Which I know you've done. Now, let's go to addition, right. We talked about attitude descriptions not being a vector group, they are actually part of the SO(3) group. You can't just take one DCM, and add another DCM. Actually, let me go back and look at this. So the definition, if we have n and your b frame, and pretend to have two right hands, okay? So if they're coincidental, exactly the same, what is the corresponding DCM description, Charles? If n and b are the same coordinate frame? >> Probably be the [INAUDIBLE] >> Right, because then this, if n1 is equal to b1, this is going to be cosine of 0, which is 1. Then everything else is orthogonal to 0, you just end up with a diagonal matrix. So the zero rotation isn't just a 000, 000, 000. That wouldn't make sense, because that's not orthognal, you can't invert it, they aren't unit lengths. So the zero rotation is an identity matrix, that's what you have to remember, all right? Good, so let's go back, actually I was here, And so, additions, yep, now we want to add several different rotations. I have three frames, the N frame, the B frame, and the R frame. Both just given through the vector sets of b 123, r 123, n 123, so it's kind of shorthand. So I'm just using the two-letter notation here, but just to see. So if C maps n to b, exactly what we have in the earlier slide. And the second line shows how b maps into r through C', so 3'. In the two-letter, this would be the RB matrix that goes from B to R. This would be to BN, that goes from N to B. So if you cascade these together, you can say, well, if R hat is equal to C' times B, I can take B and substitute it back in, and then just do that math. And you get, okay, C' times C times n. And I'm looking for the mapping from N to R directly. Well, these two matrices multiplied together, I'm just calling that answer C'', is what maps N to R. If we do the two-letter notation, to me, that's actually easier. because then this was BN, this was RB. And you get RB times BN, and you end up with RN. So the DCM allows a very simple way to add rotations. If I have the attitude from the inertial to the space station, this is the DCM. And then from the space station to the astronaut, this is the DCM. And you want to add the two orientations, you just multiply them out. If you did identities, this was always the quick thing. If people would just add DCMs, so space station, inertial are the same, astronaut and inertial all the same. Therefore, BN and RB are all identity. If you've just added two identity matrices, you end up with 2's on the diagonal, which would give you huge warning flags. And say, I can't be doing the right math, right, because I'm not ending up with an orthogonal description. Yes sir, what was your question? >> That answered my question. >> That answered your question, okay. So good, so to add orientations, this is a very fundamental way. And if you know how to go from any attitude set to DCM's, even if you don't know how to do these Cayley-Klein parameters' addition properties. If you can at least map them to DCM's, you can add them here. And that's why the DCM's very important, that's one of the reasons it's very important. So here we're doing additions, let's say we want to do subtraction. So we're going to now say, look, you know BN, and you know RN. But I'm looking for the relative attitude of body, relative to reference. So what I'm looking for is BR. How do I find this now? Go ahead. >> BN times RN transpose. >> BN times, let's just make it in two steps. If I go BN, you would need NR, right? because then you'd go from R to N, N to B, which gives you R to B directly. But we don't have NR, we have the opposite, we have the inverse of it. And like with omegas, if you have the inverse of omega B relative to A, we just flip the sign. For DCMs, if you inverse of a DCM description, we just have to transpose, or flip the two letters. So, we can say this is equivalent to BN RN transpose. So you can see, once you can do addition, you can actually also do subtraction. You just have to know, I'm not going from N to B, I need the one from B to N. Well, it's just a flip, a transposed version of it, that's it. So then you can go back and forth, and do addition and subtraction. This is a very fundamental way, there's no approximation in this, there's no small angle set. This gives you, to machine precision, always the correct answer. And so for many codes, you will see these things actually being used. But there are other ones that we use for particular coordinates, too, that are quite convenient. They sometimes exist, and sometimes it's very complicated, and you go back to DCM's. Especially when you're mixing coordinates, if your sensing is done in quarternions and your math is done in Rodrigues parameters, you need the difference between two? I'm probably just mapping them to the DCM, doing it there, and then pulling out what I need. So you'll see this process repeated a few times. Good,
