0:05

So here's a example we want to run through.

Â Here, I have some principle inertias that are 350, 300, and 400.

Â Now, just by the numbers, a little bit unfortunate, the intermediate one

Â if 50 beneath the max and is 50 above the min, so it's plus and minus 50.

Â This is why when we look at these wheels, the critical wheel speeds,

Â where things go stable unstable, it happens to be symmetric.

Â 0:31

But, generally these inertias aren't split evenly like that.

Â So generally those critical wheel speeds are not necessarily symmetric, all right.

Â This example will have them symmetric.

Â The wheel inertia is ten kilograms, just to make a lot of math easy.

Â Spacecraft is to spin about 60 rpm, about b1, so that's our Omega e1 that you have.

Â And so now, we're trying to figure out how fast do we have to spin this wheel

Â at least to guarantee linear stability of the system.

Â So you want both bracketed terms to be either positive or negative.

Â Now, those two bracketed terms could be positive, that's the Set 1.

Â That's where we have I1 must be greater than these other two terms here.

Â In that case, the left bracket and the right bracket are both positive, and

Â then we have a stable system.

Â Set 2, actually just simply reverses this, greater than becomes a less than.

Â This is the condition that makes the first and second bracket go negative and

Â that is also a stable configuration, all right.

Â So now, we want to look at what is the range of spin rates.

Â So the way I like to think of this is just look at the spin axis.

Â 1:43

We have zero.

Â And then, I can go positive spin rate and negative spin rate.

Â And we're going to put in inequality condition to start to go it has to be

Â at least to the right of this mark.

Â And it has to be at least to the left of this mark, so what's left?

Â The union or

Â intersections of these different areas, that's how I solve this stuff at least.

Â So if we're looking at this, in this problem

Â if the wheel speed is zero, is this system going to be stable?

Â 2:19

If the wheel speed is zero, there's no longer a dual spin,

Â there's simply a single rigid body.

Â And we're spinning about b1.

Â About which inertia axis are we spinning then?

Â >> It's not going to be stable.

Â >> All right, because now we have 350.

Â So b1 is an axis of intermediate inertia in this particular example, right.

Â So we know up front, zero cannot be included in my solution space.

Â I have to, if this is going to work,

Â I have to have something positive or negative.

Â And we'll see how that drops out.

Â So if we do this, here's the two inequality conditions.

Â The second one, for example here, this is one set.

Â You can say okay, I2 minus, you can bring this term over,

Â that gives you I2 minus I1, divide by this.

Â 3:08

Iw, I2 minus I1, gives you minus 350 divided by 10.

Â No, it gets you minus 50 divided by ten, gives you minus five.

Â So by bringing this over, omega hat has to be bigger than minus five.

Â If I'm just going to draw that here, so let's say,

Â here's the minus five part, that means, what did I say, less than?

Â All right, thinking too many steps ahead, okay.

Â No, it's greater than.

Â Okay, so that means we would have to be

Â somewhere greater than minus five.

Â But that's only one condition.

Â Now the other condition,

Â here I'm just going through the math a little bit more steps, right.

Â I'm bringing this condition and bringing this part over to the left hand side.

Â I1 back over to the right hand side.

Â I divide by IW2, here it's I3 minus I1.

Â You plug in the numbers, which you do something similar in the homework, so

Â I'm just not going to.

Â You know how to do this algebra.

Â Then omega hat as to be greater than plus five.

Â So if we look at this now and say, okay,

Â there's also a plus five, and here's the other point.

Â 4:36

[INAUDIBLE] Great.

Â So that union between both, right?

Â because we need both of them to be positive.

Â That's going to be here, all right?

Â Outs greater than.

Â Is that the only domain, though?

Â 4:54

Carlo, what do you think?

Â >> Will be less than minus five.

Â >> Less than minus five, now talk me through why that it's also domain correct.

Â >> Because the second set of equations doesn't give us that answer.

Â >> Yep, the first set always has greater than, greater than.

Â You've identified two points where each bracketed term flips signs.

Â Once you've found those, you've really found, instead of greater than,

Â it's all less than.

Â So instead of always being to the right of,

Â it's always going to be all to the left of.

Â And, again, the same unions.

Â So if you did that to make both brackets negative, If I pick a different color,

Â let me make blue for negative right.

Â Then you would have to be here or here and

Â the union of that is just going to be here.

Â 5:40

So you can see as expected the origin is not included in the solution space.

Â Because as Brian was pointing out,

Â this is a single rigid body spinning about axis remediated inertia.

Â We know it's not stable, but if the wheel spins up enough

Â 5:55

then at some point, you'll notice with this wheel spinning,

Â you're spinning oars is an oblate body, you're spinning by axis of max inertia.

Â That wheel by itself, the way it's defined.

Â It's always going to be stable, right?

Â So if that wheel spins fast enough, one way to think of this is the stability

Â of that wheel is going to overcome the instability of the spacecraft spin and

Â that's how we stabilize it.

Â But you have to get to some minimum amount where it's just equal and

Â then you have to be greater than that, hopefully, quite a bit greater that

Â it will increase your stiffness and your response time as well, right.

Â So either you can go positive or you can go negative and that will get you there.

Â Now, so good.

Â So in this case, I have, this is one of the wheel's speed, the critical speeds,

Â you could use or

Â it has to be bigger than that really, that's what you have to write.

Â Not equal, but probably bigger than that.

Â Or it could also be less than minus 300 in this case.

Â There's two sets, both brackets positive, both brackets negative.

Â 7:00

here I am saying, we are spinning about major axis.

Â So in this case, b1 is axis of maximum inertia and without a rotor spin, so

Â the origin, we would have to be stable, which is what I am showing.

Â If we are spinning up more about the spacecraft by definition,

Â already has a positive spin, about b1.

Â That is how we derived to all this stuff.

Â That's how picked b1.

Â And now, the wheel is also spinning about that.

Â If the spacecraft is max inertia stable, the wheel is max inertia stable and

Â they're both spinning about positive b1, it's always just going to be stable.

Â More and more stable.

Â If you start to spin it the opposite direction though,

Â even though the space craft by itself without that wheel will stable,

Â 7:47

it has the speed momentum that helps stabilize it.

Â What happens with the wheel, if it's in the opposite direction,

Â it's going to start to pull out the total momentum gets reduced, actually.

Â If this is spinning at 10 RPM positive and

Â the wheel is 100 RPM negative, you're pulling out momentum.

Â And in fact, the momentum perspective, at some point it acts like the separatrix,

Â it acts like the unstable intermediate axis motion and

Â you will drive a system unstable.

Â 8:14

But the good news is no matter what axis, principle axis,

Â you want to spin nominally, we can always find a wheel speed that will stabilize it.

Â [LAUGH] The bad news is no matter how stable the spacecraft was before you

Â touched it, once you touch it, you have the capability to drive it unstable.

Â 8:32

So make sure you have the right real speeds otherwise

Â your sponsor will not be happy.

Â At some point again, if you go fast enough,

Â then the wheel momentum will dominate and it's very, very stable.

Â Whatever the spacecraft's doing is almost noise.

Â So with all duo spinners to extremes infinity real speeds always stable.

Â But in between, there is a region, a finite zone, that is unstable.

Â So for an intermediate axis speed, which we looked at.

Â And in our problem, we had plus and

Â minus five, that was just because of the inertias, it doesn't have to be symmetric.

Â If you see something like this that excludes the origin,

Â right away I could say,

Â this must be spinning about an axis of intermediate inertia with a dual-spinner.

Â Just from the solution space.

Â And, if you have an axis of a least inertia,

Â you're spinning about this region that you shouldn't be spinning about is actually in

Â the positive spin direction and just all comes out of the mathematics then.

Â Jordan. >> Sorry,

Â I missed how can it be stable if you're spinning it less than negative five.

Â >> Both bracketed terms.

Â because the two brackets have to either be positive, and

Â that was the argument for the black hash lines.

Â But we found the two critical points.

Â If we also less than those critical points,

Â then both brackets both become negative.

Â >> Okay.

Â >> And that's why this is also a possible answer, out of the mathematics.

Â >> Thanks. >> Yeah, good.

Â 9:53

So anyway, but this gives you now a quick solution space.

Â You can do these easy homework with this yourself, to come up with it.

Â But just remember the extremums are always included.

Â The origin is only included for a max and min inertia case and

Â then you kind of look at the pattern.

Â If I see the pattern, I know right away what type of spin we're doing.

Â