Robotics & Automation Lecture 23 Closed Kinematic Chain And

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Robotics & Automation Lecture 23 Closed Kinematic Chain and Parallel Mechanisms John T. Wen November 10, 2008 JTW-RA23 RPI ECSE/CSCI 4480 Robotics I Examples of closed kinematic chains ² Constrained robot, e.g., robot in contact with environment (polishing, assembly, etc.). ² Multiple interacting robots, e.g., multi-finger grasp, multiple cooperative robots. ² Parallel mechanism, e.g., 4-bar linkage, slider-crank, Stewart-Gough Platform, Delta robot. qq 2 qq 1 qq 3 November 10, 2008Copyrighted by John T. Wen Page 1 JTW-RA23 RPI ECSE/CSCI 4480 Robotics I Kinematics Forward kinematics and inverse kinematics are the same as before: given joint positions, find the task frame, and vice versa – subject to the closed chain constraints. Usually: Serial mechanism has complicated geometry, inverse kinematics is more difficult than the forward kinematics. Parallel mechanisms usually are based on imposing kinematic constraints to simple serial mechanisms. Therefore, inverse kinematics is usually easy. The forward kinematics needs to take into account of constraints, and is more difficult to solve. November 10, 2008Copyrighted by John T. Wen Page 2 JTW-RA23 RPI ECSE/CSCI 4480 Robotics I Constrained mechanisms Consider a serial kinematic chain with n joints (q 2 Rn). Suppose the chain is constrained: f(q) = 0; f : Rn ! Rk: Then the mechanism has n ¡ k (unconstrained) DOF. Example: ² 3-DOF Planar arm with tip constrained to move along a line (2 dof) ² 4-bar linkage (1 dof) November 10, 2008Copyrighted by John T. Wen Page 3 JTW-RA23 RPI ECSE/CSCI 4480 Robotics I Greubler’s Formula Consider a planar mechanism. Let N = # of links, g = # of joints, fi = # of DOF of ith joint. DOF: 3N g Constraints: 3g ¡ ∑i=1 fi. g Net # of DOF: 3(N ¡ g) + ∑i=1 fi for spatial mechanisms (bodies are free to translate and rotate in Cartesian space), we have g Net # of DOF: 6(N ¡ g) + ∑ fi: i=1 Example: planar platform with three legs. November 10, 2008Copyrighted by John T. Wen Page 4 JTW-RA23 RPI ECSE/CSCI 4480 Robotics I Kinematics: 4-bar linkage Example rotation: R = exp(zˆq1)exp(zˆq3) = exp(zˆq2)exp(zˆq4) (1 constraint) position: p14 = exp(zˆq1)(p13 + exp(zˆq3)p34) = p12 + exp(zˆq2)p24 (2 constraints) Inverse kinematics: given R (equiv. qE = q1 + q3 = q2 + q4), find (q1;q2;q3;q4). Forward kinematics: Suppose q1 is the active joint. Given q1, find qE . Typically, inverse kinematics is easy (since each leg is simple), but forward kinematics hard (solving the constraint equations). November 10, 2008Copyrighted by John T. Wen Page 5 JTW-RA23 RPI ECSE/CSCI 4480 Robotics I Slider Crank qq2 qq1 qq3 Structure equation (kinematics + constraints): (q1+q2+q3)zˆ R03 = R01R12R23 = e p0T = R01 p12 + R01R12 p23 + R01R12R23 p3T R03 = I T y p0T = 0: November 10, 2008Copyrighted by John T. Wen Page 6.

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