Angular Velocity

Loading...

Do curso por University of Pennsylvania

Robotics: Aerial Robotics

1922 classificações

University of Pennsylvania

Robotics: Aerial Robotics

1922 classificações

Curso 1 de 6 em Specialization Robotics

How can we create agile micro aerial vehicles that are able to operate autonomously in cluttered indoor and outdoor environments? You will gain an introduction to the mechanics of flight and the design of quadrotor flying robots and will be able to develop dynamic models, derive controllers, and synthesize planners for operating in three dimensional environments. You will be exposed to the challenges of using noisy sensors for localization and maneuvering in complex, three-dimensional environments. Finally, you will gain insights through seeing real world examples of the possible applications and challenges for the rapidly-growing drone industry. Mathematical prerequisites: Students taking this course are expected to have some familiarity with linear algebra, single variable calculus, and differential equations. Programming prerequisites: Some experience programming with MATLAB or Octave is recommended (we will use MATLAB in this course.) MATLAB will require the use of a 64-bit computer.

Na lição

Geometry and Mechanics

Welcome to Week 2 of the Robotics: Aerial Robotics course! We hope you are having a good time and learning a lot already! In this week, we will first focus on the kinematics of quadrotors. Then, you will learn how to derive the dynamic equations of motion for quadrotors. To build a better understanding on these notions, some essential mathematical tools are discussed in supplementary material lectures. In this week, you will also complete your first programming assignment on 1-D quadrotor control. If you have not done so already, please download, install, and learn about Matlab before starting the assignment.

Transformations16:23

Euler Angles12:27

Axis/Angle Representations for Rotations14:06

Angular Velocity9:07

Conheça os instrutores

Vijay Kumar
Nemirovsky Family Dean of Penn Engineering and Professor of Mechanical Engineering and Applied Mechanics
School of Engineering and Applied Science

Now that we've gotten familiar with this concept of rotations and

displacements more generally,

it's time to start thinking about the rate of change of rotations.

And that leads us to the concept of an angular velocity vector.

What does it mean to differentiate a rotation and get a velocity?

We know the analogue for position vectors, you differentiate a position vector,

you get a velocity.

Now we want to take a three by three rotation matrix and differentiate that.

Well, if you take this matrix and you differentiate it, the first thing you

want to recognize is that this matrix is not just a bunch of numbers it's actually

Orthogon and there are two relationships which govern the Orthogonality.

Let's differentiate both sides of each equation,

if you do that you'll essentially use the product rule.

And you come up with two identities that relate

to the derivative of the rotation matrix and the transpose of a rotation matrix.

And this will tell you that R transpose R dot,

where R dot is a derivative of R, and R dot R transpose are both skewed symmetric.

In other words,

there's only three independent elements in these q symmetric matrices.

So rather than think in terms of the derivative of a rotation matrix,

we will think in terms of the derivative,

pre-multiplied by R transpose, or post-multiplied by R transpose.

Back to our canonical example for a rigid body with a generic position vector and

its rotated position, which we call P prime.

Once again, we'll make sure that the origins are the same, and

we'll consider the displacement from P to P prime.

Recall that we use the position vector P to denote the coordinates of the point P,

with respect to A1, A2, A3, and the vector q to denote

the coordinates of the point p prime with respect to A1, A2, A3.

I want us to remember now that the vector P is

a three by one vector that consists of coordinates of P in a body fixed frame.

Now if the rigid body is rigid and stays rigid, this vector P does not change.

The only thing that changes as a rigid body rotates is a vector q.

As the rigid body rotates, the rotation matrix changes

as a function of time, and the vector q changes as a function of time.

But the vector p stays constant.

So let's take this equation and differentiate this.

If you differentiate both sides and, again, take advantage of the fact that p

remains constant, you will see that the derivatives that are appear

are only the derivatives of q and the derivatives of R.

[NOISE] So p as we've seen before is the position in a body fixed frame.

Q dot on the other hand is the velocity in an inertial frame.

If I pre multiply both sides by the transpose of the rotation matrix

I get a familiar form on the right hand side.

On the left hand side, it's the velocity and the body fix frame, Q dot.

The velocity in the inertia frame is being transformed back to the body fix frame.

And the right hand side, the familiar quantity that I see is R transpose R dot.

And that encodes the angular velocity in a body-fixed frame.

Let's call that omega hat, and

attach a superscript b to denote that it's a body-fixed angular velocity.

Again, it is the angular velocity of the rigid body, but

we've chosen to write the components in a body-fixed frame, which is not shown here.

Another way of writing the equation relating the velocity in

inertial frame and the position in the body-fixed frame is by

rewriting p on the right-hand side in terms of R transposing q.

So now let's take a look at the quantities on both sides of this equation.

Q dot is the velocity in inertial frame.

We've now transformed it back to the body-fixed frame.

But on the righ thand side you see another familiar quantity

R dot R transpose which is a skew symmetric matrix.

This encodes the angular velocity in inertial frame.

We call that omega hat with a superscript s to denote the fact that it is

not in the body fix frame but instead it's a spacial angular velocity.

We've seen before that skewed symmetric matrices encode cross products.

So what you're seeing in these two equations is essentially

our ability to generate velocities

by taking the cross product of an angular velocity vector with a position vector.

In the first equation, it's the angular velocity in the body-fixed frame

which yields the velocity in the body-fixed frame.

In the second equation, it's the angular velocity in the inertial frame

which then yields the velocity in the inertial frame.

So we have two different representations of the same angular velocity vector.

The first one is written in terms of basis vectors on the body-fixed frame,

and the second one is written in terms of basis vectors in inertial frame.

[SOUND] So here's an exercise worth doing.

Let's consider a simple rotation, a rotation about the z axis.

This is easy to visualize, and it's also easy to write down.

Let's take the transpose of this.

And then let's differentiate R with the respect to Time.

So R dot will only depend on the derivative of theta.

So as this angle changes R changes.

And you can write R dot as a function of theta dot by simply pre multiplying

it by a matrix that has mostly zeroes except for cosine theta and sine theta.

So if you do the same computations for this very simple matrix and

derive the expressions for R transpose R dot, and

R dot R transpose you'll discover those two matrices are the same.

And this happens in this very special case where axis of rotation is constant.

In this particular case this symmetric matrix corresponds

to the 0,0,1 vector which is the z axis.

And this is something you should have expected.

If you rotate a rigid body about the z axis and the axis of rotation is constant,

then clearly, the angular velocity vector will also be along the z axis.

What happens if your rotation is obtained by composing two rotations?

In this case the rotations are about the z axis through theta.

About the x axis through phi.

Now if I do the computations I differentiate r.

I pre multiply by r transpose to get the body fixed angle of velocity

I'll get two terms.

One that depends only on the rate of change of Rz and

the second that depends only on the rate of change of Rx.

The same thing is true if I use the spatial angular velocity which is

R dot R transpose I get two terms.

One that depends on the rate of change for Rz.

And the second that depends on the rate of change of Rx.

You can see that these two expressions are different.

In both case they consist of two terms.

One depending on theta dot and the second depending on phi dot but

because the axis of rotation is not fixed the two expressions are different.

The body fixed angular velocity and

the spatial angular velocity have different expressions.

Explore nosso catálogo

Registre-se gratuitamente e obtenha recomendações, atualizações e ofertas personalizadas.

O Coursera proporciona acesso universal à melhor educação do mundo fazendo parcerias com as melhores universidades e organizações para oferecer cursos on-line.

© 2019 Coursera Inc. Todos os direitos reservados.

Baixar na App Store

Baixar no Google Play