If we make it rigid. We've already jumped ahead a little bit which is good. Sometimes just rearranging your order helps things sink in more, right? If it's rigid, what is the key differentiation? So Sheila was on the right track earlier that it's all about these derivatives, especially this little r dot. And if it's rigid, then the derivative of this vector as seen by the body frame has to be 0. So you just end up with omega cross r, right? If it's not rigid, then this little mass particle is maybe deploying, it's moving, it's wobbling, it's doing something. And you have extra turns. That's, in fact, what you'd have to include. Now you plug this in, and we only focus about this one. You can separate this also for rigid body, like this plus this other term. I'm not going to do that here. But in the end, the angle momentum about the center of mass ended up being this minus r tilde, r tilde, dm times omega. What did we call this big bracketed term here? >> Is that the inertia tensor? >> That's the inertia tensor. And I like using the word tensor versus matrix, because there's a difference. We have a position vector but then we use a matrix to define the vector components with respect to some frame. So the matrix represents a vector but the vectors you can fundamentally just add vectors and add arrows and you don't have to assign frames. Same thing here. Now in this tilde notation, this is basically the cross product operator which will be the vector equivalent of it, right? You have to make sure that r tilde. You have to take this vector, r, with respect to some frame, the B frame, as an example. This would be in the B frame, that's what you measure. Now it's apples and apples and you can do all your matrix math. You wouldn't have this in the C frame and this in the B frame. And then you'd be mixing frames and it just wouldn't work out well, all right? So we have to do the integration by picking a body frame and then we get this. And also little r is defined relative to this body's center of mass. That's how we defined little r. What if we have a different point? Maybe this thing isn't spinning about its center of mass, but it's pivoted to an end of an arm, and it's pivoting like this. All of a sudden, I need the inertia about this endpoint, not about the center point, right? Do we have to redo this integration over and over again? No, right? So there's formulas on how to do this. This, in essence, is my inertia tensor times omega. So written like this, treat this as a completely coordinate-independent formulation. When you're actually numerically evaluated in some program like MATLAB or C or something. You have to make sure it's the same frames. But when we derive controls, we'll just write it this way and not assign frames yet. That's the bookkeeping you do at the very end. So good, that was the inertia tensor. So let's wrap up some final things. What if we have a, let's just flip the order. Let's do attitudes first, just to mix it up. What if we have a different orientation? You have the inertia tensor given to you in B frames but I want the inertia tensor in frame components. I don't have to redo all those body intervals again. Once you have them, you're good. >> Do you have to left multiply and right multiply? >> Okay, so give me the two letters of each DCM. >> [INAUDIBLE] >> So it would be BE, right, or no EB, sorry, backwards. >> Won't let me erase. Okay, EB, and the next one is? >> [INAUDIBLE] >> Or B. I like to remember it this way actually. because this way if I know BE, great, then have to transpose this one. If I know EB, fine, you have to transpose that one, right? But it's basically if this is in the B frame, then it's a two-dimensional. Think of tensors like two-dimensional vectors. So instead of using the DCM once, it's a 2D thing, you have to use it twice. And you're absolutely correct, it's a left and a right multiplication. And when you bump them up like that the letters B should match up. This is all on the B frame. And then what's on the outside everything's being mapped to the E frame and you get a inertia tensor in the E frame. So a different rule of mnemonics trick that you can help to remember this stuff. Definitely use it twice. Good, so that's how we avoid, once we have this in some body fixed frame, we can easily map it to any other body fixed frame. I will use this trick several times, especially when deriving variable speed, c and g equations of motion, and so forth. So what do I mean by a principal coordinate frame? Is it Tosh? What does that mean? >> The [INAUDIBLE]? >> Mm-hm, so let's say F is principal. So what about the diagonals? >> Of the principal. >> Let's just call this I1, I2, I3. What about the off diagonals? >> They're 0. >> So, they're 0. If this is a principal frame, you have a coordinate frame such that you get a nice diagonal form. Chuck, can we always do this? Or are there assumptions on the shape that allow us to do this? It's a rigid system. >> Yes, you can always do it. >> You can always do this. That's why for analysis very often you see papers or even our own stuff. We need to start out with assuming it's a principal frame. because instead of having nine elements to track and now with symmetry it's actually six, but still it's about I can go down to three. Three principal inertias fully define the mass distribution of a system. Without loss of generality we can always do this. In the code, though, we typically solve, actually we do control problems that include the full tensor. Because in real life they will never give you the mass balance. You may start out with a frame that's perfect, and at the very last minute astronaut A and B switch places, and next thing you know your CG's offset. Crap, my inertia tensor's now fully populated, darn astronauts! So there's always something. So in the code, we want to have formulations and controls that handle any inertia tensor. But we want to also realize for analysis and we design it typically they're diagonal or nearly diagonal. The off diagonal terms might be very small. We want to have solutions that work for everything. So good, this is symmetric matrix from the definition. Could you have inertias, I1 = 800, I2 = 20, and I3 = 50. I'm just using generic units. I'm not throwing in anything particular, they're just consistent numbers. Is that possible? So we didn't discuss, I'm just curious. Everybody here has seen inertia tensors before. Are there constraints in the principal inertia values? You find lots of publications that they publish stuff, where they throw in inertias like this. And as an editor I can go, hey, wait a minute. This is a non-physical system. There are limits to how different your inertias can be. In fact there's an inequality constraint that says, the I, principal inertia + the J of principal inertia have to be greater than or equal to the kth principal inertia. So if you sum any two of them, you have to be bigger than the third, whatever the third is. And if you go, and let's just look at that. Here we go. If you look in this, this is the I1, I2, I3. These off diagonals are all 0, right? So I1 i r2 squared and 3 squared. This one is r1 squared and 3 squared. So summed up, I get r1 squared 2 squared and 2r3 squared. Compared to the third, it's just r1 squared and 2 squared plus an r3 squared. That squared term can at best be 0, which means they're equal. But if anything else, it's going to have to have something bigger, all right? So when you set up inertias, be careful that you actually use a realistic example where the sum of 82 principal inertias has to be bigger than or equal in a limiting case to the third one. So there's different kind of properties that you can pick up here. So good, we have that. Now we know how to do this. How do we go from, what's the relationship? If you have a regular 3x3 matrix and you're trying to find the principal inertias. How do you go from the 3x3 tensor matrix representation to the three principals? What was back row scratching your head. What's your name? >> Mark. >> Mark, thank you. >> Can you repeat the question please? >> Sorry? >> Can you repeat the question please? >> How do I go from a 3x3 inertia tensor representation to a diagonal? How do I find the principal values of- >> You have to find the eigenvalues and the eigenvectors. >> Exactly, right? So you have to find eigenvalues, eigenvectors of the system. And the eigenvalues,of the 3x3, they are all of the principal inertias, that's cool. And then the associated eigenvectors are the axes of that principal inertia. And are relative to the body, it turns out. This is a principal axis, and this is a principal axis, then you'd have a third one, right? Now are these eigenvectors always going to be unique? Jordan, think of a cylinder. If you get this math, I can give you a nice 3x3 matrix of a cylinder with some weird skewed frame. You find principal inertias. If it's a cylinder, what should you expect now, all of a sudden? >> Expect one to be up the axis and two to be in that, in any plane. >> Right. >> Anywhere in that plane. >> So you will have a repeated eigenvalues set orthoganol to the symmetry axis. The inertia of a cylinder about this axis or this axis is just going to be the same. So you get a repeated eigenvalue. And the plane is unique in that case, but not necessarily the values, right? So this is how these things all relate again. And these matrices are symmetric, turns out they're also symmetric definite because the principal inertias have to be positive. You would never have a negative principal inertia. It just doesn't make physical sense. So, okay, so good properties there, good! Now, the last thing we want to review is, what if we have a different point? We have the inertia tensor about c, I want the inertia tensor about a point q. How do I go from this to this, without redoing all the math, all the integration? Kevin. >> You need add, The mass of the whole body times, Just like the vector between the two points? >> Yeah, so let me just say that c relative to q, they're close. It's usually a parallel axis there, right? If you do it for 1d rotation it was always easy, it was something the inertia about the center of mass plus the distance. So if I have the inertia about the center of mass, I want the inertia about this endpoint. I do mass times distance squared, it's squared I think, that I get there. This is essentially mass times distance squared, and this is the distance between those two points that you'd have. It doesn't have to be an inertia point, it could be another point as well, like base point. Now what are some the trickeries in here? This is typically given to you in the Q. Let's say this is your spacecraft queue, Star Trek, right? Anybody remember Q? That was one of the characters in there. So in his spacecraft, Q lives in a very Q-centric world. So this Rc relative to Q is usually given to you in a Q frame. This stuff is probably given to you in the body frame. Now before I actually can do this math, what do you have to do with these quantities? >> You have to change them. Body frame to Q frame. I'm sure, but body frame to Q frame. >> So you have to map this to the Q, is that the only way to do this math? Warda, or is there another option? You could, because this isn't the Q, everything is kind of in the Q space, you'd have to make this into Q. What else could we do? This isn't a B. Is there any way you can make this in the B frame? Tony, what would you do? >> Do a rotation matrix on the R. >> You can also just say, well, so I know this in the Q frame or I can also find c relative to q in the B frame. Which is BC times R C relative to Q in the Q frame, right? If you did this now you plug these in here, and then this. Just make sure it's consistent. I didn't say we have to have the answer in B or Q components. You can do either. Well, you can do one and then map the answer back into the frame you want. That works as well, if it's a third frame that's not either C or Q or B. So you can see lots of different ways this goes. This is the general math but if this is the coordinate independent way, a coordinate frame independent way to write it. When you evaluate it, still make sure you have all the right stuff. Otherwise, nothing works. Okay, good.