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ECE 486 Lecture
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Safwan Choudhury
June 06, 2012
Education
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ECE 486 Lecture
Safwan Choudhury
June 06, 2012
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Transcript
ECE 486: Robot Dynamics and Control Practical Applications of the
Jacobian Safwan Choudhury May 31, 2012
Brief Introduction
Bipedal Locomotion
Bipedal Robot 14 DOF Lower Body
q3 q2 q1
q4 q5
q7 q6
Electromechanical Design SolidWorks + Custom Toolchain
High Performance Direct Drive Micromo DC Motors + Misumi Drivetrain
Components
Machined on Campus Engineering Machine Shop (E3)
Full Dynamic Simulations Simulink + SimMechanics + QUARC
Basic Joint Control 7DOF Leg w/ Fixed Base
The Jacobian Differential Kinematics ˙ x = J ˙ q
Computing the Jacobian Columns The “Geometric” Approach Recall
Computing the Jacobian Columns The “Geometric” Approach Revolute Joints Ji
= zi 1 ⇥ ( on oi 1) zi 1 Ji = zi 1 0 Prismatic Joints Recall
Why?
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
Motivating Example Direct Joint Control
What about complex motions? Inverse Kinematics? Other Methods?
Jacobian Inverse Control Differential Kinematics ˙ q = J 1
˙ x
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
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
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
Motivating Example Jacobian Inverse Control
Motivating Example Jacobian Inverse Control
Jacobian Transpose Control Differential Kinematics ⌧ = JT F
Gravity Compensation Center of Mass (COM) as an End Effector
Motivating Example Without Gravity Compensation
Computing the Jacobian Columns The “Geometric” Approach Recall
Center of Mass Equation Rigid Body Physics Recall xcom =
P ximi P mi
Gravity Compensation Center of Mass (COM) as an End Effector
Partial Center of Mass Rigid Body Physics
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
Jacobian Transpose Control With Gravity Compensation
Whole Body Control A Jacobian-Based Approach ˙ q = 2
6 6 4 JCOM J1 J2 J3 3 7 7 5 1 ˙ x
Independent Leg Motions Two Jacobian’s Stacked: JL + JR
Shifting Balance Three Jacobian’s Stacked: JCOM + JL + JR
Questions?