velocity kinematics - tu chemnitz
TRANSCRIPT
Velocity Kinematics
Dr.-Ing. John Nassour
Artificial Intelligence & Neuro Cognitive Systems FakultΓ€t fΓΌr Informatik
The Jacobian
13.01.2018 J.Nassour 1
13.01.2018 J.Nassour 2
Motivation
β’ Positions are not enough when commanding motors.
β’ Velocities are needed for better interaction.
β’ How fast the end-effector move given joints velocities?
β’ How fast each joint needs to move in order to guarantee a desired end-effector velocity.
13.01.2018 J.Nassour 3
Differential Motion
Base
Forward Kinematics π β π₯
Differential Kinematicsπ + πΏπ β π₯+πΏπ₯
We are interested in studying the relationship: πΏπ β πΏπ₯
Linear and angular velocities
13.01.2018 J.Nassour 4
Jointβs VelocityPrismatic Joint:
π£ = πππ = 0
Revolute Joint :
π£ = ππ Γ ππ = ππ
π is the unit vector.
ππ π£
π
π
π£
π
π
13.01.2018 J.Nassour 5
Jointβs Velocity
With more than one joint, the end effector velocities are a function of joint velocity and position:
π£π
= π( π1, π2, π1, π2)
For any number of joints:
π£π
= π( π, π)
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 6
The Jacobian
The Jacobian is a matrix that is a function of joint position, that linearly relates joint velocity to tool point velocity.
π£π
= π₯(π1, π2) π1 π2
For the linear velocity:
π£ = π₯π£ π βΊ π₯ π¦
=π₯11 π₯12
π₯21 π₯22
π1 π2
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 7
The Jacobian
The elements of the Jacobian π₯ππ can be obtained by partial differentiation of the forward kinematic equations:
π₯ π¦
=π₯11 π₯12
π₯21 π₯22
π1 π2
ππ₯
ππ‘=
ππ₯
ππ1
ππ1ππ‘
+ππ₯
ππ2
ππ2ππ‘
ππ¦
ππ‘=
ππ¦
ππ1
ππ1ππ‘
+ππ¦
ππ2
ππ2ππ‘
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 8
The Jacobian
The elements of the Jacobian π₯ππ can be obtained by partial differentiation of the forward kinematic equations:
π₯ π¦
=π₯11 π₯12
π₯21 π₯22
π1 π2
ππ₯
ππ‘=
ππ₯
ππ1
ππ1ππ‘
+ππ₯
ππ2
ππ2ππ‘
ππ¦
ππ‘=
ππ¦
ππ1
ππ1ππ‘
+ππ¦
ππ2
ππ2ππ‘
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 9
The Jacobian
In this example the forward kinematics are given by:
π₯ = π2 cos(π1)π¦ = π2 sin(π1)
Find the elements of the Jacobian π₯ππ.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 10
The Jacobian
In this example the forward kinematics are given by:
π₯ = π2 cos(π1)π¦ = π2 sin(π1)
Find the elements of the Jacobian π₯ππ.
π₯11 =ππ₯
ππ1
= βπ2 sin(π1) π₯12 =ππ₯
ππ2
= πππ (π1)
π₯21 =ππ¦
ππ1
= π2 cos(π1) π₯22 =ππ¦
ππ2
= π ππ(π1)
This is the linear velocity Jacobian.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 11
The Jacobian
In this example the forward kinematics are given by:
π₯ = π2 cos(π1)π¦ = π2 sin(π1)
Find the elements of the Jacobian π₯ππ.
π₯11 =ππ₯
ππ1
= βπ2 sin(π1) π₯12 =ππ₯
ππ2
= πππ (π1)
π₯21 =ππ¦
ππ1
= π2 cos(π1) π₯22 =ππ¦
ππ2
= π ππ(π1)
This is the linear velocity Jacobian.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 12
The Jacobian
In this example the forward kinematics are given by:
π₯ = π2 cos(π1)π¦ = π2 sin(π1)
Find the elements of the Jacobian π₯ππ.
π₯11 =ππ₯
ππ1
= βπ2 sin(π1) π₯12 =ππ₯
ππ2
= πππ (π1)
π₯21 =ππ¦
ππ1
= π2 cos(π1) π₯22 =ππ¦
ππ2
= π ππ(π1)
This is the linear velocity Jacobian.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 13
The Jacobian
In this example the forward kinematics are given by:
π₯ = π2 cos(π1)π¦ = π2 sin(π1)
Find the elements of the Jacobian π₯ππ.
π₯11 =ππ₯
ππ1
= βπ2 sin(π1) π₯12 =ππ₯
ππ2
= πππ (π1)
π₯21 =ππ¦
ππ1
= π2 cos(π1) π₯22 =ππ¦
ππ2
= π ππ(π1)
This is the linear velocity Jacobian.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 14
The Jacobian
In this example the forward kinematics are given by:
π₯ = π2 cos(π1)π¦ = π2 sin(π1)
Find the elements of the Jacobian π₯ππ.
π₯11 =ππ₯
ππ1
= βπ2 sin(π1) π₯12 =ππ₯
ππ2
= πππ (π1)
π₯21 =ππ¦
ππ1
= π2 cos(π1) π₯22 =ππ¦
ππ2
= π ππ(π1)
This is the linear velocity Jacobian.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 15
The Jacobian
In this example the forward kinematics are given by:
π₯ = π2 cos(π1)π¦ = π2 sin(π1)
Find the elements of the Jacobian π₯ππ.
π₯11 =ππ₯
ππ1
= βπ2 sin(π1) π₯12 =ππ₯
ππ2
= πππ (π1)
π₯21 =ππ¦
ππ1
= π2 cos(π1) π₯22 =ππ¦
ππ2
= π ππ(π1)
This is the linear velocity Jacobian.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 16
The Jacobian
In this example the forward kinematics are given by:
π₯ = π2 cos(π1)π¦ = π2 sin(π1)
Find the elements of the Jacobian π₯ππ.
π₯11 =ππ₯
ππ1
= βπ2 sin(π1) π₯12 =ππ₯
ππ2
= πππ (π1)
π₯21 =ππ¦
ππ1
= π2 cos(π1) π₯22 =ππ¦
ππ2
= π ππ(π1)
This is the linear velocity Jacobian.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 17
The Jacobian
In this example the forward kinematics are given by:
π₯ = π2 cos(π1)π¦ = π2 sin(π1)
Find the elements of the Jacobian π₯ππ.
π₯11 =ππ₯
ππ1
= βπ2 sin(π1) π₯12 =ππ₯
ππ2
= πππ (π1)
π₯21 =ππ¦
ππ1
= π2 cos(π1) π₯22 =ππ¦
ππ2
= π ππ(π1)
This is the linear velocity Jacobian.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 18
The Jacobian
In this example the forward kinematics are given by:
π₯ = π2 cos(π1)π¦ = π2 sin(π1)
Find the elements of the Jacobian π₯ππ.
π₯11 =ππ₯
ππ1
= βπ2 sin(π1) π₯12 =ππ₯
ππ2
= πππ (π1)
π₯21 =ππ¦
ππ1
= π2 cos(π1) π₯22 =ππ¦
ππ2
= π ππ(π1)
This is the linear velocity Jacobian.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 19
The Jacobian
The angular velocity Jacobian:
π£π
= π₯(π1, π2) π1 π2
For the angular velocity:
π = π₯π π
π = π₯1 π₯2
π1 π2
In this example: π₯1 = 1 , π₯2 = 0
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 20
The Jacobian
The angular velocity Jacobian:
π£π
= π₯(π1, π2) π1 π2
For the angular velocity:
π = π₯π π
π = π₯1 π₯2
π1 π2
In this example: π₯1 = 1 , π₯2 = 0
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 21
The Jacobian
The angular velocity Jacobian:
π£π
= π₯(π1, π2) π1 π2
For the angular velocity:
π = π₯π π
π = π₯1 π₯2
π1 π2
In this example: π₯1 = 1 , π₯2 = 0
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 22
Full Manipulator Jacobian
By combining the angular velocity Jacobian and the linear velocity Jacobian:
π£π
= π₯(π1, π2) π1 π2
π₯ π¦π
= βπ2 sin(π1) cos(π1)π2 cos(π1) sin(π1)
1 0
π1 π2
The full Jacobian is an nΓm matrix where n is the number of joints, and m is the number of variables describing motion.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 23
Full Manipulator Jacobian
Work out the linear and the angular velocities, with joint 2 extendedto 0.5 m. The arm points in the x direction. Joint 1 is rotating at 2 rad/s and joint 2 is extending at 1 m/s.
π₯ π¦π
= βπ2 sin(π1) cos(π1)π2 cos(π1) sin(π1)
1 0
π1 π2
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 24
Full Manipulator Jacobian
Work out the linear and the angular velocities, with joint 2 extendedto 0.5 m. The arm points in the x direction. Joint 1 is rotating at 2 rad/s and joint 2 is extending at 1 m/s.
π₯ π¦π
= βπ2 sin(π1) cos(π1)π2 cos(π1) sin(π1)
1 0
π1 π2
π₯ π¦π
= 0 10.5 01 0
21
=112
π₯=1 m/s ; π¦=1 m/s ; π=2 rad/s
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 25
Inverting The Jacobian
To determine the joint velocities for a given end effector velocity, weneed to invert the Jacobian:
π£π
= π₯ π π
π = π₯β1 ππ£π
13.01.2018 J.Nassour 26
Inverting The Jacobian
Find the joint velocities ( π1, π2) in terms of the end effector velocity ( π₯, π¦).
π₯ π¦
= βπ2 sin(π1) cos(π1)π2 cos(π1) sin(π1)
π1 π2
π1 π2= π₯β1 π
π₯ π¦
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 27
Inverting The Jacobian
Find the joint velocities ( π1, π2) in terms of the end effector velocity ( π₯, π¦).
π1 π2=
βπ2 sin(π1) cos(π1)π2 cos(π1) sin(π1)
β1 π₯ π¦
π1 π2=
1
βπ2
sin(π1) βcos(π1)βπ2 cos(π1) βπ2 sin(π1)
π₯ π¦
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 28
Inverting The Jacobian
Find the joint velocities ( π1, π2) in terms of the end effector velocity ( π₯, π¦).
π1 π2=
βπ2 sin(π1) cos(π1)π2 cos(π1) sin(π1)
β1 π₯ π¦
π1 π2=
1
βπ2
sin(π1) βcos(π1)βπ2 cos(π1) βπ2 sin(π1)
π₯ π¦
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 29
Inverting The Jacobian
Find the joint velocities ( π1, π2) in terms of the end effector velocity ( π₯, π¦).
π1 π2=
βπ2 sin(π1) cos(π1)π2 cos(π1) sin(π1)
β1 π₯ π¦
π1 π2=
1
βπ2
sin(π1) βcos(π1)βπ2 cos(π1) βπ2 sin(π1)
π₯ π¦
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 30
Inverting The Jacobian
Find the joint velocities ( π1, π2) in terms of the end effector velocity ( π₯, π¦).
π1 π2=
βπ2 sin(π1) cos(π1)π2 cos(π1) sin(π1)
β1 π₯ π¦
π1 π2=
1
βπ2
sin(π1) βcos(π1)βπ2 cos(π1) βπ2 sin(π1)
π₯ π¦
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 31
Inverting The Jacobian
Find the joint velocities ( π1, π2) in terms of the end effector velocity ( π₯, π¦).
π1 π2=
βπ2 sin(π1) cos(π1)π2 cos(π1) sin(π1)
β1 π₯ π¦
π1 π2=
1
βπ2
sin(π1) βcos(π1)βπ2 cos(π1) βπ2 sin(π1)
π₯ π¦
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 32
Inverting The Jacobian
Find the joint velocities ( π1, π2) in terms of the end effector velocity ( π₯, π¦).
π1 π2=
βπ2 sin(π1) cos(π1)π2 cos(π1) sin(π1)
β1 π₯ π¦
π1 π2=
1
βπ2
sin(π1) βcos(π1)βπ2 cos(π1) βπ2 sin(π1)
π₯ π¦
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 33
Inverting The Jacobian
π1 π2=
1
βπ2
sin(π1) βcos(π1)βπ2 cos(π1) βπ2 sin(π1)
π₯ π¦
The example arm points in the x Direction, with joint 2 extended to 0.5 m.
Find the joint velocities to move the end effector such that: π₯=1 m/s ; π¦=1 m/s
π1= 2 rad/s ; π2= 1 m/s
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 34
Singularities
The effect of the determinant in the inverse Jacobian example:
π1 π2=
1
βπ2
sin(π1) βcos(π1)βπ2 cos(π1) βπ2 sin(π1)
π₯ π¦
Whenever π2= 0 m, there are no valid joint velocity solutions.
Limited end effector velocities give unlimited joint velocities.
π2, π2π£
ππ1, π1
13.01.2018 J.Nassour 35
Singularities
If the determinant of a square Jacobian is zero, the manipulator cannot be controlled.
β’ It is useful to observe the determinant of the Jacobian as the robot moves to avoid singularities.
β’ Avoid configuration where the determinant approaches zero.
13.01.2018 J.Nassour 36
Jointβs VelocityPrismatic Joint:
π£ = πππ = π
Revolute Joint :
π£ = ππ Γ ππ = ππ
π is the unit vector.
ππ π£
π
π
π£
π
π
13.01.2018 J.Nassour 37
Angular Velocity
Prismatic joint gives π = 0 and revolute joint gives π = ππ
The general Jacobian for the angular velocity: π₯π = π1π§0 π2π§1 β¦ πππ§π β 1
β’ ππ is 1 if the joint is revolute and 0 if the joint is prismatic. β’ π§πis the direction of the z axis of the ith coordinate frame with
respect to the base frame.β’ π§π is the first three elements of third column of the general
transformation matrix.
A1.A2β¦Ai = T0i =
π₯π₯ π¦π₯π₯π¦ π¦π¦
π§π₯ ππ₯π§π¦ ππ¦
π₯π§ π¦π§0 0
π§π§ ππ§0 1
Example: RRP Robot
13.01.2018
Find the angular velocity Jacobianfor the arm RRP.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 38
π₯π = π1π§0 π2π§1 β¦ πππ§π β 1
π1 =? , π2 =?, π3 =?
π 20= π΄ 1 π΄ 2 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
π 10= π΄ 1 =
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
Example: RRP Robot
13.01.2018
Find the angular velocity Jacobianfor the arm RRP.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 39
π₯π = π1π§0 π2π§1 β¦ πππ§π β 1
π1 = 1 , π2 = 1, π3 = 0
π 20= π΄ 1 π΄ 2 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
π 10= π΄ 1 =
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
Example: RRP Robot
13.01.2018
Find the angular velocity Jacobianfor the arm RRP.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 40
π₯π = π1π§0 π2π§1 β¦ πππ§π β 1
π1 = 1 , π2 = 1, π3 = 0
π§0=001
, π§1=βπ 1π10
, π§2=
βπ1π 2βπ 1π 2βπ2
π 20= π΄ 1 π΄ 2 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
π 10= π΄ 1 =
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
Example: RRP Robot
13.01.2018
Find the angular velocity Jacobianfor the arm RRP.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 41
π₯π = π1π§0 π2π§1 β¦ πππ§π β 1
π1 = 1 , π2 = 1, π3 = 0
π§0=001
, π§1=βπ 1π10
, π§2=
βπ1π 2βπ 1π 2βπ2
π 20= π΄ 1 π΄ 2 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
π 10= π΄ 1 =
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
Example: RRP Robot
13.01.2018
Find the angular velocity Jacobianfor the arm RRP.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 42
π₯π = π1π§0 π2π§1 β¦ πππ§π β 1
π1 = 1 , π2 = 1, π3 = 0
π§0=001
, π§1=βπ 1π10
, π§2=
βπ1π 2βπ 1π 2βπ2
π 20= π΄ 1 π΄ 2 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
π 10= π΄ 1 =
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
Example: RRP Robot
13.01.2018
Find the angular velocity Jacobianfor the arm RRP.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 43
π₯π = π1π§0 π2π§1 β¦ πππ§π β 1
π1 = 1 , π2 = 1, π3 = 0
π§0=001
, π§1=βπ 1π10
, π§2=
βπ1π 2βπ 1π 2βπ2
π 20= π΄ 1 π΄ 2 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
π 10= π΄ 1 =
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
Example: RRP Robot
13.01.2018
Find the angular velocity Jacobianfor the arm RRP.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 44
π₯π = π1π§0 π2π§1 β¦ πππ§π β 1
π1 = 1 , π2 = 1, π3 = 0
π§0=001
, π§1=βπ 1π10
, π§2=
βπ1π 2βπ 1π 2βπ2
π 20= π΄ 1 π΄ 2 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
π 10= π΄ 1 =
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
Example: RRP Robot
13.01.2018
Find the angular velocity Jacobianfor the arm RRP.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 45
π₯π = π1π§0 π2π§1 β¦ πππ§π β 1
π1 = 1 , π2 = 1, π3 = 0
π§0=001
, π§1=βπ 1π10
, π§2=
βπ1π 2βπ 1π 2βπ2
π 20= π΄ 1 π΄ 2 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
π 10= π΄ 1 =
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
Example: RRP Robot
13.01.2018
Find the angular velocity Jacobianfor the arm RRP.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 46
π₯π = 001
βπ 1π10
000
π₯π = π1π§0 π2π§1 β¦ πππ§π β 1
π1 = 1 , π2 = 1, π3 = 0
π§0=001
, π§1=βπ 1π10
, π§2=
βπ1π 2βπ 1π 2βπ2
Example: RRP Robot
13.01.2018
Find the angular velocity of the endeffector in the configuration shown with joint 1 rotating at 2 rad/s, joint 2 rotating at -1 rad/s and joint 3 fixedAt 2 m.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 47
Example: RRP Robot
13.01.2018
Find the angular velocity of the endeffector in the configuration shown with joint 1 rotating at 2 rad/s, joint 2 rotating at -1 rad/s and joint 3 fixedAt 2 m.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 48
π₯π = 001
βπ 1π10
000
= 001
010
000
ππ₯
ππ¦
ππ§
=001
010
000
2β19
=0β12
πππ/π
The end effector is rotating 0 rad/s around x axis, -1 rad/s around y axis, and 2 rad/s around z axis.
Example: RRP Robot
13.01.2018
Find the angular velocity of the endeffector in the configuration shown with joint 1 rotating at 2 rad/s, joint 2 rotating at -1 rad/s and joint 3 fixedAt 2 m.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 49
π₯π = 001
βπ 1π10
000
= 001
010
000
ππ₯
ππ¦
ππ§
=001
010
000
2β19
=0β12
πππ/π
The end effector is rotating 0 rad/s around x axis, -1 rad/s around y axis, and 2 rad/s around z axis.
Example: RRP Robot
13.01.2018
Find the angular velocity of the endeffector in the configuration shown with joint 1 rotating at 2 rad/s, joint 2 rotating at -1 rad/s and joint 3 fixedAt 2 m.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 50
π₯π = 001
βπ 1π10
000
= 001
010
000
ππ₯
ππ¦
ππ§
=001
010
000
2β19
=0β12
πππ/π
The end effector is rotating 0 rad/s around x axis, -1 rad/s around y axis, and 2 rad/s around z axis.
Example: RRP Robot
13.01.2018
Find the angular velocity of the endeffector in the configuration shown with joint 1 rotating at 2 rad/s, joint 2 rotating at -1 rad/s and joint 3 fixedAt 2 m.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 51
π₯π = 001
βπ 1π10
000
= 001
010
000
ππ₯
ππ¦
ππ§
=001
010
000
2β10
=0β12
πππ/π
The end effector is rotating 0 rad/s around x axis, -1 rad/s around y axis, and 2 rad/s around z axis.
Example: RRP Robot
13.01.2018
Find the angular velocity of the endeffector in the configuration shown with joint 1 rotating at 2 rad/s, joint 2 rotating at -1 rad/s and joint 3 fixedAt 2 m.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 52
π₯π = 001
βπ 1π10
000
= 001
010
000
ππ₯
ππ¦
ππ§
=001
010
000
2β19
=0β12
πππ/π
The end effector is rotating 0 rad/s around x axis, -1 rad/s around y axis, and 2 rad/s around z axis.
Example: RRP Robot
13.01.2018
Find the angular velocity of the endeffector in the configuration shown with joint 1 rotating at 2 rad/s, joint 2 rotating at -1 rad/s and joint 3 fixedAt 2 m.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 53
π₯π = 001
βπ 1π10
000
= 001
010
000
ππ₯
ππ¦
ππ§
=001
010
000
2β19
=0β12
πππ/π
The end effector is rotating 0 rad/s around x axis, -1 rad/s around y axis, and 2 rad/s around z axis.
Example: RRR Robot
13.01.2018
Find the angular velocity of the end effector in the configuration shown with joint 1 rotating at 2 rad/s, joint 2 rotating at -1 rad/s and joint 3 rotating at 2 rad/s.
βThis is a typical example that can be discussed in the oral examβ
J.Nassour 54
Link 0 (fixed)
Joint 1
Link 1
Joint variable π½1
Joint 2
Link 2
Joint variable π½2
Link 3
ππ
Joint 3
Joint variable π½3
ππ
ππ
ππ
ππ
ππ ππ
ππ
ππ
ππ
ππ
ππ
Link 1= 2 mLink 2= 2 mLink 3= 2 m
Linear Velocity Jacobian
13.01.2018
The velocity of the end effector for an n link manipulator is the race of change of the origin of the end effector frame with respect to the base frame.
ππ0 =
π=1
ππππ
0
πππ ππ
J.Nassour 55
The Origin ππ0
13.01.2018
The origin of the ith reference frame
ππ0 =
π11 π12π21 π22
π13 ππ₯π23 ππ¦
π31 π320 0
π33 ππ§0 1
J.Nassour 56
Linear Velocity Jacobian
13.01.2018
ππ0 =
π=1
ππππ
0
πππ ππ
The contribution of each joint in the linear velocity of the end effector.
π₯π£π=πππ
0
πππ
Each column in the Jacobian is the rate of changes of the end effector in the base reference frame with respect to the rate of change of a joint variable qi.The ith column shows the movement of the end effector caused by ππ.
J.Nassour 57
Prismatic Joint
13.01.2018
Displacement along the axis of actuation π§πβ1:
ππ0 = πππ§πβ1
0
π£ = π§πβ10 ππ
πππ= ππβπ
π
J.Nassour 58
End effectorVelocity
Joint Velocity
Revolute Joint
13.01.2018
Displacement around the axis of actuation π§πβ1:
ππ0 = π Γ π
ππ0 = ππ π§πβ1
0 Γ ππ β ππβ1
π£ = π§πβ10 Γ ππ β ππβ1 ππ
πππ= ππβπ
π Γ ππ β ππβπ
J.Nassour 59
End effectorVelocity
Joint Velocity
Example: RRP Robot
13.01.2018
Find the linear velocity Jacobian for the arm RRP.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 60
π 10=
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
, π 20=
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
, π30 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 βπΏππ1π 2βπ 1π 2 βπΏππ 1π 2
βπ 2 00 0
βπ2 3 βπΏππ20 1
Example: RRP Robot
13.01.2018
Find the linear velocity Jacobian for the arm RRP.
For Prismatic joint:
π₯π£π= π§πβ1
0
For Revolute joint:
π₯π£π= π§πβ1
0 Γ ππ β ππβ1
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 61
π 10=
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
, π 20=
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
, π30 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 βπΏππ1π 2βπ 1π 2 βπΏππ 1π 2
βπ 2 00 0
βπ2 3 βπΏππ20 1
Example: RRP Robot
13.01.2018
Find the linear velocity Jacobian for the arm RRP.
For Prismatic joint:
π₯π£π= π§πβ1
0
For Revolute joint:
π₯π£π= π§πβ1
0 Γ ππ β ππβ1
π₯π£1= π§0
0 Γ π3 β π0
π₯π£1=
001
Γ
βπΏππ1π 2 β 0βπΏππ 1π 2 β 03 βπΏππ2 β 0
=πΏππ 1π 2βπΏππ1π 2
0
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 62
π 10=
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
, π 20=
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
, π30 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 βπΏππ1π 2βπ 1π 2 βπΏππ 1π 2
βπ 2 00 0
βπ2 3 βπΏππ20 1
Example: RRP Robot
13.01.2018
Find the linear velocity Jacobian for the arm RRP.
For Prismatic joint:
π₯π£π= π§πβ1
0
For Revolute joint:
π₯π£π= π§πβ1
0 Γ ππ β ππβ1
π₯π£1= π§0
0 Γ π3 β π0
π₯π£1=
001
Γ
βπΏππ1π 2 β 0βπΏππ 1π 2 β 03 βπΏππ2 β 0
=πΏππ 1π 2βπΏππ1π 2
0
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 63
π 10=
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
, π 20=
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
, π30 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 βπΏππ1π 2βπ 1π 2 βπΏππ 1π 2
βπ 2 00 0
βπ2 3 βπΏππ20 1
Example: RRP Robot
13.01.2018
Find the linear velocity Jacobian for the arm RRP.
For Prismatic joint:
π₯π£π= π§πβ1
0
For Revolute joint:
π₯π£π= π§πβ1
0 Γ ππ β ππβ1
π₯π£2= π§1
0 Γ π3 β π1
π₯π£2=
βπ 1π10
Γ
βπΏππ1π 2βπΏππ 1π 2βπΏππ2
=
βπΏππ1π2βπΏππ 1π2
πΏ3π 12 π 2 + πΏ3π1
2 π 2
=
βπΏππ1π2βπΏππ 1π2πΏ3 π 2
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 64
π 10=
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
, π 20=
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
, π30 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 βπΏππ1π 2βπ 1π 2 βπΏππ 1π 2
βπ 2 00 0
βπ2 3 βπΏππ20 1
Example: RRP Robot
13.01.2018
Find the linear velocity Jacobian for the arm RRP.
For Prismatic joint:
π₯π£π= π§πβ1
0
For Revolute joint:
π₯π£π= π§πβ1
0 Γ ππ β ππβ1
π₯π£2= π§1
0 Γ π3 β π1
π₯π£2=
βπ 1π10
Γ
βπΏππ1π 2βπΏππ 1π 2βπΏππ2
=
βπΏππ1π2βπΏππ 1π2
πΏ3π 12 π 2 + πΏ3π1
2 π 2
=
βπΏππ1π2βπΏππ 1π2πΏ3 π 2
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 65
π 10=
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
, π 20=
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
, π30 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 βπΏππ1π 2βπ 1π 2 βπΏππ 1π 2
βπ 2 00 0
βπ2 3 βπΏππ20 1
Example: RRP Robot
13.01.2018
Find the linear velocity Jacobian for the arm RRP.
For Prismatic joint:
π₯π£π= π§πβ1
0
For Revolute joint:
π₯π£π= π§πβ1
0 Γ ππ β ππβ1
π₯π£3= π§2
0
π₯π£3=
βπ1π 2βπ 1π 2βπ2
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 66
π 10=
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
, π 20=
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
, π30 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 βπΏππ1π 2βπ 1π 2 βπΏππ 1π 2
βπ 2 00 0
βπ2 3 βπΏππ20 1
Example: RRP Robot
13.01.2018
Find the linear velocity Jacobian for the arm RRP.
For Prismatic joint:
π₯π£π= π§πβ1
0
For Revolute joint:
π₯π£π= π§πβ1
0 Γ ππ β ππβ1
π₯π£3= π§2
0
π₯π£3=
βπ1π 2βπ 1π 2βπ2
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 67
π 10=
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
, π 20=
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
, π30 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 βπΏππ1π 2βπ 1π 2 βπΏππ 1π 2
βπ 2 00 0
βπ2 3 βπΏππ20 1
Example: RRP Robot
13.01.2018
Find the linear velocity Jacobian for the arm RRP.
For Prismatic joint:
π₯π£π= π§πβ1
0
For Revolute joint:
π₯π£π= π§πβ1
0 Γ ππ β ππβ1
ππ =
π³πππππ βπ³πππππ βππππβπ³πππππ βπ³πππππ βππππ
π π³π ππ βππ
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 68
π 10=
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
, π 20=
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
, π30 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 βπΏππ1π 2βπ 1π 2 βπΏππ 1π 2
βπ 2 00 0
βπ2 3 βπΏππ20 1
Example: RRP Robot
13.01.2018
Find the linear velocity Jacobian for the arm RRP.
For Prismatic joint:
π₯π£π= π§πβ1
0
For Revolute joint:
π₯π£π= π§πβ1
0 Γ ππ β ππβ1
ππ =
π³πππππ βπ³πππππ βππππβπ³πππππ βπ³πππππ βππππ
π π³π ππ βππ
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 69
π 10=
π1 0π 1 0
βπ 1 0π1 0
0 β10 0
0 30 1
, π 20=
π1π2 π 1π 1π2 βπ1
βπ1π 2 0βπ 1π 2 0
βπ 2 00 0
βπ2 30 1
, π30 =
π1π2 π 1π 1π2 βπ1
βπ1π 2 βπ³πππππβπ 1π 2 βπ³πππππ
βπ 2 00 0
βπ2 π βπ³πππ0 1
Example: RRP Robot
13.01.2018
Find the linear velocity of the end effector in the configuration shown with joint 1 rotation at 2 rad/sec, joint 2 rotating at -1 rad/sec, and joint 3 fixed at 2 m.
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 70
Example: RRP Robot
13.01.2018
Find the linear velocity of the end effector in the configuration shown with joint 1 rotation at 2 rad/sec, joint 2 rotating at -1 rad/sec, and joint 3 fixed at 2 m.
π₯π£ =
πΏ3π 1π 2 βπΏ3π1π2 βπ1π 2βπΏ3π1π 2 βπΏ3π 1π2 βπ 1π 2
0 πΏ3 π 2 βπ2
π1 = 0Β°, π2 = β90Β°, πΏ3 = 2 π
π₯π£ =0 0 12 0 00 β2 0
π£ =0 0 12 0 00 β2 0
2β10
=042
π/π
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 71
Example: RRP Robot
13.01.2018
Find the linear velocity of the end effector in the configuration shown with joint 1 rotation at 2 rad/sec, joint 2 rotating at -1 rad/sec, and joint 3 fixed at 2 m.
π₯π£ =
πΏ3π 1π 2 βπΏ3π1π2 βπ1π 2βπΏ3π1π 2 βπΏ3π 1π2 βπ 1π 2
0 πΏ3 π 2 βπ2
π1 = 0Β°, π2 = β90Β°, πΏ3 = 2 π
π₯π£ =0 0 12 0 00 β2 0
π£ =0 0 12 0 00 β2 0
2β10
=042
π/π
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 72
13.01.2018 J.Nassour 73
Inverting The Jacobian
β’ Analytical inverse (more DOF more Complexity)
β’ Numerical inverse
Example: RRP Robot
13.01.2018
Find the joint velocities in the configuration shown if the desired linear velocities of the end effector are: 0 m/sec on x axis 4 m/sec on y axis 2 m/sec on z axis
π₯π£ =
πΏ3π 1π 2 βπΏ3π1π2 βπ1π 2βπΏ3π1π 2 βπΏ3π 1π2 βπ 1π 2
0 πΏ3 π 2 βπ2
π1 = 0Β°, π2 = β90Β°, πΏ3 = 2 π
π₯π£ =0 0 12 0 00 β2 0
π₯π£β1 =
0 0.5 00 0 β0.51 0 0
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 75
Example: RRP Robot
13.01.2018
Find the joint velocities in the configuration shown if the desired linear velocities of the end effector are: 0 m/sec on x axis 4 m/sec on y axis 2 m/sec on z axis
π₯π£β1 =
0 0.5 00 0 β0.51 0 0
π =0 0.5 00 0 β0.51 0 0
042
=2β10
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 76
Example: RRP Robot
13.01.2018
Find the joint velocities in the configuration shown if the desired linear velocities of the end effector are: 0 m/sec on x axis 4 m/sec on y axis 2 m/sec on z axis
π₯π£β1 =
0 0.5 00 0 β0.51 0 0
π =0 0.5 00 0 β0.51 0 0
042
=2β10
Joint 3
Joint 1
Joint 2
Tool ππ
ππ
ππ
ππππππ
ππ
ππ
ππ
ππ
ππ
ππ
π³3
3π
J.Nassour 77
13.01.2018 J.Nassour 79
Example: RR Robot
ππ
ππ
π°π
π1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
End Effector
Xe
Ye
Find the end-effector velocities in function of joint velocities.
The total derivatives of the kinematics equations:
πππ =πππ(π1, π2)
ππ1ππ1+
πππ(π1, π2)
ππ2ππ2
πππ =πππ(π1, π2)
ππ1ππ1+
πππ(π1, π2)
ππ2ππ2
The Jacobian matrix represents the differential relationship between the joint displacement and the resulting end effector motion.
It comprises the partial derivatives of ππ(π1, π2) and Yπ(π1, π2)with respect to the joint displacements π1 and π2.
13.01.2018 J.Nassour 80
Example: RR Robot
ππ
ππ
π°π
π1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
End Effector
Xe
Ye
π₯π£ =
πππ(π1,π2)
ππ1
πππ(π1,π2)
ππ2ππ
π(π
1,π
2)
ππ1
πππ(π
1,π
2)
ππ2
π₯π£ = βπΌ1π 1β πΌ2π 12 βπΌ2π 12πΌ1π1 + πΌ2π12 πΌ2π12
13.01.2018 J.Nassour 81
Example: RR Robot
ππ
ππ
π°π
π1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
End Effector
Xe
Ye
π₯π£ =
πππ(π1,π2)
ππ1
πππ(π1,π2)
ππ2ππ
π(π
1,π
2)
ππ1
πππ(π
1,π
2)
ππ2
π₯π£ = βπΌ1π 1β πΌ2π 12 βπΌ2π 12πΌ1π1 + πΌ2π12 πΌ2π12
13.01.2018 J.Nassour 82
Example: RR Robot
ππ
ππ
π°π
π1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
End Effector
Xe
Ye
π₯π£ = βπΌ1π 1β πΌ2π 12 βπΌ2π 12πΌ1π1 + πΌ2π12 πΌ2π12
We can divide the 2-by-2 Jacobian into two column vectors:
π₯π£= (π₯1 , π₯2), π₯1 , π₯2 β π½2 Γ 1
We can then write the resulting end-effector velocity vector:
ππ = π₯1 . π1+ π₯2 . π2
Each column vector of the Jacobian matrix represents the end βeffector velocity generated by the corresponding joint moving when all other joints are not moving.
13.01.2018 J.Nassour 83
Example: RR Robot
ππ
ππ
π°π
π1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
End Effector
Xe
Ye
π₯π£ = βπΌ1π 1β πΌ2π 12 βπΌ2π 12πΌ1π1 + πΌ2π12 πΌ2π12
π₯1 =βπΌ1π 1 β πΌ2π 12πΌ1π1+ πΌ2π12
π₯2 =βπΌ2π 12πΌ2π12
13.01.2018 J.Nassour 84
Example: RR Robot
ππ
ππ
π°π
π1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
End Effector
π₯1 =βπΌ1π 1 β πΌ2π 12πΌ1π1 + πΌ2π12
, π₯2 =βπΌ2π 12πΌ2π12
Illustrate the column vector of the Jacobian in the space at the end-effector point.
13.01.2018 J.Nassour 85
Example: RR Robot
ππ
ππ
π°π
π1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
End Effector
π₯1 =βπΌ1π 1 β πΌ2π 12πΌ1π1 + πΌ2π12
, π₯2 =βπΌ2π 12πΌ2π12
Illustrate the column vector of the Jacobian in the space at the end-effector point.
π₯2 points in the direction perpendicular to link 2.
While π₯1 is not perpendicular to link 1 but is perpendicular to the vector (Xe,Ye).This is because π₯1 represent the endpoint velocity caused by joint 1 when joint 2 is not rotating.In other word, link 1 and 2 are rigidly connected, becoming a single rigid body of length (Xe,Ye) and π₯1 is the tip velocity of this body.
13.01.2018 J.Nassour 86
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
π₯1 =βπΌ1π 1 β πΌ2π 12πΌ1π1 + πΌ2π12
, π₯2 =βπΌ2π 12πΌ2π12
Illustrate the column vector of the Jacobian in the space at the end-effector point.
π₯2 points in the direction perpendicular to link 2.
While π₯1 is not perpendicular to link 1 but is perpendicular to the vector (Xe,Ye).This is because π₯1 represent the endpoint velocity caused by joint 1 when joint 2 is not rotating. In other word, link 1 and 2 are rigidly connected, becoming a single rigid body of length (Xe,Ye) and π₯1 is the tip velocity of this body.
π₯2
π₯1
13.01.2018 J.Nassour 87
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
π₯1 =βπΌ1π 1 β πΌ2π 12πΌ1π1 + πΌ2π12
, π₯2 =βπΌ2π 12πΌ2π12
If the two Jacobian column vectors are aligned, the end-effector can not be moved in an arbitrary direction.
This may happen for particular arm configurations when the two links are fully contracted or extracted.
These arm configurations are referred to as singular configurations.
ACCORDINGLY, the Jacobian matrix become singular at these positions.
Find out the singular configurationsβ¦
π₯2
π₯1
13.01.2018 J.Nassour 88
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
π₯1 =βπΌ1π 1 β πΌ2π 12πΌ1π1 + πΌ2π12
, π₯2 =βπΌ2π 12πΌ2π12
π₯2
π₯1
13.01.2018 J.Nassour 89
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
π₯π£ = βπΌ1π 1 β πΌ2π 12 βπΌ2π 12πΌ1π1 + πΌ2π12 πΌ2π12
The Jacobian reflects the singular configurations.When joint 2 is 0 or 180 degrees:
πππ‘ π₯π£ = detβ πΌ1 Β± πΌ2 π 1 βπΌ2π 1πΌ1 Β± πΌ2 π1 Β±πΌ2π1
= 0
π₯2
π₯1
13.01.2018 J.Nassour 90
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
Forward kinematics:
ππ = πΌ1 π1 + πΌ2 π12ππ = πΌ1 π 1 + πΌ2 π 12
π₯π£ = βπΌ1π 1 β πΌ2π 12 βπΌ2π 12πΌ1π1 + πΌ2π12 πΌ2π12
The Jacobian reflects the singular configurations.When joint 2 is 0 or 180 degrees:
πππ‘ π₯π£ = detβ πΌ1 Β± πΌ2 π 1 βπΌ2π 1πΌ1 Β± πΌ2 π1 Β±πΌ2π1
= 0
π₯2
π₯1
13.01.2018 J.Nassour 91
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
π₯2
π₯1
Work out the joint velocities π ( ππ, ππ) in terms of the end effector velocity Ve(Vx,Vy).
If the arm configuration is not singular, this can be obtained by taking the inverse of the Jacobian matrix:
π = π½β1. ππ
Note that the differential kinematics problem has a unique solution as long as the Jacobian is non-singular.
Since the elements of the Jacobian matrix are function of joint displacements, the inverse Jacobian varies depending on the arm configuration. This means that although the desired end-effector velocity is constant, the joint velocities are not.
13.01.2018 J.Nassour 92
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
π₯2
π₯1
We want to move the endpoint of the robot at a constant speed along a path starting at point βAβ on the x-axis, (+2.0, 0), go around the origin through point βBβ (+Ι, 0) and βCβ (0, +Ι), and reach the final point βDβ (0, +2.0) on the y-axis. Consider I1 = I2.
Work out joints velocities along this path.
D
13.01.2018 J.Nassour 93
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
π₯2
π₯1
We want to move the endpoint of the robot at a constant speed along a path starting at point βAβ on the x-axis, (+2.0, 0), go around the origin through point βBβ (+Ι, 0) and βCβ (0, +Ι), and reach the final point βDβ (0, +2.0) on the y-axis. Consider I1 = I2.
Work out joints velocities along this path.
The Jacobian is:
π₯π£ = βπΌ1π 1 β πΌ2π 12 βπΌ2π 12πΌ1π1+ πΌ2π12 πΌ2π12
13.01.2018 J.Nassour 94
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
π₯2
π₯1
The Jacobian is:
π₯π£ = βπΌ1π 1 β πΌ2π 12 βπΌ2π 12πΌ1π1+ πΌ2π12 πΌ2π12
The inverse of the Jacobian is:
π₯π£β1 =
1
πΌ1πΌ2π 2
πΌ2π12 πΌ2π 12βπΌ1π1β πΌ2π12 βπΌ1π 1 β πΌ2π 12
We want to move the endpoint of the robot at a constant speed along a path starting at point βAβ on the x-axis, (+2.0, 0), go around the origin through point βBβ (+Ι, 0) and βCβ (0, +Ι), and reach the final point βDβ (0, +2.0) on the y-axis. Consider I1 = I2.
Work out joints velocities along this path.
13.01.2018 J.Nassour 95
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
π₯2
π₯1
The inverse of the Jacobian is:
π₯π£β1 =
1
πΌ1πΌ2π 2
πΌ2π12 πΌ2π 12βπΌ1π1β πΌ2π12 βπΌ1π 1 β πΌ2π 12
π1 =πΌ2π12. ππ₯ + πΌ2π 12. ππ¦
πΌ1πΌ2π 2
π2 =βπΌ1π1β πΌ2π12 . ππ₯ + [βπΌ1π 1β πΌ2π 12]. ππ¦
πΌ1πΌ2π 2
We want to move the endpoint of the robot at a constant speed along a path starting at point βAβ on the x-axis, (+2.0, 0), go around the origin through point βBβ (+Ι, 0) and βCβ (0, +Ι), and reach the final point βDβ (0, +2.0) on the y-axis. Consider I1 = I2.
Work out joints velocities along this path.
13.01.2018 J.Nassour 96
π1 =πΌ2π12. ππ₯ + πΌ2π 12. ππ¦
πΌ1πΌ2π 2
π2 =βπΌ1π1 β πΌ2π12 . ππ₯ + [βπΌ1π 1 β πΌ2π 12]. ππ¦
πΌ1πΌ2π 2
D
13.01.2018 J.Nassour 97
π1 =πΌ2π12. ππ₯ + πΌ2π 12. ππ¦
πΌ1πΌ2π 2
π2 =βπΌ1π1 β πΌ2π12 . ππ₯ + [βπΌ1π 1 β πΌ2π 12]. ππ¦
πΌ1πΌ2π 2
D
13.01.2018 J.Nassour 98
Example: RR Robot
Note that the joint velocities are extremely large near the initial and the final points, and are unbounded at points A and D. These are the arm singular configurations q2=0.
π1 =πΌ2π12. ππ₯ + πΌ2π 12. ππ¦
πΌ1πΌ2π 2
π2 =βπΌ1π1β πΌ2π12 . ππ₯ + [βπΌ1π 1β πΌ2π 12]. ππ¦
πΌ1πΌ2π 2
D
13.01.2018 J.Nassour 99
Example: RR Robot
π1 =πΌ2π12. ππ₯ + πΌ2π 12. ππ¦
πΌ1πΌ2π 2
π2 =βπΌ1π1β πΌ2π12 . ππ₯ + [βπΌ1π 1β πΌ2π 12]. ππ¦
πΌ1πΌ2π 2
As the end-effector gets close to the origin, the velocity of the first joint becomes very large in order to quickly turn the arm around from point B to C. At these configurations, the second joint is almost -180 degrees, meaning that the arm is near singularity.
D
13.01.2018 J.Nassour 100
Example: RR Robot
π1 =πΌ2π12. ππ₯ + πΌ2π 12. ππ¦
πΌ1πΌ2π 2
π2 =βπΌ1π1β πΌ2π12 . ππ₯ + [βπΌ1π 1β πΌ2π 12]. ππ¦
πΌ1πΌ2π 2
This result agrees with the singularity condition using the determinant of the Jacobian:
πππ‘ π₯π£ = sin π2 = 0 for π2 = ππ, π = 0, Β±1,Β±2,β¦
D
13.01.2018 J.Nassour 101
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
π₯2
π₯1
Using the Jacobian, analyse the arm behaviour at the singular points. Consider (l1=l2=1).
The Jacobian is:
π₯π£ = βπΌ1π 1 β πΌ2π 12 βπΌ2π 12πΌ1π1+ πΌ2π12 πΌ2π12
, π₯1 =βπΌ1π 1β πΌ2π 12πΌ1π1+ πΌ2π12
, π₯2 =βπΌ2π 12πΌ2π12
For q2=0:
π₯1 =β2π 12π1
, π₯2 =βπ 1π1
The Jacobian column vectors reduce to the ones in the same direction. Note that no endpoint velocity can be generated in the direction perpendicular to the aligned arm links (singular configuration A and D).
13.01.2018 J.Nassour 102
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
π₯2
π₯1
Using the Jacobian, analyse the arm behaviour at the singular points. Consider (l1=l2=1).
The Jacobian is:
π₯π£ = βπΌ1π 1 β πΌ2π 12 βπΌ2π 12πΌ1π1+ πΌ2π12 πΌ2π12
, π₯1 =βπΌ1π 1β πΌ2π 12πΌ1π1+ πΌ2π12
, π₯2 =βπΌ2π 12πΌ2π12
For q2=π :
π₯1 =00, π₯2 =
π 1βπ1
The first joint cannot generate any endpoint velocity, since the arm is fully contracted (singular configuration B).
13.01.2018 J.Nassour 103
Example: RR Robot
ππ
ππ
π°ππ1
π2π°π
π₯2
π₯1
Using the Jacobian, analyse the arm behaviour at the singular points. Consider (l1=l2=1).
At the singular configuration, there is at least one direction is which the robot cannot generate a non-zero velocity at the end effector.
Example: RRR Robot
13.01.2018
The robot has three revolute joints that allow the endpoint to move in the three dimensional space. However, this robot mechanism has singular points inside the workspace. Analyze the singularity, following the procedure below.
J.Nassour 104
Link 0 (fixed)
Joint 1
Link 1
Joint variable π½1
Joint 2
Link 2
Joint variable π½2
Link 3
ππ
Joint 3
Joint variable π½3
ππ
ππ
ππ
ππ
ππ ππ
ππ
ππ
ππ
ππ
ππ
Link 1= 2 mLink 2= 2 mLink 3= 2 m
Step 3 Find the joint angles that make det J =0.Step 4 Show the arm posture that is singular. Show where in the workspace it becomes singular. For each singular configuration, also show in which direction the endpoint cannot have a non-zero velocity.
Step 1 Obtain each column vector of the Jacobian matrix by considering the endpoint velocity created by each of the joints while immobilizing the other joints.
Step 2 Construct the Jacobian by concatenating the column vectors, and set the determinant of the Jacobian to zero for singularity: det J =0.
13.01.2018 J.Nassour 105
Stanford Arm
π π
π6
π½π
π½π
π½π
π½π
π½π
ππ
ππ
ππ
ππ
ππ ππ
π2
ππ
ππ ππ
π
ππ
ππ
Give one example of singularity that can occur.
Whenever π½π = π , the manipulator is in a singular configuration because joint 4 and 6 line up. Both joints actions would results the same end-effector motion (one DOF will be lost).
13.01.2018 J.Nassour 106
PUMA 260
π½π
π½π
π½π
π½π
π½π π½π
ππ
ππππ
ππ
ππ
ππ
ππ
ππ
ππππ
ππ
ππ
ππππ
ππ
ππ
ππππ
ππ
ππππ
π π
ππ
π π
ππ
π π
Give two examples of singularities that can occur.
Whenever π½π = π , the manipulator is in a singular configuration because joint 4 and 6 line up. Both joints actions would results the same end-effector motion (one DOF will be lost).
Whenever π½π = βππ , the manipulator is in a singular configuration. In this situation, the arm is fully extracted. This is classed as a workspace boundary singularity.
13.01.2018 J.Nassour 107
π½π
π½π
π½π
π½π
π½π
ππ»
ππ»ππ»
ππππ
ππ
ππ
ππππ
ππ
ππ
ππ
ππ
ππ
ππ
ππ
ππ
ππ
NAO Left Arm
ππ
ππππ