1. Consider the RP planar manipulator shown below. Angle OAE is a right angle. T

Question

1. Consider the RP planar manipulator shown below. Angle OAE is a right angle. The revolute joint rotates about origin O. The length of the first linkage is a constant parameter OA a, The length of the second linkage is a variable AE-d a. (20 points) Obtain the forward kinematic equations relating the end-effecter position X, y and orientation to the joint displacements 0, and d2 b. (20 points) Given the end effector position (*, yE), derive the inverse kinematics. Your answer should define ( , ,d2 ) as functions of the ( XE , y) position.

Explanation / Answer

GIVEN RP Planar manipulator

a. From the given diagram and trigonometry
   end effector coordinates xe, ye are given by
   xe = a1*cos(theta1) - d2*sin(theta1)
   ye = a1*sin(theta1) + d2*cos(theta1)
   orientation, phie of end effector is given by
   phie = 90 + theta1

b. xe = a1*cos(theta1) - d2*sin(theta1)
   ye = a1*sin(theta1) + d2*cos(theta1)
   in matrix form

   [xe] = [cos(theta)   -sin(theta)][a1]
   [ye]   [sin(theta)   cos(theta)][d2]

   premultiplying with the inverse of trigonometric matrix
   (1/cos^2(theta) + sin^2(theta)) [cos(theta)       sin(theta)] [ xe] = [a1]
                                   [-sin(theta)   cos(theta)]   [ye]   [d2]

   hence a1 = xe*cos(theta) + ye*sin(theta)
   d2 = -xe*sin(theta) + ye*cos(theta)

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