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, F5, FE , Fcam and Fobj , are attached to the
fifth link, the end-effector, the camera, and the object, respectively.
5Tcam =

0 0 −1 2
0 −1 0 1
−1 0 0 6
0 0 0 1
 , 5TE =

0 −1 0 0
1 0 0 2
0 0 1 5
0 0 0 1
 , camTobj =

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 between FE and Fobj ,
5 marks
(b) the distance between the end-effector and the object, i.e., the distance between the
origins of FE and Fobj .
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. F0 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 along y0
direction. The axis of rotation of Joint 3 is along x0 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 34T and its inverse
4
3T .
3 marks
(d) Find the numerical transformation matrix 04T by assuming θ1 = 90
◦, θ2 = 45◦, θ3 =
−45◦, and θ4 = 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
Γ1
2Γ2
2 + 3Γ1
2Γ2 − 4Γ1Γ22 + 4 = 0
2Γ1
2Γ2
2 + Γ1
2Γ2 − 3Γ1Γ22 + 1 = 0
Please reduce the above two equations into a univariant polynomial in Γ1
4. (10 marks) Use the linear function with parabolic blends to generate a smooth trajectory
from x = 0m at t = 0s to x = 0.9m at t = 1s. The required acceleration in the blend region
is x¨ = 4.8m/s2. 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
1T =

1 0 0 0
0 1 0 0
0 0 1 d1
0 0 0 1
 , 12T =

c2 −s2 0 0
0 0 −1 0
s2 c2 0 0
0 0 0 1
 , 23T =

1 0 0 0
0 1 0 L2
0 0 1 0
0 0 0 1

The mass centres of links 1 and 2 are located in the middle of the links. Both inertia tensors
contain the principle inertia terms (Ixx, Iyy and Izz) only. The principle frame of each link
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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.
0g =
 −g0
0
 =
 −100
0

Conduct outward propagation up to frame F2 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: d1 = 1 m, L2 = 0.5 m, v1 = 0, ω2 = 0, v˙1 = 1, ω˙2 = 1,
m1 = 10 kg, m2 = 5 kg, I1xx = 0.05 kgm
2, I1yy = 0.01 kgm
2, I1zz = 0.2 kgm
2,
I2xx = 0.05 kgm
2, I2yy = 0.01 kgm
2, I2zz = 0.1 kgm
2.
3 marks
(c) Suppose an external force 3f3 (along positive x3 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: d1 = 1 m, L2 = 0.5 m, θ2 = −pi/6, v1 = 0, ω2 = 0,
v˙1 = 1, ω˙2 = 1, m1 = 10 kg, m2 = 5 kg, I1xx = 0.05 kgm
2, I1yy = 0.01 kgm
2, I1zz = 0.2
kgm2, I2xx = 0.05 kgm
2, I2yy = 0.01 kgm
2, I2zz = 0.1 kgm
2, 3f3 = 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˙ +Bθ˙ +G(θ) + JT (θ)Fe
where J is the Jacobian matrix. u and Fe 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
RX(θ) =
 1 0 00 cθ −sθ
0 sθ cθ

RY (θ) =
 cθ 0 sθ0 1 0
−sθ 0 cθ

RZ(θ) =
 cθ −sθ 0sθ cθ 0
0 0 1

Transformation matrix
i−1
i T = RX(αi−1)DX(ai−1)RZ(θi)DZ(di)
i−1
i T =

cθi −sθi 0 ai−1
sθcαi−1 cθcαi−1 −sαi−1 −sαi−1di
sθsαi−1 cθsαi−1 cαi−1 cαi−1di
0 0 0 1

Products
Let
a = axi+ ayj + azk
b = bxi+ byj + bzk
then
a.b = axbx + ayby + azbz
a× b = (aybz − azby)i+ (azbx − axbz)j + (axby − aybx)k
Jacobian matrix
For revolute joint i
0Ji =
[
0Zi × (0PE − 0PiORG)
Zi
]
For prismatic joint i
0Ji =
[
0Zi
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) = a0 + a1t+ a2t
2 + a3t
3 + a4t
4 + a5t
5
where at t = 0
x(0) = x0
x˙(0) = x˙0
x¨(0) = x¨0
and at t = tf
x(tf ) = xf
x˙(tf ) = x˙f
x¨(tf ) = x¨f
then the coefficients of the polynomial can be obtained as
a0 = x0
a1 = x˙0
a2 =
x¨0
2
a3 =
−20x0 + 20xf − (12x˙0 + 8x˙f )tf − (3x¨0 − x¨f )t2f
2t3f
a4 =
30x0 − 30xf + (16x˙0 + 14x˙f )tf − (3x¨0 − 2x¨f )t2f
2t4f
a5 =
−12x0 + 12xf − (6x˙0 + 6x˙f )tf − (x¨0 − x¨f )t2f
2t5f
Trigonometry
sin(A+B) = sinA cosB + cosA sinB
cos(A+B) = cosA cosB − sinA sinB
and
C2 = A2 +B2 − 2AB cos(θ)
when A, B and C are the three edges of a triangle and θ is the angle between edges A and B.
Homogeneous transformation matrix
A
BT =
[
A
BR
ApBorg
0T 1
]
A
BT
−1 =BA T =
[
A
BR
T −ABRTApBorg
0T 1
]
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Monash University
Final Examination
MEC4456/TRC4800
Robotics
Parabolic blends
θb = θ0 +
1
2
θ¨t2b
tb =
t
2
−
√
θ¨2t2 − 4θ¨(θf − θ0)
2θ¨
Definition of rotation matrix
A
BR =
[
AxB
AyB
AzB
]
=
xB · xA yB · xA zB · xAxB · yA yB · yA zB · yA
xB · zA yB · zA zB · zA

Fixed angles representation
A
DRXY Z(γ, β, α) =
cαcβ cαsβsγ − sαcγ cαsβcγ + sαsγsαcβ sαsβsγ + cαcγ sαsβcγ − cαsγ
−sβ cβsγ cβcγ

Euler angles representation
A
DRXY Z(α, β, γ) =
cαcβ cαsβsγ − sαcγ cαsβcγ + sαsγsαcβ sαsβsγ + cαcγ sαsβcγ − cαsγ
−sβ cβsγ cβcγ

Angular velocity and cross-product matrix:
Ω ≡ Q˙ QT =
 0 −ωz ωyωz 0 −ωx
−ωy ωx 0
 =
ωxωy
ωz
×
Propagation of Position p
i
ip =
i
iOi+1 +
i
i+1p =
i
iOi+1 +
i
i+1R
i+1
i+1p
Velocity propagation
i+1ω i+1 =
i+1
i Q
iω i + θ˙i+1
i+1zi+1
i+1vi+1 =
i+1
i Q(
iω i × iipi+1 +i vi)
Jacobian
f˙1
f˙2
...
f˙m
 =

∂f1
∂x1
∂f1
∂x2
. . . ∂f1∂xn
∂f2
∂x1
∂f2
∂x2
. . . ∂f2∂xn
...
...
. . .
...
∂fm
∂x1
∂fm
∂x2
. . . ∂fm∂xn


x˙1
x˙2
...
x˙n

Parallel-axis theorem of inertia tensor:
AI =C I +m[pTc pcI3 − pcpTc ]
Newton-Euler Laws:
f = mv˙c
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Monash University
Final Examination
MEC4456/TRC4800
Robotics
n = Icω˙ +ω × Icω
Dynamic outward iterations:
i+1
0 ω i+1 =
i+1
i R
i
0ω i + θ˙i+1
i+1zi+1
i+1
0 ω˙ i+1 =
i+1
i R
i
0ω˙ i + (
i+1
i R
i
0ω i)× (θ˙i+1 i+1zi+1) + θ¨i+1 i+1zi+1
i+1
0 O¨i+1 =
i+1
i R (
i
0O¨i +
i
0 ω˙ i × iiOi+1 +i0 ω i × (i0ω i × iiOi+1))
i
0p¨ci =
i
0 O¨i +
i
0 ω˙ i × iipci +i0 ω i × (i0ω i × iipci)
i+1Fi+1 = mi+1
i+1v˙ci+1
i+1Ni+1 =
i+1ICi+1
i+1ω˙i+1 +
i+1ωi+1 × i+1ICi+1i+1ωi+1
For prismatic joint
i+1ω˙ i+1 =
i+1
i R
iω˙ i
i+1v˙ i+1 =
i+1
i R(
iω˙ i × ipi+1 + iω i × (iω i × ipi+1) + iv i) + 2i+1ω i+1 × d˙i+1z + d¨i+1z
Dynamic inward iterations (Force and Torques):
ifi =
i fci +
i
i+1 R
i+1fi+1
ini =
i nci +
i
i+1 R
i+1ni+1 +
i pci × ifci +i Oi+1 × ii+1R i+1fi+1
τi =
i nTi
izi
For prismatic joint
τi =
i fTi
izi
Page 8 of 8 End of exam.

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