Last time we started up with that big chunk of rigid body kinematics. There are some highlights. I want to do a quick review. This is what you should be remembering. If nothing else, get everybody warmed up for the new material. Or the angles, which is the next section. But let's review these things first. The fundamentals that we had of attitude descriptions. How many coordinates do I need as a minimum to define a three dimensional orientation? What's your name? >> Ryan. >> Ryan. >> Three, minimum. >> Three, okay. If I have three. Sorry, what was your name? >> Mandar. >> Mandar, that's right. If you have three coordinates, what can you say about those attitude descriptions? >> They might go singular. >> They might? They will. [LAUGH] Might is not good enough on a prelim exam. They will, right? There's no question. If you only have three coordinates, they must go singular somewhere. Okay, what was your name? Cody, how did those singularities manifest themselves? >> I don't know. >> Mariano, how do they manifest themselves? Do you know the coordinates at that singular? >> [COUGH] Yeah, I'm sorry I'm just out of it. >> [LAUGH] Okay. Daniel? >> You get gimbal lock then one of the Euler angles isn't defined. >> Right, so there's an ambiguity in the definition and we will see that today. That's one of the ways and in particular in the differential kinematic equation, that's always the equation that relates your coordinate rates to the omegas and vice versa. That one will have zero over zero issues because all of a sudden. The easy one is think of orbits. Most of you had orbits right? So if you have your orbit plane there's an ascending node, inclination, argument periapsis. If you have no inclination you're doing ascending node and argument periapsis. There's two rotations about the same axis which you can do. It's perfectly valid. But if you have the orbit and you try to figure out what those two angles are. There is a infinity of answers. If you've shifted 60 degrees now, okay great. But that's where periapsis is but you could have 30 and 30. You could have 20 and 40. Those are the ambiguities we'll see after today as we get in into Euler angles. That's how they will have singularities. Anybody remember the other way that coordinates can go singular? So either they're ambiguous, What was the other one? They blow up to infinity, okay? So these are the things you definitely, when you show up here, I really recommend review the stuff. You have two, three minutes, go over the highlights again. because this what is hopefully going to stick, with me using it all now again. So either they go ambiguous or they go off to infinity, great. Are there ways to avoid singularities? What's your name? Brett? >> Yeah, by using more than three coordinates. >> Okay if you use four, sorry next to Brett. What was your name? >> David. >> David, if you use four coordinates are you guaranteed those coordinates are non singular? >> No, because one could be dependent of the same thing. >> Yeah, you could make some crazy set where you have two Euler Angles for some reason. And that's still not going to help you, right? So going to four enables you to have a non-singular description. And there's quaternions. And in the book also we talk about Caley-Klein parameters. Which are four dimensional complex numbers. But you have to go at least to four. Now what's the one attitude description we've already covered? Andrew? >> Direction cosine matrices? >> Is that one singular or non singular? >> Non singular? >> It's actually non singular. We didn't discuss it last time but if you look at it the whole way we derived it. There was no attitude where you couldn't get these cosines right? So now we have more than four, as David was saying, right? So no, did I get your name right? David, right. Okay, I did have David right. So we're going to more than four. How many coordinates, Jordan, do we have with the DCM? So how many constraints must you have on the system? >> Three? >> Talk me through your process. Why do you think it's three? >> Number of constraints. Although [INAUDIBLE] there would be nine because there are nine angles that relate each axis to each other. >> So you have nine constraints? So you have nine coordinates and nine constraints. That means there's zero degrees of freedom left. How many degrees of freedom do you have in general? What's the minimum number of coordinates we need to do general three dimensional attitudes? >> Three >> Three, so we do need three. And you have nine coordinates. So how many constraints must you have? >> Three. >> Dude, it's six. >> [LAUGH] Six, okay. >> Why? >> I can't think of it right now. I dont know. >> So let's think of this. Let's talk about constraints quickly, maybe that's the confusion. Let's say this is your particle, it has to move around. A point mass, Jordan, Tebow, a point mass, Tebow, has how many degrees of freedom? >> Three. >> Three, right. So you can move it anywhere in space. If you are saying this point mass has to move on this plane. Now, Tebow, how many? Right, why? >> Because it only has two dimensional space to live in. >> Right, this is a two dimensional sub-space. It has to be here. So the height, if you've got a coordinate frame on this plane, right? The height has to be 0. Or if you say this particle has to move 1 meter off the floor, you've constrained one of the coordinates and you're left with a 2 degree of freedom system. So the number of coordinates minus the number of constraints leaves you with the number of degrees of freedom. That's a fundamental thing. If you have not seen that, look it up, read it up quick. Or find a Wiki page on this or something. But this is something that I'll be talking with these terms and expecting you guys to know what I'm talking about. Hopefully, if not, raise questions again. We definitely want to get into this. So we do have six constraints. With the DCM, where do these constraints come from? I can't always find everybody's face quick enough. Was it Warda? Where do these constraints come from on the DCM? And you don't need to read, I know you can read. Give me your thoughts. That tells me roughly where you stand right now. If I draw a DCM, let me just give you this. Could you have an element that is three? >> No. I mean, yeah. But some of your, the column should be, should give us a unit vector. >> Exactly. Cause they represent actually the base vectors of the B frame. If this is the representation of BN, right? Then the rows would B1, B2, B3 and N frame components. And the columns are N1, N2, N3 in B frame components. So they all have to have unit length. You can't have numbers. They come from the definition of cosines. So you know cosine is between plus or minus 1. So it can't be bigger than norms of 1, so that can't happen, so that's no good. So okay. So you actually gave. What I gave is a bunch of stuff. because width then each being normal, and this being normalized. These are not unrelated. You could go, well, that's three. That's three. But then somebody else says, wait a minute. Base vectors, right? The first row crossed with the second row has to give you plus the third row. because they're right-handed coordinate systems. The first column crossed the second has to give you plus the third. But also the second row, the third row crossed have to give you plus the first one. And this third crossed second has to give you minus one. Well, all of a sudden I'm coming up with ten constraints. Does that mean I have a minus one degree of freedom problem? What does it mean? >> I don't know what minus one degree of freedom means. [LAUGH] >> Doesn't make sense, right? So if you end up with that like wait a minute I've way more constraints. These constraints are actually are all related in the end. You know you have nine coordinates. It's a three degree for your problem. You end up having the equivalent of six constraints. But there's many ways that they manifest themselves. Also we said the inverse of the matrix has to be the transpose. Well that imposes all kinds of constraints which embeds all thought than normal rows and columns and all the other stuff that we have, okay? So in the end it six constraints there is many ways to write them but you have to be very careful when you write the constraints the you are not just duplicating the same constraints. They have to be independent constraints to really reduce your degrees of freedom. That's what we have with the DCM so keep on it's singularity free but you do have a bunch of constraints which impacts our DR integration that we do. Let's see. What else did I have? What is the determinate of any orthogonal matrix? Back row, green shirt, what's your name? >> Ben. >> Ben, thanks. What is the determinate of any orthogonal matrix? >> Zero? >> No. Good guess, zero is always the good answer, right? >> Uh-huh >> Not quite. What was your name? >> Armzel. >> Armzel. >> Plus minus one. >> Plus minus one, precisely. So we went through that algebra. There's something any orthogonal matrix is true. Now, how do we call our rotation matrix? It's a particular type of orthogonal matrix. >> Right handed? >> We want a right-handed one, yes, but what's the mathematical term in a paper? It's something orthogonal. >> Orthonormal? >> Orthonormal just means it really has units, columns and rows. What's the word I'm looking for? >> Proper? >> Proper, right? If you hear proper orthogonal matrix, again, please review these things before class because they will go much quicker. And hopefully it will stick, and therefore you'll do better on your exam. So proper orthogonal, now what does that mean? If it's a proper orthogonal, it's to determine a plus one or minus one? >> Plus one. >> Plus one, and, yeah, right? That's actually the check for that's an easy check. You just take your deter, your DCM. And if it's orthogonal, the inverse has to be the transpose, great. But that doesn't guarantee its a, it's a rotation made. It has to also be proper orthogonal. If you take the determinant Mathlab. If hopefully plus one, then you're good. Okay, additions, substructions, how do we add DCMs? So what was your name, you light green shirt? >> Khalian. >> Khalian, okay, how do you add a DCMs if you have orientations? >> Because BN is like [BR] and [RN] we just add it to give [BN]. >> How do you add? You do an add like this? [BR] plus [BN]. >> No we multiply. >> [LAUGH] What's you answer? Or do you want a lifeline. >> No. We multiply them. >> Multiply, yeah. Don't do addition. That gives you nothing. That gives you nothing sensible. It'll give you something, but it's not sensible. So this would give you the N. Okay. So here you're given this, you're given this and I'm adding two orientations. So good. Let's do the quick flip side, if we. Next o Nathaniel in the back, what's your name? >> Nate. >> Nate. Let's say you are given [BN] and you are given [RN] but I'm trying to find the difference [BR]. How would you do that? >> Begin,take the transpose of [RN] multiply by [BN] is equivalent to the inverse. >> All right, you can post multiply both side by all transpose, that cancels here, moves it to the left hand side, and here we go. So the one, if you know the addition, you should know subtraction, its' really the same, it should, with the two rotations. If you're given BN, great. If you're given NB, well, you can relate them, right. You just transpose them, that's it. So good. So we did additions and now we also do the ordinary differential equations, the differential kinematic equations. So we had C that we wrote and we derived this and it was omega till the C and this is equivalent in 3.6 actually, you need these two letter version to do this because you need BR dot, right? Because you want to be careful what that means but BN dot here is omega BN in a 2D form and that omega BN is written in B frame components times the DCM again. So this and this are equivalent. That's a differential kinematic equation. So let's talk about this okay, if you have sounding rocket. Originally standing straight up let's say that's a reference frame so our attitude is zero motion relative to the reference for it. So, Charles, what is the DCM if B and N are identical? >> Identity. >> Identity, right? So just the identity operator. Only the diagonal cosine's a one everything else is a zero. Makes it really easy. Good. So we know the initial latitude, now you fire off the sounding rocket and you have a rate gyro on board, as so many people here do this, now you record it. Puts it on a hard drive. You pick up the sounding rockets wherever it lands and put it back in the data. You know the initial latitude. Now you know the omegas. You can integrate this and figure out precisely what was the attitude of the sounding rocket at any point of its flight. You've got exactly what you need. But again, we've measured omega BN. Once we get to the kinetic side, the second chunk, we can figure out differential equations to predict with forces and torques what happens to omega BN. But what's going to be a challenge if you numerically integrate this equation? Are there some concerns you might have? Sorry, what's your name, purple shirt near the end? >> Spencer. >> Sorry? >> Spencer. >> Spencer, thanks. Any issues you can foresee if you integrated this differential equations? These sets of differential equations? >> Deciding what frame you would differentiate them in? >> No, this is a matrix now. This is a matrix equation. You're going to get the DCM, which represents B relative to N, just in a nice 3 by 3 form for convenience. That's not an issue. We know the DCM doesn't have singularity so we should never have 0 over 0 in fact these will always be well behaved. But what's going to be the problem? Any integration method you use, is it going to be perfect, Andrew? Little errors okay. We have nine coordinates. If these errors creep in, are those errors going to satisfy all those six constraints we have? To keep us, I cannot think of people's names. Warta, right? You were talking about unit columns and unit rows, right? That was important. If you now have slide integration errors, are your rows and columns can have unit length? Probably not. So as soon as you have this singularity free description label, no problems. Now life isn't that easy. You should know better by now. You're all old enough, right. If you get one benefit, there's some old secrets you have to deal with. So, if it's a redundant system, with more than three coordinates, these coordinates must satisfy constraints. And so, a regular integrator won't know about those constraints, and you will have little errors creep in. And that means we will have to and there's processes for doing that. This matrix, because otherwise, long term you can have integration errors build up where one element might be 1.001, and that's going to cause all kinds of craziness with your transformations and things that you do, so be aware of that. There are other types of integrators. Anybody here heard of symplectic integrators? Symplectic integrators have a particular mathematical form that allows you to input constraints and say, look, I have to integrate this orbit, but my energy has to be precisely this amount. You might still have integration errors but your energy will not be violated. And you can do additive integration like that too, where you're saying, look, I have to integrate these DCMs. It has six constraints. You have to put them in some mathematical form. But you can make sure you keep it there, but it gets a lot more complicated to do that. So there's a whole world of how to integrate redundant constraint coordinate systems. Just something I wanted to highlight. But the benefit is no singularity, easy, it's a simple linear algebra math to do that, okay? Good, that was kind of wrapping up the review. Now, this is where we finished up last time with this equation. And we've discussed it, you can see how we derived it. There's one other thing I want to go through. This differential kinematic equation, C dot equal to minus omega tilde C. We derived it for the rotation matrix. It turns out this is actually true for any orthogonal matrix. And you see orthogonal matrices appearing in lots of dynamical systems, especially if you take structures, the system masked matrices that you can decompose, in the eigen norm, eigenmode decompositions. They appear lots of places. So what we can do actually, we can prove that this form that we have, that C dot is equal to minus omega tilde C. If C is orthogonal, then its time evolution of that matrix must satisfy this form. Now what does omega mean, omega tilde, if you have a 16 million degree of freedom structures problem? Good luck, I don't know. But it doesn't have the nice rigid body interpretation. But it will have the skew symmetric relationship that you can figure out. And we'll see this a little bit later when we go to higher, more than three, the people are looking at not just three dimensional latitude, but four and five dimensional latitudes as well. We'll see a little bit of hints of that. So look at this. If it's orthogonal, we said C times C transpose has to be identity, right? because C transpose is the inverse. So the matrix times its own inverse is going to be identity. Its time derivative has to be 0. That's why we get there. Or if you take the chain rule, C dot C transpose and C, C dot transpose, you just expanded it. Now you can start to plug in, we postulate and make a hypothesis and say okay, does this form hold for general n by n matrices? So we just plug that in and so here you have a C dot C transpose which gives you minus omega tilde C C transpose. This other part here, this is a little linear algebra. We have C dot transpose. Here I'm giving you C dot, not C dot transpose. If you transpose to C dot, what happens to this matrix multiplication? It flips. If you've forgotten that, remember that. It's an easy thing if you haven't seen that before, check it out for yourself. So, C dot transpose is going to be C times omega tilde, sorry, C transpose times omega tilde transpose. But, what is omit? So that gives you this part. And there's still the minus sign there. C, C transpose is identity. C, C transpose is identity. So those just vanish out of these multiplications. They have no impact, which leaves you with minus omega tilde. Now why does minus omega tilde transpose become plus omega tilde? Evan. >> So, since it's only used in cross product matrix, when you take its transpose, it's defined with a negative- >> Right, that was one of the fundamental things we reviewed at the beginning, is that the tilde matrix transpose is the same thing as minus a tilde matrix. And it relates to the vector cross product order, right, if you could flip them, there's a minus sign that appears. So you can get rid of the transpose by adding an extra minus sign, so minus minus makes it a plus. And at the end you have minus omega tilde plus omega tilde equal to zero. So this is a simple little proof. In exam, definitely fair game. Can you prove that this form holds for general orthogonal matrices? You just have to deal with the orthogonal definition, start to differentiate, plug it in, make some proper deductions and voila. So it's kind of a cool thing. And so we'll see in this class a few times higher dimensional representations as well, not just three dimensional stuff. We won't go far in that, I'll just give you a glimpse of that world. There's a whole other thing going on.
