ECE 486 Lecture

Transcript

1. ECE 486: Robot Dynamics and Control Practical Applications of the Jacobian Safwan Choudhury May 31, 2012
2. Brief Introduction
3. Bipedal Locomotion
4. Bipedal Robot 14 DOF Lower Body
5. q3 q2 q1
6. q4 q5
7. q7 q6
8. Electromechanical Design SolidWorks + Custom Toolchain
9. High Performance Direct Drive Micromo DC Motors + Misumi Drivetrain Components
10. Machined on Campus Engineering Machine Shop (E3)
11. Full Dynamic Simulations Simulink + SimMechanics + QUARC
12. Basic Joint Control 7DOF Leg w/ Fixed Base
13. The Jacobian Differential Kinematics ˙ x = J ˙ q
14. Computing the Jacobian Columns The “Geometric” Approach Recall
15. Computing the Jacobian Columns The “Geometric” Approach Revolute Joints Ji =  zi 1 ⇥ ( on oi 1) zi 1 Ji =  zi 1 0 Prismatic Joints Recall
16. Why?
17. Motivating Example q dq QUARC Visualization System Timebase Kp KP(1:7) Knee Pitch 60 Kd KD(1:7)*5 Hip Yaw 0 Hip Roll 0 Hip Pitch -30 EN 1 D2R D2R D2R D2R D2R D2R D2R Biped τ q q′ q′′ Ankle Yaw 0 Ankle Roll 0 Ankle Pitch -20 Direct Joint Control
18. Motivating Example Direct Joint Control
19. What about complex motions? Inverse Kinematics? Other Methods?
20. Jacobian Inverse Control Differential Kinematics ˙ q = J 1 ˙ x
21. Jacobian Inverse Control 1. Compute Jacobian matrix w.r.t. end effector 2. Invert the matrix (pseudoinverse if ) 3. Obtain by multiplying 4. Obtain by integrating ˙ q q = Z ˙ q q ˙ q = J 1 ˙ x n > 6
22. Motivating Example Work Space Analysis QUARC Visualization Trajectory Reference Model Configuration Joint Space Analysis Jacobian Inverse Transformation q′ = J-1x′ J x′ q q′ Jacobian Computation DQREF QREF q → x q x x dqref qref Jacobian Inverse Control
23. Motivating Example τ q q′ q′′ PD Controller q error q′ error τ control erse Transformation q′ = J-1x′ q q′ an Computation Control Torques DQREF QREF q → x q q dq dqref qref Jacobian Inverse Control
24. Motivating Example Jacobian Inverse Control
25. Motivating Example Jacobian Inverse Control
26. Jacobian Transpose Control Differential Kinematics ⌧ = JT F
27. Gravity Compensation Center of Mass (COM) as an End Effector
28. Motivating Example Without Gravity Compensation
29. Computing the Jacobian Columns The “Geometric” Approach Recall
30. Center of Mass Equation Rigid Body Physics Recall xcom = P ximi P mi
31. Gravity Compensation Center of Mass (COM) as an End Effector
32. Partial Center of Mass Rigid Body Physics
33. Jacobian Transpose Control 1. Compute the partial center of masses for each joint 2. Form the COM Jacobian matrix 3. Obtain from the basic formula 4. Obtain by multiplying J com ~ FG = m~ g ~ F ⌧G ⌧ G = JT com ~ F G
34. Jacobian Transpose Control With Gravity Compensation
35. Whole Body Control A Jacobian-Based Approach ˙ q = 2 6 6 4 JCOM J1 J2 J3 3 7 7 5 1 ˙ x
36. Independent Leg Motions Two Jacobian’s Stacked: JL + JR
37. Shifting Balance Three Jacobian’s Stacked: JCOM + JL + JR
38. Questions?