0:16

So what are the some key things we did with these polar plots?

Nathaniel, can you talk me through here?

What does this stuff actually show us,

how do we get these kind of plots, what are we doing?

>> We're showing the energy as the, and

the stability of the spinning, rigid body.

>> Okay. >> So-

>> So what's this sphere?

Let's go back to basics, you're jumping into details.

>> I don't remember the name of it.

>> Okay.

Was it Andrew?

>> No.

>> Sorry, next to Nathaniel, yes, Ick?

>> Nick. >> That was off, completely.

>> The momentum sphere.

>> It's a momentum sphere, right?

So instead of using omega one's, two's, and

three's as independent rate coordinates, we use h one's, two's, and three's.

Which is nothing but omega i's times the principal inertia, that's it, right?

But it's convenient because the momentum magnitude being constant,

not the momentum vector.

We're using a sub result, just a magnitude being constant.

It gives us a nice geometric answer,

it's a sphere, what causes these intersections on this sphere?

So we're intersecting a sphere with what?

1:29

An ellipsoid, an ellipsoid represents energy.

So your rotational kinetic energy, using the same H1 2 coordinates,

we're able to rewrite and it's a classic ellipsoidal shape.

So we've got this ellipsoid and we have to intersect, so

here I'm showing you lots of different energy levels.

Nathaniel was talking about stability.

If you hear stability,

the first thing you have to think about is stability about what?

So in these plots we identified some equilibrias, spin equilibrias.

So if you take an object and you spin it about a certain axis,

it's just going to keep on spinning, and that omega stays constant.

And all the omegas stay constants, right.

What were the equilibrias we identified for a torque free rigid body?

2:36

All right, this is a common source of confusion.

That's why I'm glad we'll be going over this again especially as we get into

control stuff.

Equilibrium just means, x dot is equal to zero, and here, x is my omega right?

So all my omega rates are zero, it's not mathematical, not mathematical.

That's an equilibrium of the system all right?

Equilibrium has nothing to do with stability.

Stability is a different argument but sometimes,

people confuse equilibrium being as stable as things turning a round.

So you've got the classical pendulum this is one of the equilibrium.

3:11

And what's the other equilibrium we have over planar pendulum?

Right, upside down.

Those are two equilibrius, now its a must, that's it.

But if I look at now stability I can talk about neighboring motions,

if you just off a little bit on this one.

Gravity will basically help stabilize and

it stays closed a little bit that thing it settles.

If we turned it upside down, if we have to use a little bit of departure,

are you going to stay close and of course not.

We know it's can deviate away, right?

So, just keep those two mathematical concepts separate,

equilibrium just means your X star is equal to 0.

If you put it in this condition it would remain there perfectly.

Stability now talks about well if I'm off a little bit and that's kind of where

the real world starts because we never have it perfectly, well do I stay close?

So of these you basically answered the two spins that would be stable

if I have a slightly more than my minimum energy condition,

I would be one of these wobbles, and you can see you stay reasonably close.

Same thing here this is the maximum energy condition,

I'm spinning about the axis of least inertia.

And if I'm off a little bit you can

see wobble lines these intersection lines are staying close.

4:58

It's not just a single spin for the intermediate energy level

that's the solution, it's actually there's these different arcs.

Now if you're on the let's say, you can see the arrows, they come in.

What can you say about how you will approach this pure spin condition?

>> Tebow, like if you're exactly on it, it'll take infinite to get there.

>> Right.

You never actually reach this point in a fine right amount of time.

RIght, so you're only asymptotically ever reach it on this energy

if you slightly off this one.

We saw that in the video last time with the astronaut spinning that key.

Right, it kind of span.

It looked like it was reasonably stable, a little bit of wabble but staying close.

The next thing you know, it flipped over, wabbled here and

then it flipped back again, and it just kept doing that.

That was one of the lines like here or here.

So when it's close to the sphere spin about the intermediate axis of inertia,

it tens to hand out there a long, long time.

And then it flips over, but you never remain there.

Whereas this one, you can hang out near a pure spin about axis of maximum inertia.

And you will stay there forever, whereas this one you will keep flipping back and

forth back and forth.

6:25

Debo.

>> There's any kind of friction you're going to go down in the energy level and

get to the separate.

>> This is all assuming a perfectly mathematically rigid

body and any real body will have some panels that will flex just a little bit,

inside there might be just a little bit of give.

Every time you have a little bit of friction you're losing mechanical energy,

and you could lose some, these are all internal forces, all right.

With internal forces I can actually change my energy state,

I can lose energy, but can you bring your energy state to 0?

With enough flapping of the panels, of fuel sloshing going on,

will your space craft ever come to rest?

7:17

>> Which one was that?

That was this one.

All right, you do [INAUDIBLE].

Right, your momentum internal forces and fuel slosh,

as complicated as equations are.

Partial differential equations and surface interactions and every stoke equations and

who knows what else.

It only internal forces and we've shown up angular momentum

of a system cannot change unless you apply external forces.

So, that's why when you're loosing energy, you will asymptotically get to

this kind of the case which you can never bring the craft to rest.

All right, so those are all quick answer questions that we've had.

So good, these little pole hole plots, they are quite common,

you see these sometimes with different analysis.

We'll see after, it's going to do a spin or similar configuration.

We're not spending too much time on one but [INAUDIBLE] classic literature,

you'll often find these kind of illustrations.

And I wanted to make sure you're familiar with that.

Here the thing we did last time was Torque-Free Motion.

These are the equations where we assume to have a body of frame that's principal,

therefore the inertia tensor is diagonal, right.

We get this nice form.

Now, if we have an axis symmetric body,

you can see from here what happens on the right hand side.

You always have differences of inertia, and if it's axis symmetric that basically

means two inertias have to be equal about like this pen,

about orthogonal axis are equal.

Therefore one of the terms is going to go to zero and

immediately we know that omega dot is zero and that's going to be a constant.

Very, and the trick how to solve this last time was,

we want to go from these equations.

We know Omega 3 is constant, and

to solve this they're now to couple first order differential equation

Omega 1 that depends on Omega 2, and Omega 2 that depends on Omega 1.

Anybody remember the trick how we got to solve these two

coupled first order differential equations?

Yeah? >> One of the equations?

>> We differentiate them again, which can seem counterintuitive.

We're trying to get rid of dots,

in fact we add dots first because it makes it easier.

By differentiating them we get basically omega 1 dot,

omega 3's a constant, you get omega 2 dot.

Now we get to plug in the coupled one that depends on omega 1, and you end up with

a differential equation that's only omega 1s, but we traded two coupled.

First order differential equations for

two uncoupled second order differential equations right.

So that was a trade off but this one is a classic spring mass system so

you can solve this very easily.

Anybody who has differential equations should know how to solve this all right,

which is what we did.

So this is a classic answer if you have an axisymmetric body And

when we do a numerical simulation, just looking at the tumble rates,

you can see one of them flat lines.

That's expected, so

the spin about the axis of symmetry is always going to be constant.

And the other two give you this nice sinusoidal response, but

if you look at the attitude, it's not easy to predict.