MR95015 ROBOTICS

. Dynamic analysis

a. Dynamic model by the Euler-Lagrange formulation

Objectives Modeling the dynamics

Comparison of different approaches

Energy based (Euler-Lagrange) Force and velocity propagation (Newton)

Kinematics deal with position and velocity

Dynamics deal with force and torque

You can go a long way with kinematics simulation

You can do more with dynamics simulation, e.g. maintain

constant pressure of the end effecter

Dynamics vs. Kinematics

Euler-Lagrange

It is based on the kinetic and potential energies in the system

The kinetic and potential energy should be written as function of the generalized coordinates (i.e. minimum number of variables required to specific the movement of the system)

With the kinetic and potential energies it is form the Lagrangian or Lagrange function Then the Lagrange equation is evaluated based on the Lagrangian

Euler-Lagrange

Lagrange steps

Determine generalized coordinates or degrees of freedom (d.o.f.)

Kinetic energy

Potential energy

Substitution into Lagrange equation

Considering the generalized forces

Degree of freedom

Number of independent position variables which

would have to be specified in order to located all

parts of the mechanism.

Generalized coordinates

1 2Suppose that with any dimensions ( , , , ), the position of a given system can be specified, then these dimensions (magnitudes)are called generalized coordinates of the system with d.o.f.,

ss q q qs

s

1 2

and its velocities ( , , , ) are called generalized velocities.

The generalized positions and velocities are the mechanical state of system,i.e. the minimum information required at time to de

sq q q

t

termine the position ofthe system at any .t dt

The force (or torque) applied to a generalized coordinate

Generalized force

Generalized coordinates: examples

1 movement1 d.o.f. 1 generalized coordinate :

x

y

m

l

cos( )a t

Generalized coordinates: examples

1

2

x

y

m1

m2

l1

l21

2

2 movement2 d.o.f.

2 generalized coordinate :

Generalized coordinates: examples

J

2 movement1 d.o.f. 1 generalized coordinates

Generalized coordinates: examples

3 movement2 d.o.f. 2 generalized coordinates

Generalized coordinates: examples

Robot with horizontal plate

7 movements4 d.o.f. 4 generalized coordinates

Kinetic energy

Energy due to motion: rotation and/or translation

Kinetic energy

Position and velocity

2 2 2 2 2

2 2 2 2 2

In a cartesian coordinate system

In cilindric coordinates

In spheric coordinates

v dx dy dz v

v dr r d dz

2 2 2 2 2 2 2 senv dr r d r d

The representation of the position and velocity depends on the coordinate system under consideration

Kinetic energy: example

x

y

m

l

cos( )a t

2 2 2

22

The coordinates of the mass are sin , cosTherefore its kinetic energy is given by

1 1 ( cos ) ( sin )2 2

2

m m

m

m

mx l y l

T mv m l l

mlT

1

2

x

y

m1

m2

l1

l2

Kinetic energy: example

1

2

2

2 2 2

211 1

2 2 2

2 1 2 2

2 1 2 2

2 22

221

The kinetic energy for mass 1 is given by

( ) , 2

For its coordinates , are given by sen sen ,

cos cos

therefore

( )2

2

m

m

m

m m m

mT l

m x yx l l

y l l

mT x y

m l

2 2 21 2 2 1 2 1 2 1 22 cos( )l l l

Kinetic energy and inertial effects (Inertia Moment)

The kinetic energy in previous slides was due to the mass effects

If the rigid body has inertial effects, then they will also generate kinetic energy, the so called inertia moment

The kinetic energy of an inertia is function of the angular velocity, and it is given by

2 21 1 2 2

where: inertia angular or rotational velocity angular position

IT I I

I

Kinetic energy and inertial effects: example

x

y

m

l

cos( )a t

2

2 2

The kinetic energy due to inertia is given by1 2

Therefore the total kinetic energy is12

I

m I

T I

T T T ml I

I

Kinetic energy and inertial effects: example

1

2

x

y

m1

m2

l1

l2

1

2

1 1 2 2

211

222

2 2 21 1 1 1

The kinetic energy due to inertia 1 is

, 2

and for inertia 2

, 2

therefore the total kinetic energy is given by

1 ( )2 2

I

I

m I m I

IT

IT

T T T T T

Im l I

222 2 2 22

1 1 2 2 1 2 1 2 1 2 2 cos( )2m l l l l

Inertia Moment

3 1

In general the angular velocity has 3 components, i.e.

:

Therefore the inertia is a tensor or a matrix of 3 3, that is given by an integral as function of the dens

T

x y z

2 211 12 13

2 221 22 23

2 231 32 33

ity distributiuon, i.e.

( , , ) V

y z xy xz I I II x y z xy z x yz dx dy dz I I I

xz yz x y I I I

Inertia Moment The tensor of inertia can be diagonalized along the principal axes,

thus along the principal axes, the principal moments are 0 0

0 00 0

such that along the p

xx

yy

zz

II I

I

11 12 13

21 22 23

31 32 33

rincipal axes of rotation0 0

0 00 0

x xx x

y yy y

zz

I I I II I I II I I z I z

Mass center

So far the masses have been concentrated at the end of the link

In general there exist the mass center which is the point in which

can be considered that all the mass is concentrated, and moreover

all the particles of the body have cero relative velocity w.r.t. the

mass center

When the mass center is considered, then its position coordinates

must be taken into account to calculate the kinetic energy

Example

1 1 1 1 1 1

2 2 21 1 1 1 1

Mass center coordinates sin( ), cos( )

Thus its kinetic energy is 1 1 2 2

c c c c

c

x l q y l q

T m l q I q

Translational motion: (Newton equation)

Rotational motion: (Euler equation)

More general: (Lagrange) Generalized coordinates + generalized forces Using variational principle

Equations of motion

( )d mvF madt

( )d I I Idt