0:04

So these are the equilibrias we're trying to identify.

Â When are - if I have equilibrium states, it's like with the pendulum, right?

Â Theta zero was an equilibrium, 180 was an equilibrium.

Â And then you can look at 180 plus deltas, right, to look at small departures.

Â With the omegas, I'm saying I have some omegas

Â equilibrium spin, and then I can look at small departures.

Â What if I'm wobbling slightly off axis?

Â Do things stay close or does things go crazy?

Â Right, that would be unstable.

Â So if you look at equilibrium conditions, we'd make the big omega dot equal to zero.

Â I've basically rewritten it, but instead of just having omega

Â one, two, threes, it's omega E1, 2 and 3s to highlight these equilibrium conditions.

Â And what came out of this was that we had to have this condition, essentially.

Â The spacecraft has to be spinning about the same axis as the wheel.

Â The other two have to be zero.

Â Then all my omega E1, 2s, and 3 dots become zero.

Â That's an equilibrium.

Â So now we're going to look at small departures.

Â So we're using linear stability analysis.

Â It is - if you haven't done too much of this, this is nothing but a basically spring mass

Â damper-like system in the end.

Â Right?

Â Once we linearize about this motion, you end up with something plus

Â cx dot plus kx, and then from then we can argue stability.

Â So the angular accelerations, if you put dots on everything, you would just have

Â your nominal spins plus the departures.

Â And we know, for our equilibriums, the nominal spins about two

Â and three are actually going to be zero.

Â This one could be none, or at the equilibrium, it has to be zero.

Â Omega E2 is going to be zero.

Â E3 is going to be zero.

Â This one is going to be a non-zero value.

Â This is going to be a non-zero.

Â So you just plug in the terms and one - there's different ways you can linearize.

Â Later on, we'll do Taylor series expansions and come up with first order terms.

Â Here, I'm just plugging it in.

Â I see delta times delta, that's second order, I drop it.

Â If you have omega E1, that's some number

Â times a delta; well, that's still first order, I keep it.

Â First order times zero; well, that's definitely zero, right?

Â And you drop it.

Â But that's basically what you do when you run through these.

Â These equilibrias are both zero and only have first order terms.

Â So you can run.

Â And in your homeworks, you've also do some linearizations.

Â If you now drop all the second order or the zero terms,

Â this is what you end up with a dual spinner.

Â So right away, this looks a little bit like the axisymmetric case we covered,

Â where we had two inertias being equal and whatever was the axis of symmetry,

Â the spin rate about that axis ended up being

Â constant, while the other ones you could wobble

Â and we had sines and cosine answers.

Â Here, in this mathematics, we have our departure.

Â This is not my nominal spin rate.

Â This is my delta omega one.

Â So if I'm supposed to be spinning at one degree per second,

Â but somebody kicked it off with one point one degree per second,

Â right, then you're delta omega one is point one degree per second.

Â This tells me how does that point one degree per second change with time.

Â And it turns out, it doesn't.

Â So if you're spinning off a dual spinner

Â and you inject it into that orbit

Â and instead of doing 360 degrees per day, if it's doing 360.1 degrees per day,

Â well, it will just keep doing that.

Â So we call this kind of a marginal stability.

Â It doesn't grow, it doesn't shrink.

Â It doesn't do anything.

Â It's just flat lines there.

Â It's not - so it doesn't recover itself if it's off.

Â It doesn't get closer again at some point like that.

Â So if you - if you weren't happy with that behavior,

Â you probably would need some active control to nudge it

Â and get it within your tolerances.

Â I have to have a drift rate that's less than one thousandth of a degree

Â per day or something, you know, to always be pointing at the Earth.

Â So - so dual spinner doesn't give you any passive stabilizing component.

Â It doesn't make it go unstable, it just -

Â whatever air you have, that's what you stick with.

Â That's one way to look at this.

Â Now the other two, again, omega E1, that's our nominal spin rate.

Â 360 per day, that's one rev per day like a geo satellite would have to do.

Â These are the departure motions

Â and you can see that my departure spin rate

Â about the second axis is a function only of delta omega three.

Â So delta omega two dot depends on delta omega three

Â and delta omega three dot depends on delta omega two.

Â The wheel speeds are just constants that we hold here.

Â So we have two coupled first order differential equations.

Â Matt, do you remember our trick we had earlier?

Â How to solve two coupled first order differential equations?

Â You take the derivative from it?

Â Exactly.

Â We want to get rid of the derivatives to find

Â - or find an answer to this, and it turns out

Â it's often easier to first differentiate again.

Â And that's what we're gonna do.

Â Because if I take the derivative of this, all of this is nothing but constant.

Â My nominal speed is a fixed value.

Â It's supposed to be doing 360 per day, right?

Â This is a fixed wheel speed; inertias are fixed.

Â So everything is a constant.

Â So I only have delta omega three dot.

Â Then, when I take this derivative, I can plug in this equation

Â and get one second order differential equation.

Â Very similar to what we did mathematically with that single spacecraft axisymmetric case.

Â So if I do this, I took its derivative, all this is constant,

Â you get this, plug in the other equation, you end up with this one.

Â Now this should look very nice because this is basically

Â - instead of delta omega two as a variable, just call it X,

Â and you end up with x double dot plus KX equal to zero.

Â And if you have any information about a spring mass system,

Â the only way this is going to be stable

Â and give you oscillatory part, is you have to have positive stiffness.

Â All right?

Â So we have to design this k.

Â This is not really a spring, this is just a mathematical coefficient

Â that comes out of this - this dynamical system.

Â So we have to pick inertias and spins and nominal things such that k

Â is going to be positive.

Â Right?

Â That's going to give us stability for the other two axes.

Â The first axis is just going to be marginal.

Â That's all you get.

Â Now, in this case, I have different ways to break it down.

Â Essentially, you have one bracketed term times a second bracketed term.

Â It has to be positive.

Â So Lucas, what can you say about the signs of the bracketed terms?

Â What must be true of them for this to be stable?

Â They have to cancel with each other.

Â How do you cancel?

Â What would you mean by canceling sines?

Â Does this first bracketed term have to be positive or negative?

Â And what the second bracketed term has to be?

Â Not sure.

Â All right.

Â We want the product of two terms to be positive.

Â So what must be true of the elements of that product?

Â They're both positive or both negative.

Â Right.

Â As long as they're the same sign, we are good.

Â Don't just jump right to well, this - if this has to

Â be positive, this has to be positive, this has to be positive.

Â That's not quite true.

Â That's a condition, but it's not the only condition.

Â Either we have positive-positive, which is stability, or positive-negative,

Â sorry, negative-negative, which is stability.

Â What you don't want to do is have positive times negative.

Â That would definitely be an unstable situation.

Â All right?

Â And that's - that's essentially, with all the math, that's what it breaks down.

Â So now, if we have some inertias

Â and we have some nominal spin rate, I want to spin about this axis at

Â two RPM, 50 RPM, whatever you're doing,

Â what allowable wheel speeds do I have to guarantee I have positive stiffness?

Â All right?

Â This is what it boils down to.

Â So we can rewrite this different ways from this parameter here, k,

Â I can factor out omega E1.

Â That means I divide this wheel speed by the nominal spacecraft spin rate

Â and that's what's omega hat.

Â You see, omega hat is just a normalized version of the wheel speed.

Â So if this is point one,

Â that means your wheel is rotating at 10 percent of the spacecraft speed.

Â If it's four, you're rotating four times faster than the spacecraft speed.

Â That's one way to look at it, it's a convenient way.

Â And then we factored out some other inertias and brought it back together.

Â This number squared is always going to be positive inertias, or hopefully positive.

Â Otherwise you have some imaginary system.

Â I'd be very interested in seeing that.

Â But then you're back to bracket times bracket.

Â But there's no more ratios.

Â It's a little bit simpler to look at.

Â And again, we need both of them to be positive or both of them to be negative.

Â Let's look at the one case.

Â We argued stability looking at pole hold plots if we don't have a wheel spinning, right?

Â So if we lock omega hat to be zero, we've got nothing but one rigid body

Â and we should be able to recover our results we saw from the pole hold plots,

Â which said without energy loss, spins about least inertia were stable,

Â spins about max inertia were stable - we've stayed close - but spins

Â about intermediate inertia were always unstable.

Â Right?

Â And we can see that if we look at this.

Â So if I take out the omega hat part, all you have is the difference between I1 and I3

Â and I1 and I2.

Â And you can readily see that if we have a spin about max inertia, if

Â B1 is a max inertia case,

Â then this difference has to be positive

Â because I1 is the biggest one, so negative

Â anything else is going to still be above zero; I1 minus

Â I2 also has to be positive, because

Â I1 is larger than I2; so positive-positive.

Â As we predicted with the pole hold plots, we're getting a stable response

Â in a linear sense.

Â And the least axis of inertia, if I1 is the smallest inertia,

Â then I1 minus I3 will be negative, I1 minus I2 will be negative

Â and we have negative-negative and we're also stable.

Â That's the axis of least inertia.

Â The skinny spin.

Â So linearly, they look stable.

Â We know already with energy loss -

Â and the pole holds actually gave us much more global

Â arguments, this is only for linearized local arguments -

Â but we regained the results that we had earlier.

Â Again, this is a classic result you should be able to do with those equations.

Â So let's look at non.

Â This is the condition we had if we do have the wheel speed.

Â So there's always two sets of conditions that comes out of this.

Â Either we pick a wheel speed such that this is bigger than zero - that means

Â I1 has to be bigger than this part, so I bring all

Â that to the right-hand side and that's what you have here -

Â or I1 has to be bigger than this part over here, which I brought to the right-hand side.

Â So you end up with two inequalities.

Â And both of them have to be satisfied, right?

Â If both of them are satisfied, you are positive-positive.

Â If Instead of greater than, if I flip the sine and say well if this one is less than,

Â that makes this negative, this is less than, that makes this negative.

Â So it's the same boundary points.

Â If you make it equal, you have two points on the wheel.

Â If you look at your wheel speed spectrum

Â - zero's in the middle, then you go left, right,

Â you will find two points that are critical,

Â where one bracketed term switches sines

Â and another point where another bracketed term switches sines.

Â Once you've identified them, you want to find out where you -

Â are you between the points, are you outside of those points?

Â What is the domain that makes it stable?

Â The beauty is we can do this regardless of the inertias.

Â So I can always actually enforce this condition.

Â