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Continuing with the Jacobian and its uses

Continuing with the Jacobian and its uses. ME 4135 R. R. Lindeke, Ph. D. Connecting the  Operator to the Jacobian. Examination of the Velocity Vector: If we consider motion to be made in UNIT TIME Then x dot which is dx / dt and dt = 1 (unit time step) then x dot = dx

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Continuing with the Jacobian and its uses

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  1. Continuing with the Jacobianand its uses ME 4135 R. R. Lindeke, Ph. D.

  2. Connecting the  Operator to the Jacobian • Examination of the Velocity Vector: • If we consider motion to be made inUNIT TIME • Then xdot which is dx/dtand dt = 1 (unit time step) then xdot= dx • Similarly for ydot, zdot, and the ’sthey are: dy, dz and x, y, and z respectively

  3. These data then can build the operator Populate it with the outtakes from the DdotVector – which was found from: J*Dqdot

  4. Using these two ideas: • Forward Motion in Kinematics: • Given Joint Velocities and Positions • Find Jacobian (a function of Joint positions) & T0n • Compute Ddot, finding di’s and i’s – in unit time • Use the di’s and i’s to build  • With  and T0ncompute new T0n • Apply IKS which gets new Joint Positions • Which builds new Jacobian and new Ddot •  and so on

  5. Most Common use of Jacobian is to Map Motion Singularities • Singularities are defined as: • Configurations from which certain directions of motion are unattainable • Locations where bounded (finite) TCP velocities may correspond to unbounded (infinite) joint velocities • Locations where bounded gripper forces & torques may correspond to unbounded joint torques • Points on the boundary of manipulator workspaces • Points in the manipulator workspace that may be unreachable under small perturbations of the link parameters • Places where a unique solution to the inverse kinematic solution does not exist (No solutions or multiple solutions)

  6. Finding Singularities: • They exist wherever the Determinate of the Jacobian vanishes: • Det(J) = 0 • As we remember, J is a function of the Joint positions so we wish to know if there are any combinations of these that will make the determinate equal zero • And then try to avoid them!

  7. Finding the Jacobian’s Determinate • We will decompose the Jacobian by Function: • J11 is the Arm Joints contribution to Linear velocity • J22 is the Wrist Joints contribution to Angular Velocity • J12 is the (secondary) contribution of the ARM joints on angular velocity • J12 is the (secondary) contribution of the WRIST joints on the linear velocity • Note: Each of these is a 3X3 matrix in a full function robot

  8. Finding the Jacobian’s Determinate • Considering the case of the Spherical Wrist: • J12: • Of course O3, O4, O5 are a single point so if we ‘choose’ to solve the Jacobian (temporally) at this (wrist center) point then J12 = 0!

  9. Finding the Jacobian’s Determinate • With this simplification: • Det(J) = Det(J11)Det(J22) • The device will be singular, then, whenever either Det(J11) or Det(J22) equals 0 • These “separated” Singularities would be considered ARM Singularities or Wrist Singularities, respectively

  10. Lets Compute the ARM Singularities for a Spherical Device • From Earlier: • To solve “Expand by Minors” along 3rd row

  11. Let’s Compute the ARM Singularities for a Spherical Device After simplification: the 1st term is zero; The second term is d32*C2*C22; The 3rd term is d32*C2*S22

  12. Lets Compute the ARM Singularities for a Spherical Device This is the ARM determinate, it would be zero whenever Cos(2) = 0 (90 or 270)

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