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

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

Theta zero was an equilibrium, 180 was an equilibrium.

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

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

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

What if I'm wobbling slightly off axis?

Do things stay close or does things go crazy?

Right, that would be unstable.

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

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

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

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

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

The other two have to be zero.

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

That's an equilibrium.

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

So we're using linear stability analysis.

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

damper-like system in the end.

Right?

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

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

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

your nominal spins plus the departures.

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

and three are actually going to be zero.

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

Omega E2 is going to be zero.

E3 is going to be zero.

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

This is going to be a non-zero.

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

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

Here, I'm just plugging it in.

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

If you have omega E1, that's some number

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

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

And you drop it.

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

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

So you can run.

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

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

this is what you end up with a dual spinner.

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

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

the spin rate about that axis ended up being

constant, while the other ones you could wobble

and we had sines and cosine answers.

Here, in this mathematics, we have our departure.

This is not my nominal spin rate.

This is my delta omega one.

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

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

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

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

And it turns out, it doesn't.

So if you're spinning off a dual spinner

and you inject it into that orbit

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

well, it will just keep doing that.

So we call this kind of a marginal stability.

It doesn't grow, it doesn't shrink.

It doesn't do anything.

It's just flat lines there.

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

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

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

you probably would need some active control to nudge it

and get it within your tolerances.

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

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

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

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

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

That's one way to look at this.

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

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

These are the departure motions

and you can see that my departure spin rate

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

So delta omega two dot depends on delta omega three

and delta omega three dot depends on delta omega two.

The wheel speeds are just constants that we hold here.

So we have two coupled first order differential equations.

Matt, do you remember our trick we had earlier?

How to solve two coupled first order differential equations?

You take the derivative from it?

Exactly.

We want to get rid of the derivatives to find

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

it's often easier to first differentiate again.

And that's what we're gonna do.

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

My nominal speed is a fixed value.

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

This is a fixed wheel speed; inertias are fixed.

So everything is a constant.

So I only have delta omega three dot.

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

and get one second order differential equation.

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

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

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

Now this should look very nice because this is basically

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

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

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

the only way this is going to be stable

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

All right?

So we have to design this k.

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

that comes out of this - this dynamical system.

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

is going to be positive.

Right?

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

The first axis is just going to be marginal.

That's all you get.

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

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

It has to be positive.

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

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

They have to cancel with each other.

How do you cancel?

What would you mean by canceling sines?

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

And what the second bracketed term has to be?

Not sure.

All right.

We want the product of two terms to be positive.

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

They're both positive or both negative.

Right.

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

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

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

That's not quite true.

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

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

sorry, negative-negative, which is stability.

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

That would definitely be an unstable situation.

All right?

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

So now, if we have some inertias

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

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

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

All right?

This is what it boils down to.

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

I can factor out omega E1.

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

and that's what's omega hat.

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

So if this is point one,

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

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

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

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

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

Otherwise you have some imaginary system.

I'd be very interested in seeing that.

But then you're back to bracket times bracket.

But there's no more ratios.

It's a little bit simpler to look at.

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

Let's look at the one case.

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

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

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

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

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

about intermediate inertia were always unstable.

Right?

And we can see that if we look at this.

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

and I1 and I2.

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

B1 is a max inertia case,

then this difference has to be positive

because I1 is the biggest one, so negative

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

I2 also has to be positive, because

I1 is larger than I2; so positive-positive.

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

in a linear sense.

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

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

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

That's the axis of least inertia.

The skinny spin.

So linearly, they look stable.

We know already with energy loss -

and the pole holds actually gave us much more global

arguments, this is only for linearized local arguments -

but we regained the results that we had earlier.

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

So let's look at non.

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

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

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

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

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

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

So you end up with two inequalities.

And both of them have to be satisfied, right?

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

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

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

So it's the same boundary points.

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

If you look at your wheel speed spectrum

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

you will find two points that are critical,

where one bracketed term switches sines

and another point where another bracketed term switches sines.

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

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

What is the domain that makes it stable?

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

So I can always actually enforce this condition.