0:24

If the omega is not just prescribed or

Â measured, if you really have a free tumbling body.

Â You kick off the satellite from the rocket.

Â What happens as it gets released?

Â I need to be able to mathematically predict how omega will evolve, and

Â that means I need to find differential equations for omega.

Â And what we're going to use always, for translation,

Â we use f equals ma essentially.

Â That's what it boils down to, that's one of the forms.

Â Here, for rotation, we always go back to H dot equal to L.

Â And so, here, H is taken about the center of mass.

Â So, we know H dot equal to L holds.

Â And L is the torque.

Â About of all the extra influences about the center of mass of the system.

Â 1:05

So okay if we know it's the inertia derivative of H that is equal to

Â this equal to this torque.

Â So we have to be now a little bit careful in how we derive this.

Â I've chosen to take a body relative derivative of this plus now,

Â I have two omega b, relative to n, crossed with that h vector again, right.

Â The transport theorem isn't just for positions and velocities and so forth.

Â It's true for any vectorial quantity that you're differentiating,

Â as seen by a rotating frame.

Â Angular momentum counts as also a vector.

Â So this is true.

Â Now why did we choose to do the body frame derivative of this?

Â 2:21

Jordan. >> Finishes.

Â >> Why? >> Constant in time.

Â >> So even if this thing is tumbling and spinning and

Â going through space as seen by you inertial observers.

Â The mass distribution keeps changing.

Â This point is here, it's up, it's left, it's flat, it's all over the place right?

Â That's why doing this derivative on any inertia tensers inertia frame

Â you can do it.

Â But it becomes a time varying quantity and you have to track all this stuff.

Â And it gets harder.

Â Whereas the derivative of the inertia tensor as seen by the body frame

Â is trivial if it's rigid.

Â Because in the mass distribution, as seen by myself, if I'm rigid and

Â I'm rotating and doing stuff, that mass distribution is exactly the same.

Â And this is going to vanish.

Â If you had a deployable mass,

Â you slowly having your solar panels from your cube sat and doing something.

Â Then some servos that are moving them out.

Â Well then you would have body relative derivatives and that's the point

Â where you include extra terms, all of a sudden, into these equations.

Â That's what gives you those extra stuff.

Â So good, so that term vanished.

Â Now here, this dot, Lewis, implies what type of derivative?

Â >> It implies an inertial derivative.

Â >> Here it's a body frame derivative.

Â So how did I go from a body frame derivative to inertial derivative?

Â >> Make the vectors consistent between the two.

Â >> Right, this isn't omega of b relative to n.

Â So therefore, the b framed derivative and

Â the n framed derivative have to be the same.

Â That's the one vector where you find this kind of stuff,

Â where the derivatives between the two frames is the same if it's

Â the angular velocity between the two frames.

Â Because the omega cross product term always goes to 0.

Â Exactly.

Â So, if you want to be careful, show all these steps.

Â In an exam don't just go from here and that's clearly this.

Â I'm, you can be missing points.

Â I want to see all these sub steps.

Â Argue every case as you understand, this is why this goes to zero here.

Â And argue correctly why this other one goes to this, right?

Â So, good.

Â So, the body around this derivative is just going to be i*omega-dot.

Â Plus this cross-product term.

Â So, we're almost there.

Â 4:22

Now, you put this together and you get this differential equation.

Â I put this cross-product, H=Ic, so

Â I have omega cross is the same thing as omega two the h.

Â That's essentially how we got to this last step.

Â I took it over to the right hand side, so

Â I have i omega that is equal to minus omega til the i omega plus l c.

Â That's the classic orders rotational equations of motion.

Â This is what predicts now,

Â in your integration, how does my omega evolve in time.

Â And for a rigid body, tumbling like this, there's two aspects.

Â One is, this is a gyroscopic.

Â And you notice, it's omega and omega, so it's omega squared, is essence.

Â 5:23

Why did this become so hard so quickly sometimes?

Â Well, we'll talk about different ways to write this.

Â This form, we have a nice, generic i.

Â So, we can have a principle coordinate frame as we talked about earlier.

Â Or we can have any body fixed frame.

Â This can be a regular 3 by 3 matrix representation and

Â it will work, all right.

Â So, this is a very general format.

Â So in your integration, I would recommend this is the equation you should implement

Â because that way you're solving always the most general case and

Â the easier cases become sub classes of that.

Â If you pick, sorry Jordan?

Â >> Quick notational question, you have sub scripts on l and h but

Â you don't have them on i?

Â >> Yeah.

Â Is that purposeful?

Â Or wouldn't you [CROSSTALK].

Â >> Just lazy.

Â >> Same point?

Â >> I was just lazy.

Â >> Okay.

Â >> If you want to keep track, I do recommend.

Â If you want to keep track of them,

Â this has to be the inertia about point C in this case.

Â If you wanted to keep track of that.

Â For these kind of rigid body dynamics,

Â people very quickly drop them and it's just implied.

Â But it's a subtlety.

Â If you have multi body stuff and you take in moments and torques about you know,

Â there's a different body fix point, there's a docking port point.

Â Definitely keep track of those suckers,

Â because it's easy to get it mixed up when you're adding these things.

Â So here I just dropped it out of convenience,

Â this is, you see this a lot actually in the notations,

Â they don't declare themselves as being lazy, but [SOUND] I don't care.

Â Okay, so we have this.

Â Now, let's say we have a principal corded frame.

Â As we were hearing earlier,

Â that means all the matrix representation of this tensor will just have diagonals.

Â Lots of off diagonals go to zero.

Â And the same thing here.

Â So, omega I omega will just become i1 omega 1, i2 omega 2, i3 omega 3.

Â Really simple.

Â And then that 10 to the omega 2,

Â you carry out this math and you up with these equations.

Â So, these are equivalent to these equations,

Â this just assumes I've chosen principal axes of the body.

Â Whereas this allows for any body fixed axises.

Â So, yes integration wise this is what you use.

Â But, in analysis we often find it much easier to just look at this.

Â And in fact we will be using these equations a lot here coming up.

Â The torque free motion and discuss different things.

Â So, you can see the classic kind of results and L, L 1 2 3,

Â is basically, this is all taken in a body fixed frame that's principle.

Â So, these are the same vector components with respect to that frame And then we go.

Â 7:52

So, you can see now for example,

Â if you have a axis symmetric body, we talked earlier about different shapes.

Â because if you have a cylinder, a rocket body, a lot of things in space tend to be

Â barrels because everything fits in a rocket.

Â Well the rocket itself is kind of a barrel shape, it has an axis of symmetry.

Â In that case two principle inertias will always be equal.

Â So since in here the gyroscopics are all these differences of two principle

Â inertias you know right away [INAUDIBLE] metric bodies one of these gyroscopic

Â terms is going to vanish right and that's an insight to have.

Â I'll see this in a few cases that we'll look at.

Â Now let's talk about how to integrate these things.

Â One question I have for you is, does,

Â 8:37

we've derived differential nematic equation, just MRPs.

Â We know sigma odd is equal to one over four b times is omega vector right.

Â So omega appears defiantly going to need that.

Â Otherwise we just need sigma Here, I need omega, omega, the inertia,

Â it's a rigid body so with relative to B frame, this is a fixed inertia tensor.

Â And then, you have external torques.

Â 9:43

Andre?

Â >> I'd say simultaneously.

Â >> Because?

Â But you delivered the question.

Â >> I know. [LAUGH] >> Hopefully you're good enough at by now where you know

Â I'm asking you a trick question.

Â And B, the gut feeling probably is I know simultaneous should definitely work.

Â Separating, now you need to argue a defense why was that valid, right.

Â There's simple answers, definitely.

Â I would also do it simultaneously, but why.

Â Is it ever okay to just integrate omegas?

Â 10:14

And then do that.

Â Daniel. >> Don't have torques.

Â >> Okay. Torque free motion.

Â If I have torque free motion, I don't have my that works only here.

Â If I don't have the torque, absolutely.

Â You could just integrate omega.

Â Now in my codes, let me just say upfront, I always do it simultaneously.

Â Just, that's the most general answer, right.

Â And I don't want to write code for these cases when I have no torque or

Â here I have torque.

Â If I do them simultaneously, it doesn't matter if I have torque or not.

Â But absolutely, you could do that but let's say we do have torque.

Â 11:02

Need to have a situation where your angular rates aren't coupled to

Â your actual angle, itself, right?

Â >> Well, yes, the angular accelerations we're worried about.

Â It's omega dot.

Â Does omega dot couple with the angles?

Â This will never make it couple with angles.

Â This never makes it couple with angles.

Â The only term that could make us couple with attitudes Is this.

Â My example with thrusters.

Â If I'm firing pairs of thrusters that produce a torque and

Â spin it, does that matter now on the altitude?

Â 11:51

>> Gradient.

Â Spencer.

Â >> Solo radiation pressure.

Â >> Solo radiation pressure.

Â Right. Daniel,

Â talking about atmospheric drag, those effects.

Â Gravity gradients.

Â There's all these kind of disturbances, often environmental.

Â All of a sudden, you need the attitude.

Â And then, whoa, now it matters.

Â When we do another thing [INAUDIBLE] as humans,

Â we will be developing feedback control, and even the simplest form the torque,

Â we come up with a thruster control torque, which is minus a gain times attitude error

Â minus a gain time rate error or something very simplistic like that, right?

Â And all of a sudden, through our algorithm, we're coupling in our attitude.

Â because we want to point here, not just come to rest over here, right?

Â And the thruster firings will be a function of attitude.

Â So yes, so when you look at these things and it's a good PHD prelim oral questions.

Â 12:41

There are ways we can one way decouple them and solve omega separately and

Â doing that parallel as Andre was saying.

Â Absolutely your gut feeling is right.

Â That's probably the way it always will work and

Â in fact that's how you should program it.

Â And the way attitude couples into it, thought, is only through this part,

Â which could be environmental effects like drag, SRP, gravity gradient or

Â also we could external torques.

Â That's our control solution, all right?

Â That's what's producing these things.

Â And that's where attitude couples in as well.

Â