Now let's see, good, we've got about 20 minutes. Let's go through some examples. When you solve these problems, this is really how I want you. It's kind of a cookbook formula. Those in 3200 have already seen this a little bit, but it's how do you approach kinematics. This is purely kinematics. I'm not throwing in forces, some torques, some mass. I'm just saying I have a position vector or some vectorial quantity. I want to take its first derivative. I want to take its second derivative. But I've written things in terms of rotating frames. How do I do this, right? So let's just work out some simple one. This is an example, very classical. Just looking at planar motion. So here's an E frame with e1, e2 and then e3 is pointing out of the board, right? And this is a vector r. And this is a particle P that I'm tracking. So I'm giving you actually quite a bit of information here. What I want to do is, the question is what is the natural derivative of r? Let me be explicit. This is the point O, the origin. Let me get rid of E. That's O. You know what? Let me get rid of this as well. I'll define this separately. So you have this given. Some frame. Some vector. It's planar motion that we're looking at and it'll make the math a little bit easier to do it here quickly. And we can go through this. So the first step that you have to do is write vector. What’s the vector? If I could write nicely, that we are looking for? And in this case, we need the motion of P relative to this point O. If you want to be explicit, you could write P relative to O, but I'm going to use an r shorthand just so I don't have a lot of P relative to Os floating around, all right? That's what I need. What's the laziest way? Laziness is a virtue here. What's the easiest, laziest way that we can write this r vector to go from O to P? Daniel. What do you think? >> [INAUDIBLE] >> Sorry, [INAUDIBLE] ahead of you. >> [INAUDIBLE] >> What's the easiest, laziest way to write a vector r that goes from point O to point P? >> [INAUDIBLE] >> Well, but how do we write that mathematically? So using E ray coordinates? You're not lazy enough. I need you more lazy. What's the easier way to write this? >> [INAUDIBLE] >> Yeah, if you just write there's a length. That's just going to be r, some length here. Let's just call it L. We're just making up this problem. There's a certain distance, d or L. Now what did you call the other one, Chuck? >> [INAUDIBLE] >> Okay, not what I would've called it, but that's good. That makes it live and real. So you're always trying to trick me. r hat, so r hat has to be unit direction vector, but it's basically saying, hey, that point P is 4 meters in that direction, that's it. You've got a question? No, okay, just you're raising your hand. [LAUGH] That's probably the easiest way. But now r hats, if we want to use this, r hat is going to be a rotating vector. So if we want to use the transport theorem to use the derivatives, to get the derivatives, we have to know what's the angular velocity between these two frames. So step two, is get the angular velocities. No, I'm sorry. I'm mixing this scene up, hold on. Step two and three are really correlated. To get the angular velocities, that's going to be step three. Get all omegas basically. But to get the omegas, we have to have full frame definitions. Earlier we're talking about, could we have a frame around the velocity vector, and that's not really possible. You can lock in one direction but the rotation about that, there's an infinity of possible frames that could do it, right? So we have to, here too, I only have one vector. So you're going to have to define two other vectors to fully set up this frame. So one is defined here. Where would you want to put the other vectors? What would make life easy? Need to be orthogonal, we know that right-handed, right, unit length, all this kind of stuff. Jordan, what do you think? >> Direction out the board, the same for both frames. >> Okay, so here you want to, now I'm going to define e3 is out of the board, all right? And I need a vector here out of the board. Good, so we can share e3 between two frames. Where is the third vector going to go? [INAUDIBLE] It's r. >> Up? >> That way. >> Could it also go down? >> Yep, unless you had three were into the board. >> No, okay, I'm glad you said that. Let's draw what I agree. This is the easiest way for me typically. What do we want to call this vector? We need a name. >> Theta hat. >> Theta hat, okay. See, I would've had a frame P, with p1, p2, p3, which would've been much easier. But no, you guys have to get complicated on me. Okay, so here we go. We have r hat, theta hat, and e3. I told you you could mix names, and you're proving me right. Okay, so we can do this. So we have to define the frames. And how many frames do you need is really directly related to how many omegas you're going to need. If you have all the different vectors that you need to get from A to B, B to C, D to E, you may need many frames. And then for, you'll need many omegas. In this case, it's all planar motion. I agree with Jordan, you could have the third one skewed, but why? Then you have all these weird, orthogonal angles to do. So this would work. So let's define this frame. Point P. What is the first axis now, Jordan? >> r hat. >> r hat, theta hat and e3. You really should write down these frames. Because the ordering is going to be important, especially when you guys get creative as you are, not just doing p1, 2, 3, as I would have done. But you are. We're going to do r hat, theta hat, and e3. Now I'm going to go back quickly to Jordan's earlier comment. He said it had to be up in this direction. Is there any way I could have had theta hat point down here and still define this P frame to be right-handed? >> [INAUDIBLE] >> No, we're keeping e3, we're not touching e3, we're not touching r hat. I'm only flipping theta hat. What must change in this definition to make it right-handed? She used the right hand, right-handed, not left left-handed. Sorry? >> [INAUDIBLE] >> If you would flip the definition and say my first vector is theta hat, in which case that would be here, my second vector is r, then the third one is out of the board. So you could actually do it and make it right-handed but now you're really making things confusing. [LAUGH] So whatever is easiest. It all gets you to the right answer. It's just names, and it's good, in the problems, to mix it up. because I shouldn't confuse you because instead of point body frame B, I made body frame T. And now all hell broke loose, you know? And then that means you didn't understand the stuff, you were just plugging in formulas. Matt? >> [INAUDIBLE] It also has an order when you put them into matrix form. because then in the P frame, I understand the first scalar in that prebound one is times the first vector direction. And then it matters as well. So let me do tensors and so forth, it did all, they didn't matters. >> [INAUDIBLE] to the problem? >> No. Laziness is my convention and it typically gets where we come from what's here I would use r1, that's my first one and then I build everything around it and that tends to make my life easier. But there's no real convention about this. So good, so write the vectors. We get from here to here. The easiest way typically is from here to here and I just define a frame that goes, well I need to go two meters that a way. And that away r hat, right? And then we need to flesh it out, Jordan called the other directions theta hat and e3 hat, and the rest gets there. Now we need omegas. We only need one omega, we only have two frames, right? Omega is the angular rate between two frames, so I only need one, and let's just find one. I'm going to get omega P relative to E. What is that going to be? This is planar motion. In this case. What do you think, Tebo? >> And so omega is theta dot where theta is the angle between r and e one and with an arrow in the upwards direction. >> Thank you, that was my next question. Right. You put your thumb along e three, curl your fingers, that would be a positive angular rotation, perfect, we got theta now that's so we need theta dot and what's the axis? >> Three. >> D 3 that's it. When you think of angular velocities as simply magnitude times the direction. That's how I break them and down two. I go okay this frame and this frame, what's going on? Well, they're rotating about this axis and in these problems we're solving right now, they're often rotating about some common axis. Here we're rotating both frames about E3 and that makes it a lot simpler. If it's a 4D, 3D tumble, you will see in chapter three how we handle those omegas. It's just a little bit more bookkeeping. What we need, right? So good. So let's write this out. We've written position, gotten frames, and now we differentiate. So the problem statement is that I'm looking for inertial derivatives as I'm assuming e here is defined as an inertial frame. Everybody nod there heads. So Evan, what does the inertial frame mean? [NOISE] >> The non rotating one. >> If I'm not rotating, am I guaranteed that this is a non >> That is an inertial frame. >> No. I suppose. >> What defines a frame to be inertial. >> Non accelerating >> Non accelerating. If you're rotating, there's centrifugal accelerations immediately .If something's rotating default boom, not inertial, right? But if you're accelerator and going faster and faster and faster, you are not an inertial frame, right? Could you be traveling at a constant speed? Yes. because in that case, there's zero acceleration. So you don't have to be a station. Some people say inertial frame means stationary frame. If you say that in the prelim, I'm going to raise all kinds of flags and say, wait a minute. Give me more details. It doesn't have to be stationary, it just has to be non-accelerating, that's what it boils down to. So hopefully it's something you've already heard of. So now, we're looking for r dot, that's defined to be the, I'll make it explicit. That's d N, that's dt of r Would you like to differentiate this directly as seen by an end frame. Hopefully the answer is no, because out hat doesn't. It varies the time, it's going to be none zero. We'll have to figure this things out. Maybe take vector components, I will introduce all of the signs and co-signs, and I'm way too lazy to do that, okay? So, as seen by what frame is the derivative of this right hand side, could be really easy. What do you think? Nick. Is it Nickle, no Nickles, Nick. I'm getting the name wrong. Yes, right over here. Trevor. I'm way off, sorry. >> So as seen by what frame is the right hand derivative is going to be very easy to do? >> Say the P frame. >> Yeah. Why? because right now, you're just lucky. 50/50, I mean, that's pretty good odds. >> [LAUGH] >> Why the p-frame? >> because the vector r is moving according [INAUDIBLE] Essentially, it's the body frame that we're dealing with? It's wishy washy. If I have this derivative of this term, the p frame derivative of this term, which one of these derivatives is going to go to zero? >> R Hat. >> R Hat. That's why. Right? Because immediately, with respect to what frame or all these crazy rotating parts not going to matter. That's essentially what we're doing with the transport theorem. You're always choosing a frame where it's almost like inertial. It just means hey, I can just treat this vector stuff as fixed things and not worry about them. That's the trick. That's the essence of the transport theorem. So that means if I write this out, I have a ddt(r) times r hat and it's being very explicit right now. You could do. You could skip a lot of these steps and put them together. Times DDT of R hat there. [INAUDIBLE] by the P frame. So this whole thing is. Shit. I shouldn't have. I messed it up. Let me just erase this. So the p derivative is going to be this, right? Then should I write a p over here on the dt of R? >> [INAUDIBLE] >> You can. >> [INAUDIBLE] >> So what is a p frame derivative of a scalar? [INAUDIBLE] I would say if you're doing a derivative of a scalar, it's just a time derivative. Don't put frames in there, because I've seen too many people, especially this homework, this is one problem, I think it's 3.6 that you'll be going through, that's really fun. Where all these little subtleties matter and all of a sudden people put transport theorems on scalars and have omegas cross the scalars and doing all kinds of crazy stuff that makes absolutely no sense. So if you just have a scalar you just doing a time derivative. I would say just write just a time derivative, that's way more rigorous. I agree, if this is composed of vectors and to vectors you could take derivative c anything. But this is just a time derivative, so here we said this is going to go to 0, so you only really just have r. R hat. And r is just a scalers such as a time derivative of that times r hat. Now to get this derivative, I'm going to do the p frame derivative which is r hat r. R hat. What do I have to add here to make, this is the P frame derivative. I need the R frame derivative. They're not the same, right? What do I have to add to make this complete? Sorry, yep, right there, Matt, thank you. >> [INAUDIBLE] >> Which is? >> [INAUDIBLE] >> Good, so that was my next question; which letters go here? So your saying q with respect to E, no, P. Must be P because we have a P-frame, crossed with r that we do there, yep. because we took the P-frame derivative here, so we need omega P relative to E. Again, that thing's just placeholders with the letters. Yes, Marion? >> [INAUDIBLE] >> We can, give me a better name? >> [INAUDIBLE] >> [LAUGH] >> Better is relative. >> [LAUGH] >> But we can, okay, so so far, we didnt need it yet. So we'll make it a q-hat, and we'll make this a q. Absolutely, so now we have to compute this. So the first part is r-dot r-hat + omega is theta-dot E3 x with r, which is r r-hat. So the first part is still the same, right? Now this crossed product, the scalars you can bunch up together. And just have r theta-dot, what is E3 x r? Now with these names, see, it's the 3rd vector crossed with the 1st gives you 2nd, right, and plus the 2nd. Once you have the right term, make sure it's plus and minus. In this case, it's plus. So in this case, the newly-christened q-hat appears. Again, I would've used P1, 2, and 3, but that's just me. And that's it, that is our inertial derivative of this. And you can see, we've solved it in a very quoted frame agnostic way. We're just saying, you have to somehow know r-hat is and q-hat is, and they're orthogonal. And so if you have this frame defined in this problem, this is the theta angle, now you can do your sines and cosines. And put it in MATLAB, and compute an actual matrix representation in the n-frame, the b-frame, whatever frame you want. But we're only doing the frames when we need to, at the very, very end. So in the problem statements, I often say, find the inertial derivative. It does not mean put every component of this vector into the inertial frame. If anything, you're really going to aggravate me. I want you, in the homework, to use rotating frames. If I see lots of sines and cosines, I'm probably just going to get my red pen out and slashing off points. Use rotating frames, that's the point. These are really simple, boring problems, all right? The purpose here is practice how to use rotating frames. Follow these steps, get through this stuff, and then you can come up with some way to write it. In some of the problems, there's ambiguities. I don't always give you the e-frames. There's offsets you can put in, so everybody's answers might be slightly different. And that's perfectly fine, all right? But this is the steps. We only assign corner frames when we absolutely have to. And that's probably at the very end, if you want to get actual numerical answers to compute something than using all these states, okay? Any questions about this? Let's just set up some other problems that relate a little bit to homework. So in this case, this was simple. If you did have to get inertial acceleration, let's just talk through that. So we have our dot, All right? And the problem statement part B says give me the inertial acceleration of this. What are the steps that we have to do in this case? What do you think? Yes, go ahead, Kyle, right? >> [INAUDIBLE] >> Casey, okay, I was off, Casey. >> [INAUDIBLE] >> What frame would you differentiate this to make life easiest? >> [INAUDIBLE] >> The P-frame that we had, right? because then r-hat, q-hat are all fixed, right, when you do this, +. Which omega, Casey, do you need here? >> [INAUDIBLE] >> E in this case. >> [INAUDIBLE] >> E was the n-one, crossed with the vector, itself, and you carry it out, all right? So let's look at different problems. I'm just going to quickly show you some, we got two minutes to show quick highlights and try to mix things up. Let's say we have a position vector that is a a1-hat because it's a frame, a 1, 2, and 3. Somebody was nicely lazy. And b b1-hat, there's a b-frame, b1, 2, and 3. Now I have to get a derivative of this. As seen by what frame would you choose to differentiate this? What was your name again, Matt? >> Yeah. >> You kind of rolled your eyes, so that means something startled you here. >> [INAUDIBLE] >> Do you? Differentiation is a linear operator, right? So what does that mean? How can you solve this? How can you be extra lazy and make this really easy? >> [INAUDIBLE] >> Yes, exactly, so if your vector has components in this frame and components in that frame, pick all the ones that are in one frame. So I would say this part is going to be a A-frame derivative +. Which omega do we need to use here? because I need to find an inertial derivative in the end, so what omega would go here? >> A with respect to n. >> A with respect to n crossed with the vector itself. And then this part would become a B-frame derivative, all right, of this stuff + omega B with respect to n crossed with this stuff again, right? So just remember that because with that, no matter how complicated this stuff looks, you can always, this is all the q, this is all the p, this is all the s, chunk them together. But you have to have figured out the proper omegas. And that's again, note, omega's a vector. So if you have omega an, and omega ba, you could add them to get omega bn, or something. You have to do the proper vector math to find these things. So that's good, one last problem, yeah. What if, Let's just put us into orbit. Here's the spacecraft, and this is the orbit frame. Here that's radial, tangential, and orbit normal, ih, all right? That's there, so that's an orbit frame, defined this way, {ir, i theta, ih}. One homework problem in particular deals with this. And that is you have the position vector here, r. No, let's say, forget this one. There's another thing moving around here that's a point, P. And you write this vector relative to the orbit. There's another satellite, an astronaut floating relative to the space station. I want to know how does the astronaut's position vary as seen by the space station, O. What type of time derivative are you taking now of this vector if that's what the problem statement says? How does this vector vary as seen by O? Do we take an N derivative, Andrew? >> [INAUDIBLE] >> Which one? >> [INAUDIBLE] >> Yeah, space station one, right? So check for that. Most of the problems always ask for inertial, inertial, inertial derivative, inertial velocity, inertial acceleration. But there's some that, all of a sudden, things are twisting and rotating and you're on this Ferris wheel or something. And it says, hey, how does this vectorial quantity change as seen by an observer in this other frame. In that case, you're picking an O-frame. Transport theorem still applies, you just do the same stuff. Yes, Matt? >> [INAUDIBLE] >> Yes, you may have to flip the, >> [INAUDIBLE] You have the rotation rates relative to [CROSSTALK] >> Yeah, because you may have this and say, look, I can easily take the derivative in N for some reason. Maybe this is given in N-frame stuff. But then you would need omega N relative to O crossed with that just to complete the transport theorem. Again, these letter are perfectly interchangeable. Just look out for what the problem statement says. So if it's asking for inertial derivative or a-frame derivative, it's just how you differentiate it. It's differentiated as seen by these observers. You can use mixed frames. Doesn't have to be all in the e-frame or the b-frame. You can mix the frames unless, if I need specifics, and I don't think any of these homeworks ask for it, I would say, hey, express your answer in terms of e-frame components. That means at some point you have to do your sines and cosines and map everything into one frame. But I don't typically here. I'm really trying to encourage you to use rotating frames. That's the whole purpose of this, okay? We're a few minutes over, but that's good. We'll pick up here Tuesday. Start these homeworks, come back with good questions. Let's see where you get stuck. As you apply it, that's always when all the little intricacies come in.