Artificial Intelligence & Neuro Cognitive Systems Fakultät für Informatik Velocity Kinematics The Jacobian Dr.-Ing. John Nassour 13.01.2018 J.Nassour 1 Motivation • Positions are not enough when commanding motors. • Velocities are needed for better interaction. • How fast the end-effector move given joints velocities? • How fast each joint needs to move in order to guarantee a desired end-effector velocity. 13.01.2018 J.Nassour 2 Differential Motion Forward Kinematics 휃 → 푥 Differential Kinematics 휃 + 훿휃 → 푥+훿푥 Base We are interested in studying the relationship: 훿휃 ↔ 훿푥 Linear and angular velocities 13.01.2018 J.Nassour 3 Joint’s Velocity Prismatic Joint: 푞 푣 = 푞 푘 푘 푣 휔 = 0 푣 Revolute Joint : 휔 푣 = 푞 푘 × 푟 푞 휔 = 푞 푘 푘 푟 푘 is the unit vector. 13.01.2018 J.Nassour 4 Joint’s Velocity With more than one joint, the end effector velocities are a function of joint velocity and position: 휔 푞1, 푞1 푣 푣 푞2, 푞2 = 푓(푞 , 푞 , 푞 , 푞 ) 휔 1 2 1 2 For any number of joints: 푣 = 푓(푞 , 푞) 휔 13.01.2018 J.Nassour 5 The Jacobian The Jacobian is a matrix that is a function of joint position, that linearly relates joint velocity to tool point velocity. 휔 푣 푞 1 푞1, 푞1 푣 = 풥(푞1, 푞2) 푞2, 푞2 휔 푞2 For the linear velocity: 푥 풥11 풥12 푞1 푣 = 풥푣 푞 ⟺ = 푦 풥21 풥22 푞2 13.01.2018 J.Nassour 6 The Jacobian The elements of the Jacobian 풥푖푗 can be obtained by partial differentiation of the forward kinematic equations: 휔 푥 푞 풥11 풥12 1 푞1, 푞1 푣 = 푞2, 푞2 푦 풥21 풥22 푞2 푑푥 휕푥 푑푞 휕푥 푑푞 = 1 + 2 푑푡 휕푞1 푑푡 휕푞2 푑푡 푑푦 휕푦 푑푞 휕푦 푑푞 = 1 + 2 푑푡 휕푞1 푑푡 휕푞2 푑푡 13.01.2018 J.Nassour 7 The Jacobian The elements of the Jacobian 풥푖푗 can be obtained by partial differentiation of the forward kinematic equations: 휔 푥 푞 풥11 풥12 1 푞1, 푞1 푣 = 푞2, 푞2 푦 풥21 풥22 푞2 푑푥 휕푥 푑푞 휕푥 푑푞 = 1 + 2 푑푡 휕푞1 푑푡 휕푞2 푑푡 푑푦 휕푦 푑푞 휕푦 푑푞 = 1 + 2 푑푡 휕푞1 푑푡 휕푞2 푑푡 13.01.2018 J.Nassour 8 The Jacobian In this example the forward kinematics are given by: 푥 = 푞 cos(푞 ) 2 1 휔 푦 = 푞2 sin(푞1) 푞1, 푞1 푣 푞2, 푞2 Find the elements of the Jacobian 풥푖푗. 13.01.2018 J.Nassour 9 The Jacobian In this example the forward kinematics are given by: 푥 = 푞 cos(푞 ) 2 1 휔 푦 = 푞2 sin(푞1) 푞1, 푞1 푣 푞2, 푞2 Find the elements of the Jacobian 풥푖푗. 휕푥 휕푥 풥11 = = −푞2 sin(푞1) 풥12 = = 푐표푠(푞1) 휕푞1 휕푞2 휕푦 휕푦 풥21 = = 푞2 cos(푞1) 풥22 = = 푠푖푛(푞1) 휕푞1 휕푞2 This is the linear velocity Jacobian. 13.01.2018 J.Nassour 10 The Jacobian In this example the forward kinematics are given by: 푥 = 푞 cos(푞 ) 2 1 휔 푦 = 푞2 sin(푞1) 푞1, 푞1 푣 푞2, 푞2 Find the elements of the Jacobian 풥푖푗. 휕푥 휕푥 풥11 = = −푞2 sin(푞1) 풥12 = = 푐표푠(푞1) 휕푞1 휕푞2 휕푦 휕푦 풥21 = = 푞2 cos(푞1) 풥22 = = 푠푖푛(푞1) 휕푞1 휕푞2 This is the linear velocity Jacobian. 13.01.2018 J.Nassour 11 The Jacobian In this example the forward kinematics are given by: 푥 = 푞 cos(푞 ) 2 1 휔 푦 = 푞2 sin(푞1) 푞1, 푞1 푣 푞2, 푞2 Find the elements of the Jacobian 풥푖푗. 휕푥 휕푥 풥11 = = −푞2 sin(푞1) 풥12 = = 푐표푠(푞1) 휕푞1 휕푞2 휕푦 휕푦 풥21 = = 푞2 cos(푞1) 풥22 = = 푠푖푛(푞1) 휕푞1 휕푞2 This is the linear velocity Jacobian. 13.01.2018 J.Nassour 12 The Jacobian In this example the forward kinematics are given by: 푥 = 푞 cos(푞 ) 2 1 휔 푦 = 푞2 sin(푞1) 푞1, 푞1 푣 푞2, 푞2 Find the elements of the Jacobian 풥푖푗. 휕푥 휕푥 풥11 = = −푞2 sin(푞1) 풥12 = = 푐표푠(푞1) 휕푞1 휕푞2 휕푦 휕푦 풥21 = = 푞2 cos(푞1) 풥22 = = 푠푖푛(푞1) 휕푞1 휕푞2 This is the linear velocity Jacobian. 13.01.2018 J.Nassour 13 The Jacobian In this example the forward kinematics are given by: 푥 = 푞 cos(푞 ) 2 1 휔 푦 = 푞2 sin(푞1) 푞1, 푞1 푣 푞2, 푞2 Find the elements of the Jacobian 풥푖푗. 휕푥 휕푥 풥11 = = −푞2 sin(푞1) 풥12 = = 푐표푠(푞1) 휕푞1 휕푞2 휕푦 휕푦 풥21 = = 푞2 cos(푞1) 풥22 = = 푠푖푛(푞1) 휕푞1 휕푞2 This is the linear velocity Jacobian. 13.01.2018 J.Nassour 14 The Jacobian In this example the forward kinematics are given by: 푥 = 푞 cos(푞 ) 2 1 휔 푦 = 푞2 sin(푞1) 푞1, 푞1 푣 푞2, 푞2 Find the elements of the Jacobian 풥푖푗. 휕푥 휕푥 풥11 = = −푞2 sin(푞1) 풥12 = = 푐표푠(푞1) 휕푞1 휕푞2 휕푦 휕푦 풥21 = = 푞2 cos(푞1) 풥22 = = 푠푖푛(푞1) 휕푞1 휕푞2 This is the linear velocity Jacobian. 13.01.2018 J.Nassour 15 The Jacobian In this example the forward kinematics are given by: 푥 = 푞 cos(푞 ) 2 1 휔 푦 = 푞2 sin(푞1) 푞1, 푞1 푣 푞2, 푞2 Find the elements of the Jacobian 풥푖푗. 휕푥 휕푥 풥11 = = −푞2 sin(푞1) 풥12 = = 푐표푠(푞1) 휕푞1 휕푞2 휕푦 휕푦 풥21 = = 푞2 cos(푞1) 풥22 = = 푠푖푛(푞1) 휕푞1 휕푞2 This is the linear velocity Jacobian. 13.01.2018 J.Nassour 16 The Jacobian In this example the forward kinematics are given by: 푥 = 푞 cos(푞 ) 2 1 휔 푦 = 푞2 sin(푞1) 푞1, 푞1 푣 푞2, 푞2 Find the elements of the Jacobian 풥푖푗. 휕푥 휕푥 풥11 = = −푞2 sin(푞1) 풥12 = = 푐표푠(푞1) 휕푞1 휕푞2 휕푦 휕푦 풥21 = = 푞2 cos(푞1) 풥22 = = 푠푖푛(푞1) 휕푞1 휕푞2 This is the linear velocity Jacobian. 13.01.2018 J.Nassour 17 The Jacobian In this example the forward kinematics are given by: 푥 = 푞 cos(푞 ) 2 1 휔 푦 = 푞2 sin(푞1) 푞1, 푞1 푣 푞2, 푞2 Find the elements of the Jacobian 풥푖푗. 휕푥 휕푥 풥11 = = −푞2 sin(푞1) 풥12 = = 푐표푠(푞1) 휕푞1 휕푞2 휕푦 휕푦 풥21 = = 푞2 cos(푞1) 풥22 = = 푠푖푛(푞1) 휕푞1 휕푞2 This is the linear velocity Jacobian. 13.01.2018 J.Nassour 18 The Jacobian The angular velocity Jacobian: 푣 푞1 = 풥(푞1, 푞2) 휔 휔 푞 2 푞1, 푞1 푣 푞2, 푞2 For the angular velocity: 휔 = 풥휔푞 푞1 휔 = 풥1 풥2 푞2 In this example: 풥1 = 1 , 풥2 = 0 13.01.2018 J.Nassour 19 The Jacobian The angular velocity Jacobian: 푣 푞1 = 풥(푞1, 푞2) 휔 휔 푞 2 푞1, 푞1 푣 푞2, 푞2 For the angular velocity: 휔 = 풥휔푞 푞1 휔 = 풥1 풥2 푞2 In this example: 풥1 = 1 , 풥2 = 0 13.01.2018 J.Nassour 20 The Jacobian The angular velocity Jacobian: 푣 푞1 = 풥(푞1, 푞2) 휔 휔 푞 2 푞1, 푞1 푣 푞2, 푞2 For the angular velocity: 휔 = 풥휔푞 푞1 휔 = 풥1 풥2 푞2 In this example: 풥1 = 1 , 풥2 = 0 13.01.2018 J.Nassour 21 Full Manipulator Jacobian By combining the angular velocity Jacobian and the linear velocity Jacobian: 휔 푣 푞 1 푞1, 푞1 푣 = 풥(푞1, 푞2) 푞2, 푞2 휔 푞2 −푞 sin(푞 ) cos(푞 ) 푥 2 1 1 푞 = 1 푦 푞2 cos(푞1) sin(푞1) 푞 휔 1 0 2 The full Jacobian is an n×m matrix where n is the number of joints, and m is the number of variables describing motion. 13.01.2018 J.Nassour 22 Full Manipulator Jacobian Work out the linear and the angular velocities, with joint 2 extended to 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. 휔 푞1, 푞1 푣 푞2, 푞2 −푞 sin(푞 ) cos(푞 ) 푥 2 1 1 푞 = 1 푦 푞2 cos(푞1) sin(푞1) 푞 휔 1 0 2 13.01.2018 J.Nassour 23 Full Manipulator Jacobian Work out the linear and the angular velocities, with joint 2 extended to 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. 휔 푞1, 푞1 푣 푞2, 푞2 −푞 sin(푞 ) cos(푞 ) 푥 2 1 1 푞 = 1 푦 푞2 cos(푞1) sin(푞1) 푞 휔 1 0 2 푥 0 1 2 1 푦 = 0.5 0 = 1 1 휔 1 0 2 푥 =1 m/s ; 푦 =1 m/s ; 휔=2 rad/s 13.01.2018 J.Nassour 24 Inverting The Jacobian To determine the joint velocities for a given end effector velocity, we need to invert the Jacobian: 푣 = 풥 푞 푞 휔 푣 푞 = 풥−1 푞 휔 13.01.2018 J.Nassour 25 Inverting The Jacobian Find the joint velocities (푞1 , 푞2 ) in terms of the end effector velocity (푥 , 푦 ). 휔 푥 −푞2 sin(푞1) cos(푞1) 푞1 푞1, 푞1 푣 = 푞2, 푞2 푦 푞2 cos(푞1) sin(푞1) 푞2 푞 푥 1 = 풥−1 푞 푞2 푦 13.01.2018 J.Nassour 26 Inverting The Jacobian Find the joint velocities (푞1 , 푞2 ) in terms of the end effector velocity (푥 , 푦 ). 휔 −1 푞 , 푞 푞1 −푞2 sin(푞1) cos(푞1) 푥 1 1 푣 = 푞2, 푞2 푞2 푞2 cos(푞1) sin(푞1) 푦 푞 1 sin(푞 ) −cos(푞 ) 푥 1 = 1 1 푞2 −푞2 −푞2 cos(푞1) −푞2 sin(푞1) 푦 13.01.2018 J.Nassour 27 Inverting The Jacobian Find the joint velocities (푞1 , 푞2 ) in terms of the end effector velocity (푥 , 푦 ). 휔 −1 푞 , 푞 푞1 −푞2 sin(푞1) cos(푞1) 푥 1 1 푣 = 푞2, 푞2 푞2 푞2 cos(푞1) sin(푞1) 푦 푞 1 sin(푞 ) −cos(푞 ) 푥 1 = 1 1 푞2 −푞2 −푞2 cos(푞1) −푞2 sin(푞1) 푦 13.01.2018 J.Nassour 28 Inverting The Jacobian Find the joint velocities (푞1 , 푞2 ) in terms of the end effector velocity (푥 , 푦 ).