0:06

So if we have a vector r which is probably derived to transport theorem,

Â r1, b1, right.

Â This all looks so simple and it gets confusing until you really own this.

Â If you write this out and say, okay that's equivalent to this

Â matrix representation r1, r2, r3 in b frame components.

Â When we derived the transport theorem we went through this and said, okay.

Â If I just take the b frame derivative of this r vector.

Â Sorry, r, then you treat b1, 2, 3 as constants, right?

Â because [INAUDIBLE] by yourself your right is always your right.

Â And all you have is r1 dot b1 plus r2

Â dot b2 plus r3 dot b3, all right.

Â If you put this in matrix form this would just be r1 dot r2 dot r3 dot

Â 1:05

So you can see from here if all you did was took this matrix.

Â And did a differentiation operator on this matrix of these three scalars.

Â Take the time derivative of does this vary, this vary, this vary.

Â Because this matrix represents a vector in the b frame.

Â And you're just taking derivatives of the scalar part, not the vectorial.

Â That means you're implicitly keeping the vectorial parts b1, 2, and 3 constant.

Â That means you're taking a b frame derivative of this, all right.

Â And that's the key that we had in that problem 3.6, from where that goes.

Â So that means this is equivalent

Â to just taking d/dt of r in the b frame.

Â If this is the matrix representation of the vector r b frame component.

Â And then the homework when you do this This is the term that you start out with.

Â For you it's a del omega in b frame components,

Â treated as a matrix, del omega is omega minus br, omega r.

Â You can add the actual letters if you want to make it explicit.

Â 2:18

And then starting today,

Â just take time derivatives of the matrix equation, all right?

Â And you don't get the same thing if you do just this as part a.

Â Because you haven't taken the same type of derivative of the vectorial quantity.

Â Part a, we did inertial derivatives.

Â Part b implied you're taking body frame derivatives.

Â So at the end you have to say, well, what I found so

Â far is just a body frame derivative.

Â To map it to inertial, you have to add the omega product term.

Â And so hopefully, that'll get you there.

Â But be consistent.

Â I don't want to see any homeworks with people doing transport theorems on DCNs.

Â And doing a b frame derivative of br.

Â What on earth does that mean?

Â It's just a time derivative of that stuff, okay?

Â So, really be very, very careful with your notation.

Â This is where you're practicing.

Â If you get this down, and you do it well,

Â the rest the class will be that much easier for you.

Â 3:11

There was one that I wanted to highlight quickly that came up,

Â when are things constant?

Â So if I'm talking about, this is a disk.

Â This is one wheel a, one wheel b and there is a person in a and so forth.

Â Those wheels are all constant radii.

Â If it's not constant typically highlight it,

Â there was one problem that has a telescoping tube.

Â That length L should be time varying,

Â the other one was the ball in the rotating tube.

Â So there the position coordinate r varies with time because

Â the ball can slide in and out of that tube.

Â There you have to treat it.

Â The other problem we had was this one where you had a disk rolling on a ring.

Â And you have to relate the angle phi to this angle theta.

Â Because I just want it in terms of one of the angles in the end.

Â And you have to realize that this disk is actually rolling on this ring

Â without slip.

Â And that no slip condition is what you use to relate this to your coordinates.

Â Okay Jordan?

Â >> I'm a little confused on how we define these vetrics things?

Â >> Okay. >> So if I have vetrics b hat equals dcm

Â of bn vectrics n hat.

Â 4:23

So are these vectrix, vectrice?

Â I don't know how you would, what's the plural?

Â >> I don't know either.

Â [LAUGH] >> [LAUGH]

Â >> Are these defined in frames?

Â because I would think if say you're- >> No.

Â >> Okay.

Â >> You see this is where we're starting to mix.

Â We're using matrix, linear algebra math, to do vector math.

Â But in the end, that's just b1, b2, b3.

Â Where does b1 really point?

Â Well, I can solve the problem without ever specifying that.

Â As you can do orbit problem, where you basically come up with,

Â this is the orbit position.

Â But where does this ir vector actually point?

Â Well, now you have to have kept track of the attitude of that frame and

Â then you can resolve it and compute it.

Â >> So in this form, I would not put numbers into the DCM?

Â I would just kind of use it as a matrix.

Â >> This is, no, no, this is correct.

Â because let's go back at look at the DCM.

Â No, that's a good question,

Â because that means you're starting to look at the details.

Â When we defined b1 we took the direction cosines.

Â And you ended up having cosine alpha 1 1

Â b 1 + cosine alpha 1 2 b 2 + cosine alpha 1 3 b 3.

Â >> [INAUDIBLE] ends right, because- >> Yeah, thanks.

Â [COUGH] I'm not trying to trick you I just messed up.

Â Okay, ends, ends, ends, perfect right?

Â So we've written that out and so these are numbers and

Â they in fact become the first row of this bn matrix.

Â This is how you're going to map one frame to another.

Â In chapter one, one frame and another,

Â typically we always define them in nice ways.

Â But they only differ by one angle.

Â That makes it easy.

Â And then if you have to project one into the other it's a bunch of sines and

Â cosines across two of the axes.

Â This gives you the complete 3D version essentially.

Â >> But what I'm confused about, isn't then b1 in the end frame?

Â because we've defined it in terms of n unit vectors?

Â >> What is a vector?

Â 6:22

>> B1 is a vector, has a unit direction and has a unit magnitude and

Â some direction.

Â In relation to the n-frame base vectors,

Â it's going to be this magnitude times the direction.

Â This magnitude and times the direction, and this magnitude times this direction.

Â I still don't know where b1 actually points,

Â unless I know how n-frame plants, all right?

Â Relative to, and I could say it's going to be this way but

Â if you don't know where n is then you have more math to do to track that.

Â But this is still a vector equation in the end that you have of this.

Â because the fundamental quantities are magnitude times direction.

Â Magnitude times direction, magnitudes times direction.

Â So if the derivatives of this, you have to specify this with respect to what, frame.

Â That's why for the vetrics if I'm taking a time derivative

Â of vetrics I have to specify it's an inertial derivative.

Â because that way I know inside, b1, 2, and 3, taking inertial derivatives of them.

Â Or am I doing body frame derivatives of them?

Â It's just a way to use linear algebra to group a bunch

Â of vector equations together.

Â You're typically very familiar with taking linear algebra and

Â grouping together a bunch of scalar equations, all right?

Â Three equations scalar three unknowns,

Â you know how to group them together matrix form, invert that matrix and you solve it.

Â Now we can do the same thing with vectors as well.

Â It's just some subtleties we have to observe, especially when differentiating.

Â 7:40

Okay, we don't use this too much.

Â But it's a nice fundamental thing when we define it.

Â because once this kind of makes sense from here derive a lot of

Â the different properties.

Â And it's a nice compact way to derive these equations without writing out

Â all the components all the time.

Â Okay, good.

Â So at this stage we always review last time.

Â Andrew, without looking at your notes, what did we do?

Â 8:19

>> It's a sequence of rotation.

Â >> Right, and that's the key.

Â How many sequences do we use for 3D attitude?

Â >> Three.

Â >> Three, you might see in robotics six Euler angle-like quantities

Â just because they have six joints on that robot manipulator.

Â Fine.

Â But in general, free tumbling,

Â this is a three degree problem; we need three angles.

Â And we do this as a sequence, right?

Â It's not that the spacecraft has to actually move like a robot and

Â doing this kind of thing.

Â 9:14

So that's one set of Euler angles.

Â How many other sets?

Â There are other sequences that we can do.

Â >> Eight?

Â >> Not quite.

Â >> 12.

Â >> 12.

Â Why 12?

Â Yes.

Â So the first axis you can do one, two or three.

Â So we are not doing rotations about some axis skewed with the original frame.

Â You only have three options.

Â Either your first, your second or your third base vector.

Â That's how these things are typically defined.

Â So you're doing a 3, a 2 and a 1.

Â The first one you got three options, but the second one,

Â you can only do one of the other two axes, not the same axis again.

Â Otherwise you end up getting silly stuff.

Â So, 12 sets, these 12 sets, how do we break them down?

Â There are groupings [INAUDIBLE].

Â There are groupings we have, these Euler angles.

Â 10:19

>> Broadly symmetric or asymmetric.

Â >> Symmetric and asymmetric.

Â Would 3-2-1 be symmetric set or an asymmetric set?

Â >> Asymmetric.

Â >> It's an asymmetric set, okay.

Â So a symmetric would be 3-1-3, good.

Â Why does it matter if it's symmetric or asymmetric?

Â 10:47

>> It impacts, and next to David.

Â Let me just go down the row.

Â Don't know your name anymore.

Â >> Russel.

Â >> Russel, thank you, yes.

Â >> The singularities are different.

Â >> Right, now what makes an Euler angle set go singular?

Â Which of the three angles is a troublemaker?

Â 11:21

And if it's asymmetric, like a 3-1-3, again it's the second one,

Â now what singularities, Maurice, do we have?

Â >> For which one?

Â >> 3-1-3, a symmetric set?

Â >> One, for the second one, right?

Â >> It's the second angle.

Â But at what angles do you go singular?

Â >> Minus 90 and 90.

Â >> That's with an asymmetric set.

Â That's what Russell just said.

Â >> 0 and 1.

Â >> All right, that's the other option, good.

Â >> [LAUGH] >> So if orbit inclinations,

Â that's the easy way I remember this.

Â Orbit inclination, you've got your ascending node.

Â Inclination argument it's a 3-1-3 sequence.

Â Zero inclination you do two rotations about the three axis and you can do it.

Â But you can imagine the inverse mapping is going to have all kinds of mathematical

Â issues because of ambiguities.

Â There are infinity of angles that sum up to be that one sum that you need.

Â So symmetric, asymmetric, right away I can say well,

Â that's where the trouble is going to be.

Â I don't even have to look at the math.

Â I know right away where this stuff goes.

Â