Monash University
Final Examination
MEC4456/TRC4800
Robotics
There are 6 questions for a total of 85 marks. Attempt all questions. If you need to make
assumption, please write it down clearly. An equation sheet is attached to the end of this
paper.
1. (7 marks) In a 6-DOF robotic system, a camera is attached to the fifth link of the robot.
The sixth link is the end-effector. The camera observes an object and determines its frame
relative to the camera’s frame. Four frames,F
5
,F
E
,F
cam
andF
obj
, are attached to the
fifth link, the end-effector, the camera, and the object, respectively.
5
T
cam
=
00−1 2
0−101
−1006
0001
,
5
T
E
=
0−1 0 0
100 2
001 5
000 1
,
cam
T
obj
=
0 0 1 2
1 0 0 3
0 1 0 5
0 0 0 1
Using the above information, determine
(a) the transformation matrix betweenF
E
andF
obj
,
5 marks
(b) the distance between the end-effector and the object, i.e., the distance between the
origins ofF
E
andF
obj
.
2 marks
Figure 1: Shoulder and Elbow Joints of Baxter Robot
2. (20 marks) The schematic diagram of the left shoulder and elbow of Baxter robot is shown
in Fig 1.F
0
is the ground frame. The axes of rotation, i.e. the z-axes of Joints 1 to 4 are
as illustrated. At the home position, the axes of rotation of Joints 2 and 4 are alongy
0
direction. The axis of rotation of Joint 3 is alongx
0
direction. For the shoulder and elbow
joints:
(a) Assign coordinate frames in Fig. 1.
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Monash University
Final Examination
MEC4456/TRC4800
Robotics
5 marks
(b) Find the DH parameters.
8 marks
(c) Find transformation matrix
3
4
Tand its inverse
4
3
T.
3 marks
(d) Find the numerical transformation matrix
0
4
Tby assumingθ
1
= 90
◦
,θ
2
= 45
◦
,θ
3
=
−45
◦
, andθ
4
= 90
◦
.
4 marks
3. (12 marks) In deriving the solution of the inverse kinematics of a general 3-dof robot, the
following equations are formulated
Γ
1
2
Γ
2
2
+ 3Γ
1
2
Γ
2
−4Γ
1
Γ
2
2
+ 4 = 0
2Γ
1
2
Γ
2
2
+ Γ
1
2
Γ
2
−3Γ
1
Γ
2
2
+ 1 = 0
Please reduce the above two equations into a univariant polynomial in Γ
1
4. (10 marks) Use the linear function with parabolic blends to generate a smooth trajectory
fromx= 0m att= 0s tox= 0.9m att= 1s. The required acceleration in the blend region
is ̈x= 4.8m/s
2
. Write the equations for the whole trajectory.
Figure 2: PR Manipulator
5. (26 marks) Consider a PR manipulator as shown in Fig. 2. The transformation matrices
are given as follows
0
1
T=
1 0 00
0 1 00
0 0 1d
1
0 0 01
,
1
2
T=
c
2
−s
2
00
00−1 0
s
2
c
2
00
0001
,
2
3
T=
1 0 00
0 1 0L
2
0 0 10
0 0 01
The mass centres of links 1 and 2 are located in the middle of the links. Both inertia tensors
contain the principle inertia terms (I
xx
, I
yy
and I
zz
) only. The principle frame of each link
Page 3 of 8Please go on to the next page. . .
Monash University
Final Examination
MEC4456/TRC4800
Robotics
is parallel to the DH frame and located at the mass centre. Use the following gravitational
acceleration g for analytical and numerical solutions.
0
g=
−g
0
0
=
−10
0
0
Conduct outward propagation up to frame F
2
and inward propagation.
(a) Derive the analytical solution for outward propagation and show all velocities, accel-
erations, inertia forces, and inertia moments.
10 marks
(b) With the analytical solution for outward propagation, compute the numerical solution
with the following parameters:d
1
= 1 m,L
2
= 0.5 m,v
1
= 0,ω
2
= 0, ̇v
1
= 1, ̇ω
2
= 1,
m
1
= 10 kg,m
2
= 5 kg,I
1xx
= 0.05 kgm
2
,I
1yy
= 0.01 kgm
2
,I
1zz
= 0.2 kgm
2
,
I
2xx
= 0.05 kgm
2
,I
2yy
= 0.01 kgm
2
,I
2zz
= 0.1 kgm
2
.
3 marks
(c) Suppose an external force
3
f
3
(along positivex
3
axis) is applied at the tip of the
manipulator; all other components of the external force and moment are zero. Derive
the analytical solution for inward propagation and present all relevant elements (f,n,
τ).
10 marks
(d) With the analytical solution for inward propagation, compute the numerical solution
with the following parameters:d
1
= 1 m,L
2
= 0.5 m,θ
2
=−π/6,v
1
= 0,ω
2
= 0,
̇v
1
= 1, ̇ω
2
= 1,m
1
= 10 kg,m
2
= 5 kg,I
1xx
= 0.05 kgm
2
,I
1yy
= 0.01 kgm
2
,I
1zz
= 0.2
kgm
2
,I
2xx
= 0.05 kgm
2
,I
2yy
= 0.01 kgm
2
,I
2zz
= 0.1 kgm
2
,
3
f
3
= 1 N.
3 marks
6. (10 marks) The dynamics of a robot in joint space is given by
u=M(θ)
̈
θ+C(θ,
̇
θ)
̇
θ+B
̇
θ+G(θ) +J
T
(θ)F
e
whereJis the Jacobian matrix.uandF
e
are joint torques and force on the end-effector,
respectively. Derive the dynamics of this robot in task space.
END OF EXAM
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Monash University
Final Examination
MEC4456/TRC4800
Robotics
EQUATION SHEET
Rotation matrices
R
X
(θ) =
100
0cθ−sθ
0sθ cθ
R
Y
(θ) =
cθ0sθ
010
−sθ0cθ
R
Z
(θ) =
cθ−sθ0
sθ cθ0
001
Transformation matrix
i−1
i
T=R
X
(α
i−1
)D
X
(a
i−1
)R
Z
(θ
i
)D
Z
(d
i
)
i−1
i
T=
cθ
i
−sθ
i
0a
i−1
sθcα
i−1
cθcα
i−1
−sα
i−1
−sα
i−1
d
i
sθsα
i−1
cθsα
i−1
cα
i−1
cα
i−1
d
i
0001
Products
Let
a=a
x
i+a
y
j+a
z
k
b=b
x
i+b
y
j+b
z
k
then
a.b=a
x
b
x
+a
y
b
y
+a
z
b
z
a×b= (a
y
b
z
−a
z
b
y
)i+ (a
z
b
x
−a
x
b
z
)j+ (a
x
b
y
−a
y
b
x
)k
Jacobian matrix
For revolute joint i
0
J
i
=
[
0
Z
i
×(
0
P
E
−
0
P
iORG
)
Z
i
]
For prismatic joint i
0
J
i
=
[
0
Z
i
0
]
Lagrange dynamics
The Lagrangian
L(q, ̇q) =k(q, ̇q)−u(q)
The equation of motion
d
dt
∂L
∂ ̇q
−
∂L
∂q
=τ
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Monash University
Final Examination
MEC4456/TRC4800
Robotics
Polynomial for trajectory generation
For a quintic polynomial
x(t) =a
0
+a
1
t+a
2
t
2
+a
3
t
3
+a
4
t
4
+a
5
t
5
where att= 0
x(0) =x
0
̇x(0) = ̇x
0
̈x(0) = ̈x
0
and att=t
f
x(t
f
) =x
f
̇x(t
f
) = ̇x
f
̈x(t
f
) = ̈x
f
then the coefficients of the polynomial can be obtained as
a
0
=x
0
a
1
= ̇x
0
a
2
=
̈x
0
2
a
3
=
−20x
0
+ 20x
f
−(12 ̇x
0
+ 8 ̇x
f
)t
f
−(3 ̈x
0
− ̈x
f
)t
2
f
2t
3
f
a
4
=
30x
0
−30x
f
+ (16 ̇x
0
+ 14 ̇x
f
)t
f
−(3 ̈x
0
−2 ̈x
f
)t
2
f
2t
4
f
a
5
=
−12x
0
+ 12x
f
−(6 ̇x
0
+ 6 ̇x
f
)t
f
−( ̈x
0
− ̈x
f
)t
2
f
2t
5
f
Trigonometry
sin(A+B) = sinAcosB+ cosAsinB
cos(A+B) = cosAcosB−sinAsinB
and
C
2
=A
2
+B
2
−2ABcos(θ)
when A, B and C are the three edges of a triangle andθis the angle between edges A and B.
Homogeneous transformation matrix
A
B
T=
[
A
B
R
A
p
Borg
0
T
1
]
A
B
T
−1
=
B
A
T=
[
A
B
R
T
−
A
B
R
T A
p
Borg
0
T
1
]
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Monash University
Final Examination
MEC4456/TRC4800
Robotics
Parabolic blends
θ
b
=θ
0
+
1
2
̈
θt
2
b
t
b
=
t
2
−
√
̈
θ
2
t
2
−4
̈
θ(θ
f
−θ
0
)
2
̈
θ
Definition of rotation matrix
A
B
R=
[
A
x
B
A
y
B
A
z
B
]
=
x
B
·x
A
y
B
·x
A
z
B
·x
A
x
B
·y
A
y
B
·y
A
z
B
·y
A
x
B
·z
A
y
B
·z
A
z
B
·z
A
Fixed angles representation
A
D
R
XY Z
(γ,β,α) =
cαcβ cαsβsγ−sαcγ cαsβcγ+sαsγ
sαcβ sαsβsγ+cαcγ sαsβcγ−cαsγ
−sβcβsγcβcγ
Euler angles representation
A
D
R
XY Z
(α,β,γ) =
cαcβ cαsβsγ−sαcγ cαsβcγ+sαsγ
sαcβ sαsβsγ+cαcγ sαsβcγ−cαsγ
−sβcβsγcβcγ
Angular velocity and cross-product matrix:
Ω
Ω
Ω≡
̇
Q Q
T
=
0−ω
z
ω
y
ω
z
0−ω
x
−ω
y
ω
x
0
=
ω
x
ω
y
ω
z
×
Propagation of Positionp
i
i
p=
i
i
O
i+1
+
i
i+1
p=
i
i
O
i+1
+
i
i+1
R
i+1
i+1
p
Velocity propagation
i+1
ω
ω
ω
i+1
=
i+1
i
Q
i
ω
ω
ω
i
+
̇
θ
i+1
i+1
z
i+1
i+1
v
i+1
=
i+1
i
Q(
i
ω
ω
ω
i
×
i
i
p
i+1
+
i
v
i
)
Jacobian
̇
f
1
̇
f
2
.
.
.
̇
f
m
=
∂f
1
∂x
1
∂f
1
∂x
2
...
∂f
1
∂x
n
∂f
2
∂x
1
∂f
2
∂x
2
...
∂f
2
∂x
n
.
.
.
.
.
.
.
.
.
.
.
.
∂f
m
∂x
1
∂f
m
∂x
2
...
∂f
m
∂x
n
̇x
1
̇x
2
.
.
.
̇x
n
Parallel-axis theorem of inertia tensor:
A
I=
C
I+m[p
T
c
p
c
I
3
−p
c
p
T
c
]
Newton-Euler Laws:
f=m
̇
v
c
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Monash University
Final Examination
MEC4456/TRC4800
Robotics
n=I
c
̇
ω
ω
ω+ω
ω
ω×I
c
ω
ω
ω
Dynamic outward iterations:
i+1
0
ω
ω
ω
i+1
=
i+1
i
R
i
0
ω
ω
ω
i
+
̇
θ
i+1
i+1
z
i+1
i+1
0
̇
ω
ω
ω
i+1
=
i+1
i
R
i
0
̇
ω
ω
ω
i
+ (
i+1
i
R
i
0
ω
ω
ω
i
)×(
̇
θ
i+1
i+1
z
i+1
) +
̈
θ
i+1
i+1
z
i+1
i+1
0
̈
O
i+1
=
i+1
i
R(
i
0
̈
O
i
+
i
0
̇
ω
ω
ω
i
×
i
i
O
i+1
+
i
0
ω
ω
ω
i
×(
i
0
ω
ω
ω
i
×
i
i
O
i+1
))
i
0
̈
p
ci
=
i
0
̈
O
i
+
i
0
̇
ω
ω
ω
i
×
i
i
p
ci
+
i
0
ω
ω
ω
i
×(
i
0
ω
ω
ω
i
×
i
i
p
ci
)
i+1
F
i+1
=m
i+1
i+1
̇v
ci+1
i+1
N
i+1
=
i+1
I
Ci+1
i+1
̇ω
i+1
+
i+1
ω
i+1
×
i+1
I
Ci+1
i+1
ω
i+1
For prismatic joint
i+1
̇
ω
ω
ω
i+1
=
i+1
i
R
R
R
i
̇
ω
ω
ω
i
i+1
̇
v
v
v
i+1
=
i+1
i
R
R
R(
i
̇
ω
ω
ω
i
×
i
p
p
p
i+1
+
i
ω
ω
ω
i
×(
i
ω
ω
ω
i
×
i
p
p
p
i+1
) +
i
v
v
v
i
) + 2
i+1
ω
ω
ω
i+1
×
̇
d
i+1
z+
̈
d
i+1
z
Dynamic inward iterations (Force and Torques):
i
f
i
=
i
f
ci
+
i
i+1
R
i+1
f
i+1
i
n
i
=
i
n
ci
+
i
i+1
R
i+1
n
i+1
+
i
p
ci
×
i
f
ci
+
i
O
i+1
×
i
i+1
R
i+1
f
i+1
τ
i
=
i
n
T
i
i
z
i
For prismatic joint
τ
i
=
i
f
T
i
i
z
i
Page 8 of 8End of exam.
