So here's a example we want to run through. Here, I have some principle inertias that are 350, 300, and 400. Now, just by the numbers, a little bit unfortunate, the intermediate one if 50 beneath the max and is 50 above the min, so it's plus and minus 50. This is why when we look at these wheels, the critical wheel speeds, where things go stable unstable, it happens to be symmetric. But, generally these inertias aren't split evenly like that. So generally those critical wheel speeds are not necessarily symmetric, all right. This example will have them symmetric. The wheel inertia is ten kilograms, just to make a lot of math easy. Spacecraft is to spin about 60 rpm, about b1, so that's our Omega e1 that you have. And so now, we're trying to figure out how fast do we have to spin this wheel at least to guarantee linear stability of the system. So you want both bracketed terms to be either positive or negative. Now, those two bracketed terms could be positive, that's the Set 1. That's where we have I1 must be greater than these other two terms here. In that case, the left bracket and the right bracket are both positive, and then we have a stable system. Set 2, actually just simply reverses this, greater than becomes a less than. This is the condition that makes the first and second bracket go negative and that is also a stable configuration, all right. So now, we want to look at what is the range of spin rates. So the way I like to think of this is just look at the spin axis. We have zero. And then, I can go positive spin rate and negative spin rate. And we're going to put in inequality condition to start to go it has to be at least to the right of this mark. And it has to be at least to the left of this mark, so what's left? The union or intersections of these different areas, that's how I solve this stuff at least. So if we're looking at this, in this problem if the wheel speed is zero, is this system going to be stable? Brian. >> They'd take a look at the inequalities again. >> No, no. If the wheel speed is zero, there's no longer a dual spin, there's simply a single rigid body. And we're spinning about b1. About which inertia axis are we spinning then? >> It's not going to be stable. >> All right, because now we have 350. So b1 is an axis of intermediate inertia in this particular example, right. So we know up front, zero cannot be included in my solution space. I have to, if this is going to work, I have to have something positive or negative. And we'll see how that drops out. So if we do this, here's the two inequality conditions. The second one, for example here, this is one set. You can say okay, I2 minus, you can bring this term over, that gives you I2 minus I1, divide by this. Iw, I2 minus I1, gives you minus 350 divided by 10. No, it gets you minus 50 divided by ten, gives you minus five. So by bringing this over, omega hat has to be bigger than minus five. If I'm just going to draw that here, so let's say, here's the minus five part, that means, what did I say, less than? All right, thinking too many steps ahead, okay. No, it's greater than. Okay, so that means we would have to be somewhere greater than minus five. But that's only one condition. Now the other condition, here I'm just going through the math a little bit more steps, right. I'm bringing this condition and bringing this part over to the left hand side. I1 back over to the right hand side. I divide by IW2, here it's I3 minus I1. You plug in the numbers, which you do something similar in the homework, so I'm just not going to. You know how to do this algebra. Then omega hat as to be greater than plus five. So if we look at this now and say, okay, there's also a plus five, and here's the other point. This is the domain that makes the second bracket go to B positive. What is the actual solution space then for stability? [INAUDIBLE] Great. So that union between both, right? because we need both of them to be positive. That's going to be here, all right? Outs greater than. Is that the only domain, though? Carlo, what do you think? >> Will be less than minus five. >> Less than minus five, now talk me through why that it's also domain correct. >> Because the second set of equations doesn't give us that answer. >> Yep, the first set always has greater than, greater than. You've identified two points where each bracketed term flips signs. Once you've found those, you've really found, instead of greater than, it's all less than. So instead of always being to the right of, it's always going to be all to the left of. And, again, the same unions. So if you did that to make both brackets negative, If I pick a different color, let me make blue for negative right. Then you would have to be here or here and the union of that is just going to be here. So you can see as expected the origin is not included in the solution space. Because as Brian was pointing out, this is a single rigid body spinning about axis remediated inertia. We know it's not stable, but if the wheel spins up enough then at some point, you'll notice with this wheel spinning, you're spinning oars is an oblate body, you're spinning by axis of max inertia. That wheel by itself, the way it's defined. It's always going to be stable, right? So if that wheel spins fast enough, one way to think of this is the stability of that wheel is going to overcome the instability of the spacecraft spin and that's how we stabilize it. But you have to get to some minimum amount where it's just equal and then you have to be greater than that, hopefully, quite a bit greater that it will increase your stiffness and your response time as well, right. So either you can go positive or you can go negative and that will get you there. Now, so good. So in this case, I have, this is one of the wheel's speed, the critical speeds, you could use or it has to be bigger than that really, that's what you have to write. Not equal, but probably bigger than that. Or it could also be less than minus 300 in this case. There's two sets, both brackets positive, both brackets negative. So if I'm giving you some problem like this though, here I am saying, we are spinning about major axis. So in this case, b1 is axis of maximum inertia and without a rotor spin, so the origin, we would have to be stable, which is what I am showing. If we are spinning up more about the spacecraft by definition, already has a positive spin, about b1. That is how we derived to all this stuff. That's how picked b1. And now, the wheel is also spinning about that. If the spacecraft is max inertia stable, the wheel is max inertia stable and they're both spinning about positive b1, it's always just going to be stable. More and more stable. If you start to spin it the opposite direction though, even though the space craft by itself without that wheel will stable, it has the speed momentum that helps stabilize it. What happens with the wheel, if it's in the opposite direction, it's going to start to pull out the total momentum gets reduced, actually. If this is spinning at 10 RPM positive and the wheel is 100 RPM negative, you're pulling out momentum. And in fact, the momentum perspective, at some point it acts like the separatrix, it acts like the unstable intermediate axis motion and you will drive a system unstable. But the good news is no matter what axis, principle axis, you want to spin nominally, we can always find a wheel speed that will stabilize it. [LAUGH] The bad news is no matter how stable the spacecraft was before you touched it, once you touch it, you have the capability to drive it unstable. So make sure you have the right real speeds otherwise your sponsor will not be happy. At some point again, if you go fast enough, then the wheel momentum will dominate and it's very, very stable. Whatever the spacecraft's doing is almost noise. So with all duo spinners to extremes infinity real speeds always stable. But in between, there is a region, a finite zone, that is unstable. So for an intermediate axis speed, which we looked at. And in our problem, we had plus and minus five, that was just because of the inertias, it doesn't have to be symmetric. If you see something like this that excludes the origin, right away I could say, this must be spinning about an axis of intermediate inertia with a dual-spinner. Just from the solution space. And, if you have an axis of a least inertia, you're spinning about this region that you shouldn't be spinning about is actually in the positive spin direction and just all comes out of the mathematics then. Jordan. >> Sorry, I missed how can it be stable if you're spinning it less than negative five. >> Both bracketed terms. because the two brackets have to either be positive, and that was the argument for the black hash lines. But we found the two critical points. If we also less than those critical points, then both brackets both become negative. >> Okay. >> And that's why this is also a possible answer, out of the mathematics. >> Thanks. >> Yeah, good. So anyway, but this gives you now a quick solution space. You can do these easy homework with this yourself, to come up with it. But just remember the extremums are always included. The origin is only included for a max and min inertia case and then you kind of look at the pattern. If I see the pattern, I know right away what type of spin we're doing.