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Differential Kinematic Equation.

Â This is really the one that relates, again,

Â the omega vector to my coordinate rates.

Â What we measure is angular velocity in omega.

Â What we use in dynamics is omega.

Â But when we integrate I have to somehow integrate the attitude.

Â So here I'm using the DCM as an attitude description.

Â So I have to come up with DCM rates.

Â I have to have a three by three matrix which I can integrate and

Â then find the new attitude, all right.

Â And how do we get that?

Â So that's what we're looking for.

Â Again this DCM is just a three by three by choice.

Â Nobody said we had to write those nine numbers into a three by three form.

Â 0:47

We've chosen it because that three by three formulation's very convenient,

Â it's orthogonal, has organized properties, it makes our life easier.

Â But it's basically just nine numbers we've stacked into a three by three.

Â This derivative does not have any frame dependencies.

Â It's just cosine, cosine to scalars.

Â As we said earlier the derivatives of scalars is just a time derivative

Â of scalar.

Â There's not as seen by some frame.

Â We talked about energy and mass in this case.

Â Same thing here.

Â So in homeworks, I can promise you there will be problem 3.6, many of you will be

Â tempted, it's so, I want to put a letter here, but it's rubbish.

Â Don't do it, resist temptation.

Â DCMs, you're taking derivative,

Â it's just a straight matrix derivative of these terms.

Â That's what we're talking about here.

Â So we have to relate it to omega.

Â Omega, this is omega, this BN, this will be omega BN.

Â And this is what we had earlier, the omega one, two, three B frame components.

Â This is what we have to map it together.

Â Now, I'm going to show you this relationship here.

Â This is one of your homeworks, I think.

Â You have to prove this identity.

Â So if I'm looking at individual vectors, the transport theorem holds.

Â There's an I here.

Â This vector is the Ith component of a vectrix which is really just BI hat in

Â essence, right?

Â And the inertial derivative of BI hat is the same thing as the body frame

Â derivative of BI hat plus omega B/N cross that vector again,

Â classic transport theorem.

Â And I can do that for every one of my vectrix components.

Â And if you do that, you end up, there is a matrix equivalent thing of this that says,

Â the derivative of all three components can be written as minus omega tilde b hat.

Â Now what is this omega tilde?

Â In vector math here, if x crossed y, is a cross product, one vector crossed another.

Â There's a matrix equivalent operator which is called a tilde operator,

Â sometimes called a cross product operator.

Â Some people write it with a matrix with the word cross symbol sometimes.

Â I always have it with the matrix with a tilde, that little squiggly mark,

Â on top of it.

Â That times this.

Â This you want to prove to yourself.

Â Just take x as x 1, 2, 3.

Â y as y 1, 2, 3.

Â Do your classic way of doing cross products.

Â Look at the answer.

Â Or take the x 1, 2, 3, fill up this skew symmetric times y 1, 2, 3 and

Â you should get exactly the same answer.

Â 3:01

The tilde matrix is skew symmetric.

Â So you have three vector components and they fill the upper diagonal and

Â lower diagonal part.

Â But they have opposite signs and this sign definition is heavily used but

Â not universally used.

Â You will find some papers that flip the sign definition, just look out for that.

Â It really screws up other parts, but this is one of the most commonly used form.

Â So check this out.

Â r tilde, x tilde y, if you see that you should think it's nothing but x cross y.

Â That's why if you have x tilde x, what would that have to be?

Â Yeah?

Â >> Zero.

Â >> Zero, exactly.

Â A vector cross x tilde x, knowing that represents a vector crossed with a vector.

Â If it's x and x it has to be zero.

Â And you can plug in the numbers and you'll see a bunch of x 3 minus x.

Â Everything cancels out.

Â Good, so go through this math, define this stuff.

Â But you can see the inertial derivative of this in the body frame,

Â we know this will go to zero.

Â Here, we don't just get omega cross the vectrix,

Â we get minus omega cross the vectrix.

Â 4:02

So it looks almost like the transport theorem if we're working

Â on a vectrix instead of vectors as we have here.

Â But there's a sign flip.

Â And it comes out of the linear algebra.

Â So once you plug in the vectors B 1, 2, 3, write things in components,

Â you'll be able to prove this to yourself.

Â I'm not, this is basic linear algebra vector math.

Â This is one of your homeworks.

Â But given this definition, that's what we need to define the DCM rotation matrix.

Â And we'll get this and then that'll be a good stopping point.

Â So if we have B is equal to C times n.

Â That's the vevtrix, direction cosine times the vectrix.

Â If I take time derivatives of this.

Â On the left-hand side this is a matrix of vectors.

Â So here I do have to specify a frame.

Â I'm taking an inertial derivative.

Â That's going to be the inertial derivative of this set because these components

Â time a vectrix, a vectors, it's still a vector frame.

Â Then chain rule, the c varies with time and the n might vary with time.

Â In this case, no, I see by the n frame, n 1, 2, 3 are fixed.

Â That's going to go to 0 with the C, and

Â this is where I drop the n because it's just a derivative of components.

Â It's like in the homework when you have l times a2, all right.

Â The derivative of l is just a derivative of the scalar,

Â either it varies with time or it doesn't.

Â And here it does.

Â So we keep that.

Â This part drops out, that gives you C dot times n hat.

Â That's just from the basic definition and chain rule.

Â The next thing is too, we also, earlier, this is what you'll do in your homework,

Â the inertial derivative of the b vectrix is minus omega tilde b.

Â If this has to be the same we can plug in this definition, drop it in there.

Â 5:39

And if you do that then you have minus Omega tilde, C, N hat,

Â this all comes together.

Â This has to be equal to the same derivative here so this gets pulled down.

Â So if this times N hat is equal to this times N hat.

Â You can group that together, and

Â then this bracketed term times N hat has to go to zero.

Â This is the classic math argument.

Â This has to be true for any set of N hats.

Â You can't pick a particular frame which happens to makes this math go to zero,

Â this has to be true for any frame.

Â So the only way that happens is this bracketed has to individually go to zero.

Â And while we have derived the differential kinematic

Â equation that you need to integrate.

Â So C dot is equal to minus omega tilde C.

Â Or if you want to write this out in the two letter notation I'll do that.

Â That would be BN dot ends up

Â being minus BN tilde times BN.

Â Now this B vector, when you take vector components and construct this tilde

Â matrix, with respect to what frame do we have to take vector components?

Â 6:54

If you go look at that If you go back,

Â look at the definition we had here when we created this stuff.

Â And when you derive this stuff actually,

Â you will always pick Omega bn to be expressed in B1, B2 and B3.

Â So this differential kinematic equation, you can make a note, this,

Â if you have BN dot then minus BN tilde, BN has to be expressed in B frame components.

Â Otherwise things are flipped.

Â Again, it seems trivial now, wait until you have six frames and

Â everything gets relative.

Â This is why you really want to be on top of these details.

Â And make sure you get the right stuff.

Â >> So if you expressed your omega in the inertial frame,

Â would you right-hand multiply instead of left-hand?

Â >> There's ways to flip and come up with the equivalent definitions, yes.

Â If you use this definition that's what's required.

Â There's ways to rewrite this where you could have other components and

Â kind of build that in.

Â Yup, so just once you know the basic tools there's lots of combinations in how you

Â could rewrite this into terms of other stuff, and

Â come up with an analytic answer.

Â But good, so we're just two minutes over.

Â So we'll stop here.

Â We'll pick it up with a brief review of DCMs, and

Â then we're going to go into the next set of attitude coordinates.

Â Things will get more.

Â