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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

Continuing with Jacobian and its uses

ME 4135 – Slide Set 7

R. R. Lindeke, Ph. D.

connecting the operator to the jacobian
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
these data then can build the operator
These data then can build the operator:

Populate it with the outtakes from the Ddot Vector – which was found from: J*Dqdot

using these two ideas
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
most common use of jacobian is to map motion singularities
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)
finding singularities
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!
finding the jacobian s determinate
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
finding the jacobian s determinate8
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’)
finding the jacobian s determinate9
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
lets compute the arm singularities for a spherical device
Lets Compute the ARM Singularities for a Spherical Device
  • From Earlier efforts we found that:
  • To solve lets “Expand by Minors” along 3rd row
lets compute the arm singularities for a spherical device11
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

lets compute the arm singularities for a spherical device cont
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)