Transcript Continuing with Jacobian and its uses – Slide Set 7 ME 4135

Continuing with Jacobian
and its uses
ME 4135 – Slide Set 7
R. R. Lindeke, Ph. D.
Connecting the  Operator to the Jacobian
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Examination of the Velocity Vector:
If we consider motion to be made in
UNIT 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 z
respectively
  
x
 
  
 y

  
D z 
 
 x
 y 
 
 z 
These data then can build the operator:
 0

z


  y

 0
 z
0
y
 x
x
0
0
0
dx 

dy

dz 

0
Populate it with the outtakes from the Ddot
Vector – which was found from: J*Dqdot
Using these two ideas:
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Forward Motion in Kinematics:
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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
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Singularities are defined as:
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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:
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They exist wherever the Determinate of the
Jacobian vanishes:
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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
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We will decompose the Jacobian
by Function:
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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
 J 11 J 12 
J 

 J 21 J 22 
Finding the Jacobian’s Determinate
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Considering the case of the Spherical Wrist:
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J12:  Z3   On  O3  Z 4   On  O4  Z 5   On  O5  
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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!
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This really states that On= O3= O4= O5 (which is
a computation convenience but not a ‘real
Jacobian’)
Finding the Jacobian’s Determinate
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With this simplification:
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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
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From Earlier efforts we found that:
 d3S1C 2 d3C1S 2 C1C 2 


J 11  d3C1C 2 d3S1S 2 S1C 2



0
d3C 2
S 2 
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To solve lets “Expand by Minors” along 3rd
row
Lets Compute the ARM Singularities for a
Spherical Device
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J 11  0   d3C1S 2  S1C 2    d3S1S 2  C1C 2 
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 S 2   d S1C 2  d S1S 2    d C1S 2  d C1C 2  
 d3C 2   d3S1C 2  S1C 2    d3C1C 2  C1C 2 
3
3
3
After simplification: the 1st term is zero;
The second term is d32S22C2;
The 3rd term is d32C22C2
3
Lets Compute the ARM Singularities for a
Spherical Device, cont.
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J 11  0  d C2 C2  d S 2 C2
2
3
2
J 11  d C2  C2  S 2
2
3
J 11  d C2
2
3
2
2
3
2
2

This is the ARM determinate,
it would be zero whenever
Cos(2) = 0 (90 or 270)