque equations. If we go and look at this differential equations, this is in a vectorial form, but really if you think of it in the scalar components, how many differential equations does this line, this boxed line give me? Three. Three. How many states does this system have? Three? Well, you have omegas. There's attitude, but for attitude, we have the differential kinematic equations, right? So attitude is always covered, so let's just ignore the attitude part. But this differential equation, you could solve for omega_dot. Great. But what is Omega_dot? Right? Per wheel, you don't know what that angular acceleration is gonna be. What is gamma_double_dot? We don't know that either. So there's two extra degrees of freedom in this system. Each mechanical device provided one additional degree of freedom because we have a single-axis rotation going from body to gimbal, gimbal to wheel. So this only gives you three differential equations. We need five, and that's because we have one device. If we have five of these devices, I need way more, but then this kind of becomes repetitive. So we have let's say we have four VSCMGs, this is still the equation you would use. You would need this with summations over all the devices and I'll show you how to do that. But then we still need what's called the motor torque equations per device. Because,ultimately, what you would simulate is, hey, I'm putting in one newton-meter on that spin axis for the reaction wheel and half a newton-meter on the gimbal axis. What's the dynamic response gonna be? All right. So we need these extra equations which we don't have yet. So that's what we're gonna look at next. How do we find these motor torque equations? And essentially, we go back to the board and say, look, we're looking at H_dot equal to L. If you think of it as a free-body diagram â€“ let me just draw that out. If we think of this as a â€“ you've got your wheel. Right. And here's the spin axis, and the wheel is free to rotate about g_s. There's your rotation. Now Omega. So this axis is a motor â€“ I know there's a motor torque that makes it accelerate about this axis. If you have zero voltage with the motor broke, it would be a free-spinning wheel as, you know, it would spin up, spin down. There's no motor torque applied. This is very similar to that [inaudible] problem you did with the cylinder and that rotisserie grill architecture. Right? And it was the b-b_prime axis. That was the axis where it was free to spin. So if we had no motor torque, this would be a free-spin axis. With the motor torque, you get to control what torque is applied. All right. So we know this is where our motor torque acts, but this is just one axis. You will have two other axes that are orthogonal to this. And as the spacecraft is tumbling and moving â€“ remember, this axis has a particular orientation in the craft â€“ the structure has to actually produce these other two torques. So, we can do H_dot equal to L, about the center of mass of this wheel now, which is what we kind of derived earlier already, and we can then use that to relate and come up with a scalar equation that we need. So that's what I'm showing you here. So if you do this, H_dot equal to L â€“ the dynamical system is just the wheel â€“ the only thing that interfaces with the wheel is the motor torque in one axis and then the structural torques along the other two. That's what makes sure this wheel doesn't just flip sideways relative to the gimbal frame, right? As you gimbal it, you're specifying two of the three dimensions of the wheel. It's the spin that's actually free. So we can do this, and you've derived this already, we had this result earlier when we derived the full equations of motion, so I get to rewrite it again. And the trick now is we have to recognize here that H_dot equal to L, that torque that's acting on just the wheel, the torque that's along the g_s axis must be what I call u_s. That's the spin-axis motor torque. The other two I'm calling tau. Those are the structural torques that keep this wheel aligned relative to that. If you needed to know what the structural torques were, you're worried about breaking those hinges and bearings and stuff, you can actually go back here and then do your simulation and then, postprocessing, plug in all these states. And all these g_t's and g_g's are gonna be the structural torques that are acting on the system. So this approach actually gives you the internal torques, if you want. Now the equations of motion, I don't need them. So what we do need to do is what is along the g_s axis, that must be equal. So here, you can see we've got only this part. This part must be equal to the motor torque, and that's where I find my motor torque equation here. It's essentially the motor torque is the wheel inertia times the angular acceleration plus these two other terms. If we ignore gimbaling, how big is your spacecraft acceleration compared to wheel acceleration? What do you think? About the same size? An order of magnitude difference? Two, three orders of magnitude difference? How fast the spacecraft rotate, you think? What's a fast speed with before they declare emergencies? Three degrees a second? Two degrees, three degrees, just a couple of degrees per second, doesn't take much. You know, how fast does the wheel go? Maximum speed, 5-6,000 RPM, way faster. So it's accelerations are gonna well exceed what the spacecraft. So some people actually just go, you know what, this is my motor torque equation, which really makes your life simpler because it decouples this equation completely from the attitude states and the gimbal states. But it's not the true thing. Momentum won't be preserved if you do that. This is the full answer that you have. And in fact, if you're gimbaling at the same time, the gimbal rate does cross-couple back into this torque. There's a gyroscopic torque that you have to account for. And so this is the full motor torque equation that you have. Right? So now we have this extra differential equation we needed. We had earlier one for omega_dot. With this one, if I throw it in, I can solve for this one. But you can see it's also coupled to this one. So you get linear systems of equations you have to solve simultaneously. Or you could take this â€“ I'll show you in a moment â€“ we can take this and back-substitute it in, and this will give us the reaction where only creations of motion that I'll show you â€“ the classic results. So, different ways you can formulate it. So that's one extra differential equation. We also need it for the gamma_double_dot. That needs the motor torque u_g. What's that torque there? There, the dynamical system I'm using, it's not just the gimbal frame, but inside the gimbal frame is a wheel. And the torques I'm worrying about is outside of the gimbal frame, so the gimbal frame to body is where the motor torque is attached between gimbal and body. So, inside of it, the dynamical system is actually both the wheel with the gyroscopics and the frame with its gyroscopics. So we computed these h_dot's earlier. If you sum them up, this is what you get. And now, this stuff about the gimbal axis actually has to be equal to the motor torque. The other two are the structural torques being produced. So it's the same kind of a process, right? Two structural. This is the motor torque here that's really spinning this combined â€“ as you rotate the frame, you're also rotating the wheel. It's a combined system. And you can come up, you can solve for this in the end and then that's what you get. So if you drop this and ignore other terms, you could roughly say that torque is equal to inertia times angular acceleration, but it's a pretty crude approximation. This gives you the full coupling. And you can see, if you have a torque, you can get your angular acceleration for the gimbal, but also you get the spacecraft angular acceleration coupling into it. So, either you back-substitute or you solve it as a linear system of equations. But now we actually have five differential equations and given torques acting on the system, we can solve for everything.