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So what we want to do next is actually look at motor torque equations.

Â If we go and look at this differential equations,

Â this is in a vectorial form,

Â but really if you think of it in the scalar components,

Â how many differential equations does this line,

Â this boxed line give me?

Â Three.

Â Three.

Â How many states does this system have?

Â Three?

Â Well, you have omegas. There's attitude,

Â but for attitude, we have the differential kinematic equations, right?

Â So attitude is always covered,

Â so let's just ignore the attitude part.

Â But this differential equation,

Â you could solve for omega_dot.

Â Great. But what is Omega_dot?

Â Right? Per wheel, you don't know what that angular acceleration is gonna be.

Â What is gamma_double_dot?

Â We don't know that either.

Â So there's two extra degrees of freedom in this system.

Â Each mechanical device provided one additional degree of freedom because we have

Â a single-axis rotation going from body to gimbal, gimbal to wheel.

Â So this only gives you three differential equations.

Â We need five, and that's because we have one device.

Â If we have five of these devices,

Â I need way more, but then this kind of becomes repetitive.

Â So we have let's say we have four VSCMGs,

Â this is still the equation you would use.

Â You would need this with summations over all the devices

Â and I'll show you how to do that.

Â But then we still need what's called the motor torque equations per device.

Â Because,ultimately, what you would simulate is, hey,

Â I'm putting in one newton-meter on that spin axis for

Â the reaction wheel and half a newton-meter on the gimbal axis.

Â What's the dynamic response gonna be? All right.

Â So we need these extra equations which we don't have yet.

Â So that's what we're gonna look at next.

Â How do we find these motor torque equations?

Â And essentially, we go back to the board and say, look,

Â we're looking at H_dot equal to L. If you think of it

Â as a free-body diagram â€“ let me just draw that out.

Â If we think of this as a â€“ you've got your wheel.

Â Right. And here's the spin axis,

Â and the wheel is free to rotate about g_s.

Â There's your rotation.

Â Now Omega.

Â So this axis is a motor â€“ I know there's

Â a motor torque that makes it accelerate about this axis.

Â If you have zero voltage with the motor broke,

Â it would be a free-spinning wheel as,

Â you know, it would spin up, spin down.

Â There's no motor torque applied.

Â This is very similar to that [inaudible] problem you did with

Â the cylinder and that rotisserie grill architecture.

Â Right? And it was the b-b_prime axis.

Â That was the axis where it was free to spin.

Â So if we had no motor torque,

Â this would be a free-spin axis.

Â With the motor torque, you get to control what torque is applied. All right.

Â So we know this is where our motor torque acts,

Â but this is just one axis.

Â You will have two other axes that are orthogonal to this.

Â And as the spacecraft is tumbling and moving â€“ remember,

Â this axis has a particular orientation in

Â the craft â€“ the structure has to actually produce these other two torques.

Â So, we can do H_dot equal to L,

Â about the center of mass of this wheel now,

Â which is what we kind of derived earlier already,

Â and we can then use that to relate and come up with a scalar equation that we need.

Â So that's what I'm showing you here.

Â So if you do this, H_dot equal to L â€“

Â the dynamical system is just the wheel â€“ the only thing that

Â interfaces with the wheel is the motor torque in

Â one axis and then the structural torques along the other two.

Â That's what makes sure this wheel doesn't just flip

Â sideways relative to the gimbal frame, right?

Â As you gimbal it, you're specifying two of the three dimensions of the wheel.

Â It's the spin that's actually free.

Â So we can do this, and you've derived this already,

Â we had this result earlier when we derived the full equations of motion,

Â so I get to rewrite it again.

Â And the trick now is we have to recognize here that H_dot equal to L,

Â that torque that's acting on just the wheel,

Â the torque that's along the g_s axis must be what I call u_s.

Â That's the spin-axis motor torque.

Â The other two I'm calling tau.

Â Those are the structural torques that keep this wheel aligned relative to that.

Â If you needed to know what the structural torques were,

Â you're worried about breaking those hinges and bearings and stuff,

Â you can actually go back here and then do your simulation and then,

Â postprocessing, plug in all these states.

Â And all these g_t's and

Â g_g's are gonna be the structural torques that are acting on the system.

Â So this approach actually gives you the internal torques, if you want.

Â Now the equations of motion, I don't need them.

Â So what we do need to do is what is along the g_s axis,

Â that must be equal.

Â So here, you can see we've got only this part.

Â This part must be equal to the motor torque,

Â and that's where I find my motor torque equation here.

Â It's essentially the motor torque is the wheel inertia times

Â the angular acceleration plus these two other terms.

Â If we ignore gimbaling,

Â how big is your spacecraft acceleration compared to

Â wheel acceleration? What do you think?

Â About the same size?

Â An order of magnitude difference?

Â Two, three orders of magnitude difference?

Â How fast the spacecraft rotate, you think?

Â What's a fast speed with before they declare emergencies?

Â Three degrees a second?

Â Two degrees, three degrees,

Â just a couple of degrees per second, doesn't take much.

Â You know, how fast does the wheel go?

Â Maximum speed, 5-6,000 RPM, way faster.

Â So it's accelerations are gonna well exceed what the spacecraft.

Â So some people actually just go,

Â you know what, this is my motor torque equation,

Â which really makes your life simpler because it decouples this equation

Â completely from the attitude states and the gimbal states. But it's not the true thing.

Â Momentum won't be preserved if you do that.

Â This is the full answer that you have.

Â And in fact, if you're gimbaling at the same time,

Â the gimbal rate does cross-couple back into this torque.

Â There's a gyroscopic torque that you have to account for.

Â And so this is the full motor torque equation that you have.

Â Right? So now we have this extra differential equation we needed.

Â We had earlier one for omega_dot.

Â With this one, if I throw it in,

Â I can solve for this one.

Â But you can see it's also coupled to this one.

Â So you get linear systems of equations you have to solve simultaneously.

Â Or you could take this â€“ I'll show you in a moment â€“

Â we can take this and back-substitute it in,

Â and this will give us

Â the reaction where only creations of motion that I'll show you â€“ the classic results.

Â So, different ways you can formulate it.

Â So that's one extra differential equation.

Â We also need it for the gamma_double_dot.

Â That needs the motor torque u_g.

Â What's that torque there?

Â There, the dynamical system I'm using,

Â it's not just the gimbal frame,

Â but inside the gimbal frame is a wheel.

Â And the torques I'm worrying about is outside of the gimbal frame,

Â so the gimbal frame to body is where the motor torque is

Â attached between gimbal and body.

Â So, inside of it, the dynamical system is actually both

Â the wheel with the gyroscopics and the frame with its gyroscopics.

Â So we computed these h_dot's earlier.

Â If you sum them up, this is what you get.

Â And now, this stuff about the gimbal axis actually has to be equal to the motor torque.

Â The other two are the structural torques being produced.

Â So it's the same kind of a process, right?

Â Two structural.

Â This is the motor torque here that's really spinning

Â this combined â€“ as you rotate the frame,

Â you're also rotating the wheel.

Â It's a combined system.

Â And you can come up, you can solve for this in the end

Â and then that's what you get.

Â So if you drop this and ignore other terms,

Â you could roughly say that torque is equal to inertia times angular acceleration,

Â but it's a pretty crude approximation.

Â This gives you the full coupling.

Â And you can see, if you have a torque,

Â you can get your angular acceleration for the gimbal,

Â but also you get the spacecraft angular acceleration coupling into it.

Â So, either you back-substitute or you solve it as a linear system of equations.

Â But now we actually have five differential equations

Â and given torques acting on the system,

Â we can solve for everything.

Â