WHEEL MOBILE ROBOTS
LECTURE– 02 - KINEMATICS
dr inż. Tomasz Buratowski
Faculty of Mechanics and Robotics
Department of Robotics and Mechatronics
• Kinematics of arbitrarily selected point belonging to a
particular section of the system can be analyzed with the use of so-
called kinematic equations
• Each movable part must be connected with
Mobile robots - kinematics
2
Kinematics modeling of the two-wheels
mobile robot
3
Robot consists of:
•
chassis (4)
•
drive system (1&2)
•
idler wheel (3)
Kinematics modeling of the two-wheels
mobile robot
4
Proper coordinate system was connected with particular parts
of the model:
• chassis (4) and x
4
y
4
z
4
coordinate system (CS) with the
origin in the mass center of the part
• x
1
y
1
z
1
and x
2
y
2
z
2
CSs are connected with driving systems
with origins in points B and C
• x
0
y
0
z
0
CS is stationary coordinate system and is base
frame of reference
Kinematic description of the model is based on equations
for characteristic points of the robot and assumes that it
travels with constant velocity of the point A V
A
Kinematics modeling of the two-wheels
mobile robot
Kinematic equation for point H has form of:
5
H
H
H
T
r
,
0
,
4
Kinematics modeling of the two-wheels
mobile robot
6
Transformation matrix of the frame of
reference x
4
y
4
z
4
to x
0
y
0
z
0
for point H has
form of:
1
0
0
0
1
0
0
)
sin(
0
)
cos(
)
sin(
)
cos(
0
)
sin(
)
cos(
1
3
3
,
0
,
4
r
l
y
l
x
T
A
A
H
1
0
0
4
l
H
Positioning vector of the point H with
reference to the x
4
y
4
z
4
has form of:
Kinematics modeling of the two-wheels
mobile robot
After evaluation of the formula
we obtained equation of motion of the point
H written in two equivalent forms:
H
H
H
T
r
,
0
,
4
1
)
sin(
)
cos(
4
1
3
3
l
r
l
y
l
x
r
A
A
H
1
)
sin(
)
cos(
1
4
1
3
3
l
r
l
y
l
x
z
y
x
A
A
H
H
H
7
Kinematics modeling of the two-wheels
mobile robot
In the next step the equation for the circular
path are presented:
4
1
))
cos(
1
(
)
sin(
l
r
z
R
y
R
x
H
H
H
After comparison of the equation of motion
with path formulae we get:
)
sin(
)
cos(
1
(
)
cos(
)
sin(
3
3
l
y
R
l
x
R
A
A
8
Kinematics modeling of the two-wheels
mobile robot
Differentiating with respect to time equations
for point H
)
sin(
)
cos(
1
(
&
)
cos(
)
sin(
3
3
l
y
R
l
x
R
A
A
we obtain equation for speed of the point H
along x and y axis
0
)
cos(
)
sin(
3
R
l
x
A
0
)
sin(
)
cos(
3
R
l
y
A
9
Kinematics modeling of the two-wheels
mobile robot
Next characteristic point is point A for which
constant speed was assumed and projection
of speed vectors on to X and Y axis of the
base reference frame was made.
0
)
cos(
A
A
v
x
0
)
sin(
A
A
v
y
10
Kinematics modeling of the two-wheels
mobile robot
To fully describe parameters of motion of our
structure we need angular velocities and
rotation angles of the wheels (1&2).
Those velocities are derived from equations
for velocity of the tangent points between
wheel and ground (assuming no slip)
K
K
K
T
T
r
4
,
1
,
0
,
4
for wheel 1:
11
Kinematics modeling of the two-wheels
mobile robot
Transformation matrix of the frame of
reference x
4
y
4
z
4
to x
0
y
0
z
0
for point K has
form of:
1
0
0
0
1
0
0
)
cos(
0
)
cos(
)
sin(
)
sin(
0
)
sin(
)
cos(
1
1
1
,
0
,
4
r
l
y
l
x
T
A
A
K
12
Kinematics modeling of the two-wheels
mobile robot
Transformation matrix of the frame of
reference x
1
y
1
z
1
to x
4
y
4
z
4
for point K has
form of:
1
0
0
0
0
0
)
sin(
)
cos(
0
1
0
0
0
0
)
cos(
)
sin(
1
1
1
1
4
,
1
T
13
Kinematics modeling of the two-wheels
mobile robot
Positioning vector for the point K with
respect to x
1
y
1
z
1
reference frame has form
of:
1
0
)
sin(
)
cos(
1
1
1
1
r
r
K
After differentiating of equation of motion for
the point K we get:
K
K
K
T
T
v
4
,
1
,
0
,
4
14
Kinematics modeling of the two-wheels
mobile robot
After substitution of all part of the equation
of motion for point K we get equations for
velocities for point K
0
0
)
sin(
)
sin(
)
cos(
)
cos(
0
1
1
1
1
1
1
l
r
y
l
r
x
z
y
x
A
A
K
K
K
15
Kinematics modeling of the two-wheels
mobile robot
Having in mind the fact that V
K
= 0 (no-slip
condition) in scalar form we get:
)
cos(
)
cos(
1
1
1
l
r
x
A
)
sin(
)
sin(
1
1
1
l
r
y
A
16
Kinematics modeling of the two-wheels
mobile robot
For the second wheel we proceed in similar
manner.
The equation for motion of the point L
(tangent point between 2 wheel and the
ground) was derived:
L
L
L
T
T
r
4
,
2
,
0
,
4
17
Kinematics modeling of the two-wheels
mobile robot
Transformation matrix of the frame of
reference x
4
y
4
z
4
to x
0
y
0
z
0
for point L has
form of:
1
0
0
0
1
0
0
)
cos(
0
)
cos(
)
sin(
)
sin(
0
)
sin(
)
cos(
2
1
1
,
0
,
4
r
l
y
l
x
T
A
A
L
18
Kinematics modeling of the two-wheels
mobile robot
Transformation matrix of the frame of
reference x
2
y
2
z
2
to x
4
y
4
z
4
for point L has
form of:
1
0
0
0
0
0
)
sin(
)
cos(
0
1
0
0
0
0
)
cos(
)
sin(
2
2
2
2
4
,
2
T
19
Kinematics modeling of the two-wheels
mobile robot
Positioning vector for the point L with respect
to x
2
y
2
z
2
reference frame has form of:
After differentiating of equation of motion for
the point L we get:
1
0
)
sin(
)
cos(
2
2
2
2
r
r
L
L
L
L
T
T
v
4
,
2
,
0
,
4
20
Kinematics modeling of the two-wheels
mobile robot
After substitution of all part of the equation
of motion for point L we get equations for
velocities for point L
0
0
)
sin(
)
sin(
)
cos(
)
cos(
0
1
2
2
1
2
2
l
r
y
l
r
x
z
y
x
A
A
L
L
L
21
Kinematics modeling of the two-wheels
mobile robot
Having in mind the fact that V
L
= 0 (no-slip
condition) in scalar form we get:
)
cos(
)
cos(
1
2
2
l
r
x
A
)
sin(
)
sin(
1
2
2
l
r
y
A
22
Kinematics modeling of the two-wheels
mobile robot
Summing up, we obtained system of
equations describing motion of the two-
wheeled robot:
0
3
3
2
2
2
5
r
v
l
A
calculation of:
• displacement
• velocity
• acceleration
• angle of rotation
• angular velocities
• solving simple and inverse
kinematic problems
calculation of:
• kinematic parameters of
idler wheel
23
0
)
cos(
)
cos(
0
)
cos(
)
cos(
0
)
sin(
0
)
cos(
0
)
sin(
)
cos(
0
)
cos(
)
sin(
1
2
2
1
1
1
3
3
l
r
x
l
r
x
v
y
v
x
R
l
y
R
l
x
A
A
A
A
A
A
A
A
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
Based on previosly derived equations of
motion for our robot we conduct simulation
of inverse kinematic problem for proposed
path:
24
Start-up
Straight line movement
Arc movement
Deceleration and stop
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
We assume working conditions and physical
features of the robot:
• Point A velocity v
A
=0.3 [m/s],
• Start-up 2 [s],
• Straight line movement 0.5 [s],
• Acr movement 4.7 [s] with R=1.5 [m],
• Angle of chassis rotation 0
-54
,
• Straight line movement 0.5 [s],
• Deceleration and stop 2 [s].
25
l
1
[m]
l
3
[m]
l
4
[m]
l
5
[m]
r
1
[m]
r
2
[m]
r
3
[m]
0.163
0.07
0.2
0.07
0.0825
0.0825
0.035
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
26
Speed and trajectory generator model
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
27
Output from speed and trajectory generator model
v
A
[m/s]
Sygnał o trajektorii
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
28
Model of inverse kinematic solver (as input there is trajectory from
previously discused model)
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
29
Output for the angle of rotation of the chassis and its angular velocity
[rad]
[rad/s
]
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
30
Output for the angle of rotation of wheels 1,2 and 3
1
[rad]
2
[rad]
3
[rad]
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
31
Output for the angular speed of wheels 1,2 and 3
1
[rad/s]
2
[rad/s]
3
[rad/s]
Simulation of the simple kinematic
problem with the use of Matlab Simulink
software
32
Based on the results from inverse kinematic
problem we can check correctness of our results
by plotting the path as a result of simple kinematic
problem.
fig. Model solving simple kinematics problem
Simulation of the simple kinematic
problem with the use of Matlab Simulink
software
33
Fig. Output from simple kinematics solver - path
As a result we obtain previosly defined path for inverse
kinematics problem. That means our model is made in a
proper way.
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software (straight line path)
34
Start-up
Straight line movement
Deceleration and stop
Lets assume different path
– straight line
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
35
Output from speed and trajectory generator model
v
A
[m/s]
trajectory signal
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
36
Output for the angle of rotation of wheels 1,2 and 3
1
[rad]
2
[rad]
3
[rad]
Simulation of the inverse kinematic
problem with the use of Matlab Simulink
software
37
Output for the angular speed of wheels 1,2 and 3
Simulation of the simple kinematic
problem with the use of Matlab Simulink
software
38
Fig. Output from simple kinematics solver - path
As a result we obtain previosly defined path for inverse
kinematics problem. That means our model is made in a
proper way.
THANK YOU
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