T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Mobile Robot
Kinematic Structure for Control Issues
Lecture 2
Jeong-Yean Yang
2020/10/22
1
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Wheel from Stone Age
• Maya did NOT use wheels.
– Even pulley and tools.
2
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Two Wheeled Robot
3
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Two wheeled robot has some problems
4
• Falling down…
• Why it is unstable?
Stability Problem with wheels
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Two wheeled robot has some problems
• It is stable, but angular position control has some
noise.
• Why?
5
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Two Wheeled Robot?
No, One wheel Spherical Robot
• Balance control is required.
• But, very fast and low power consumption
6
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Spherical Robot
7
• Tilt feedback is required.
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
For Mechanics Analysis
• F is derived from Linear Momentum
• Moment, M is derived from Angular Momentum
– H= r x L = r x mv ( x is a cross product)
• Moment is often called Torque.
8
0
0
lim
lim
t
t
L
m v
ma
F
t
t
0
0
lim
lim
(
= )
t
t
H
m v
r
r ma
r F
M or
T
t
t
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Statics Equilibrium
• Force equilibrium
– Sum of all external forces should be zero
– If all force sum is zero, there is no movement by an
acceleration
• Moment equilibrium
– Sum of all external moments should be zero
– If moment is zero, there is NO rotation.
9
0
F
ma
0
M
r F
0
F
ma
M
r F
I
dynamics
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Angular Velocity and Acceleration
• Angle
• Angular velocity
• Angular Acceleration
10
I
I
I
( )
t
( )
d
t
dt
2
2
( )
d
t
dt
( )
x
x t
v
x
a
x
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
V = w x r
• V = =w x r (Cross product)
• A =
11
x
y
z
r
w
0
0
0
x
y
z
x
y
x
y
z
x
y
i
j
k
i
j
k
w
w
w
r
j
r
i
r
r
r
r
r
𝑟𝑥𝜛
𝑟𝑦𝜛
dr
dt
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
r
r
r
x
re
x
re
r e
x
r
r
e
r
r
e
Remind
ˆ
ˆ r=const.
ˆ
0
ˆ
0
r
x
re
x
r e
x
r
e
2
2
d r
r
dt
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Moment of Inertia, I
• What is Momentum?
– Conservation of Momentum: L = mv = const.
– Time Differentiation of L
–
– Angular Momentum, H = ?
– Time Differentiation of H=?
12
dL
dmv
dv
L
m
ma
dt
dt
dt
H
r L
r mv
m r v
m r
w r
=m
d
d
dr
dv
H
r L
r mv
v
r
dt
dt
dt
dt
Too complex.. T_T
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Angular Moment of Inertia
(for Rigid Body)
• Remind
• Because it is a Rigid body,
– r is a constant value.
• Simplification
13
=m
d
dr
dv
H
r L
r mv
v
r
dt
dt
dt
0
r
=
d
dv
dw
H
r L
r mv
r m
mr
r
mr
r
dt
dt
dt
i i
i
i
H
m r
r
Torque on Rigid body.
Remind that all is
i
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Angular Momentum of Inertia
• Rotation of Particles on Origin.
14
i i
i
i
H
m r
r
2
0
0
(
)
0
0
0
0
0
0
r
r
i
i
i
i
i
i
i
i
i
e
e
k
e
e
k
r
r
r
r
r
r
k
r
r
r
2
2
2
2
i i
i
m
i i
i
m
H
m r
r dm
I
I
m r
r dm
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Wheel Dynamics
15
I,
m,a
=
T
w
N
f
2
2
1)
2)
0
1
3)
2
1
2
x
y
c
ma
f
ma
W
N
I
mr
T
rf
mr
T
rf
o
2
2
2
4)
1
2
2
3
x
a
r
mr
f
mr
T
mr
T
mr
When T=const., if r then but w
𝛼
vs
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Dynamic Model of Two wheeled Robot
16
I,M
m
+
=
I,M
m
+
mg
Mg
c
M
r F
I
0
y
y
F
ma
mg
Mg
R
f
R
I:rotating inertia, L: length, m, M: each mass
f: friction, R: reaction force, r: radius of disc
L
'
(
)
x
x
F
ma
Ma
ma
Mr
m
r
L
f
r
2
c
I
mL
rf
o
c
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
System Dynamics: Two wheeled Robot
• If we cancel f with eq1,
17
2
2
*
*
* 2
(
)
0
0
( )
1
( )
T( )
dynamics with Two Poles
c
I
mL
Mr
mr r
L
I
I
T
Laplace Transform
s
G s
s
I s
system
Double
Pole.
Remind that
Poles on jw axis mean
Marginal stable
Zero because there is no Torque Input
Add Torque Input
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Double Pole has No Damping
18
mx
F
Given system even without Damping
and Feedback control
d
p
d
e
x
x
F
K e
K e
PD Feedback Control
(
)
(
)
p
d
p
d
d
d
d
p
p
d
p
d
mx
K e
K e
K
x
x
K
x
x
mx
K x
K x
K x
mx cx
kx
F
K x
D control changes system dynamics
(Stiffness, Damping)
Remind that PD does not change on
inertia mx
What is Double Pole? Remind mx’’=F
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
19
Disturbances on Marginal Stable System
I,M
m
+
=
I,M
m
+
mg
Mg
f
R
L
r
o
D<<1
D is very small.
'
(
)
x
x
F
ma
Ma
ma
Mr
m r
L
D
f
o
M
r F
I
2
c
I
mL
rf
DL
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
20
'
(
)
x
x
F
ma
Ma
ma
Mr
m r
L
f
D
2
c
I
mL
rf
DL
*
(
)
I
L
r D
Without no Torque Input
*
* 2
( )
1
( )
T( )
I
T
s
G s
s
I s
Stability of Torque Input
Stability of Disturbances
*
* 2
(
)
( )
(
)
( )
I
L
r D
s
L
r
D s
I s
( )
( )
T( )
( )
s
s
s
D s
Disturbances at L+r
is bigger than
Torque input
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
L
Self Balancing with Small Movement
21
I,M
+
=
Mg
f
R
L
r
o
mg
Small
disturbance
D
q
L
3
1
sin
(0
.....) by Taylor series.
3!
q<<1
sin
L
q
L
q
q
if
L
q
Lq
*
*
* 2
2
sin
2
2
Q( )
D(s)
M
I
D r
mgL
q
I
I q
mgLq
D r
r
s
I s
mgL
Small Movement Linear Assumption
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Root Locus of Self Balancing
22
* 2
Q( )
2
D(s)
s
r
I s
Lmg
wj
When w=0 ( no velocity),
It is very stable
When w is increasing. ( oscillation)
It is stable, too.
But, a Roly-Poly can fall with Strong force
What is wrong in our model?...
Think the assumption that sin q = q
If q >> 1, the assumption fails.
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Self Balancing with Large Movement
23
*
*
* 2
2
2
* 4
* 2
2
sin
sin
2
Q( )
2
1
D( )
D( )
1
Q( )
M
I
D r
mgL
q
I
I q
mgL
q
D r
L
s
I s
mg
r
s
s
s
s
s
I s
I s
mgL
rlocus(tf([1 0 1], [ 10 0 10 0 5]));
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Real Axis
Im
a
g
A
x
is
If w increase, system becomes oscillatory.
T&C LAB-AI
Dept. of Intelligent Robot Eng. MU
Robotics
Self Balancing
with Small or Large Movement
24
*
*
* 2
2
sin
2
2
Q( )
D(s)
M
I
D r
mgL
q
I
I q
mgLq
D r
r
s
I s
mgLs
Small Movement =
Linear Assumption
Always
Stable!
Large Movement
= Non linear Eq.
*
*
2
* 4
* 2
2
sin
sin
2
D( )
1
Q( )
M
I
D r
mgL
q
I
I q
mgL
q
D r
s
s
s
I s
I s
mgL
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Real Axis
Im
a
g
A
x
is
Stability wrt
Inputs
