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Newton-Euler Equations

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Robotics: Aerial Robotics

Université de Pennsylvanie

4.5 (2,058 notes) | 81K étudiants inscrits

Cours 1 de 6 dans Robotique Spécialisation

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.

Visualiser le programme de cours

Compétences que vous apprendrez

Motion Planning, Robotics, Quadcoper, Matlab

Avis

4.5 (2,058 notes)

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CG

Feb 12, 2016

Well balanced mix of theory and practical applicability. Explanation of the material is also very good.\n\nThe assignments are nicely built on the taught material to stimulate understanding.

AJ

Nov 22, 2017

Exceptional Material, not only are the concept are explained well, the supplementary material are provided for all possible requirement along side. Advance course with minimum prerequisite

À partir de la leçon

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.

Formulation3:18

Newton-Euler Equations6:05

Principal Axes and Principal Moments of Inertia6:03

Quadrotor Equations of Motion5:13

Supplementary Material: State-Space Form4:32

Supplementary Material: Getting Started With the First Programming Assignment3:22

Enseigné par

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

Transcription

Let's remind ourselves what these equations look like for a single particle. If you have a single particle of mass m, and you know the resultant force acting on this particle. Then the acceleration is just obtained by Newton's second law, the total force equals mass times acceleration a. What does this equation look like when you get to a system of particles and then a rigid body? Well, let's start with a system of particles. Before we can define Newton's Second Law for a system of particles, we must first define the center of mass. The picture on the right hand side shows you a set of particles. P1, P2, and so on. Let's assume that we can write position vectors P1, P2, Pi, for each of these particles. Let's denote the mass of the particle by M sub I. We can now compute the average position vector, by weighting each position vector with the appropriate mass. So the equation you see here, mi times pi, is essentially the weighted sum of all the position vectors. You divide that by the total mass, m, and that gives you a new position vector. So this position vector defines the center of mass. It turns out that the center of mass for a system of particles, S, behaves in exactly the same way as a single particle would have behaved. If it had been located at the center of mass. So instead of writing f equals ma for a system of particles, you would write f equals m times the acceleration of the center of mass. And that's essentially what you see in this equation. You take the net force, sum up all the individual forces to get F. And that is equated then to the total mass M times the derivative of the velocity of the center of mass. Here the notation requires a little bit of explanation. The superscript A refers to the fact that you're computing all these quantities in an inertial frame A. The superscript C, refers to the fact that you're computing the velocity of the center of mass. And remind yourself that the right hand side is essentially the rate of change of linear momentum. So if I call L the Linear Momentum, then F the total force is just equal to the Rate of change of Linear Momentum. This is Newton's second law for a system of particles. Once again the derivative of each of these quantities, must be performed in an inertial frame in order for this equation to be valid. And by the way, this is also true for a rigid body. If you think about it, a rigid body is nothing but a set of infinite particles, all glued together rigidly. So if its valid for a set of particles, it must also be valid for a collection of infinite particles. So that was Newton's second law. What's the equivalent for rotational motions? Let's derive the rotational equations of motion for a rigid body. For linear motion we consider the linear momentum of the rigid body and it's derivative. For rotational motion, the analogous quantity to consider is the angular momentum. So if you look at this equation, H is the angular momentum of the rigid body with the origin C, the center of mass, in the inertial frame A. So we want to compute the rate of change of angular momentum of the rigid body B, and that equals the net moment applied to the rigid body. Once again the differentiation must be done in an inertial frame. In order to perform the calculation on the left hand side, we have to replace H by something we can easily measure. It turns out that the angular momentum is nothing but the inertia times the angular velocity. Now, these computations have to be done in three dimensions. The angular momentum is a three dimensional vector. The angular velocity is a three dimensional vector. The quantity in between is the inertial tensor, whose components can be written as a three by three matrix. Here the subscript c denotes the fact that you measured the components with c as the origin. Remember c is the center of mass. The angular velocity is also obtained in the inertial the leading superscript A captures this. And the trailing superscript B, tells you this is the angular velocity of the rigid body B that you are measuring. Finally, N is the net moment. Take all the external forces, compute their moments, and add these moments to all external couples or torques. Once again. The trailing subscript C tells you that you're computing moments with C as an origin.

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