0:18

You're excluded, sorry.

Â You're out of the club.

Â Okay, vector differentiation.

Â Let me give you a simple vector.

Â The pen, my shoulder.

Â This is about one meter to my right, okay, that's it.

Â So back row, yes, yes, sir.

Â What's your name?

Â >> Casey.

Â >> Casey, thank you, Casey.

Â So now the question is,

Â what's the derivative, the time derivative of this vector?

Â And this vector goes from my shoulder to this panel.

Â That's it, so my hand is going front right, that is the vector.

Â So what's the time derivative of this vector?

Â 0:54

>> Well, if you've defined it in the initial frame,

Â it's the timed derivative of each component.

Â >> Is what?

Â >> Under derivitaves of each component?

Â >> And what are they, zero or non-zero?

Â Let me break that down.

Â >> Typically non-zero.

Â >> Okay. Are there other ways you can define this?

Â >> Yes, if you had a moving coordinate frame, you can use the,

Â I forgot the name of the transformation law.

Â I forgot what it's called.

Â >> We'll get to it, yep.

Â >> But yeah. >> So what moving specifically?

Â >> Non inertial >> Okay,

Â if it's a non-inertial coordinate frame, what's the derivative?

Â 1:51

Take a guess, you've got 50/50, do you feel lucky?

Â >> Since you're moving around with those >> Going to guess non-zero.

Â >> Non-zero, okay.

Â Everybody agree?

Â So, you're saying, even with a rotating frame and inertial frame, in both cases,

Â this time derivative is going to be non-zero.

Â That basically summarizes it.

Â 2:13

Why not? You're shaking your head.

Â Forgot your name already.

Â >> Matt. >> Matt, thank you.

Â >> It depends on what frame you're talking about.

Â I can set a frame in my own body.

Â >> Who gave it to you?

Â The inertial and the rotating phase?

Â Not a specific frame.

Â >> I put a frame right on your body that's moving around, with all your motions.

Â >> And you're being very specific about keeping it-

Â >> Guess it's getting tired.

Â >> [INAUDIBLE] your chest.

Â So you would have a zero.

Â So you did get to the right, almost there.

Â >> Inertially, you guys are great inertial observers.

Â You're not moving, you're not rotating, you're lock fixed focused on me, right?

Â This is so exciting.

Â So we're sitting here and I'm moving around as.

Â So as seen by you guys, this vector is actually changing.

Â It's not changing magnitude,

Â my arm isn't going bigger or smaller this distance, but it's changing direction.

Â And so there would be an inertia as seen by inertial observers this changes, but

Â if we have a frame that's body fixed, we do this.

Â This pen is to my right one meter.

Â 3:19

This pen is still to my right one meter.

Â All right, so, as seen by body fixed observer,

Â all of a sudden, this vector doesn't appear to do it.

Â So >> Let's summarize.

Â My question was, what is the time derivative of this vector?

Â Is it zero or non-zero?

Â So what is the answer in the end?

Â What do you think?

Â You're looking confused at me.

Â 3:53

The answer is, my question is rubbish.

Â My question makes absolutely no sense, I'm just tricking you guys.

Â If you take a I'm guaranteed to ask you this question in some form or

Â manner to see if you are actually paying attention, right?

Â If you ask somebody, what's the time of a vector?

Â It's meaningless.

Â 4:11

You're not, you're being ambiguous, right?

Â You have to be more precise and say "Look, I need a time derivative as seen by this

Â frame." Now I can talk about how does this vector evolve.

Â Cause you are pointing out,

Â theres different observers that will see this vector evolving very differently, and

Â therefore the time derivatives will be very differently.

Â And that's all what this next section is about, how do we put this in mathematics,

Â how do we write these things?

Â And as, Matt, right?

Â I'm slowly getting it, as Matt is pointing out, the nice, we love zeroes, right?

Â We're all lazy, because in dynamics, something,

Â something complicated times zero, yes!

Â All that something goes away, right?

Â And we don't need to keep track of it >> So that's beautiful and

Â we'll find it many, many times.

Â So we often seek frames with a spec to which these vectors don't move

Â because then it's much easier to take these derivates.

Â 5:05

But then you can't Comes at a penalty, right?

Â because the body frame derivative is not the same as the inertial derivative, and

Â in the equations of motions, we're talking about,

Â Kate was talking about the physics last time, right?

Â f equals ma, h not equal to l.

Â All those derivatives are inertial derivatives, not body frame derivatives.

Â So we are going to have to do both.

Â 5:23

But do it in a way that makes our life easier.

Â Okay?

Â So, let's look at this.

Â The first example, this is all in the textbook as well.

Â But it's a very simple example where I'm looking at a body fixed vector.

Â So r doesn't change just by the body.

Â Just think of it as a shishkabob or a rotisserie chicken or something.

Â you've got something with a fixed rotation axis and this very blobby looking chicken.

Â And it's slowly rotating about it, and

Â looking at where's the right wing on that chicken as it goes around the [INAUDIBLE].

Â And pretend it's really tied tightly so

Â this thing doesn't change, doesn't morph and change with time, right?

Â That's essentially what we're looking at here.

Â And if you want to compute the Velocity.

Â You can do very, very basic math here with dot products cross products.

Â The magnitude, you can see, that's the projected radius.

Â This point is going to move on a circle about this axis between a A and B,

Â and this circle will have the radius R times sine theta,

Â because this has a length R, and that's the projection onto that axis, so

Â this is the opposite, it's the right-hand triangle.

Â Very basic stuff.

Â I'm not going to go through details.

Â You guys can do this on your own.

Â And then to get to direction, if this is the angular velocity,

Â and that's the position vector, the velocity vector will be pointing

Â right down here And that's going to be orthogonal to on and orthogonal to omega.

Â So with the right order and

Â normalizing it, I get the direction times the magnitude, that gets me the vector.

Â Really, that's it.

Â 6:48

What you can do here is omega cross r, if you look at the cross product rule,

Â it's the magnitude of the first, the magnitude of the other,

Â times sine of the angle between them.

Â So this term is actually the same as this term, just kind of drops out.

Â And so the initial derivative of this body fixed points is nothing but

Â omega crossed r, that's it.

Â That's the one thing I want to remember from the slide right?

Â A body fixed derivative, the inertial derivative of a body fixed axis,

Â and ours here is a body fixed axis, is just going to be the angular velocity of

Â that body relative to inertial crossed with that vector again.

Â 7:23

That's it.

Â good. So, that's easy.

Â Basic, basic geometry.

Â So, let's derive this thing,

Â so Kyle, you were talking about the, I call it the transport theorem.

Â Different textbooks have cross-product rules or other names for it sometimes, but

Â in my education system it was called the transport theorem.

Â That's the name I use, but you might find it under other names.

Â That's what we're going to derive here now.

Â You'll see how with a Simple rules

Â will understand how do we get to this wonderful property.

Â 7:51

We have a vector and writing R.

Â And sometimes people are very bias and think R has to only be a position vector.

Â Not true, or it could be really anything.

Â All I need here is a vector, and I've broken it up along three orthogonal axes.

Â But this is still the magnitude, direction, magnitude,

Â direction, all this works nicely.

Â So b1, 2, 3 could be a rotating frame.

Â Like a body fixed frame that we're looking at.

Â Now I need the angular velocity of this b frame relative to inertial, so

Â I'm being very explicit.

Â Omega B relative to N like we saw before omega1b1, omega2b2,

Â omega3b3 using the same angular velocity vector, that's the position vector.

Â Now, if we are taking the initial derivative of

Â r, Differentiation, so a linear processor.

Â If you have the derivative of a plus b, you can take the derivative of

Â a plus the derivative of b and put them back together, right?

Â So if you have a lot of terms, it's just bookkeeping.

Â So let's look at the first term.

Â If this vector, the position of r1 there is the time, so I'm sorry,

Â if r1, the magnitude varies with time, and this whole frame's rotating,

Â then both of these terms could actually vary as seen by inertial observer,

Â and we're looking for inertial derivative.

Â So, if you take the chain rule, and just differentiate r1b1,

Â you're going to get the derivative of r1 times b1 plus r1 times in this case

Â the inertial derivative of b1.

Â We said if this is my body fixed axis I need the inertial derivative,

Â this one's going to be non-zero.

Â And I have to figure this thing out and it gets a little bit more complicated.

Â So this rule will show you how to get around that.

Â 9:31

So instead of thinking we know we want the inertial derivatives let's do a side step.

Â And as Matt was pointing out earlier,

Â we have body frames with respect to which these b123s are fixed.

Â And their time derivatives are always going to be 0.

Â That's kind of the motivation.

Â So if I took the same vector and

Â I differentiated as seen by the b frame, that's what this left superscript means.

Â 9:56

As seen by, it doesn't mean r has to be written in b frame components,

Â it often is, but it's not required, I just need the derivative, as seen by a b frame.

Â That means you're basically sitting on this ferris wheel at that location,

Â strapped in, and as you're rotating, how does this position change as seen by you,

Â a rotating observer?

Â Different ways to write that, but that's my mathematical notation,

Â left superscript ddt, b framed derivative of r.

Â So in this case, b1, b2, and b3,

Â as Matt was pointing out, those are body fixed vectors.

Â My right is always my right.

Â My front is always my front and my up is always my up.

Â No matter what I'm doing with my orientations.

Â So if I'm taking the derivative as seen by here,

Â I will only have the non-zero scalar parts.

Â So if this pen isn't at a fixed location but it's slowly moving relative to me,

Â going up and down, all right, those would be the r., r.., and r....

Â As seen by me, a rotating observer, this other astronaut is slowly drifting ahead,

Â wobbling going off, probably drunk.

Â Who does such motions?

Â Okay, so this is what we take advantage of, right?

Â As seen by ourselves, the b frame,

Â any b frame fixed vectors derivatives must be 0 by definition.

Â So, that's really nice, we like this, but we want an inertial derivative.

Â 11:27

So, if we did the inertial derivative, that's the chain rule I was

Â describing earlier, if you dissect the derivative r1b1,

Â you will have r1.b1+ r1 times the inertial derivative of b1.

Â because now, we have to treat b1 as a time varying quantity again, right?

Â My right axis is varying with time because I keep wobbling all over the place, and

Â I'm not drunk by the way.

Â Okay so, here we have to go through this, same thing for 2, same thing for 3.

Â So, just a little bit of bookkeeping, that's what we have to do.

Â So now, let's recognize these things.

Â b1 is a body fixed frame and I have to take its inertial derivative.

Â In that rotisserie problem that I showed you, I had a body fixed point.

Â 12:08

And I had to get it to inertial derivative, and then with a little

Â geometry you quickly found that it was simply r., was simply omega crossed the r.

Â In our case it's not r., it's b1., the innertial derivative of b1.

Â 12:31

I'm just using that identity that we derived earlier.

Â So for body fixed vectors,

Â the natural derivative it's simply omega cross that vector again.

Â So good, we could actually put this back together because we put these definitions

Â in here, you going to get an omega cross b1, omega cross b2, omega b3.

Â It's an omega cross, omega cross, omega cross,

Â which you can take out to the left-hand side.

Â I'll just do this math quickly just to show you.

Â So you end up with r1 times

Â Omega BN and b1, right?

Â So plus r2 omega BN,

Â no b2, plus r3,

Â omega BN cross b3.

Â All these omegas, you can factor out to the left.

Â These are just scalars.

Â If you have a scalar times vector math, that scalar you can put anywhere inside.

Â It doesn't really change the answer.

Â So I can put the scalar in here and then factor out the omega, which is common, so

Â you get omega BN, r1 b1.

Â Same thing here, omega bn gets factored out.

Â So what's left is r2b2 and then r3b3.

Â And what is this sum in the end?

Â r, exactly, so we're back to r.

Â And there's a cross product.

Â So in this case all that complicated right-hand side is simply omega cross r.

Â 14:09

Whoa, it jumped ahead of me.

Â Here we go.

Â So this is going to be nothing but omega cross r, which is what I have here.

Â On the left-hand side you can see and only I have this quantity and

Â I have the 1 rate, the 2 rate and the 3 rate times b 1, 2, 3.

Â This we saw on this side.

Â That is actually the definition of the p frame derivative of r, right?

Â That's where you are treating as b1, 2 and 3 as being fixed.

Â 14:42

And now you are just taking the magnitudes of the vector parts and

Â see do they vary with chart.

Â Does the length change does the four backward parts change and so forth.

Â That's it, so that becomes nothing but the b frame derivative of r.

Â And this is how we relate the two, the n frame and the b frame.

Â 15:04

In the b frame, we have a lot of things like omega that's defined.

Â We measure it in the b frame.

Â The inertia tensor, if this is your spacecraft flying around,

Â it has a certain mass distribution.

Â But as seen by you guys, the tip is up, the tip is sideways, the tip is down.

Â The mass changes with time.

Â And that's a real pain to keep track of.

Â So, if we do these inertia transfers, if you write them all and differentiate

Â the mass seen by the body frame, the mass if it is rigid, stays the same.

Â And it'll make life so much easier.

Â So you will see this appearing over and over in different systems.

Â So, that pulls down to this.

Â 15:38

A small equation, but a big slide for it.

Â Why? because you're going to use this

Â over and over.

Â Once you become one with this formula and understand it, all those silly sines and

Â cosine rules you had to learn as a freshman go out of the window.

Â This is it. You really just need this formula.

Â Write everything in the b frame and

Â I'll go through some examples here today to show you how we apply this now.

Â And make life a little bit easier.

Â So when I write this formula, I have Bs here and I have Ns.

Â 16:05

In my derivation did anything require B to be a body-fixed frame or just a B frame

Â and N to be an inertial frame, could N be another orbit frame, for example?

Â Yeah, there was nothing specific,

Â it's just two letters, it was the angular velocity.

Â I need omega of that frame relative to the other.

Â So in this formula, please treat B and N as placeholders.

Â 16:33

If you need the derivative as seen by the Q frame.

Â And you're taking instead, the derivative as seen by the L frame, but

Â you need omega Q relative to L and then you can apply it.

Â Right, you can just substitute those variables and you'll be good.

Â 17:01

People, something, what- >> [INAUDIBLE]

Â >> Well, you could, actually.

Â That's something in the homework you have to do.

Â I asked you to get inertial velocity.

Â So you do these rotating frame derivatives and

Â then cross products to get the answers in a nice compact way.

Â [COUGH] Excuse me.

Â But if you have to get accelerations, you do the same thing over.

Â So you could use velocity vectors.

Â But what other vectors have you seen so far?

Â 17:47

What about kinetic energy?

Â Do we use the transport theorem on kinetic energy?

Â What's your name?

Â Sorry, yes, purple shirt?

Â >> Spencer.

Â >> Spencer, what do you think?

Â Kinetic energy, do we apply transport theorem on kinetic energy?

Â 18:25

But you bring up a good point, Spencer.

Â Let me just write that out.

Â So let's look at energy of a particle.

Â T = m times v dotted with

Â itself / 2, right?

Â This is your inertial velocity, I've one my r dot already.

Â Now, if I need my power equation, I'm differentiating kinetic energy,

Â which is a scalar.

Â So in that case, let me go to the next slide, I just have d/dt (T).

Â If you're taking derivatives of scalars, don't throw frames in there.

Â You don't get extra points, you just get me upset.

Â because now, it means, you just don't know what you're doing.

Â You're just throwing letters everywhere, right?

Â Scalars, mass isn't simple example, a rocket is firing or

Â I'm eating three cheesecakes a day and gaining 2 pounds a day.

Â If I do that on Earth or I do that on the Moon, the Moon's rotating relative to us.

Â Sadly, it makes absolutely no difference,

Â I'm still gaining about one kilo a day, it doesn't matter.

Â So your mass rate is just a scalar quantity and it doesn't matter on

Â the observer with respect to our frame, it's just a time derivative.

Â 19:42

So now we need d/dt, and as Spencer was talking about,

Â m/2 times the velocity squared, which I'm writing as an inner product.

Â But it's based, the scalar quantity energy is based on vectorial quantities.

Â So let's say it's constant mass, this rocket is not eating cheesecakes, so

Â we're doing fine, but the velocities are changing.

Â So I'm being very explicit, I have d/dt, v dotted with v.

Â 20:15

What is that going to be?

Â because all of a sudden now, I have to differentiate vs, and

Â I'm writing time derivatives.

Â I told you earlier, if you do that in the prelim, I'm yelling at you.

Â Going that's bogus, this doesn't mean anything, right?

Â 21:24

>> [INAUDIBLE] >> You could put it in the B frame.

Â >> Not B, V.

Â >> V?

Â >> [INAUDIBLE] >> Well,

Â how do you center a frame around the velocity?

Â Velocity is a single vector.

Â So you would get one axis, but you would be missing a full frame definition.

Â So that would be a challenge there.

Â But really, you can put any frame here.

Â So let's just get away from B and N and I can say I have to pick a frame, but

Â lemme just pick the Q frame just to make it different, why not, right?

Â I can do that.

Â 21:59

And now, like we did earlier, you have to expand.

Â This is something times something product rule differentiation.

Â You would have m / 2 times the Q frame

Â derivative of v dotted with v + v dotted with

Â the Q frame derivative of v here, right?

Â And now, I need to figure out, what the Q frame derivative of this v?

Â 22:24

So hopefully, you've expressed v in Q frame

Â components if you have v as being L times Q3 hat.

Â That's what happens to be the velocity,

Â the Q frame derivative of Q3 is just going to be 0.

Â And you need to now know does L change with time or not?

Â So now, you can start putting it all back together again, all right?

Â But you can pick any frame, which is kind of the cool thing.

Â You could also pick the inertial frame, you could pick the body frame,

Â you could pick a whole different frame, it has nothing to do with the system.

Â And in every time, you better get the same answer.

Â 23:00

So I just want to throw this at you, we're not going to use it much early on.

Â But as we get later on into the course, we'll be talking about this again.

Â Well, we need dot, we have all of these scalar quantities and

Â I need the time derivatives out of them.

Â But the scalar quantities are dot products between two vectors.

Â The one thing you can't do is if you're doing the derivatives here,

Â you can't just say well and differentiating this as seen by Q.

Â 23:28

That would give you rubbish, right?

Â Once you go here and you've chosen this time derivative of this scalar

Â quantity composed of vectorial sub terms.

Â Once you pick a frame, you have to be consistent that everything in that math

Â has to be QQQ or BBB or NNN, so you can't mix and match within this stuff.

Â But on the outside the differentiation of a scaler, you can do anything you wish,

Â which is great, that's going to save us.

Â Especially, when we get to CMG math, and then we can always pick the frame that

Â makes our life the easiest and gets the answers the quickest.

Â 24:03

lot's of subtleties on how this comes together, it's really for vectors.

Â Don't put these frame dependent derivatives on scalars.

Â And if it's a scalar that depends on vectors once you get this stage,

Â you can choose whatever makes it work.

Â 24:26

Just comments, we do a lot of inertial derivatives in this material.

Â because f equals ma h dot equal to l, all those derivatives from physics,

Â those are all inertial derivatives.

Â It gets very tedious to write this out, DDT is seen by inertial frame.

Â So there's a definition that, unless I say otherwise, if you see a dot

Â over a vectorial quantity, it's implied to be an inertial derivative.

Â And then if it's something else, we'll specify.

Â Sometimes, we use a lot of body frames and use a prime.

Â Dot is inertial, prime is a body frame derivative.

Â We use some short hands as we go through it, but

Â that's kind of a typical thing, so good.

Â