Now, let's look at this. What we're going to start to do now is, talk about stability. And as soon as we talk about stability, we always talk about small departures all right? This is an equilibrium and this is an equilibrium. Equilibrium has nothing to do with stability. Equilibrium just means, if I put it particular side of state. Theta is equal to 180 degrees, if I am measuring this is zero, this would be 180. What happens if I have small motions than stability?. Equilibrium just means, if you this at this particular state, your state rates will go to 0. X go to 0. And as we did with the spin stability of a single rigid body, we're going to look at for what conditions now will this vanish, if I did n't have any of this. If I only have this, You can see, for what conditions will omega 1, 2, and 3 dot go to zero? One case is of course trivial, as you make all your rates 0. That not exactly a spinning condition, it's going to have a trivial solution. If you have something else though, the only way you can make with general inertias, the only way you can make all of three terms go to zero Is how? Andre? >> By axisymmetric. >> It doesn't have to be axis, we can look at axisymmetric. If it's axisymmetric about one of the axis, this could go to 0 which says, omega 1 is constant, right. And then, you have to make these two go to 0 if omega 1 is a non 0 value. The only way these can vanish, these are not the same inertias cause it's not the spherical symmetry, is omega 3 and omega 2 would have to be zero, right? That's one way, yes. But for general inertias, if the inertia differences are never zero, how'd you make all three of them go to zero without having all three omegas go to zero. Just to big omega and [INAUDIBLE]. >> Well, we don't have big omega right now. Let's ignore the real. If you just have single rigid body. Spencer. >> Would you spin it about the axis of maximum inertia and then, have 0 spinning about the other two axes? >> You could. In fact, you could spin it about any of the principal axes. because if omega 1 is nonzero, let's say omega 1 was axis of maximum inertia. Then omega 1 is nonzero, this is nonzero. But Omega 2s and 3s is 0. I got 0 times 0, something finite times 0, which is also 0 0 again something times 0, 0 right it all vanishes. This is where we identify quickly from the mathematics now the three equilibrium spins for general inertia is appears spin about to b1, b2 or b3. Which also what we saw graphically in the pole hole plots. And we discussed several different ways, right? Those are the three spin equilibrius. But in real life, we don't spin spacecraft around intermediate axis or leased axis because we have to worry about any anticipation and chaotic behavior. So, it really limits your design to just spin about axis of max inertia. Geostationary satellites tend to be tall, those Boeing 501s I believe were these big tubular shapes and they want to spin so they're always pointing at the earth. So, they want to spin about the least axis of inertia which we know is an issue with energy loss, and always be pointing at the earth. And so, this dual spinner allows us to spin actually about any axis. I could even spin with the dual spin about axis of intermediate inertia. And with the right spin conditions which we'll develop I can make it stable. So this is a way to provide passive stability without the dual spinner, you could only long term spin about axis of max inertia. With the wheel once we add this stuff. We look at the mathematics now. I can show you that we have other equilibrias. And conditions where if we pick the right wheel speeds, I can make any one of them stable without feedback. So rigid body, we have to spin about B1, 2, and 3. Those are the equilibrias. Now, let's do the same exercise with a wheel. And we're going to say this wheel with a dual spinner, your nominal case is the wheel is moving at a constant rate. That means omega dot vanishes. Now, I want to find an equilibrium, where is omega 1 dot, 2 dot, and 3 dot 0? And I'm going to have Non 0 omega 1, 2s, or 3s. I don't want everything to be stationary. That's back to the trivial answer, but I also want to have a non 0 omega. How can we make that work? >> [COUGH] >> All right, because you kind of basically go through deductions. Like Sherlock Holmes, you deduct what can't be possible. Whatever is left, however improbably, must be the answer, right? So if we start out here, this is already zero because it's a constant wheel speed. The only way to make this zero, and again, we're looking at general inertia still, in this case, is either to omega 2 or 3 has to be zero. Okay, let's pick omega 2. Right? Start the deduction. If omega 2 is zero, this is going to be 0 and this is a finite wheel speed times 0 it's all going to be 0. And you go hey, I like this. Okay? But now, omega 3. If this is going to go to 0, you could make omega 1, 0. But if we don't want all three of them to be 0. If you make omega 1 0, omega 3 has to be 0. This term goes to 0. Omega 1 being 0 times something. If omega 3 is not 0, then you have nonzero times nonzero. You don't get a 0 second derivative of the omegas. A zero derivative over the second Omega. So, that wouldn't work. So, instead what you can see is you make Omega two zero. That vanishes, cancels this, makes this all zero. The only way you can make this one zero is to take Omega three that's zero. Then even if Omega one is non zero. It doesn't matter. And the wheel can be non-zero, and it doesn't matter. They're both multiplied times something zero. So, that's going to be the equilibrium. You notice the first axis is also, so I can have an omega one that's non-zero. But that one axis is also the axis that's a spin axis of the wheel. That's where this came from. About any other axis, you can't really make this work I could have picked omega three, went through the same logic, you end up with the same answer with this stuff. So, instead of being able to have spin equlibrias about any of the principle axes of the space craft with a dual spinner, you're now too constrained to only have an equilibria about the spin axis of the wheel. If I have a tumble about omega 2 and not omega 1 and 3 I end up here with cross coupling terms that will give me non 0 derivatives right that wouldn't be an equilibrium spin it would be more of a general tumble that you would have. So, yes Ansel? >> You were talking about how this was passive but you would need some kind of. Big or bigger right. If you have to actually carry this out. >> You would spin it up. And we'll get to the spin up part in a few slides, probably on Tuesday next week. But here it's just once we've axed the spin condition. And then, you just have a little motor that keeps it spinning at a fixed rate. There's no external feedback that says look at the stars, look at the horizon, look at everything and stabilize me. It's just holding it at a fixed speed. That's it. >> Big omegadot as zero. >> Yes, so the dual spinner, the nominal case is big omega dot to zero.