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Continuing with Jacobian and its uses

Continuing with Jacobian and its uses. ME 4135 – Slide Set 7 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 : dt = t = 1 Then x dot (which is dx/dt) – is dx

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Continuing with Jacobian and its uses

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  1. Continuing with Jacobian and its uses ME 4135 – Slide Set 7 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: dt = t = 1 • Then xdot (which is dx/dt) – is dx • Similarly for ydot, zdot, and the ’s.They are: dy, dz and x, y, and zrespectively

  3. These data then can build the operator: Populate it with the outtakes from the Ddot Vector – 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 T0n compute new T0n • Apply IKS to new T0n 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 problem 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 • J21 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! • This really states that On= O3= O4= O5 (which is a computation convenience but not a ‘real Jacobian’)

  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 efforts we found that: • To solve lets “Expand by Minors” along 3rd row

  11. Lets Compute the ARM Singularities for a Spherical Device After simplification: the 1st term is zero; The second term is d32S22C2; The 3rd term is d32C22C2

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

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