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Differential Kinematic Equation.

This is really the one that relates, again,

the omega vector to my coordinate rates.

What we measure is angular velocity in omega.

What we use in dynamics is omega.

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

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

So I have to come up with DCM rates.

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

then find the new attitude, all right.

And how do we get that?

So that's what we're looking for.

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

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

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

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

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

This derivative does not have any frame dependencies.

It's just cosine, cosine to scalars.

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

of scalar.

There's not as seen by some frame.

We talked about energy and mass in this case.

Same thing here.

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

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

Don't do it, resist temptation.

DCMs, you're taking derivative,

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

That's what we're talking about here.

So we have to relate it to omega.

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

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

This is what we have to map it together.

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

This is one of your homeworks, I think.

You have to prove this identity.

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

There's an I here.

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

essence, right?

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

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

classic transport theorem.

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

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

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

Now what is this omega tilde?

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

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

sometimes called a cross product operator.

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

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

on top of it.

That times this.

This you want to prove to yourself.

Just take x as x 1, 2, 3.

y as y 1, 2, 3.

Do your classic way of doing cross products.

Look at the answer.

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

you should get exactly the same answer.

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The tilde matrix is skew symmetric.

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

lower diagonal part.

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

not universally used.

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

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

So check this out.

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

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

Yeah?

>> Zero.

>> Zero, exactly.

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

If it's x and x it has to be zero.

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

Everything cancels out.

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

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

we know this will go to zero.

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

we get minus omega cross the vectrix.

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

on a vectrix instead of vectors as we have here.

But there's a sign flip.

And it comes out of the linear algebra.

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

you'll be able to prove this to yourself.

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

This is one of your homeworks.

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

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

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

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

If I take time derivatives of this.

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

So here I do have to specify a frame.

I'm taking an inertial derivative.

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

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

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

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

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

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

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

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

either it varies with time or it doesn't.

And here it does.

So we keep that.

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

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

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

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

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

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

this all comes together.

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

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

You can group that together, and

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

This is the classic math argument.

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

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

this has to be true for any frame.

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

And while we have derived the differential kinematic

equation that you need to integrate.

So C dot is equal to minus omega tilde C.

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

That would be BN dot ends up

being minus BN tilde times BN.

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

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

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If you go look at that If you go back,

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

And when you derive this stuff actually,

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

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

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

Otherwise things are flipped.

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

everything gets relative.

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

And make sure you get the right stuff.

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

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

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

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

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

kind of build that in.

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

could rewrite this into terms of other stuff, and

come up with an analytic answer.

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

So we'll stop here.

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

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

Things will get more.