Example: RR Robot. Illustrate the column vector of the Jacobian in the space at the end-effector point.

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1 Forward kinematics: X e = c 1 + c 12 Y e = s 1 + s 12 = s 1 s 12 c 1 + c 12, = s 12 c 12 Illustrate the column vector of the Jacobian in the space at the end-effector point. points in the direction perpendicular to link 2. End Effector While is not perpendicular to link 1 but is perpendicular to the vector (X e,y e ). This is because 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 (X e,y e ) and is the tip velocity of this body. J.Nassour 85

2 Forward kinematics: X e = c 1 + c 12 Y e = s 1 + s 12 = s 1 s 12 c 1 + c 12, = s 12 c 12 Illustrate the column vector of the Jacobian in the space at the end-effector point. points in the direction perpendicular to link 2. While is not perpendicular to link 1 but is perpendicular to the vector (X e,y e ). This is because 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 (X e,y e ) and is the tip velocity of this body. J.Nassour 86

3 Forward kinematics: X e = c 1 + c 12 Y e = s 1 + s 12 = s 1 s 12 c 1 + c 12, = s 12 c 12 If the two Jacobian column vectors are aligned, the endeffector can not be moved in an arbitrary direction. This may happen for particular arm configurations when the two links are fully contracted or etracted. These arm configurations are referred to as singular configurations. ACCORDINGLY, the Jacobian matri become singular at these positions. Find out the singular configurations J.Nassour 87

4 Forward kinematics: X e = c 1 + c 12 Y e = s 1 + s 12 = s 1 s 12 c 1 + c 12, = s 12 c 12 J.Nassour 88

5 Forward kinematics: X e = c 1 + c 12 Y e = s 1 + s 12 J v = s 1 s 12 s 12 c 1 + c 12 c 12 The Jacobian reflects the singular configurations. When joint 2 is 0 or 180 degrees: det J v = det ± s 1 s 1 ± c 1 ± c 1 = 0 J.Nassour 89

6 Forward kinematics: X e = c 1 + c 12 Y e = s 1 + s 12 J v = s 1 s 12 s 12 c 1 + c 12 c 12 The Jacobian reflects the singular configurations. When joint 2 is 0 or 180 degrees: det J v = det ± s 1 s 1 ± c 1 ± c 1 = 0 J.Nassour 90

7 Work out the joint velocities q (, ) in terms of the end effector velocity V e (V, V y ). If the arm configuration is not singular, this can be obtained by taking the inverse of the Jacobian matri: q = J 1. Ve Note that the differential kinematics problem has a unique solution as long as the Jacobian is non-singular. Since the elements of the Jacobian matri 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. J.Nassour 91

8 We want to move the endpoint of the robot at a constant speed along a path starting at point A on the -ais, (+2.0, 0), go around the origin through point B (+ɛ, 0) and C (0, +ɛ), and reach the final point D (0, +2.0) on the y-ais. Consider =. Work out joints velocities along this path. D J.Nassour 92

9 We want to move the endpoint of the robot at a constant speed along a path starting at point A on the -ais, (+2.0, 0), go around the origin through point B (+ɛ, 0) and C (0, +ɛ), and reach the final point D (0, +2.0) on the y-ais. Consider =. Work out joints velocities along this path. The Jacobian is: J v = s 1 s 12 s 12 c 1 + c 12 c 12 J.Nassour 93

10 We want to move the endpoint of the robot at a constant speed along a path starting at point A on the -ais, (+2.0, 0), go around the origin through point B (+ɛ, 0) and C (0, +ɛ), and reach the final point D (0, +2.0) on the y-ais. Consider =. Work out joints velocities along this path. The Jacobian is: J v = s 1 s 12 s 12 c 1 + c 12 c 12 The inverse of the Jacobian is: J v 1 = 1 s 2 c 12 s 12 c 1 c 12 s 1 s 12 J.Nassour 94

11 We want to move the endpoint of the robot at a constant speed along a path starting at point A on the -ais, (+2.0, 0), go around the origin through point B (+ɛ, 0) and C (0, +ɛ), and reach the final point D (0, +2.0) on the y-ais. Consider =. Work out joints velocities along this path. The inverse of the Jacobian is: J v 1 = 1 s 2 c 12 s 12 c 1 c 12 s 1 s 12 = c 12. V + s 12. Vy s 2 = c 1 c 12. V + [ s 1 s 12 ]. Vy s 2 J.Nassour 95

12 D = c 12. V + s 12. Vy s 2 = c 1 c 12. V + [ s 1 s 12 ]. Vy s J.Nassour 96

13 D = c 12. V + s 12. Vy s 2 = c 1 c 12. V + [ s 1 s 12 ]. Vy s J.Nassour 97

14 = c 12. V + s 12. Vy s 2 = c 1 c 12. V + [ s 1 s 12 ]. Vy s 2 D Note that the joint velocities are etremely large near the initial and the final points, and are unbounded at points A and D. These are the arm singular configurations = J.Nassour 98

15 = c 12. V + s 12. Vy s 2 = c 1 c 12. V + [ s 1 s 12 ]. Vy s 2 D 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 J.Nassour 99

16 = c 12. V + s 12. Vy s 2 = c 1 c 12. V + [ s 1 s 12 ]. Vy s 2 D This result agrees with the singularity condition using the determinant of the Jacobian: det J v = sin = 0 for = kπ, k = 0, ±1, ±2, J.Nassour 100

17 Using the Jacobian, analyse the arm behaviour at the singular points. Consider (l 1 =l 2 =1). The Jacobian is: J v = s 1 s 12 s 12 c 1 + c 12 c 12, = s 1 s 12 c 1 + c 12, = s 12 c 12 For =0: = 2s 1 2c 1, = s 1 c 1 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). J.Nassour 101

18 Using the Jacobian, analyse the arm behaviour at the singular points. Consider (l 1 =l 2 =1). The Jacobian is: J v = s 1 s 12 s 12 c 1 + c 12 c 12, = s 1 s 12 c 1 + c 12, = s 12 c 12 For =π: = 0 0, = s 1 c 1 The first joint cannot generate any endpoint velocity, since the arm is fully contracted (singular configuration B). J.Nassour 102

19 Using the Jacobian, analyse the arm behaviour at the singular points. Consider (l 1 =l 2 =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. J.Nassour 103

20 Joint variable θ 2 Joint 2 z 1 Joint variable θ 1 Link 0 (fied) z 0 y 1 0 Eample: RRR Robot 1 Link 1 Link 2 Joint 1 Joint 3 Joint variable θ 3 z 2 y 2 2 Link 1= 2 m Link 2= 2 m Link 3= 2 m Link 3 z 3 y 3 3 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. Step 1 Obtain each column vector of the Jacobian matri 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. 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 J.Nassour 104

21 Stanford Arm Give one eample of singularity that can occur. z 6 y 6 Whenever θ 5 = 0, 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). 6 θ 5 θ 4 z 3 z 5 θ 6 z 4 d 6 d 3 θ 2 1 z 0 z 1 θ 1 z 2 2 d J.Nassour 105

22 PUMA 260 z 1 y 1 1 θ 2 a 2 z 0 θ 3 Give two eamples of singularities that can occur. Whenever θ 5 = 0, 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 θ 3 = 90, the manipulator is in a singular configuration. In this situation, the arm is fully etracted. This is classed as a workspace boundary singularity. 0 d 2 d 4 y 3 d 6 θ 1 z 3 θ 4 3 y 2 y 4 z a θ z y 5 z 5 θ 6 y 6 z J.Nassour 106 4

23 NAO Left Arm z 1 y 1 z θ 1 θ 2 θ 3 y 2 2 z T z 2 z 3 y 3 T y T θ 5 3 θ 4 y 4 y z J.Nassour 107 z

24 NAO Right Arm J.Nassour 108