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Inverse Kinematics

Inverse Kinematics. Problem: Input: the desired position and orientation of the tool Output: the set of joints parameters. Workspaces. Dextrous workspace – the volume of space which the robot end-effector can reach with all orientations

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Inverse Kinematics

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  1. Inverse Kinematics Problem: Input: the desired position and orientation of the tool Output: the set of joints parameters

  2. Workspaces • Dextrous workspace – the volume of space which the robot end-effector can reach with all orientations • Reachable workspace – the volume of space which the robot end-effector can reach in at least one orientation • If L1=L2then the dextrous space = {origin} and the reachable space = full disc of radius 2L1 • If then the dextrous space is empty and the reachable space is a ring bounded by discusses with radiuses |L1- L2| and L1+L2 • The dextrous space is a subset of the reachable space

  3. Solutions • A manipulator is solvable if an algorithm can determine the joint variables. The algorithm should find all possible solutions. • There are two kinds of solutions: closed-form and numerical (iterative) • Numerical solutions are in general time expensive • We are interested in closed-form solutions: • Algebraic Methods • Geometric Methods

  4. Algebraic Solution • Kinematics equations of this arm: • The structure of the transformation:

  5. Algebraic Solution (cont.) • We are interested in x, y, and (of the end-effector) • By comparison of the two matrices above we obtain: • And by further manipulations: and ……

  6. Algebraic Solution by Reduction to Polynomial • The actual variable is u :

  7. Example 1

  8. Kinematic Equations of The Arm

  9. Target • By comparison we get:

  10. Kinematic Equations - Solution

  11. Example 2 (L1=0)

  12. Example 2 (cont.)

  13. Example 2 (cont.)

  14. Example 2 (cont.)

  15. y L2 L1 x Geometric Solution IDEA: Decompose spatial geometry into several plane geometry problems 2 Applying the “law of cosines”: x2+y2=l12+l22  2l1l2cos(1802)

  16. Geometric Solution (II) y Then:   The LoC gives: l22 = x2+y2+l12 - 2l1 (x2+y2) cos  So that cos  = (x2+y2+l12 - l22 )/2l1 (x2+y2) We can solve for 0   180, and then 1= x

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