So if we have a vector r which is probably derived to transport theorem, r1, b1, right. This all looks so simple and it gets confusing until you really own this. If you write this out and say, okay that's equivalent to this matrix representation r1, r2, r3 in b frame components. When we derived the transport theorem we went through this and said, okay. If I just take the b frame derivative of this r vector. Sorry, r, then you treat b1, 2, 3 as constants, right? because [INAUDIBLE] by yourself your right is always your right. And all you have is r1 dot b1 plus r2 dot b2 plus r3 dot b3, all right. If you put this in matrix form this would just be r1 dot r2 dot r3 dot So you can see from here if all you did was took this matrix. And did a differentiation operator on this matrix of these three scalars. Take the time derivative of does this vary, this vary, this vary. Because this matrix represents a vector in the b frame. And you're just taking derivatives of the scalar part, not the vectorial. That means you're implicitly keeping the vectorial parts b1, 2, and 3 constant. That means you're taking a b frame derivative of this, all right. And that's the key that we had in that problem 3.6, from where that goes. So that means this is equivalent to just taking d/dt of r in the b frame. If this is the matrix representation of the vector r b frame component. And then the homework when you do this This is the term that you start out with. For you it's a del omega in b frame components, treated as a matrix, del omega is omega minus br, omega r. You can add the actual letters if you want to make it explicit. And then starting today, just take time derivatives of the matrix equation, all right? And you don't get the same thing if you do just this as part a. Because you haven't taken the same type of derivative of the vectorial quantity. Part a, we did inertial derivatives. Part b implied you're taking body frame derivatives. So at the end you have to say, well, what I found so far is just a body frame derivative. To map it to inertial, you have to add the omega product term. And so hopefully, that'll get you there. But be consistent. I don't want to see any homeworks with people doing transport theorems on DCNs. And doing a b frame derivative of br. What on earth does that mean? It's just a time derivative of that stuff, okay? So, really be very, very careful with your notation. This is where you're practicing. If you get this down, and you do it well, the rest the class will be that much easier for you. Okay, good, any other questions on homeworks? There was one that I wanted to highlight quickly that came up, when are things constant? So if I'm talking about, this is a disk. This is one wheel a, one wheel b and there is a person in a and so forth. Those wheels are all constant radii. If it's not constant typically highlight it, there was one problem that has a telescoping tube. That length L should be time varying, the other one was the ball in the rotating tube. So there the position coordinate r varies with time because the ball can slide in and out of that tube. There you have to treat it. The other problem we had was this one where you had a disk rolling on a ring. And you have to relate the angle phi to this angle theta. Because I just want it in terms of one of the angles in the end. And you have to realize that this disk is actually rolling on this ring without slip. And that no slip condition is what you use to relate this to your coordinates. Okay Jordan? >> I'm a little confused on how we define these vetrics things? >> Okay. >> So if I have vetrics b hat equals dcm of bn vectrics n hat. So are these vectrix, vectrice? I don't know how you would, what's the plural? >> I don't know either. [LAUGH] >> [LAUGH] >> Are these defined in frames? because I would think if say you're- >> No. >> Okay. >> You see this is where we're starting to mix. We're using matrix, linear algebra math, to do vector math. But in the end, that's just b1, b2, b3. Where does b1 really point? Well, I can solve the problem without ever specifying that. As you can do orbit problem, where you basically come up with, this is the orbit position. But where does this ir vector actually point? Well, now you have to have kept track of the attitude of that frame and then you can resolve it and compute it. >> So in this form, I would not put numbers into the DCM? I would just kind of use it as a matrix. >> This is, no, no, this is correct. because let's go back at look at the DCM. No, that's a good question, because that means you're starting to look at the details. When we defined b1 we took the direction cosines. And you ended up having cosine alpha 1 1 b 1 + cosine alpha 1 2 b 2 + cosine alpha 1 3 b 3. >> [INAUDIBLE] ends right, because- >> Yeah, thanks. [COUGH] I'm not trying to trick you I just messed up. Okay, ends, ends, ends, perfect right? So we've written that out and so these are numbers and they in fact become the first row of this bn matrix. This is how you're going to map one frame to another. In chapter one, one frame and another, typically we always define them in nice ways. But they only differ by one angle. That makes it easy. And then if you have to project one into the other it's a bunch of sines and cosines across two of the axes. This gives you the complete 3D version essentially. >> But what I'm confused about, isn't then b1 in the end frame? because we've defined it in terms of n unit vectors? >> What is a vector? >> The direction and magnitude. >> B1 is a vector, has a unit direction and has a unit magnitude and some direction. In relation to the n-frame base vectors, it's going to be this magnitude times the direction. This magnitude and times the direction, and this magnitude times this direction. I still don't know where b1 actually points, unless I know how n-frame plants, all right? Relative to, and I could say it's going to be this way but if you don't know where n is then you have more math to do to track that. But this is still a vector equation in the end that you have of this. because the fundamental quantities are magnitude times direction. Magnitude times direction, magnitudes times direction. So if the derivatives of this, you have to specify this with respect to what, frame. That's why for the vetrics if I'm taking a time derivative of vetrics I have to specify it's an inertial derivative. because that way I know inside, b1, 2, and 3, taking inertial derivatives of them. Or am I doing body frame derivatives of them? It's just a way to use linear algebra to group a bunch of vector equations together. You're typically very familiar with taking linear algebra and grouping together a bunch of scalar equations, all right? Three equations scalar three unknowns, you know how to group them together matrix form, invert that matrix and you solve it. Now we can do the same thing with vectors as well. It's just some subtleties we have to observe, especially when differentiating. Okay, we don't use this too much. But it's a nice fundamental thing when we define it. because once this kind of makes sense from here derive a lot of the different properties. And it's a nice compact way to derive these equations without writing out all the components all the time. Okay, good. So at this stage we always review last time. Andrew, without looking at your notes, what did we do? >> We went over Euler angles. >> Good. Kevin. What are Euler angles? >> It's a sequence of rotation. >> Right, and that's the key. How many sequences do we use for 3D attitude? >> Three. >> Three, you might see in robotics six Euler angle-like quantities just because they have six joints on that robot manipulator. Fine. But in general, free tumbling, this is a three degree problem; we need three angles. And we do this as a sequence, right? It's not that the spacecraft has to actually move like a robot and doing this kind of thing. It's just a mathematical description on how to get to the final path. Good. So all your angles are a sequence. So let's see, so I had last row. What was your name again? >> Spencer. >> Spencer, thank you. How many sets of Euler angles are there? >> What do you mean by that? Sets of rotations? >> Yeah, so let's say we have pitch roll. What kind of a sequence of rotations was the pitch roll? >> 3-2-1. >> Right. So that's one set of Euler angles. How many other sets? There are other sequences that we can do. >> Eight? >> Not quite. >> 12. >> 12. Why 12? Yes. So the first axis you can do one, two or three. So we are not doing rotations about some axis skewed with the original frame. You only have three options. Either your first, your second or your third base vector. That's how these things are typically defined. So you're doing a 3, a 2 and a 1. The first one you got three options, but the second one, you can only do one of the other two axes, not the same axis again. Otherwise you end up getting silly stuff. So, 12 sets, these 12 sets, how do we break them down? There are groupings [INAUDIBLE]. There are groupings we have, these Euler angles. How do we break up the Euler angles? >> Broadly symmetric or asymmetric. >> Symmetric and asymmetric. Would 3-2-1 be symmetric set or an asymmetric set? >> Asymmetric. >> It's an asymmetric set, okay. So a symmetric would be 3-1-3, good. Why does it matter if it's symmetric or asymmetric? Was it David? Why does it matter if it's symmetric or asymmetric? >> I'm not sure. >> It impacts, and next to David. Let me just go down the row. Don't know your name anymore. >> Russel. >> Russel, thank you, yes. >> The singularities are different. >> Right, now what makes an Euler angle set go singular? Which of the three angles is a troublemaker? Second, all right? It's always the second angle. So what's different, Russell, between symmetric and asymmetric sets? On an asymmetric set like this, for what second angle does your description have issues? >> Plus or minus. >> Plus or minus 90. Exactly. And that's true for all the symmetrics. And if it's asymmetric, like a 3-1-3, again it's the second one, now what singularities, Maurice, do we have? >> For which one? >> 3-1-3, a symmetric set? >> One, for the second one, right? >> It's the second angle. But at what angles do you go singular? >> Minus 90 and 90. >> That's with an asymmetric set. That's what Russell just said. >> 0 and 1. >> All right, that's the other option, good. >> [LAUGH] >> So if orbit inclinations, that's the easy way I remember this. Orbit inclination, you've got your ascending node. Inclination argument it's a 3-1-3 sequence. Zero inclination you do two rotations about the three axis and you can do it. But you can imagine the inverse mapping is going to have all kinds of mathematical issues because of ambiguities. There are infinity of angles that sum up to be that one sum that you need. So symmetric, asymmetric, right away I can say well, that's where the trouble is going to be. I don't even have to look at the math. I know right away where this stuff goes.