So, good! Now what we want to start is vector differentiation. How many here have taken 3200, I think many of you actually have had it with me. Okay, I'm not going to ask you. You're excluded, sorry. You're out of the club. Okay, vector differentiation. Let me give you a simple vector. The pen, my shoulder. This is about one meter to my right, okay, that's it. So back row, yes, yes, sir. What's your name? >> Casey. >> Casey, thank you, Casey. So now the question is, what's the derivative, the time derivative of this vector? And this vector goes from my shoulder to this panel. That's it, so my hand is going front right, that is the vector. So what's the time derivative of this vector? >> Well, if you've defined it in the initial frame, it's the timed derivative of each component. >> Is what? >> Under derivitaves of each component? >> And what are they, zero or non-zero? Let me break that down. >> Typically non-zero. >> Okay. Are there other ways you can define this? >> Yes, if you had a moving coordinate frame, you can use the, I forgot the name of the transformation law. I forgot what it's called. >> We'll get to it, yep. >> But yeah. >> So what moving specifically? >> Non inertial >> Okay, if it's a non-inertial coordinate frame, what's the derivative? >> It's still a vector, right, a derivative? Is it zero or non-zero? >> Well you move it around with it, but. >> Yeah, is it zero or non-zero, right? Take a guess, you've got 50/50, do you feel lucky? >> Since you're moving around with those >> Going to guess non-zero. >> Non-zero, okay. Everybody agree? So, you're saying, even with a rotating frame and inertial frame, in both cases, this time derivative is going to be non-zero. That basically summarizes it. So, everybody agree with that answer? Why not? You're shaking your head. Forgot your name already. >> Matt. >> Matt, thank you. >> It depends on what frame you're talking about. I can set a frame in my own body. >> Who gave it to you? The inertial and the rotating phase? Not a specific frame. >> I put a frame right on your body that's moving around, with all your motions. >> And you're being very specific about keeping it- >> Guess it's getting tired. >> [INAUDIBLE] your chest. So you would have a zero. So you did get to the right, almost there. >> Inertially, you guys are great inertial observers. You're not moving, you're not rotating, you're lock fixed focused on me, right? This is so exciting. So we're sitting here and I'm moving around as. So as seen by you guys, this vector is actually changing. It's not changing magnitude, my arm isn't going bigger or smaller this distance, but it's changing direction. And so there would be an inertia as seen by inertial observers this changes, but if we have a frame that's body fixed, we do this. This pen is to my right one meter. This pen is still to my right one meter. All right, so, as seen by body fixed observer, all of a sudden, this vector doesn't appear to do it. So >> Let's summarize. My question was, what is the time derivative of this vector? Is it zero or non-zero? So what is the answer in the end? What do you think? You're looking confused at me. >> Zero. >> Zero. What do you think Chuck? It depends on where you're at. >> You're getting closer. The answer is, my question is rubbish. My question makes absolutely no sense, I'm just tricking you guys. If you take a I'm guaranteed to ask you this question in some form or manner to see if you are actually paying attention, right? If you ask somebody, what's the time of a vector? It's meaningless. You're not, you're being ambiguous, right? You have to be more precise and say "Look, I need a time derivative as seen by this frame." Now I can talk about how does this vector evolve. Cause you are pointing out, theres different observers that will see this vector evolving very differently, and therefore the time derivatives will be very differently. And that's all what this next section is about, how do we put this in mathematics, how do we write these things? And as, Matt, right? I'm slowly getting it, as Matt is pointing out, the nice, we love zeroes, right? We're all lazy, because in dynamics, something, something complicated times zero, yes! All that something goes away, right? And we don't need to keep track of it >> So that's beautiful and we'll find it many, many times. So we often seek frames with a spec to which these vectors don't move because then it's much easier to take these derivates. But then you can't Comes at a penalty, right? because the body frame derivative is not the same as the inertial derivative, and in the equations of motions, we're talking about, Kate was talking about the physics last time, right? f equals ma, h not equal to l. All those derivatives are inertial derivatives, not body frame derivatives. So we are going to have to do both. But do it in a way that makes our life easier. Okay? So, let's look at this. The first example, this is all in the textbook as well. But it's a very simple example where I'm looking at a body fixed vector. So r doesn't change just by the body. Just think of it as a shishkabob or a rotisserie chicken or something. you've got something with a fixed rotation axis and this very blobby looking chicken. And it's slowly rotating about it, and looking at where's the right wing on that chicken as it goes around the [INAUDIBLE]. And pretend it's really tied tightly so this thing doesn't change, doesn't morph and change with time, right? That's essentially what we're looking at here. And if you want to compute the Velocity. You can do very, very basic math here with dot products cross products. The magnitude, you can see, that's the projected radius. This point is going to move on a circle about this axis between a A and B, and this circle will have the radius R times sine theta, because this has a length R, and that's the projection onto that axis, so this is the opposite, it's the right-hand triangle. Very basic stuff. I'm not going to go through details. You guys can do this on your own. And then to get to direction, if this is the angular velocity, and that's the position vector, the velocity vector will be pointing right down here And that's going to be orthogonal to on and orthogonal to omega. So with the right order and normalizing it, I get the direction times the magnitude, that gets me the vector. Really, that's it. What you can do here is omega cross r, if you look at the cross product rule, it's the magnitude of the first, the magnitude of the other, times sine of the angle between them. So this term is actually the same as this term, just kind of drops out. And so the initial derivative of this body fixed points is nothing but omega crossed r, that's it. That's the one thing I want to remember from the slide right? A body fixed derivative, the inertial derivative of a body fixed axis, and ours here is a body fixed axis, is just going to be the angular velocity of that body relative to inertial crossed with that vector again. That's it. good. So, that's easy. Basic, basic geometry. So, let's derive this thing, so Kyle, you were talking about the, I call it the transport theorem. Different textbooks have cross-product rules or other names for it sometimes, but in my education system it was called the transport theorem. That's the name I use, but you might find it under other names. That's what we're going to derive here now. You'll see how with a Simple rules will understand how do we get to this wonderful property. We have a vector and writing R. And sometimes people are very bias and think R has to only be a position vector. Not true, or it could be really anything. All I need here is a vector, and I've broken it up along three orthogonal axes. But this is still the magnitude, direction, magnitude, direction, all this works nicely. So b1, 2, 3 could be a rotating frame. Like a body fixed frame that we're looking at. Now I need the angular velocity of this b frame relative to inertial, so I'm being very explicit. Omega B relative to N like we saw before omega1b1, omega2b2, omega3b3 using the same angular velocity vector, that's the position vector. Now, if we are taking the initial derivative of r, Differentiation, so a linear processor. If you have the derivative of a plus b, you can take the derivative of a plus the derivative of b and put them back together, right? So if you have a lot of terms, it's just bookkeeping. So let's look at the first term. If this vector, the position of r1 there is the time, so I'm sorry, if r1, the magnitude varies with time, and this whole frame's rotating, then both of these terms could actually vary as seen by inertial observer, and we're looking for inertial derivative. So, if you take the chain rule, and just differentiate r1b1, you're going to get the derivative of r1 times b1 plus r1 times in this case the inertial derivative of b1. We said if this is my body fixed axis I need the inertial derivative, this one's going to be non-zero. And I have to figure this thing out and it gets a little bit more complicated. So this rule will show you how to get around that. So instead of thinking we know we want the inertial derivatives let's do a side step. And as Matt was pointing out earlier, we have body frames with respect to which these b123s are fixed. And their time derivatives are always going to be 0. That's kind of the motivation. So if I took the same vector and I differentiated as seen by the b frame, that's what this left superscript means. As seen by, it doesn't mean r has to be written in b frame components, it often is, but it's not required, I just need the derivative, as seen by a b frame. That means you're basically sitting on this ferris wheel at that location, strapped in, and as you're rotating, how does this position change as seen by you, a rotating observer? Different ways to write that, but that's my mathematical notation, left superscript ddt, b framed derivative of r. So in this case, b1, b2, and b3, as Matt was pointing out, those are body fixed vectors. My right is always my right. My front is always my front and my up is always my up. No matter what I'm doing with my orientations. So if I'm taking the derivative as seen by here, I will only have the non-zero scalar parts. So if this pen isn't at a fixed location but it's slowly moving relative to me, going up and down, all right, those would be the r., r.., and r.... As seen by me, a rotating observer, this other astronaut is slowly drifting ahead, wobbling going off, probably drunk. Who does such motions? Okay, so this is what we take advantage of, right? As seen by ourselves, the b frame, any b frame fixed vectors derivatives must be 0 by definition. So, that's really nice, we like this, but we want an inertial derivative. So, if we did the inertial derivative, that's the chain rule I was describing earlier, if you dissect the derivative r1b1, you will have r1.b1+ r1 times the inertial derivative of b1. because now, we have to treat b1 as a time varying quantity again, right? My right axis is varying with time because I keep wobbling all over the place, and I'm not drunk by the way. Okay so, here we have to go through this, same thing for 2, same thing for 3. So, just a little bit of bookkeeping, that's what we have to do. So now, let's recognize these things. b1 is a body fixed frame and I have to take its inertial derivative. In that rotisserie problem that I showed you, I had a body fixed point. And I had to get it to inertial derivative, and then with a little geometry you quickly found that it was simply r., was simply omega crossed the r. In our case it's not r., it's b1., the innertial derivative of b1. And it simply going to be omega cross that vector again in this case. So the ib vector is going to be omega cross the bi. I'm just using that identity that we derived earlier. So for body fixed vectors, the natural derivative it's simply omega cross that vector again. So good, we could actually put this back together because we put these definitions in here, you going to get an omega cross b1, omega cross b2, omega b3. It's an omega cross, omega cross, omega cross, which you can take out to the left-hand side. I'll just do this math quickly just to show you. So you end up with r1 times Omega BN and b1, right? So plus r2 omega BN, no b2, plus r3, omega BN cross b3. All these omegas, you can factor out to the left. These are just scalars. If you have a scalar times vector math, that scalar you can put anywhere inside. It doesn't really change the answer. So I can put the scalar in here and then factor out the omega, which is common, so you get omega BN, r1 b1. Same thing here, omega bn gets factored out. So what's left is r2b2 and then r3b3. And what is this sum in the end? r, exactly, so we're back to r. And there's a cross product. So in this case all that complicated right-hand side is simply omega cross r. Whoa, it jumped ahead of me. Here we go. So this is going to be nothing but omega cross r, which is what I have here. On the left-hand side you can see and only I have this quantity and I have the 1 rate, the 2 rate and the 3 rate times b 1, 2, 3. This we saw on this side. That is actually the definition of the p frame derivative of r, right? That's where you are treating as b1, 2 and 3 as being fixed. And now you are just taking the magnitudes of the vector parts and see do they vary with chart. Does the length change does the four backward parts change and so forth. That's it, so that becomes nothing but the b frame derivative of r. And this is how we relate the two, the n frame and the b frame. We often like to take this. In the b frame, we have a lot of things like omega that's defined. We measure it in the b frame. The inertia tensor, if this is your spacecraft flying around, it has a certain mass distribution. But as seen by you guys, the tip is up, the tip is sideways, the tip is down. The mass changes with time. And that's a real pain to keep track of. So, if we do these inertia transfers, if you write them all and differentiate the mass seen by the body frame, the mass if it is rigid, stays the same. And it'll make life so much easier. So you will see this appearing over and over in different systems. So, that pulls down to this. A small equation, but a big slide for it. Why? because you're going to use this over and over. Once you become one with this formula and understand it, all those silly sines and cosine rules you had to learn as a freshman go out of the window. This is it. You really just need this formula. Write everything in the b frame and I'll go through some examples here today to show you how we apply this now. And make life a little bit easier. So when I write this formula, I have Bs here and I have Ns. In my derivation did anything require B to be a body-fixed frame or just a B frame and N to be an inertial frame, could N be another orbit frame, for example? Yeah, there was nothing specific, it's just two letters, it was the angular velocity. I need omega of that frame relative to the other. So in this formula, please treat B and N as placeholders. If you need the derivative as seen by the Q frame. And you're taking instead, the derivative as seen by the L frame, but you need omega Q relative to L and then you can apply it. Right, you can just substitute those variables and you'll be good. Same thing here with r, we tend to think this has to be a position vector. So give me quickly some other vectors we can throw in here, what else is going to come up. People, something, what- >> [INAUDIBLE] >> Well, you could, actually. That's something in the homework you have to do. I asked you to get inertial velocity. So you do these rotating frame derivatives and then cross products to get the answers in a nice compact way. [COUGH] Excuse me. But if you have to get accelerations, you do the same thing over. So you could use velocity vectors. But what other vectors have you seen so far? Go ahead. >> The angular force? >> Angular force was omega. At some points, we'll see between angular velocity derivatives. Actually in homework one, there's one of them that's really fun. You're going to love me with that one. And angular velocity could be there. Moving forward, angular momentum is going to go in there very quickly. What about kinetic energy? Do we use the transport theorem on kinetic energy? What's your name? Sorry, yes, purple shirt? >> Spencer. >> Spencer, what do you think? Kinetic energy, do we apply transport theorem on kinetic energy? Why? >> [INAUDIBLE] >> Not quite precise enough. What's the key thing about energy that doesn't fit to this formula? >> Scalar or vector? >> It's a scalar, right? This is really for vectorial stuff. But you bring up a good point, Spencer. Let me just write that out. So let's look at energy of a particle. T = m times v dotted with itself / 2, right? This is your inertial velocity, I've one my r dot already. Now, if I need my power equation, I'm differentiating kinetic energy, which is a scalar. So in that case, let me go to the next slide, I just have d/dt (T). If you're taking derivatives of scalars, don't throw frames in there. You don't get extra points, you just get me upset. because now, it means, you just don't know what you're doing. You're just throwing letters everywhere, right? Scalars, mass isn't simple example, a rocket is firing or I'm eating three cheesecakes a day and gaining 2 pounds a day. If I do that on Earth or I do that on the Moon, the Moon's rotating relative to us. Sadly, it makes absolutely no difference, I'm still gaining about one kilo a day, it doesn't matter. So your mass rate is just a scalar quantity and it doesn't matter on the observer with respect to our frame, it's just a time derivative. So now we need d/dt, and as Spencer was talking about, m/2 times the velocity squared, which I'm writing as an inner product. But it's based, the scalar quantity energy is based on vectorial quantities. So let's say it's constant mass, this rocket is not eating cheesecakes, so we're doing fine, but the velocities are changing. So I'm being very explicit, I have d/dt, v dotted with v. What is that going to be? because all of a sudden now, I have to differentiate vs, and I'm writing time derivatives. I told you earlier, if you do that in the prelim, I'm yelling at you. Going that's bogus, this doesn't mean anything, right? So we have to pick a frame. If we talk about time derivatives of vectors, you have to say look, I need the derivative as seen by something. Which frame do we have to put here? Tony, what do you think? >> Inertial [INAUDIBLE] >> Inertial, everybody agree? Good, I see some nodding heads. Does anybody have a different frame we could put here? Guess it's still way too shy, I think you guys have to loosen up. On the left hand side, do you care with respect to which frame you're taking a derivative if it's a scalar? No, if you don't care on the left-hand side, why do you care all of a sudden on the right-hand side? >> [INAUDIBLE] >> You could put it in the B frame. >> Not B, V. >> V? >> [INAUDIBLE] >> Well, how do you center a frame around the velocity? Velocity is a single vector. So you would get one axis, but you would be missing a full frame definition. So that would be a challenge there. But really, you can put any frame here. So let's just get away from B and N and I can say I have to pick a frame, but lemme just pick the Q frame just to make it different, why not, right? I can do that. And now, like we did earlier, you have to expand. This is something times something product rule differentiation. You would have m / 2 times the Q frame derivative of v dotted with v + v dotted with the Q frame derivative of v here, right? And now, I need to figure out, what the Q frame derivative of this v? So hopefully, you've expressed v in Q frame components if you have v as being L times Q3 hat. That's what happens to be the velocity, the Q frame derivative of Q3 is just going to be 0. And you need to now know does L change with time or not? So now, you can start putting it all back together again, all right? But you can pick any frame, which is kind of the cool thing. You could also pick the inertial frame, you could pick the body frame, you could pick a whole different frame, it has nothing to do with the system. And in every time, you better get the same answer. So I just want to throw this at you, we're not going to use it much early on. But as we get later on into the course, we'll be talking about this again. Well, we need dot, we have all of these scalar quantities and I need the time derivatives out of them. But the scalar quantities are dot products between two vectors. The one thing you can't do is if you're doing the derivatives here, you can't just say well and differentiating this as seen by Q. But for some wild, crazy reason, I want to differentiate this as seen by B. That would give you rubbish, right? Once you go here and you've chosen this time derivative of this scalar quantity composed of vectorial sub terms. Once you pick a frame, you have to be consistent that everything in that math has to be QQQ or BBB or NNN, so you can't mix and match within this stuff. But on the outside the differentiation of a scaler, you can do anything you wish, which is great, that's going to save us. Especially, when we get to CMG math, and then we can always pick the frame that makes our life the easiest and gets the answers the quickest. So you can see the simple cross product rule, lot's of subtleties on how this comes together, it's really for vectors. Don't put these frame dependent derivatives on scalars. And if it's a scalar that depends on vectors once you get this stage, you can choose whatever makes it work. Good, okay, one more slide here. Just comments, we do a lot of inertial derivatives in this material. because f equals ma h dot equal to l, all those derivatives from physics, those are all inertial derivatives. It gets very tedious to write this out, DDT is seen by inertial frame. So there's a definition that, unless I say otherwise, if you see a dot over a vectorial quantity, it's implied to be an inertial derivative. And then if it's something else, we'll specify. Sometimes, we use a lot of body frames and use a prime. Dot is inertial, prime is a body frame derivative. We use some short hands as we go through it, but that's kind of a typical thing, so good.