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Ingeniería mecatrónicaproblemas a resolverTRANSCRIPT
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MR95015 ROBOTICS
. Dynamic analysis
a. Dynamic model by the Euler-Lagrange formulation
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Objectives Modeling the dynamics
Comparison of different approaches
Energy based (Euler-Lagrange) Force and velocity propagation (Newton)
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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
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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
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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
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Degree of freedom
Number of independent position variables which
would have to be specified in order to located all
parts of the mechanism.
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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
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The force (or torque) applied to a generalized coordinate
Generalized force
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Generalized coordinates: examples
1 movement1 d.o.f. 1 generalized coordinate :
x
y
m
l
cos( )a t
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Generalized coordinates: examples
1
2
x
y
m1
m2
l1
l21
2
2 movement2 d.o.f.
2 generalized coordinate :
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Generalized coordinates: examples
J
2 movement1 d.o.f. 1 generalized coordinates
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Generalized coordinates: examples
3 movement2 d.o.f. 2 generalized coordinates
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Generalized coordinates: examples
Robot with horizontal plate
7 movements4 d.o.f. 4 generalized coordinates
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Kinetic energy
Energy due to motion: rotation and/or translation
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Kinetic energy
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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