So, we have wrapped up the first week.

First week is really pretty much review.

You have seen this stuff in some form or another.

Some of you may not have seen this thing of

a transport theorem but you have probably seen the cross product-like rules,

may not quite applied this way.

So when you are doing homeworks,

please be sure to use the transport theorem.

These are not very complicated homeworks.

It is all about practicing the principles of this stuff.

And so, some general guidelines already gave you, right?

Your position vector, then you know,

you do the easiest way possible, we do the laziest.

I mean, after introducing new frames and in this homework,

in particular, you find from the inertial frame,

we do have frame that is a one rotation about one of the axis points out to the disk,

and then from the disk, we rotate again,

that is a new frame and sometimes I show you frames,

sometimes I do not show you frames.

That is where you get to get creative and call Fred and Harry and that away,

whatever you want to do, you know.

They make the greatest fun, life fun.

But when you have that you need angular velocities.

So Brian, how many angular velocities do you have to determine?

[inaudible]

Okay,

let's say we have multiple frames,

let's say I have three frames.

How many angular velocities would you need?

[inaudible] Maybe.

Is it possible you might only need two?

[inaudible] Yeah, depends on a problem statement, right.

If you have B relative to A,

A relative to N you might need B relative to N as well,

depends how you formulated it.

So just get into it.

But really if you have three frames and you have

worked out the geometry with your right hand hopefully

what is a positive rotation about which axes are we going,

about how much are we going and you have two of the three,

do you really have to go through the geometry to get the third one?

What was your name again?

Sorry? Varadhan.

How do you find the third one?

[inaudible]. Exactly, angular velocities are vectors and this will

be a stark contrast of what we are starting today which is attitude descriptions.

Which we often write as a set of three_by_one numbers,

yaw, pitch, and roll,

you will see today the direction cosign stuff where we

have line numbers and we choose to arrange them in a certain way.

Well, attitude descriptions are not vectors.

You cannot just add yaw, pitch, roll plus yaw, pitch,

roll to get the overall stuff.

There is different math we have to do.

Angular velocities though, are vectors which is

cool because then once you know how to do

translation: I know my position relative to this person,

this person relative to that person and back to me.

You know two you can do the vectors and add them,

subtract them, just have a good notation.

So depending even if you have two frames it is usually like N+1.

That's kind of what you have to find the rest you can do through summations,

subtractions, and then you can go there,

and then the last step is always differentiate.

What you want to do is avoid as much as possible signs and cosigns,

when you leave things in rotating frames,

it is much easier it is way more compact more compact means is less chances

of missing a sign and mislabelling something dropping a distance,

an I becomes a J who knows.

But at some point and in some of these homeworks,

you will have to do a projection specially if you get a three dimensional.

The planar ones tend to be pretty simple. They all work out.

But if you do 3D, let us just kind of talk through that. That we have.

I am just going to draw up a frame.

So here's a frame I'm going to call this b1, b2, b3.

We have modern technologies, so let me use color.

I have a different frame and this is the same.

And I am going to call this one e1, e2 and e3.

Now these two frames only differ by a single axis rotation.

If you need a cross-product of

e1_cross_e3 if there are vectors within

the same frame as long as you specify the order of what is first,

second, third; you should know first cross second gives third right,

third cross first gives second,

the first cross third gives you minus the second and so forth.

That's all the typical cross-product rules you can do, that is easy.

But sometimes especially when it is 3D you come

across systems where you can do all the stuff,

this is all in b, this is all in the e frame,

but there is one term that ends up being,

I will just make up something,

I need e1 crossed with b3.

They are not orthogonal vectors right.

They are not in the same frame.

So this cross-product isn't quite trivial and we have to figure out what it is.

So here when you do it at this step,

I have to either project the e1_hat vector into

b-frame components or the b3_hat vector here into e-frame components right.

Because then you get everything in the e or everything into b.

Which way to go in the homework?

Doesn't matter. Both will give you a right answer,

just a different expression of writing that same answer right.

So how do we do this?

So in this homework right now these things typically

you're always defining frames that only differ by one axis.

If this is my first frame,

this is my second frame,

and the trick is to get the right projections and it is

easier to take this out of 3D into planar.

The up axis of my thumb is the same.

So in this case b2_hat is equal to e3_hat. That's easy.

It's the other two sets of vectors that we have to really figure out.

So the trick is you want to look down the axis that you're rotating about to go from

one frame to another and then you can draw these rotations undistorted.

So I am going to do that.

So my viewpoint is going to be looking down here

and then you can draw this any which way you want.

Let's say I have a rotation here that's

positive theta and then from here to here that's positive theta.

It's the same rotation angle.

So if I wanted to do that I'm going to look down,

twist it to make my life a little bit easier.

Let me say that b3, b1,

and then in this case b2 is sticking out of the board.

So b1, b2, b3 still right handed.

I am just looking down here taking this thing flipping it around. That's it.

Now I need the e frame and I'll use the color again just to distinguish.

So if the angle theta is 0,

e1 should b3 and then I'm rotating positive upward.

So that would be somewhere here.

That would be e1 and then e2 here is in the same plane but orthogonal,

pretend that's a straight line.

And then we have e3 sticking out of the board again.

All right. That makes sense.

So now you have drawn this,

this was all 3D.

Now you can just come back to a plane because these two axes are shared, that is easy.

Now we can go either way,

we can choose to write an e-frame vector in b-frame components or

a b-frame vector in e-frame components.

Kailey, what you want to do?

I want e to b.

OK. So let us pick e2.

And really we have to write it in b-frame though it is something b1,

something b2, plus something b3. That should be a 2, it's hard to see. There we go.

What you do something is going to be 0?

b3 or b2 should be zero.

Because that is the axis that comes orthogonal to this plane.

Right. So this one is going to be big too easy.

OK, that is the easy one.

Now here is the angle theta.

So essentially what we are doing here geometrically is you are doing projections.

How much is this vector projecting onto the b1 axis?

And that is how far we have to go,

so from here we have to go this far on the b1 direction.

All these hats are unit vectors.

When you draw these things out there all have length 1.

Right. And this of course with a projection makes it a right hand triangle.

Very, very trivial stuff.

So in this case what is this length?

How far do we have to move in the b1 direction?

Cosine theta.

Cosine theta, right.

That's the cosine of it and this would be the sine of it.

Now do you have to move in the positive b1 or

negative b1 direction to go from here to here.

Positive.

Right. So that's it.

We throw it in. Now we do the same thing here.

We do the projection of e2 onto b3,

so we have to go this far sine theta in the b3 direction.

But is it in the positive or negative b3 direction?

Negative.

Negative, in this case. So there would be a minus in front of that. And that's it.

So now if I plug this in,

this math would simply be

cosθb1_minus_sinθb3_crossed_b3.

What happens with b3 crossed

itself? Zero. We like zero, zero is good, zero is your friend.

b1_cross_b3.

What's that going to give us?

Sheila?

b1_cross_b3?

b2.

Positive or negative?

Negative.

Negative, actually. OK, good.

So -cosθ b2.

That's what this is.

This has become like that.

So now we did the projection where we absolutely needed it and

everywhere else we are using rotating frames which really keeps your life easier.

In this lecture we are going to start to get into 3D descriptions.

This is going to allow us to do more general.

You know, I need components from e into some other frame and so with the d,

c, m, we will see how to do this in general three dimensions,

but for the homework one in chapter one.

This is typically what you need,

use it as needed.

Yes, sir?

I think you've left a few things in there.

Because at the top you've got e1_cross_b3 then at the bottom you define e2.

I think that's where you've got the cos and sine.

Did I miss it? Did I really screw it up?

So much for a live example.

b3, b1, e1, e2, this is all correct, right?

I was thinking of this in terms of the algebraic definition of

the cross-products just the magnitude of each time the

sine of the angle between them and

then use right hand to figure out which direction it's going in.

So should not we end up with a sine at the end of this, [inaudible] cosine?

It depends on how the angles define,

you can see because we will have an e2_cross_b3,

the angle is actually 90_plus_theta and you put that into a trick function.

There is always wonderful trigger identities where sine

of 90 plus an angle is probably the same thing as a cosine or minus a cosine.

That's its relationships.

We're crossing here now.

And thanks for pointing out, this should have been a 2 not a 1.

We changed our example live,

but it is e2_crossed_e3 so this angle is 90_plus_theta.

I am still looking at e1_cross_b3.

I am thinking the angle between them is just theta.

No, we are doing this cross this.

So it's this angle [inaudible].

Yes. And that's where that sine can be written as a minus cosine.

That actually brings up another question I had,

which is looking at this I find it much

easier than kind of decomposing e into b coordinates.

I find it easier just to use that definition of

sine theta and then use right hand and curl rule.

It works.

[inaudible] downside of doing that.

No, it will give you the same answer,

different pass. Everybody has different way.

Some people have different ways of doing cross-product rule,

somebody it out into a matrix and do all the stuff, that's how they remember it.

I remember more the sequence of numbers.

There is no one right right way to do this.

Just want to make sure there was not some good reason

that you know about because you know where we are going.

No. If it is this simple,

there is really anything that works to get you there.

And if it is more complicated 3D that is what we are doing next.

We are going spend the next few weeks talking about different attitude descriptions and

how we get this rotation matrix and that is how you always map it back and forth.

And that is where I lose simple rules it is just I look at the math

and I compose this matrix and that is what we do.

But yeah so in just different ways,

you know some people have trouble thinking of

the geometric side and how do I get these angles in a 3D.

So the trick I found over the years is this helps a lot of people.

But if you see the direct definition,

absolutely go for it.

There is lots of ways you can do the simple math.

I just do not want anybody stumbling on this stuff and getting all wrong answers.

You need to know basic trig,

calculus, vector math that kind of thing.

OK. Any questions on homework?

Yes, sir?

One more thing. So while I was doing this assignment there

were a few times where I had a rotating frame so that I wanted to share an axis.

So say I had an inertial frame n1, n2, n3,

and then I have got rotating frame er,

er theta and it shares the n3 axis in the opposite direction of rotation.

Is it improper to define my e frame as er, eθ-n3?

So let me just write this up. So the question is n1, n2,

n3 defined in a classic way.

And then you would have er, eθ.

It is like a planar motion.

And the third one you could have e3 or you can just say n3. Is that's what you're saying?

Right.

Absolutely.

But one more complication is that because of the right hand rule the direction of

rotation of the e frame makes that n3 negative in e coordinate system.

Well n3 can only be one way.

So you want to define it as a minus this.

Yeah because it's equivalent.

I do not typically do that because

that minus sign can maybe confuse me but if I did it e3,

I would write somewhere e3 is -n3.

So if I have to flip it,

I have got the mapping right there but

there is fundamentally nothing wrong with defining,

you have defined three unit direction vectors just in a negative.

It is not that, it is the opposite of that way.

Basically that is what you are defining, to go that way.

But that shares the n3 and maybe that

makes your algebra and that is how you like to solve it.

Absolutely. There is lots of little nuances

here and everybody as you go through this stuff you should look at this and go,

"Hey, what really works for me.

How is my mind thinking? Do I like Trig?

Do I like to geometry?

Do I like this drawing vectors?" Whatever works for you.

You will get there. Any other questions right now?

I have a small one, it is just after you have done your algebra,

is that okay if your answer

is not all in the same reference frame?Some people don't like it.

I prefer that. Let me make that clear again.

All these problems that I'm giving you to say find the something something derivative.

If you are taking a frame derivative

doesn't mean you have to map everything into a frame.

you are taking an n frame derivative.

It doesn't have to be mapped into the n frame.

I can take the derivative of this pen relative to my shoulder.

I think the [inaudible] of this vector as seen by

the n frame and express it all using just rotating frames.

In fact, it makes it much easier.

It is much more compact.

That is where it is good to stop especially in an exam.

If you continue on for three pages doing bunch of sines and cosines,

some projections I'm like, "Well, that's a lot of work."

But I bet somewhere you have a whole paragraph written while you're running out of time

and this is impossible to do and I am not going to shed any tears.

I'm sorry. You should know what you're doing.

I'm just asking for the answer and it can be written in

vectorial form doesn't have to be orthogonal base vectors, we can mix vectors.

It really makes our life easier when we get through this stuff.

So this is kind of a practice round because we will be using this throughout.

If I needed you to express the answer in that I would say find the inertial

derivative and express the answer in c-frame components.

Now it doesn't mean up front step 1 put it in the c frame and then differentiate.

You can still do whatever is easiest.

It is just a very answer once you

get it because that is in the c frame, this is in the b, okay,

I am going to map that now into the c. That's when you do it at the very last step.

So still solve it whatever is easiest.

But in none of the homeworks do I actually ask you to

express the vectors in a particular frame.

So that leaves a lot of leeway in how this can be

solved and some frames you can come up with your own names.

So the [inaudible] will have to adjust for that ingredient,

obviously it won't be one answer that fits everything.

Well good. So I'm getting all good questions.

Appreciate you guys jumping on homework,

trying to get down with vectors, notations,

make sure these things are clear because if something is not,

this is every lecture we can have little,

"Hey, any questions about this?"

And if you are confused probably a sixth of the people are confused.

Please do speak up. This is where I'm getting stuck.

Let's fix that, let us move on, the rest of the lecture make that much more sense.

If you are distant students, or non-campus students who just cannot make

this lecture and you listen to the lecture and you have a question,

"Wait a minute, you know that thing about that angle I did not quite understand.

Could you explain that again?"

Let me know before the next lecture because then

the next review when I am going over this material again,

I can just incorporate those questions typically and that has worked out pretty well.

Okay. If it is something very particular just email me

or [inaudible] either of

us will try to respond right away, as soon as we can. Good, that's it for review.

So, we just basically

finished chapter one pretty quickly.

You got through some examples.

Hopefully, this warm up, get everyone in the same notation,

should not be too exciting yet.

Chapter 2, I would say skim through that you will

have some brief homework problems with Chapter 2. It's basic dynamics.

We are going to jump later on directed to the Rigid Body version of that,

that we haven't seen the stuff in Chapter 2 [inaudible] are made for a particle,

for system, for continuum.

You might want to glance through that and read to that as well.

So look at that as your reading assignment.

What we're doing now is we're jumping to Chapter 3.

Optional Review: Angular Velocities, Coordinate Frames, and Vector Differentiation

Kinematics: Describing the Motions of Spacecraft

Meet the Instructors