0:25

>> You use super particle theorem on this center of mass because it's rigid, so

Â the center of mass shouldn't be changing position.

Â >> True, if you worry about the translational part.

Â At this stage of the class, really as we're talking about rigid bodies,

Â we're going to start just dropping the translational, for most of it.

Â If not I will make it very explicit.

Â So if I'm talking about the equations of motion,

Â I'm really just talking about the rotational side of it.

Â So how do we find the rotational side of the equation?

Â What are we actually looking for?

Â 0:51

What's the variable?

Â Yeah, we're actually looking for omega dot, right?

Â That's the, omega is already a rate information so

Â omega dot is the acceleration.

Â That's your angular acceleration.

Â Once we have it we can integrate it.

Â 1:07

So whereas translation have f equals ma,

Â you go straight from forces to acceleration of your position description,

Â in attitude we have this kinetics and kinematics always broken up.

Â We have somehow added to coordinate rates related to omega, and then an omega dot

Â relates to forces and torques and everything acting on it, right.

Â So you have this clear split that you have to do.

Â So here we're looking for the second part, the kinetics.

Â What's the fundamental property, Robert, that we use to drive this?

Â 2:00

>> because- >> Ansel, can you help him?

Â >> It's torque-free?

Â >> If it's torque-free, but I didn't mention torque-free motion.

Â So let's assume it has torque, what do we need then?

Â How do we get to an omega dot?

Â 2:20

Mariel, which principles do we use to get equations of motion?

Â >> Like H equals the derivative of H time,

Â or plus, omega cross.

Â >> Okay, you've taken the derivative of it already, but

Â what is the inertial derivative of the angular momentum of the system?

Â 2:46

>> The right hand side of- >> Yeah, L.

Â >> L, there we go, right.

Â So like f equals ma, this is good for

Â a continuum if we have center of mass, or, and here implied if I don't write a sub c,

Â it's always implied to be center of mass in this notation, right?

Â So we can use H=L, the key features is, this dot has to be an inertial derivative,

Â like f equals ma, those a's have to be the inertial acceleration.

Â And this L is the net sum of all the external ports acting on the system,

Â which right now we're just writing as L.

Â Later on, that could be gravity gradient torques, it could be drag torques,

Â it could be control torques with thrusters, or something like that.

Â Many ways to break it up, good.

Â So let's review quickly, this is going to be a good, what is H of a rigid body?

Â All right, once we said this thing wasn't blob and jello and flexing, it's solid,

Â what did H become?

Â Was it, Brett?

Â 3:54

>> Is it the inertia times- >> Yes, inertia times what?

Â >> The regular rate?

Â >> Yes, and this is the 3D version of that.

Â i omega, right?

Â Review this stuff, otherwise, hour and

Â 15 minutes you're not going to have time to remember such things, okay?

Â We're getting up to that exam point.

Â So if H is this, this is the inertia tensor, right, we talked about that.

Â This is written as is in a coordinate frame independent way, you just have to

Â make sure if this is in the B frame, this needs to be also in the same B frame.

Â And this gives you the answer in a rotating frame and the B frame but

Â you could pick any frame you wish.

Â Good, now we need to do H dot.

Â So Mariel actually already got us on this track.

Â We need an inertial derivative and

Â we choose to take a body frame derivative of H.

Â I won't go all the details, we did that last time.

Â This is just a review.

Â Why do we choose to do a body frame derivative?

Â Kevin?

Â 5:28

And that's not H dot, let me just, scratch that.

Â We're just doing, we're choosing to do the body frame derivative of H.

Â So Kevin, why do the body frame derivative of I omega easier?

Â >> Because then the derivative of the inertia matrix is going to be zero.

Â >> Exactly, and see by the body, every point to here, if you just lock your

Â arms out and you start rotating, as you're rotating that arm is always to your left.

Â The mass distribution, all those moments you computed,

Â and you got the inertia tenser, they're fixed, as seen by the body.

Â As seen by you guys as I tumble an object, the mass distribution is clearly changing.

Â So if you did the inertial derivative directly on this,

Â you'd have lots of extra terms.

Â You'd have to map it to the inertial frame, DCN, DCN dots,

Â DCN dots r minus omega til DCs.

Â And it all looks much more complicated, but it's the same simple answer.

Â This ends up being, as seen by the body observer, it's 0.

Â So that's why we do this, what happens to this derivative?

Â Anybody remember?

Â What happens to that, yes?

Â >> It's just omega dot, because it's the same inertial frame.

Â >> Because this is really, if I write omega, it's short for

Â omega B relative to N.

Â We're typically dealing with B and N for

Â the next several weeks so this becomes much easier.

Â The B and the N frame derivative are the same, the cross product always drops out.

Â Good, so if you do all this stuff, then H dot as Mariel was pointing out,

Â really ends up giving you I omega dot plus I have to do the transport theorem so

Â I have an omega crossed H and H is I omega, and that's equal to L.

Â There we go.

Â 7:39

That's two frames.

Â >> Yeah, this is really written in a very frame agnostic way again.

Â We haven't picked anything, we just solved it, like we did earlier with positions and

Â angular velocities.

Â They're just vectors, we can start adding,

Â subtracting them without ever specifying where does EL really point.

Â Once you want to get a numerical answer,

Â well you have to resolve the geometry and have all the right stuff.

Â And so this is kind of the same way, this is a very, like a vector.

Â We have vectors and tensors but they're written in a frame agnostic way.

Â And when you do the math, just make sure everything's in the same frame.

Â Especially when we get to the control side, we'll be doing this a lot.

Â This is how we'll be deriving our controls, but

Â when you code it, you have to go, wait a minute.

Â This will make a R, that's coming in,

Â that's in the R frame and this stuff is in the B frame.

Â Well then implied, you're going to have to do some translations.

Â Matt?

Â 8:38

And basically what we do is, we take the body frame derivative of H, but

Â I need the inertial derivative, so you need omega crossed H as well.

Â And omega cross, H is I omega,

Â and omega cross H is the same thing as omega tilde I omega.

Â That's the form we typically see it.

Â >> Cool, thank you.

Â >> And then that's L.

Â So you can see, in exam this is something that shouldn't take you more than three

Â minutes if you know what you're doing.

Â But there's lots of subtleties, if you just jump to the answer,

Â the answer's in the equation sheet you guys have.

Â Just giving me the answer will give you zero points in the exam.

Â It's all about the path and the details.

Â So make sure every time we go through this, hopefully more and

Â more will sink in.

Â Make sure you get this stuff.

Â So, good, now we have very specific frames that make your inertia tensor diagonal.

Â Trevor, what do we call such frames?

Â >> Principle frames.

Â >> Was principle coordinate frames,

Â are there an infinity of principle coordinate frames?

Â You say no.

Â And you slow down?

Â [LAUGH] >> Well, for a particular application, for

Â a particular rigid body, it'll be the same.

Â 9:54

>> Mandara, you think so?

Â Why?

Â >> because we have one axis as they symmetry axis,

Â we can have the other two in the plane orthogonal too.

Â >> Okay but if it's a cube.

Â >> Then all of them are- >> See, all of a sudden,

Â ooh, wait a minute.

Â >> [LAUGH] >> If it's a cube what happens to

Â my principle inertias?

Â 10:21

Hm, right, all these easy questions at 8:00 in the morning.

Â >> [LAUGH] >> I need more coffee.

Â Okay, so if they're all the same, why does it make you hesitate all of the sudden?

Â 10:38

>> Yeah, then you have to kind of,

Â the eigenvectors form different ways that you can map to there.

Â You have to take a closer look at it.

Â If it's a body, let's go to a simpler body.

Â This one has three distinct inertias.

Â It's definitely rigid, right?

Â 11:15

>> Right, but if you take the moment of inertia around one of those

Â principal axis, negative moment of inertia on the other one.

Â >> Right, for every body, it doesn't actually depend on the shape,

Â every body has at least some principal axis that will diagonalize it.

Â And maybe an infinity, if it's a cylinder, that's kind of what you guys are thinking

Â of, if it's a cylinder, there's definitely an infinity of ways that you have it.

Â But if you have even three discrete ones like this, or it's something very

Â amorphous it will still have three principal inertias, principal axis.

Â But when you pick those eigenvectors, you remember wait a minute,

Â I can have a positive direction but who says this isn't your principle axis?

Â Maybe you want a frame the points towards the tail because that's where your docking

Â porter is or you want to port towards, find the points towards the nose

Â because this is a flight frame and you kind of worry about where you're flying.

Â There's immediately, well there's two options and then there's another one and

Â so there's a kind of permutation you can go through,

Â there's an infinity of principle frames that a general body would have.

Â But there's definitely a finite set because you can always rearrange what's

Â your first, what's your second, what's your third.

Â And plus or minus, just always keep it right-handed, right?

Â So yeah, there are several that we can pick and

Â so if we look at that, we went through that actually.

Â If this is diagonal, these simplify down to those three scalar equations we saw,

Â and then we can do that.

Â Now, the last question we had we covered was integration.

Â If you want to integrate the complete motion we have something and

Â we give it a spin and it tumbles.

Â 12:51

How do we now numerically solve this?

Â Tosh, help us out, how do we now write an integrator for

Â a complete 3D attitude dynamics?

Â >> So you'd look at your omega dot and [INAUDIBLE].

Â >> But besides omega dot, what else do you need?

Â What describes the full attitude state of a rigid body?

Â 13:38

>> There's no other answer but MRPs, I would say.

Â >> Not MRPs, it was a good answer, definitely.

Â >> [LAUGH] >> Thanks for sucking up, I appreciate it.

Â >> [LAUGH] >> So you need an attitude measure, right?

Â 13:51

Just like if you have f equals ma,

Â spring mass to amperes system, think back to that form.

Â Okay, we've got x double dots, that's nice.

Â But I have, every integrator writes these things back into first order form.

Â So you define that x1 is equal to x, and x2 is equal to x dot.

Â And then x1 dot is equal, and you just do that classic stuff.

Â We have to do the same thing for attitude, it's just we have different equations.

Â So we still need the attitude measure,

Â that was the integrator you wrote earlier in these classes, right?

Â So you had something for MRPs or quaternion or

Â whatever you wish to, some attitude measure.

Â So that means you have a state vector that is going to be now a,

Â if it's an MRP it's six by one.

Â If it's a quaternion it's a seven by one, right,

Â this is what we have to propagate forward.

Â Otherwise this is very much what you did before, you have your current states.

Â 14:51

So if x dot is equal to an f function, depending on these, all right.

Â Then the first order one would just give me what's the derivative of these states.

Â And then xn + 1 is xn

Â + k1 times delta t.

Â Or you can do a fourth order on the cut out or

Â whatever you wish to integrate, right?

Â It's the same integration that you're doing in those problems,

Â you're just expanding now, now we have to keep track of attitude and rates.

Â Now there was some discussion that we had regarding, could we integrate

Â attitude separately from our kinetics equation, from our omega dot equation?

Â So, Nick, do you remember the arguments?

Â How does attitude actually couple into this differential equation?

Â 15:58

It's not, >> I guess I don't

Â understand the question.

Â >> How does attitude couple into this equation?

Â Does this differential equation depend on my instantaneous attitude?

Â >> No, well, I was confident.

Â >> It was very confident, I give you kudos for that.

Â It was completely wrong, but it was very confident, absolutely.

Â 16:21

Why does attitude couple into this, where, right?

Â because omega that's just another state, that's not the coupling itself.

Â It's a derivative of some stuff.

Â The interia tensor, well, that doesn't really,

Â I don't have different mass because I point differently.

Â It's a rigid body, it doesn't matter on the orientation.

Â The only thing left is L, How does attitude couple into L?

Â What are some physical examples of torques?

Â Daniel? >> Some torques are added to it

Â independently.

Â >> Yes. >> Drag, SRP.

Â >> Drag, SRP, gravity, gradience, or even our feedback control,

Â you will see very much, that's where it couples.

Â So while there's instances where you could just solve for omega first and

Â then solve for the attitude as you did kind of.

Â I just gave you the omegas essentially in earlier homeworks, right?

Â When you code it, I highly encourage just to avoid that,

Â just give the general answer, it's simple enough to do.

Â And you want to solve your attitude and your omegas all simultaneously.

Â 17:16

Later on in class we can even add other stuff like reaction wheels or,

Â that quickly moves to 6010, the follow on class.

Â Matt? >> [COUGH] Excuse me,

Â I have a question about that equation on the last page.

Â 17:27

So when you were just asking that question I was thinking that

Â you were pointing towards the I's because this is an inertial derivative and

Â the inertial frame I will change.

Â But are these is constant because we derived it in the body frame?

Â >> These are just tensors.

Â 17:45

I mean if you say constant I have to ask constant as seen by what frame?

Â The inertia is not a matrix like a DCM,

Â where it's just a three by three grouping of cosines that we picked, right?

Â We could have picked something else.

Â This is really fundamentally a tensor.

Â Which means just like a vector it,

Â when you numerically evaluate it you have to pick a coordinate frame.

Â 18:06

So I just said look this been a completely vectorial thing I can take the body frames

Â derivative of this and what happens then is,

Â well the body frame derivative of H, and H is I omega.

Â You break it up into two parts, right?

Â Here, I can immediately say this part is going to go to zero, which simplifies it.

Â And this derivative body frame is the same thing as the inertial derivative,

Â as we discussed.

Â So this H B frame derivative becomes nothing but I omega dot, that's this part.

Â 18:37

Could then, to get the inertial derivative, I have to add omega cross H,

Â which is equivalent to here.

Â So this gives me the answer but

Â I've never actually picked a particular courting frame.

Â 18:59

Often we have, that's why we've chosen that frame.

Â Here, we haven't chosen the B frame derivative because we have a particular

Â coordinate frame, we've chosen a B frame derivative because as seen by the B frame,

Â the mass distribution for a rigid body is constant and that derivative goes to zero.

Â >> I think this was my confusion, no matter was I is, it's constant.

Â >> It's constant, exactly.

Â For any shape with asymmetries and

Â whatever's going on, this is always true, right?

Â The only thing is, it's rigid.

Â That's the single rigid body and now we have this in a very general way.

Â You can see this looks suspiciously simple in undergraduate stuff we go through this

Â very, very quickly.

Â But then all this, there's all these layers of details that are on top of

Â this that take a little bit more time to sink in, so, work with this.

Â