0:07

Last time we started up with that big chunk of rigid body kinematics.

Â There are some highlights.

Â I want to do a quick review.

Â This is what you should be remembering.

Â If nothing else, get everybody warmed up for the new material.

Â Or the angles, which is the next section.

Â But let's review these things first.

Â The fundamentals that we had of attitude descriptions.

Â How many coordinates do I need

Â as a minimum to define a three dimensional orientation?

Â What's your name? >> Ryan.

Â >> Ryan.

Â >> Three, minimum.

Â >> Three, okay.

Â If I have three.

Â 0:50

>> They might go singular.

Â >> They might? They will.

Â [LAUGH] Might is not good enough on a prelim exam.

Â They will, right?

Â There's no question.

Â If you only have three coordinates, they must go singular somewhere.

Â Okay, what was your name?

Â 1:26

Do you know the coordinates at that singular?

Â >> [COUGH] Yeah, I'm sorry I'm just out of it.

Â >> [LAUGH] Okay.

Â Daniel?

Â >> You get gimbal lock then one of the Euler angles isn't defined.

Â >> Right, so there's an ambiguity in the definition and we will see that today.

Â That's one of the ways and in particular in the differential kinematic equation,

Â that's always the equation that relates your coordinate rates to the omegas and

Â vice versa.

Â That one will have zero over zero issues because all of a sudden.

Â The easy one is think of orbits.

Â Most of you had orbits right?

Â So if you have your orbit plane there's an ascending node, inclination,

Â argument periapsis.

Â 2:21

But if you have the orbit and you try to figure out what those two angles are.

Â There is a infinity of answers.

Â If you've shifted 60 degrees now, okay great.

Â But that's where periapsis is but you could have 30 and 30.

Â You could have 20 and 40.

Â Those are the ambiguities we'll see after today as we get in into Euler angles.

Â That's how they will have singularities.

Â Anybody remember the other way that coordinates can go singular?

Â 2:48

They blow up to infinity, okay?

Â So these are the things you definitely, when you show up here,

Â I really recommend review the stuff.

Â You have two, three minutes, go over the highlights again.

Â because this what is hopefully going to stick, with me using it all now again.

Â So either they go ambiguous or they go off to infinity, great.

Â Are there ways to avoid singularities?

Â What's your name?

Â Brett? >> Yeah,

Â by using more than three coordinates.

Â >> Okay if you use four, sorry next to Brett.

Â What was your name?

Â >> David. >> David, if you use four coordinates

Â are you guaranteed those coordinates are non singular?

Â 3:26

>> No, because one could be dependent of the same thing.

Â >> Yeah, you could make some crazy set where you have two Euler Angles for

Â some reason.

Â And that's still not going to help you, right?

Â So going to four enables you to have a non-singular description.

Â And there's quaternions.

Â And in the book also we talk about Caley-Klein parameters.

Â Which are four dimensional complex numbers.

Â But you have to go at least to four.

Â Now what's the one attitude description we've already covered?

Â Andrew?

Â >> Direction cosine matrices?

Â >> Is that one singular or non singular?

Â >> Non singular?

Â >> It's actually non singular.

Â We didn't discuss it last time but if you look at it the whole way we derived it.

Â There was no attitude where you couldn't get these cosines right?

Â 4:05

So now we have more than four, as David was saying, right?

Â So no, did I get your name right?

Â David, right.

Â Okay, I did have David right.

Â So we're going to more than four.

Â How many coordinates, Jordan, do we have with the DCM?

Â 4:40

>> Number of constraints.

Â Although [INAUDIBLE] there would be nine because there are nine angles that relate

Â each axis to each other.

Â >> So you have nine constraints?

Â So you have nine coordinates and nine constraints.

Â That means there's zero degrees of freedom left.

Â 5:18

>> So let's think of this.

Â Let's talk about constraints quickly, maybe that's the confusion.

Â Let's say this is your particle, it has to move around.

Â A point mass, Jordan, Tebow, a point mass, Tebow, has how many degrees of freedom?

Â 5:33

>> Three.

Â >> Three, right.

Â So you can move it anywhere in space.

Â If you are saying this point mass has to move on this plane.

Â Now, Tebow, how many?

Â Right, why?

Â >> Because it only has two dimensional space to live in.

Â >> Right, this is a two dimensional sub-space.

Â It has to be here.

Â So the height, if you've got a coordinate frame on this plane, right?

Â The height has to be 0.

Â Or if you say this particle has to move 1 meter off the floor, you've constrained

Â one of the coordinates and you're left with a 2 degree of freedom system.

Â So the number of coordinates minus the number of constraints leaves you with

Â the number of degrees of freedom.

Â That's a fundamental thing.

Â If you have not seen that, look it up, read it up quick.

Â Or find a Wiki page on this or something.

Â But this is something that I'll be talking with these terms and

Â expecting you guys to know what I'm talking about.

Â Hopefully, if not, raise questions again.

Â We definitely want to get into this.

Â So we do have six constraints.

Â With the DCM, where do these constraints come from?

Â 7:04

But some of your, the column should be, should give us a unit vector.

Â >> Exactly. Cause they

Â represent actually the base vectors of the B frame.

Â If this is the representation of BN, right?

Â Then the rows would B1, B2, B3 and N frame components.

Â And the columns are N1, N2, N3 in B frame components.

Â So they all have to have unit length.

Â You can't have numbers.

Â They come from the definition of cosines.

Â So you know cosine is between plus or minus 1.

Â So it can't be bigger than norms of 1, so that can't happen, so that's no good.

Â So okay.

Â So you actually gave.

Â What I gave is a bunch of stuff.

Â because width then each being normal, and this being normalized.

Â These are not unrelated.

Â You could go, well, that's three.

Â That's three.

Â But then somebody else says, wait a minute.

Â Base vectors, right?

Â The first row crossed with the second row has to give you plus the third row.

Â because they're right-handed coordinate systems.

Â The first column crossed the second has to give you plus the third.

Â But also the second row,

Â the third row crossed have to give you plus the first one.

Â And this third crossed second has to give you minus one.

Â Well, all of a sudden I'm coming up with ten constraints.

Â Does that mean I have a minus one degree of freedom problem?

Â 8:16

What does it mean?

Â >> I don't know what minus one degree of freedom means.

Â [LAUGH] >> Doesn't make sense, right?

Â So if you end up with that like wait a minute I've way more constraints.

Â These constraints are actually are all related in the end.

Â You know you have nine coordinates.

Â It's a three degree for your problem.

Â You end up having the equivalent of six constraints.

Â But there's many ways that they manifest themselves.

Â Also we said the inverse of the matrix has to be the transpose.

Â Well that imposes all kinds of constraints which embeds all thought than normal rows

Â and columns and all the other stuff that we have, okay?

Â So in the end it six constraints there is many ways to write them but

Â you have to be very careful when you write the constraints the you are not just

Â duplicating the same constraints.

Â They have to be independent constraints to really reduce your degrees of freedom.

Â That's what we have with the DCM so keep on it's singularity free but

Â you do have a bunch of constraints which impacts our DR integration that we do.

Â 9:15

Back row, green shirt, what's your name?

Â >> Ben.

Â >> Ben, thanks.

Â What is the determinate of any orthogonal matrix?

Â >> Zero?

Â >> No.

Â Good guess, zero is always the good answer, right?

Â >> Uh-huh >> Not quite.

Â What was your name?

Â >> Armzel. >> Armzel.

Â >> Plus minus one.

Â >> Plus minus one, precisely.

Â So we went through that algebra.

Â There's something any orthogonal matrix is true.

Â Now, how do we call our rotation matrix?

Â It's a particular type of orthogonal matrix.

Â 10:14

And hopefully it will stick, and therefore you'll do better on your exam.

Â So proper orthogonal, now what does that mean?

Â If it's a proper orthogonal, it's to determine a plus one or minus one?

Â >> Plus one.

Â >> Plus one, and, yeah, right?

Â That's actually the check for that's an easy check.

Â You just take your deter, your DCM.

Â And if it's orthogonal, the inverse has to be the transpose, great.

Â But that doesn't guarantee its a, it's a rotation made.

Â It has to also be proper orthogonal.

Â If you take the determinant Mathlab.

Â If hopefully plus one, then you're good.

Â Okay, additions, substructions, how do we add DCMs?

Â 10:49

So what was your name, you light green shirt?

Â >> Khalian.

Â >> Khalian, okay, how do you add a DCMs if you have orientations?

Â >> Because BN is like [BR] and [RN] we just add it to give [BN].

Â >> How do you add?

Â You do an add like this?

Â [BR] plus [BN].

Â >> No we multiply.

Â >> [LAUGH] What's you answer?

Â Or do you want a lifeline.

Â >> No.

Â We multiply them.

Â >> Multiply, yeah.

Â Don't do addition.

Â That gives you nothing.

Â That gives you nothing sensible.

Â It'll give you something, but it's not sensible.

Â 11:41

Next o Nathaniel in the back, what's your name?

Â >> Nate. >> Nate.

Â Let's say you are given [BN] and you are given [RN] but

Â I'm trying to find the difference [BR].

Â How would you do that?

Â >> Begin,take the transpose of

Â [RN] multiply by [BN] is equivalent to the inverse.

Â >> All right, you can post multiply both side by all transpose,

Â that cancels here, moves it to the left hand side, and here we go.

Â So the one, if you know the addition, you should know subtraction,

Â its' really the same, it should, with the two rotations.

Â If you're given BN, great.

Â If you're given NB, well, you can relate them, right.

Â You just transpose them, that's it.

Â So good.

Â So we did additions and now we also do the ordinary differential equations,

Â the differential kinematic equations.

Â So we had C that we wrote and we derived this and

Â it was omega till the C and this is equivalent in 3.6 actually,

Â you need these two letter version to do this because you need BR dot, right?

Â Because you want to be careful what that means but BN dot here is

Â omega BN in a 2D form and

Â that omega BN is written in B frame components times the DCM again.

Â So this and this are equivalent.

Â That's a differential kinematic equation.

Â 13:06

So let's talk about this okay, if you have sounding rocket.

Â Originally standing straight up let's say that's a reference frame so

Â our attitude is zero motion relative to the reference for it.

Â So, Charles, what is the DCM if B and N are identical?

Â >> Identity.

Â >> Identity, right?

Â So just the identity operator.

Â Only the diagonal cosine's a one everything else is a zero.

Â Makes it really easy.

Â Good.

Â So we know the initial latitude, now you fire off the sounding rocket and

Â you have a rate gyro on board, as so many people here do this, now you record it.

Â Puts it on a hard drive.

Â You pick up the sounding rockets wherever it lands and put it back in the data.

Â You know the initial latitude.

Â Now you know the omegas.

Â You can integrate this and figure out precisely what was the attitude

Â of the sounding rocket at any point of its flight.

Â You've got exactly what you need.

Â But again, we've measured omega BN.

Â Once we get to the kinetic side, the second chunk,

Â we can figure out differential equations to predict with forces and

Â torques what happens to omega BN.

Â 14:31

>> Deciding what frame you would differentiate them in?

Â >> No, this is a matrix now.

Â This is a matrix equation.

Â You're going to get the DCM, which represents B relative to N,

Â just in a nice 3 by 3 form for convenience.

Â 15:15

To keep us, I cannot think of people's names.

Â Warta, right?

Â You were talking about unit columns and unit rows, right?

Â That was important.

Â If you now have slide integration errors, are your rows and

Â columns can have unit length?

Â 15:34

Probably not.

Â So as soon as you have this singularity free description label, no problems.

Â Now life isn't that easy.

Â You should know better by now.

Â You're all old enough, right.

Â If you get one benefit, there's some old secrets you have to deal with.

Â So, if it's a redundant system, with more than three coordinates,

Â these coordinates must satisfy constraints.

Â And so, a regular integrator won't know about those constraints, and

Â you will have little errors creep in.

Â And that means we will have to and there's processes for doing that.

Â This matrix, because otherwise, long term you can have integration errors build up

Â where one element might be 1.001, and that's going to cause all kinds of

Â craziness with your transformations and things that you do, so be aware of that.

Â There are other types of integrators.

Â Anybody here heard of symplectic integrators?

Â Symplectic integrators have a particular mathematical form that allows you to

Â input constraints and say, look, I have to integrate this orbit, but

Â my energy has to be precisely this amount.

Â You might still have integration errors but your energy will not be violated.

Â And you can do additive integration like that too, where you're saying, look,

Â I have to integrate these DCMs.

Â It has six constraints. You have to put them in some

Â mathematical form.

Â But you can make sure you keep it there, but

Â it gets a lot more complicated to do that.

Â So there's a whole world of how to integrate redundant constraint

Â coordinate systems.

Â Just something I wanted to highlight.

Â But the benefit is no singularity, easy,

Â it's a simple linear algebra math to do that, okay?

Â 17:22

This differential kinematic equation, C dot equal to minus omega tilde C.

Â We derived it for the rotation matrix.

Â It turns out this is actually true for any orthogonal matrix.

Â And you see orthogonal matrices appearing in lots of dynamical systems, especially

Â if you take structures, the system masked matrices that you can decompose,

Â in the eigen norm, eigenmode decompositions.

Â 17:46

They appear lots of places.

Â So what we can do actually, we can prove that this form that we have,

Â that C dot is equal to minus omega tilde C.

Â If C is orthogonal,

Â then its time evolution of that matrix must satisfy this form.

Â Now what does omega mean,

Â omega tilde, if you have a 16 million degree of freedom structures problem?

Â Good luck, I don't know.

Â But it doesn't have the nice rigid body interpretation.

Â But it will have the skew symmetric relationship that you can figure out.

Â And we'll see this a little bit later when we go to higher, more than three,

Â the people are looking at not just three dimensional latitude, but four and

Â five dimensional latitudes as well.

Â We'll see a little bit of hints of that.

Â So look at this.

Â If it's orthogonal, we said C times C transpose has to be identity, right?

Â because C transpose is the inverse.

Â So the matrix times its own inverse is going to be identity.

Â Its time derivative has to be 0.

Â That's why we get there.

Â Or if you take the chain rule, C dot C transpose and

Â C, C dot transpose, you just expanded it.

Â Now you can start to plug in, we postulate and make a hypothesis and

Â say okay, does this form hold for general n by n matrices?

Â So we just plug that in and so

Â here you have a C dot C transpose which gives you minus omega tilde C C transpose.

Â This other part here, this is a little linear algebra.

Â We have C dot transpose.

Â Here I'm giving you C dot, not C dot transpose.

Â If you transpose to C dot, what happens to this matrix multiplication?

Â 19:21

It flips.

Â If you've forgotten that, remember that.

Â It's an easy thing if you haven't seen that before, check it out for yourself.

Â So, C dot transpose is going to be C times omega tilde,

Â sorry, C transpose times omega tilde transpose.

Â But, what is omit?

Â So that gives you this part.

Â And there's still the minus sign there.

Â C, C transpose is identity.

Â C, C transpose is identity.

Â So those just vanish out of these multiplications.

Â They have no impact, which leaves you with minus omega tilde.

Â Now why does minus omega tilde transpose become plus omega tilde?

Â Evan.

Â >> So, since it's only used in cross product matrix,

Â when you take its transpose,

Â it's defined with a negative- >> Right,

Â that was one of the fundamental things we reviewed at the beginning,

Â is that the tilde matrix transpose is the same thing as minus a tilde matrix.

Â And it relates to the vector cross product order, right,

Â if you could flip them, there's a minus sign that appears.

Â So you can get rid of the transpose by adding an extra minus sign, so

Â minus minus makes it a plus.

Â And at the end you have minus omega tilde plus omega tilde equal to zero.

Â So this is a simple little proof.

Â In exam, definitely fair game.

Â Can you prove that this form holds for general orthogonal matrices?

Â You just have to deal with the orthogonal definition, start to differentiate,

Â plug it in, make some proper deductions and voila.

Â So it's kind of a cool thing.

Â And so we'll see in this class a few times

Â higher dimensional representations as well, not just three dimensional stuff.

Â We won't go far in that, I'll just give you a glimpse of that world.

Â There's a whole other thing going on.

Â