ME 4135Fall 2011R. R. Lindeke, Ph. D.

Robot Dynamics – The Action of a Manipulator When Forced

We will examine two approaches to this problem

Euler – Lagrange Approach:– Develops a “Lagrangian Function” which relates Kinetic

and Potential Energy of the manipulator thus dealing with the manipulator “As a Whole” in building force/torque equations

Newton – Euler Approach:– This approach tries to separate the effects of each link

by writing down its motion as a linear and angular motion. But due to the highly coupled motions it requires a forward recursion through the manipulator for building velocity and acceleration models followed by a backward recursion for force and torque

Euler – Lagrange approach

Employs a Denavit-Hartenberg structural analysis to define “Generalized Coordinates” as general structural models.

It provides good insight into controller design related to STATE SPACE

It provides a closed form interpretation of the various components in the dynamic model:– Inertia– Gravitational Effects– Friction (joint/link/driver)– Coriolis Forces relating motion of one link to coupling effects

of other link motion– Centrifugal Forces that cause the link to ‘fly away’ due to

coupling to neighboring links

Newton-Euler Approach

A computationally ‘more efficient’ approach to force/torque determination

It starts at the “Base Space” and moves forward toward the “End Space” computing trajectory, velocity and acceleration

Using this forward velocity information it computes forces and moments starting at the “End Space” and moving back to the “Base Space”

Defining the Manipulator Lagrangian:

( , ) ( , ) ( )

( , )

( )

L q q T q q U qhereT q q

U q

Kinetic energy of themanipulator

Potential energy of the manipulator

Generalized Equation of Motion of the Manipulator:

1, ,i i ni i

i

dF L q q L q qdt q q

is a link of the manipulator

Fi is the Generalized Force acting on Link i

Starting Generalized Equation Solution

We begin with focus on the Kinetic energy term (the hard one!)

Remembering from physics: K. Energy = ½ mV2

Lets define for the Center of Mass of a Link ‘K’:

k

k

as Linear Velocity as Angular Velocity

Rewriting the Kinetic Energy Term:

1

,2

T Tn K K K K K K

K

m DT q q

mK is Link Mass DK is a 3x3 Inertial Tensor of Link K about its center of mass

expressed WRT the base frame which characterizes mass distribution of a rigid object

Focusing on DK: Looking at a(ny) link

For this Link: DC is its Inertial Tensor About it Center of Mass

In General:

2 2

2 2

2 2

V V V

C KV V V

V V V

y z dV xy dV xz dV

D m xy dV x z dV yz dV

xz dV yz dV x y dV

Defining the terms:

The Diagonal terms at the “Moments of Inertia” of the link

The three distinct off diagonal terms are the Products of Inertia

If the axes used to define the pose of the center of mass are aligned with the x and z axes of the link defining frames (i-1 & i) then the products of inertia are zero and the diagonal terms form the “Principal Moments of Inertia”

Continuing after this simplification:

2 2

2 2

2 2

0 0

0 0

0 0

V

C KV

V

y z dV

D m x z dV

x y dV

If the Link is a Rectangular Rod (of uniform mass):

2 2

2 2

2 2

0 012

0 012

0 012

C K

b c

a cD m

a b

This is a reasonable approximation for many arms!

If the Link is a Thin Cylindrical Shell of Radius r and length L:

2

2 2

2 2

0 0

0 02 12

0 0 2 12

C K

r

r LD m

r L

Some General Link Shape Moments of Inertia:

From: P.J. McKerrow, Introduction to Robotics, Addison-Wesley, 1991.

We must now Transform each link’s Dc

Dc must be defined in the Base Space To add to the Lagrangian Solution for kinetic energy (we will call it DK):

Where: DK = [0RK*Dc *(0RK)T] Here 0RK is the rotational sub-matrix defining the Link

frame K (at the end of the link!) to the base space -- thinking back to the DH ideas

TK K KD

Defining the Kinetic Energy due to Rotation (contains DK)

0 0

. .2

. .2

TK K K

TTK K C K K

DK E

R D RK E

Completing our models of Kinetic Energy:

Remembering:

1

,2

T Tn K K K K K K

K

m DT q q

Velocity terms are from Jacobians: We will define the velocity terms as parts of a “slightly” –

(mightily) – modified Jacobian Matrix:

AK is linear velocity effect BK is angular velocity effect I is 1 for revolute, 0 for prismatic

joint types

1

1

1 0 1

0 ( )( )

( )0

KK

KK K

K K

c c A qq qJ qB qZ Z

Velocity Contributions of all links beyond K are ignored (this could be up to 5 columns!)

Focusing on in the modified jacobian

This is a generalized coordinate of the center of mass of a link

It is given by: 01 ( )

:

,0,0,1

KK K

K

c H T q chere

c

K

is a vector from frame k(at the end of link K) to the Center of Mass of Link K

land is: 2

Kc

A Matrix that essentially strips off the bottom row of the solution

Re-Writing K. Energy for the ARM:

1

,2

TK K K Kn K K

K

A q m A q B q D B qT q q

Factoring out the Joint Velocity Terms

1

,2

T TT K K T K Kn K K

K

q A m A q q B D B qT q q

1,2

n T TK K K KK K

T K

A m A B D BT q q q q

Simplifies to:

Building an Equation for Potential Energy:

1

1

( ) ( )

( ) ( )

( ) ( )

nT

K KK

n

K KK

T

U q m g c q

g

c q m c q

U q g c q

is acceleration due to gravity andIntroducing a new term:

leads to:

This is a weighted sum of the centers of mass of the links of the manipulator

Generalized coordinate of centers of mass (from earlier)

Finally: The Manipulator Lagrangian:

( , ) ( , ) ( )L q q T q q U q

1, ( )2

n T TK K K KK K

T TK

A m A B D BL q q q q g c q

Which means:

Introducing a ‘Simplifying’ Term D(q):

1

{ }n T TK K K K

K KK

D q A q m A q B q D q B q

1, [ ] ( )2T TL q q q D q q g c q

Then:

Considering “Generalized Forces” in robotics:

We say that a generalized force is an residual force acting on a arm after kinetic and potential energy are removed!?!*!

The generalized forces are connected to “Virtual Work” through “Virtual Displacements” (instantaneous infinitesimal displacements of the joints q), a Displacement that is done without physical constraints of time

Generalized Forces on a Manipulator

We will consider in detail two (of the readily identified three):

Actuator Force (torque) →

Frictional Effects →

Tool Forces →

1TW q

2TW b q q

0ToolF in generalwill be taken

Examining Friction – in detail

Defining a Generalized Coefficient of Friction for a link:

( )Kqv d s d

k K K K K Kb q b q SGN b b b e

C. Viscous Friction

C. Dynamic Friction

C. Static Friction

Combining these components of Virtual Work:

1 2

TW W W b q q

F b q

leads to the manipulator Generalized Force:

Building a General L-E Dynamic Model

Remembering:

1, ,i i ni i

dF L q q L q qdt q q

i

is a link of manipulator

Starting with this term

Partial of Lagrangian w.r.t. joint velocity

, ,

i

L q q T q qq q

1

n

ij jj

D q q

It can be ‘shown’ that this term equals:

Completing the 1st Term:

1

, n

ij jji

L q qd d D q qdt q dx

This is found to equal:

Completing this 1st term of the L-E Dynamic Model:

1 1 1

n n nij

ij j k jj k j k

D qD q q q q

q

Looking at the 2nd Term:

, ,i i i

L q q T q q U qq q q

31 1

1

( )

( )2

n nkj

k j nk j ji

k j kik j i

D qq q

qg m A q

This term can be shown to be:

Before Summarizing the L-E Dynamical Model we introduce:

A Velocity Coupling Matrix (4x4)

A ‘Gravity’ Loading Vector (nx1)

1 1 , ,2ikj ij kj

k i

C q D q D q i j k nq q

for

3

1

nj

i k j kik j i

h q g m A q

The L-E (Torque) Dynamical Model:

1 1 1

n n ni

i ij j kj k j i ij k j

D q q C q q q h q b q

Inertial Forces

Coriolis & Centrifugal

Forces

Gravitational Forces Frictional

Forces