0:06

Now the next thing, the last thing we need for this,

Â the current homework set two, but is the differential kinematic equation.

Â The kinematic differential equation.

Â This we had for the DCM already.

Â 0:24

What was the DCM, Charles?

Â What was the DCM, differential kinematic equation?

Â What were we talking about?

Â >> I don't remember kind of end.

Â >> How does the DCM rate relate to Omega?

Â 0:47

What was your name again?

Â >> Ryan. >> Ryan.

Â >> [INAUDIBLE] >> Yes so c dot was minus omega tilde c.

Â That's what we saw.

Â I just want to bring it up quick.

Â I'm not going to slide back again.

Â That was it.

Â That related, because you may have the initial altitude,

Â you know the initial orientation, so you'd have C at the time T zero.

Â If you measure omega, which we often do with rate gyros,

Â you can now integrate the altitude and

Â figure out where was that sounding rocket as it took off and spiral and did stuff.

Â That's what we're doing using the DCM.

Â The DCM was non-singular but also is highly redundant, all right.

Â We had nine coordinates for a three degree feeding problem.

Â So Jordan, if we have nine coordinates, three to meet the freedom,

Â how many constraints do we have?

Â >> Six.

Â >> Thank you, okay.

Â Six, exactly.

Â So here we're doing the same thing but we have to relate yaw,

Â pitch, roll rates to the omega.

Â 1:47

And so it's a pretty straightforward process.

Â The key to solving this is remembering what Euler angles are.

Â They're a sequence of rotations.

Â So yaw rate really means this first rotation, how is that changing with time?

Â Time stamp one, we had 30 degrees, next time stamp we had 40 degrees.

Â So one second apart, you're increasing your yaw angle at 10 degrees per second.

Â That's about this first access, right?

Â The second one is the pitch.

Â If pitch was 10 and one second later, it's 11, that means you did your yaw and

Â the increasing it very quickly that the pitch angle's only varying

Â at one degree per second, all right.

Â But that's about this intermediate axis and then finally the roll.

Â How much is the role rate?

Â How much is the roll angle changing?

Â That's your roll rate.

Â It's angular velocities of one frame to this intermediate frame,

Â which we call the prime.

Â Then the pitch, we went to the double prime.

Â And then the final rotation gets you to b, all right?

Â That was that kind of four cartoon sequence I showed you.

Â That's what we're doing here.

Â That's what these are.

Â And what we have to do is relate them to this omega.

Â Omega is the instantaneous angular velocity of this frame,

Â of b relative to n.

Â Omega is b relative to n.

Â Without writing all the letters all the time.

Â So we have those rates.

Â How do they relate to here?

Â 3:30

So we went from the N-frame, where B and N are identical.

Â Then we do a three rotation first, the yaw.

Â That gives me the prime frame.

Â This was the prime frame, the first intermediate frame.

Â Then the second rotation, we rotate about 2, so that would be the 2B 2 prime.

Â 3:47

That gets us here and then the last rotation, this is the double prime frame.

Â We're rotating about the one axis so

Â B1 double prime and B1 aren't the same in that sequence.

Â So that's where the roll is just about here.

Â 4:01

This is where you start out from.

Â You have to get this omega right, it's a three to one sequence.

Â So the first rotation's about to inertial 3.

Â Then the second one is going to be about this one, b2 prime.

Â Or it could be b2 double prime.

Â They are the same.

Â because you are rotating about that.

Â And then the last one is always in the b frame because that gets

Â you to the b frame.

Â All right, whatever the sequence is.

Â But that's the yaw, pitch, roll.

Â And you notice, we talked about what is vectors earlier, this is a vector.

Â Magnitude times the direction, magnitude times the direction,

Â magnitude times the direction.

Â 4:35

And to relate yaw rate, pitch rate, roll rate to the omega, if you go back, we

Â already have the same vector written out this way but along b1, 2, 3 components.

Â To define your differential kinematic equations,

Â fundamentally what it breaks down to is once you have this equation,

Â 4:52

you have to translate everything into b frame components.

Â So we have to map this b2 prime, which is the same thing as b2 double prime,

Â into b frame.

Â And we have to map n3 into the b frame.

Â Now you have everything is something times b1, something times b2,

Â something times b3.

Â And those somethings have to be equal to omega 1 b1, omega 2 b2, omega 3 b3.

Â That's the sequence.

Â So we're going to do this.

Â b2 prime, again, same thing as b2 double prime.

Â The trick to get to that one is the double prime frame to the final frame,

Â we did a 3, 2, 1.

Â You look down your final rotation axis, you can draw that.

Â If you don't see how the cosine, sines are, I would just draw that out, and

Â then you can see how to map b double prime to the b frame.

Â And it's just a 1D rotation, kind of what you did in chapter 1 homework.

Â That's where you come up with this kind of a relationship.

Â 5:54

Okay, let's just do this.

Â 3D thinking can sometimes, with the best of efforts, like,

Â I just don't see it right now.

Â Okay, let's do this.

Â So let's see.

Â We're looking now b1 and then we need b2, b3 okay.

Â So I'm going to just draw this out.

Â So if I look down I'm going, This is,

Â b1 is out of the board so I'm looking down b1.

Â So this would have to be b2 double prime, b3 double prime.

Â You did this sequence, I'm looking down this axis.

Â So that's my current 2 and 3 and then I'm doing the roll to get to the final body.

Â So I want to use a different color.

Â Let's go red.

Â And all right, that's the final roll angle.

Â And we're going with roll.

Â We're going from that second intermediate frame to the final frame, so

Â that's the sequence.

Â 7:16

double prime because that second rotation from here to here.

Â This is b2 prime, b2 double prime.

Â And they're rotating about the two axis.

Â They're identical.

Â Here, this one, whoop, let me get the other labels in.

Â 7:54

What's this distance going to be, from here to here?

Â >> b2 prime, wait, what, sorry.

Â Could you do that again?

Â [LAUGH] What's the distance.

Â From here to here, how far do you have to travel.

Â 8:17

>> No, B1 will be all the way to the end because b2 hat is a unit vector.

Â You've gone too far.

Â So the thing to remember when you're doing projections, these are orthogonal sets.

Â So this axis has to be orthogonal to this.

Â B2 and b3 are orthogonal.

Â That's a right hand triangle.

Â What's the length of B2 double prime there, Mariel.

Â 8:40

>> That's one.

Â >> That's one.

Â Exactly, so you're on the right track.

Â This is the full one.

Â These other two have to be less.

Â So if this is one, what's this distance.

Â >> Very sorry, so that would be cosine phi.

Â And then the other one would be sine phi.

Â >> Good, okay, let me just write that in >> This is cosine phi and

Â that's sine phi, okay.

Â We're almost done Mariel.

Â So now B2 had to go from here to this point,

Â you go cosine phi in the plus B2 or minus B2 direction.

Â Plus, right, exactly.

Â And that's the last thing we have to check.

Â So that would be a cosine phi B2 hat, good.

Â So, Tony, next to that.

Â She's already given you the distance.

Â This is how far you have to travel.

Â But in which direction do you have to travel to get from here to here.

Â 9:27

>> From B2 to the B2 hat.

Â >> No, from this point.

Â >> You want- >> To this point.

Â We've gone from here to here.

Â We have to now figure out to here, right.

Â We're trying to write B2 double prime in terms of B2 and B3 base vectors.

Â >> You would subtract sine phi.

Â >> Now we have to go this distance.

Â Since phi, but we're going to be going in the minus B3 direction.

Â 9:54

And that's it. Now

Â you've done the corner transformation to get from one frame to another.

Â Here I've done the direction cosines.

Â I've taken the projections.

Â There's many different ways to do it.

Â And really any of them are fine for the homework if you're doing this.

Â You could have just written the DCM.

Â You could have said, well this is one rotation from this frame to this frame.

Â And you could look up and one into positive phi rotation.

Â And that would have given you the correct way to map

Â one set of coordinates into another set of coordinates.

Â That could work as well.

Â So this is kind of, if you don't see the phi geometry,

Â you can always draft the mass of math of M1, M2, and M3.

Â In fact, we use that in the last step.

Â Here I typically do this because I can derive this usually quicker than looking,

Â grab my book, find the right page, do that.

Â This is such a simple thing.

Â I just have to, that's the code, that's a plus minus.

Â It's either cosine and sine and one of them is negative usually, or possibly.

Â That's what I have to check for.

Â The other one N3, now we have to map N3 into the final body.

Â That's doing a lot of rotations.

Â I don't do that in my head.

Â There's no single nice projection to get there.

Â So, what do we have to do.

Â This is a coordinate transformation.

Â And the way I got these components is, I simply looked up the DCM formula.

Â 11:08

That's even on the equation sheet you have for an exam.

Â I'm giving you some of those already, like a three, two, one.

Â And then you have to know how to use it.

Â So if you've already derived the three, two, one.

Â Then I know, I put the vectrix that maps b frame vectrix into n frame,

Â or N frame into B.

Â That's a DCM.

Â I have it in terms of order angles, so I just look it up.

Â 11:28

Let's go back to that.

Â I see some.

Â Here's the three, two, one that we'd have.

Â What happens here is we have to get the N3.

Â This maps N1, 2, 3 into B1, 2, 3.

Â I need N3, though, the opposite.

Â What do you have to do to this DCM.

Â >> [INAUDIBLE] >> Transpose it, exactly.

Â because once you have NB [COUGH], then it's going to be N1, 2,

Â 3 times the transpose of this matrix equal to B1, 2, 3.

Â Now I need the third one.

Â I need N3.

Â I don't need N1, N2.

Â You can find them but they are not useful in this math.

Â I just need the N3 part.

Â So the N3, where do I find now the three vector components that I need.

Â 12:21

It's going to be the third column, right.

Â If you transpose this,

Â the third column becomes the first row of the first, the third row.

Â The third column becomes the third row.

Â And therefore, N3 will be this much times B1.

Â This much times B2, this much times B3.

Â Or, in one of your homeworks now,

Â I'm asking you to prove this BN was B1,2,3 in N frame components transpose.

Â Now we've got the rows are B1, 2, 3 or the columns are N1, 2, 3.

Â If you remember that definition, you could look at this and

Â go well I know this is B1, 2, and 3.

Â This is N1, N2, N3.

Â Lots of different ways to get there.

Â Lots of different little tricks.

Â But that's all we have to do.

Â People stumble over the mathematics of this.

Â But really just recognize the hardest thing, to get the differential kinematic

Â equations and it is not hard, really is this part.

Â 13:13

Get the correct sequence.

Â Then once you have this you just have to figure out well that intermediate

Â frame how do i map it to B.

Â That first access, how do I map it to B.

Â It's just coordinatetransformations.

Â 13:33

That's one that I'm asking you to do in this homework too.

Â I mean you've look at the answers in the Appendix if you wish.

Â You don't get points for the answer.

Â If you just give any answer, you get zero points.

Â It's really easy grading, you know.

Â Show me the path, show me how you get this, that's what you have to go and do.

Â Good, we did that.

Â Now how do you wrap this up.

Â Here, if you plug this in and you plug this in, and then compare to here.

Â This N3 goes here.

Â And this B2 prime, B2 double prime.

Â Same thing, goes here.

Â And then you collect all equal terms.

Â Give me all the B1 components.

Â They have to be equal to omega one.

Â Right, because this three orthogonal axes and all the B2 components that you end up

Â with have to be a omega two and all the B3 components have to be omega 3.

Â Well that give me three equations, the three unknowns, and how do we solve that.

Â Linear algebra, right, we put them in a matrix form.

Â That's what I've done here.

Â This what you do when you derive this on your own, really, this last step.

Â It's just a matter of identifying that's where that term goes, this term,

Â this term, that's it.

Â That shouldn't be hard.

Â But this relates now, if you have ordered angle rates into omega.

Â We use that sometimes but it's not the most common form.

Â Typically we need to inverse.

Â I have omega because i measured it from rate gyros.

Â Or as in the next section, let me get into the kinetics,

Â we find differential equations and how to solve for omega.

Â Now I'm giving you that.

Â What is the corresponding attitude.

Â We need the inverse of this and there's no easy way around this.

Â You basically go Mathematica and ask it to invert it.

Â And it'll be so happy to do that for you.

Â You can try to do it by hand but life is too short really.

Â 15:14

That's all we need, the inverse of this matrix.

Â Find a symbolic manipulator, invert it and get here.

Â In an exam I would never ask you to do this inverse because this is really why.

Â Give it to a computer, it will do it for you.

Â Now, when we get this inverse, you can look at here.

Â Are there any mathematical issues with this differential kinematic expression.

Â 15:37

Looks good, nothing bad happens.

Â Let's look at the inverse.

Â When you do the inverse, there's this one factor you can fact,

Â you can take outside, one over cosign theta.

Â Where is this differential equation going to have issues now, Evan.

Â >> Beta equals zero or 180 Theta's equal to 0.

Â >> Right, 90.

Â Plus or minus 90.

Â >> 90. Plus or minus 90.

Â Right because at 90, cosine theta is 0.

Â Minus 90 equals n theta 0.

Â So this is the pitch angle for a 3-2-1 sequence.

Â So it's an asymmetric sequence and we promised it's a second angle that's

Â a troublemaker and it's going to be a plus or minus 90.

Â This is the way you see it mathematically manifested.

Â And if you had any other asymmetric sequence these terms would look different

Â but these is always going to the same.

Â You'll always have the one over the cos sign if it's an asymmetric sequence.

Â And if it's a symmetric sequence they'll be a one overside.

Â That would be a singular for that but that, so that's what the mathemat,

Â that's what ambiguity of these coordinates Manifest himself.

Â And all a sudden, the differential equations have stuff like 0 over 0.

Â And your numerical integrator will have a lot of trouble with that.

Â So that's how these singularities manifest themselves.

Â 16:46

We typically write this big matrix is the b matrix so

Â its what maps omega into your coordinate rates.

Â I'm going to use the same notations for other [INAUDIBLE] coordinates.

Â We'll just have omega that maps mrp's, crp's,

Â [INAUDIBLE] different coordinate set.

Â But that's a differential kinematic equation.

Â 17:11

Good.

Â Here's the same if we do a 3-1-3 sequence.

Â So you would derive that the same way.

Â He did a 3 rate for your big omega rate, your inclination rate.

Â And then a sending node rate.

Â And you do all this quotum protections.

Â Always should be a frame at the end.

Â To group terms you end up with this.

Â This term always looks good, going from [INAUDIBLE] rates to omega.

Â But you invert that metrics analytically, you'll find a one over sine, as we expect.

Â So here at zero angle or 180 degrees angle, we're dividing by zero.

Â And you have these issues with the ambiguity.

Â And that's a sensitivity question.

Â These actually manifest themselves very much in orbits.

Â So if you look at the classic orbit elements, node, inclination,

Â They're called singular, they're actually a singular coordinate axis set.

Â Because if you have an equatorial plane, or an almost equatorial or an almost.

Â Circular stuff too.

Â There's all kinds of ambiguities in how these angles are defined.

Â And all of a sudden you're can very drastically put

Â your positions on in just this much.

Â But you're working right around that singularity.

Â So in attitude terms,

Â this is how it manifests itself and you see it in lots of different fields.

Â 18:35

>> This one?

Â >> No, for the 3, 2, 1.

Â >> This one?

Â >> About going to the, yeah.

Â So for the B prime, B hat two prime.

Â >> Yeah prime would be this sequence, that's the first intermediate step.

Â >> Okay so.

Â >> Instead of calling it e or d it's still b and just along my way to b right?

Â So that's why I use b prime, it's just the convenience.

Â Mm-hm!

Â >> A line for b half.

Â One. Line is b half.

Â One is not different.

Â One will be b half, one double prime.

Â >> Here we would have all double prime.

Â Yes.

Â And you could put those on this lines if you wish.

Â And to go add those again.

Â >> So then you would write.

Â >> So all the Ps in this side are double primes.

Â All the bs here are single primes.

Â And the final bs are just bs.

Â Now, that is the final frame, once you finish the third rotation.

Â So if you look here, for example, we did a three.

Â So we are rotating about threes.

Â And three is equal to b three prime, actually.

Â b three prime and three are the same because you're just doing that rotation.

Â But now we are doing a three, a two sequence, so

Â we are rotating about two, so it would be 2 prime, and B2 double prime.

Â You can see they are the same direction.

Â That doesn't change.

Â B1 Prime was here, and leave it in there as a thin line,

Â that's really the direction of the original B1 prime.

Â [SOUND] And now that we're rotating out to the double prime, so

Â that's where that moves up.

Â And the b3 prime moves to here, to b3 double prime.

Â >> So then, would you write b-hat-1 double prime

Â in the equation of the angular velocity?

Â >> You could, they're identical.

Â Yes.

Â So if you go look at here, you could have single prime or you could do double prime.

Â They are the same thing.

Â 20:27

That's why here, if you looked at this one, I'll still write that expressively.

Â b one prime,

Â b one double prime, because that second rotation is about b two not b one.

Â The second rotation is about two,

Â so the axis is in variant between those two frames.

Â >> Okay, yeah, I was just confused- >> You're on the right track.

Â >> I was just confused with the first and the last.

Â >> Okay, so work this through, come back with questions.

Â I also have office hours tomorrow if needed.

Â We'll go from there.

Â 21:02

So some final comments, something we've already reviewed really.

Â It's the geometric singularities.

Â If it's a asymmetric set it's 90 degrees.

Â Otherwise it's zero 180.

Â Always a second angle.

Â But you can also see from these singularities we're never further than

Â a 90 degree rotation away From a singularity.

Â So they don't linearize that well.

Â It doesn't take that much before woo, you really have to go for

Â the full nonlinear stuff.

Â This we'll compare and to other coordinates as we're getting into,

Â that'll linearize way better, or we can be much further away from a singularity.

Â And just something to keep in mind as we go through this.

Â 21:56

So symmetric angles.

Â They are direct addition properties.

Â So instead of taking all these coordinates.

Â I'm mapping them to the DCM.

Â That's a 3 by 3 times a 3 by 3.

Â Gets another 3 by 3.

Â And then you don't even use all of them, right?

Â We only use five of those elements.

Â One for the second.

Â And then two sets of twos for the first and the third.

Â That's a lot of wasted math, really.

Â 22:25

But we have to use spherical trigonometry.

Â This is kind of an elegant result.

Â Mach and Schuster showed this in his paper on the survey of attitude parameters.

Â It's kind of a very elegant paper.

Â So I just want to use a few minutes to explain what is spherical trigonometry.

Â You're all used to King/Ian trigonometry.

Â Right?

Â So in a flat plane all your inner angles have to add up to

Â be 180 in the end, right, have to.

Â 23:09

Here's the North Pole.

Â So someday travels from the North Pole all the way south to the equator.

Â Then they want to travel along the equator.

Â They're taking locally a 90 degree left turn.

Â And now they walk along way And then they swim

Â >> Dodge crocodiles.

Â Who knows. They get all the way across.

Â 90 degrees across the globe.

Â And I take another left.

Â 23:48

All right, you start out here.

Â That's first, 92nd, 93rd.

Â So all of these inner angles are actually right angles.

Â So 90 + 90 is 180, but.

Â Net ninety that's 270.

Â So on a sphere, spherical trigonometry, the inner angles don't add up to 180.

Â They add up to other quantities.

Â This is kind of one of the extremes.

Â But that's what we're looking at.

Â These are the inner angles.

Â You're taking your triangles and you painted it on a unit sphere.

Â That's one way to look at it alright?

Â So this travel, we talk really about the angles.

Â This is the center of the earth.

Â There's angles here that you will see.

Â So how far have you traveled?

Â Or there are angles here, or there's these arc distances.

Â So this length would have a 90 degree distance on that unit sphere.

Â You went from here to here.

Â So everything in terms of angles on these triangles, spherical triangles.

Â 24:45

So if you go back here, A is really this angle.

Â How far have you traveled there.

Â B B is this angle and C is this angle, right?

Â So those are the distances you've traveled on this sphere and

Â then when you get to a corner point on your triangle working on this sphere,

Â how much have you rotated to go from this line to this line?

Â So what's that inner angle?

Â And I'm using capital letters for the inner angles and small letters for

Â the outer, that's the outer side.

Â It's like the length of a triangle, but the length here is in time of

Â angular distances because we're working on those units here.

Â 25:32

And these are the ones that will be used from this.

Â Well next time we'll derive these properties.

Â This is something that's useful for the next homework assignment actually.

Â But I'm going to end there.

Â This was just a kind of a quick introduction.

Â Something to get you thinking about this.

Â And we'll pick up on Thursday, okay?

Â