0:16

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

Â Nathaniel, can you talk me through here?

Â What does this stuff actually show us,

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

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

Â the stability of the spinning, rigid body.

Â >> Okay. >> So-

Â >> So what's this sphere?

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

Â >> I don't remember the name of it.

Â >> Okay.

Â Was it Andrew?

Â >> No.

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

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

Â >> The momentum sphere.

Â >> It's a momentum sphere, right?

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

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

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

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

Â not the momentum vector.

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

Â It gives us a nice geometric answer,

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

Â So we're intersecting a sphere with what?

Â 1:29

An ellipsoid, an ellipsoid represents energy.

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

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

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

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

Â Nathaniel was talking about stability.

Â If you hear stability,

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

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

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

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

Â And all the omegas stay constants, right.

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

Â 2:36

All right, this is a common source of confusion.

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

Â control stuff.

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

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

Â That's an equilibrium of the system all right?

Â Equilibrium has nothing to do with stability.

Â Stability is a different argument but sometimes,

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

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

Â 3:11

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

Â Right, upside down.

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

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

Â if you just off a little bit on this one.

Â Gravity will basically help stabilize and

Â it stays closed a little bit that thing it settles.

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

Â are you going to stay close and of course not.

Â We know it's can deviate away, right?

Â So, just keep those two mathematical concepts separate,

Â equilibrium just means your X star is equal to 0.

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

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

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

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

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

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

Â Same thing here this is the maximum energy condition,

Â I'm spinning about the axis of least inertia.

Â And if I'm off a little bit you can

Â see wobble lines these intersection lines are staying close.

Â 4:58

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

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

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

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

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

Â >> Right.

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

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

Â if you slightly off this one.

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

Â Right, it kind of span.

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

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

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

Â That was one of the lines like here or here.

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

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

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

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

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

Â forth back and forth.

Â 6:25

Debo.

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

Â get to the separate.

Â >> This is all assuming a perfectly mathematically rigid

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

Â inside there might be just a little bit of give.

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

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

Â With internal forces I can actually change my energy state,

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

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

Â will your space craft ever come to rest?

Â 7:17

>> Which one was that?

Â That was this one.

Â All right, you do [INAUDIBLE].

Â Right, your momentum internal forces and fuel slosh,

Â as complicated as equations are.

Â Partial differential equations and surface interactions and every stoke equations and

Â who knows what else.

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

Â of a system cannot change unless you apply external forces.

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

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

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

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

Â you see these sometimes with different analysis.

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

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

Â you'll often find these kind of illustrations.

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

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

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

Â therefore the inertia tensor is diagonal, right.

Â We get this nice form.

Â Now, if we have an axis symmetric body,

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

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

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

Â about orthogonal axis are equal.

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

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

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

Â we want to go from these equations.

Â We know Omega 3 is constant, and

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

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

Â Anybody remember the trick how we got to solve these two

Â coupled first order differential equations?

Â Yeah? >> One of the equations?

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

Â We're trying to get rid of dots,

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

Â By differentiating them we get basically omega 1 dot,

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

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

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

Â First order differential equations for

Â two uncoupled second order differential equations right.

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

Â you can solve this very easily.

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

Â which is what we did.

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

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

Â you can see one of them flat lines.

Â That's expected, so

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

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

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

Â