1:20

Okay,

Â let's say we have multiple frames,

Â let's say I have three frames.

Â How many angular velocities would you need?

Â [inaudible] Maybe.

Â Is it possible you might only need two?

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

Â If you have B relative to A,

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

Â depends how you formulated it.

Â So just get into it.

Â But really if you have three frames and you have

Â worked out the geometry with your right hand hopefully

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

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

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

Â What was your name again?

Â Sorry? Varadhan.

Â How do you find the third one?

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

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

Â Which we often write as a set of three_by_one numbers,

Â yaw, pitch, and roll,

Â you will see today the direction cosign stuff where we

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

Â Well, attitude descriptions are not vectors.

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

Â roll to get the overall stuff.

Â There is different math we have to do.

Â Angular velocities though, are vectors which is

Â cool because then once you know how to do

Â translation: I know my position relative to this person,

Â this person relative to that person and back to me.

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

Â subtract them, just have a good notation.

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

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

Â subtractions, and then you can go there,

Â and then the last step is always differentiate.

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

Â when you leave things in rotating frames,

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

Â of missing a sign and mislabelling something dropping a distance,

Â an I becomes a J who knows.

Â But at some point and in some of these homeworks,

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

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

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

Â I am just going to draw up a frame.

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

Â We have modern technologies, so let me use color.

Â I have a different frame and this is the same.

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

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

Â If you need a cross-product of

Â e1_cross_e3 if there are vectors within

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

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

Â third cross first gives second,

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

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

Â But sometimes especially when it is 3D you come

Â across systems where you can do all the stuff,

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

Â but there is one term that ends up being,

Â I will just make up something,

Â I need e1 crossed with b3.

Â They are not orthogonal vectors right.

Â They are not in the same frame.

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

Â So here when you do it at this step,

Â I have to either project the e1_hat vector into

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

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

Â Which way to go in the homework?

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

Â just a different expression of writing that same answer right.

Â So how do we do this?

Â So in this homework right now these things typically

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

Â If this is my first frame,

Â this is my second frame,

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

Â easier to take this out of 3D into planar.

Â The up axis of my thumb is the same.

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

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

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

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

Â So I am going to do that.

Â So my viewpoint is going to be looking down here

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

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

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

Â It's the same rotation angle.

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

Â twist it to make my life a little bit easier.

Â Let me say that b3, b1,

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

Â So b1, b2, b3 still right handed.

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

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

Â So if the angle theta is 0,

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

Â So that would be somewhere here.

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

Â pretend that's a straight line.

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

Â All right. That makes sense.

Â So now you have drawn this,

Â this was all 3D.

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

Â Now we can go either way,

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

Â a b-frame vector in e-frame components.

Â Kailey, what you want to do?

Â I want e to b.

Â OK. So let us pick e2.

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

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

Â What you do something is going to be 0?

Â b3 or b2 should be zero.

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

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

Â OK, that is the easy one.

Â Now here is the angle theta.

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

Â How much is this vector projecting onto the b1 axis?

Â And that is how far we have to go,

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

Â All these hats are unit vectors.

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

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

Â Very, very trivial stuff.

Â So in this case what is this length?

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

Â Cosine theta.

Â Cosine theta, right.

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

Â Now do you have to move in the positive b1 or

Â negative b1 direction to go from here to here.

Â Positive.

Â Right. So that's it.

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

Â We do the projection of e2 onto b3,

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

Â But is it in the positive or negative b3 direction?

Â Negative.

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

Â So now if I plug this in,

Â this math would simply be

Â cosÎ¸b1_minus_sinÎ¸b3_crossed_b3.

Â What happens with b3 crossed

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

Â b1_cross_b3.

Â What's that going to give us?

Â Sheila?

Â b1_cross_b3?

Â b2.

Â Positive or negative?

Â Negative.

Â Negative, actually. OK, good.

Â So -cosÎ¸ b2.

Â That's what this is.

Â This has become like that.

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

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

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

Â This is going to allow us to do more general.

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

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

Â but for the homework one in chapter one.

Â This is typically what you need,

Â use it as needed.

Â Yes, sir?

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

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

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

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

Â So much for a live example.

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

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

Â the cross-products just the magnitude of each time the

Â sine of the angle between them and

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

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

Â It depends on how the angles define,

Â you can see because we will have an e2_cross_b3,

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

Â There is always wonderful trigger identities where sine

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

Â That's its relationships.

Â We're crossing here now.

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

Â We changed our example live,

Â but it is e2_crossed_e3 so this angle is 90_plus_theta.

Â I am still looking at e1_cross_b3.

Â I am thinking the angle between them is just theta.

Â No, we are doing this cross this.

Â So it's this angle [inaudible].

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

Â That actually brings up another question I had,

Â which is looking at this I find it much

Â easier than kind of decomposing e into b coordinates.

Â I find it easier just to use that definition of

Â sine theta and then use right hand and curl rule.

Â It works.

Â [inaudible] downside of doing that.

Â No, it will give you the same answer,

Â different pass. Everybody has different way.

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

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

Â I remember more the sequence of numbers.

Â There is no one right right way to do this.

Â Just want to make sure there was not some good reason

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

Â No. If it is this simple,

Â there is really anything that works to get you there.

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

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

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

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

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

Â But yeah so in just different ways,

Â you know some people have trouble thinking of

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

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

Â But if you see the direct definition,

Â absolutely go for it.

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

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

Â You need to know basic trig,

Â calculus, vector math that kind of thing.

Â OK. Any questions on homework?

Â Yes, sir?

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

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

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

Â and then I have got rotating frame er,

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

Â Is it improper to define my e frame as er, eÎ¸-n3?

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

Â n3 defined in a classic way.

Â And then you would have er, eÎ¸.

Â It is like a planar motion.

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

Â Right.

Â Absolutely.

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

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

Â Well n3 can only be one way.

Â So you want to define it as a minus this.

Â Yeah because it's equivalent.

Â I do not typically do that because

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

Â I would write somewhere e3 is -n3.

Â So if I have to flip it,

Â I have got the mapping right there but

Â there is fundamentally nothing wrong with defining,

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

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

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

Â But that shares the n3 and maybe that

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

Â Absolutely. There is lots of little nuances

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

Â "Hey, what really works for me.

Â How is my mind thinking? Do I like Trig?

Â Do I like to geometry?

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

Â You will get there. Any other questions right now?

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

Â is that okay if your answer

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

Â I prefer that. Let me make that clear again.

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

Â If you are taking a frame derivative

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

Â you are taking an n frame derivative.

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

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

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

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

Â In fact, it makes it much easier.

Â It is much more compact.

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

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

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

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

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

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

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

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

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

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

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

Â derivative and express the answer in c-frame components.

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

Â You can still do whatever is easiest.

Â It is just a very answer once you

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

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

Â So still solve it whatever is easiest.

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

Â express the vectors in a particular frame.

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

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

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

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

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

Â Appreciate you guys jumping on homework,

Â trying to get down with vectors, notations,

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

Â this is every lecture we can have little,

Â "Hey, any questions about this?"

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

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

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

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

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

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

Â Could you explain that again?"

Â Let me know before the next lecture because then

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

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

Â Okay. If it is something very particular just email me

Â or [inaudible] either of

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

Â So, we just basically

Â finished chapter one pretty quickly.

Â You got through some examples.

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

Â should not be too exciting yet.

Â Chapter 2, I would say skim through that you will

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

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

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

Â for system, for continuum.

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

Â So look at that as your reading assignment.

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

Â