3: DCM Properties

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來自 University of Colorado Boulder 的課程

Kinematics: Describing the Motions of Spacecraft

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University of Colorado Boulder

Kinematics: Describing the Motions of Spacecraft

55 個評分

課程 1（共 4 門，Specialization Spacecraft Dynamics and Control）

The movement of bodies in space (like spacecraft, satellites, and space stations) must be predicted and controlled with precision in order to ensure safety and efficacy. Kinematics is a field that develops descriptions and predictions of the motion of these bodies in 3D space. This course in Kinematics covers four major topic areas: an introduction to particle kinematics, a deep dive into rigid body kinematics in two parts (starting with classic descriptions of motion using the directional cosine matrix and Euler angles, and concluding with a review of modern descriptors like quaternions and Classical and Modified Rodrigues parameters). The course ends with a look at static attitude determination, using modern algorithms to predict and execute relative orientations of bodies in space. After this course, you will be able to... * Differentiate a vector as seen by another rotating frame and derive frame dependent velocity and acceleration vectors * Apply the Transport Theorem to solve kinematic particle problems and translate between various sets of attitude descriptions * Add and subtract relative attitude descriptions and integrate those descriptions numerically to predict orientations over time * Derive the fundamental attitude coordinate properties of rigid bodies and determine attitude from a series of heading measurements

從本節課中

Rigid Body Kinematics I

This module provides an overview of orientation descriptions of rigid bodies. The 3D heading is here described using either the direction cosine matrix (DCM) or the Euler angle sets. For each set the fundamental attitude addition and subtracts are discussed, as well as the differential kinematic equation which relates coordinate rates to the body angular velocity vector.

2: Directional Cosine Matrices: Definitions18:54

3: DCM Properties7:16

4: DCM Addition and Subtraction5:37

5: DCM Differential Kinematic Equations8:10

Optional Review: Tilde Matrix Properties2:28

Optional Review: Rigid Body Kinematics and DCMs21:09

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Hanspeter Schaub
Glenn L. Murphy Chair of Engineering, Professor
Department of Aerospace Engineering Sciences

How do we identify some properties of this here?

Let's look at the determinant of the DCM.

And the trick on how to evaluate this is we don't look at the determinant of C

directly, but I'm going to look at C times C transposed.

Since C transposed is to inverse, I'm really doing C times its inverse,

which has to be identity, which is trivial.

The determine of identity, everybody here should be able to do it in their sleep

without lots of let me thinks, [INAUDIBLE] C, one.

That's it.

So we know the right hand side.

Let's expand the left hand side.

Since C is a square matrix of [INAUDIBLE] properties that determinant of A times B

is the same thing as determinant of A times the determinant of B,

which we can use here.

So good, we got that, I'mm broking\g it up.

And the determinant of C and its inverse,

actually, doing an inverse of a matrix doesn't change its determinant value.

So that's cool.

So we can really group these.

There's nothing, but the determinant of C squared has to be equal to one.

So that's a fundamental property that we can derive right here.

And that it means the determinant of C has to be the square root of one.

Which is plus minus one.

So, up to here, this is just dealing with an orthogonal matrix.

The only assumption we made was that C's orthogonal satisfies that C,

C transpose is equal to I.

That's it.

We made no statement about it being a representation of a right-handed

coordinate frame relative to each other, right?

Orthogonal matrices appear in lots of places, not just in attitude description.

So a general orthogonal matrix must have a determinant of plus or minus one.

If it has a determinant of two, it's not an orthogonal matrix.

Somehow you're scaling the vectors bigger than they should be.

But it's plus minus one.

If it's doing a mirroring, it ends up being a minus one operation.

Or if it's a left handed system, it'll give you a minus one..

So for us,

dealing with attitude descriptions, we're expecting a right-handed coordinate frame

mapped onto a right-handed coordinate frame to stay just in sync.

And that means we have to have a plus one determinant.

And that is called a proper rotation or a proper orthogonal matrix.

If you see the word,

proper in front of orthogonal, that's what we're talking about.

Just think right handed.

But mathematically, it means the determinant is plus one.

So if you just take any orthogonal looking matrix and

do a determinant operation math lab, and it gives you plus one on the three by

three, that's a representation of a right handed system.

That would work.

Any questions on this?

Proper orthogonal.

It's a simple derivation.

Yes, this is something that you should be able to do in exam.

Blank sheet of paper.

Some simple rules to get here.

This is the logic and this is where we end up.

Coordinate transformation.

This is probably the most practical use of DCNs.

We talked in homework one, we try to stay away from coordinate transformations,

do everything in rotating frames, but at some point you just can't help it.

You need to do a vector here, crossed with a skewed vector here.

How does that work out?

And if we have nice problems like in chapter one where they only differ by one

axis, there's easy ways to do it.

But sometimes it's the fall on 3D geometry.

How do I map one to another.

That's where we use this.

So there's different ways to write it out.

I'll give you this note but I'm going to back to an example and it might be just

an alternate way, you got this slide I'm showing an alternate way how to get it.

But the v vector can be written as something b1, something b2, something b3.

Or the v vector can also be written as something times n1, n2, n3, right?

There's an infinity of ways to break this vector

into three sets of orthogonal components.

So every one of those cases we can use the vectrix forms,

where the components are written as a one by three, or a three by one in this case,

transposed with a vectrix.

And you can do that, and these vectors have to be the same representation, so

you can put this around.

But in essence,

what this quickly proves to you, and I'm sure you've seen this before.

So if you go through this math If this the N frame vectrex components,

the N vectrex is nothing but C transposed to B vectrex.

And this also has to be equal to, this is the same v as here.

Therefore, this part has to be equal to this part.

And you can invert it back over, and you come up with

the B frame components are the C matrix times the N frame components.

Essentially, right.

That's this notation.

Let's go back to the notation we're using in class a little bit.

If we want to write a vector, (2, 4, 6).

I don't want to express it in eframe components.

Tony, how do I do that?

This right now is just a matrix.

How do I make it represent a vector in eframe components?

>> Multiply the e vector.

>> Well you could.

But just notationally, what do I have to add to, so

this is two times e one, four times e two.

>> [INAUDIBLE] >> Yep, that's it.

Right, that's what we use.

Easy.

And let's say I want an answer in T, just change things up.

I'm curious, what are these, if I could write.

What are the three vector components along the T one, two, and three axis, right?

So what this formula that I gave you showed is you can use the DCM but

using the two letter notation, which two letters I have to throw here?

>> T.

>> T and?

>> E. >> E.

All right.

So the T, E matrix maps from

you're going from the E frame and you end up with a T frame, right?

That's how we defined the C matrix, we went from N to B.

We had N frame components, and we ended up with b1, b2, b3.

We went from an N vectrex and ended up with a B vectrex.

It's all equivalent.

Yes.

>> Suppose you had another mapping from E to B.

So you had a DCM for E B.

>> Okay.

>> Can you go from T to B by just stacking those together?

Exactly.

And we'll get to that very next.

Let me just get to the slide and I'll show that then.

But this is the basic definition.

I hope everybody here has used the DCM rotation matrix transformation

Each one that we used in your notation.

To map vector components from one frame onto another, and

how to cascade them together.

We'll get to that very next, but that's it.

So with the two letter notation.

If I have a vector expressed in E.

I need the T E matrix to get that vector in T frame components.

If you ended up with a-.

Let's say BQ and

this is R and B frame, right.

And you claim in a paper that's RQ,

this should now raise lots of warning bells, this doesn't make sense.

What you'd have to do is transpose it and BQ transposed becomes QB.

So with this two-letter thing, you can look at, okay, this vector's given in this

frame, this is given in that, this maps from here to here.

All these letters should cascade and map up together,

hopefully, letting you keep track of it very, very simply.

Means errors.

So that's vector coordinate transformations, we.

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