L7 - Differential Motion and Jacobians 2 V1
-
Upload
zulfadhli1988 -
Category
Documents
-
view
219 -
download
0
Transcript of L7 - Differential Motion and Jacobians 2 V1
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
1/12
Continuing with Jacobian andits uses
U
niversitiKualaLumpurMalaysiaFranceInstitute
Originally prepared by: Prof Engr Dr Ishkandar Baharin
Head of Campus & Dean
UniKL MFI
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
2/12
Connecting the Operator to theJacobian
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 zrespectively
x
y
z
x
y
D z
=
U
niversitiKualaLumpurMalaysiaFranceInstitute
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
3/12
These data then can build the
operator:
0
0
0
0 0 0 0
z y x
z x y
y x z
d
d
d
=
Populate it with the outtakes from the DdotVector which was found from: J*Dqdot
U
niversitiKualaLumpurMalaysiaFrance
Institute
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
4/12
Using these two ideas: Forward Motion in Kinematics:
Given Joint Velocities and Positions
Find Jacobian (a function of Joint positions) & T0n
Compute Ddot, finding dis and is in unit time
Use the dis and is to build With and T0
n compute new T0n
Apply IKS to new T0n which gets new Joint Positions
Which builds new Jacobian and new Ddot and so on
U
niversitiKualaLumpurMalaysiaFrance
Institute
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
5/12
Most Common use of Jacobian is to MapMotion Singularities
Singularities are defined as: Configurations from which certain directions of motion are
unattainable
Locations where bounded (finite) TCP velocities may correspondto unbounded (infinite) joint velocities
Locations where bounded gripper forces & torques maycorrespond 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 solutionto the inverse kinematic problemdoes not exist (No solutions or multiple solutions)
U
niversitiKualaLumpurMal
aysiaFrance
Institute
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
6/12
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 anycombinations of these that will make thedeterminate equal zero
And then try to avoid them!
U
niversitiKualaLumpurMalaysiaFrance
Institute
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
7/12
Finding the JacobiansDeterminate
We will decompose theJacobian by Function:
J11 is the Arm Jointscontribution to Linear velocity
J22 is the Wrist Jointscontribution to AngularVelocity
J21 is the (secondary)contribution of the ARM jointson 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
11 12
21 22
J JJ J J
=
U
niversitiKualaLumpurMalaysiaFrance
Institute
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
8/12
Finding the Jacobians Determinate
Considering the case of the SphericalWrist:
J12:
Of course O3, O4, O5 are a single point soif we choose to solve the Jacobian
(temporally) at this (wrist center) pointthen J12 = 0! This really states that On= O3= O4= O5 (which
is a computation convenience but not a real
Jacobian)
( ) ( ) ( )3 3 4 4 5 5n n nZ O O Z O O Z O O
U
niversitiKualaLumpurMalaysiaFrance
Institute
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
9/12
Finding the Jacobians Determinate
With this simplification: Det(J) = Det(J11)Det(J22)
The device will be singular then whenevereither Det(J11) or Det(J22) equals 0
These separated Singularities would beconsidered ARM Singularities or WRISTSingularities, respectively
U
niversitiKualaLumpurMalaysiaFrance
Institute
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
10/12
Lets Compute the ARM Singularities for aSpherical Device
From Earlier efforts we found that:
To solve lets Expand by Minors along3rd row
3 3
3 3
3
1 2 1 2 1 2
11 1 2 1 2 1 2
0 2 2
d S C d C S C C
J d C C d S S S C
d C S
=
U
niversitiKualaLumpurMalaysiaFrance
Institute
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
11/12
Lets Compute the ARM Singularities for aSpherical Device
( ) ( ){ }
( ) ( ){ }( ) ( ){ }
3 3
3 3 3
3 3 3 3
11 0 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2
2 1 2 1 2 1 2 1 2
J d C S S C d S S C C
d C d S C S C d C C C C
S d S C d S S d C S d C C
=
+
After simplification: the 1st term is zero;
The second term is d32S2
2C2;
The 3rd term is d32C2
2C2U
niversitiKualaLumpurMalaysiaFrance
Institute
-
8/9/2019 L7 - Differential Motion and Jacobians 2 V1
12/12
Lets Compute the ARM Singularities for a
Spherical Device, cont.
( )
2 2 2 2
3 2 2 3 2 2
2 2 2
3 2 2 2
23 2
11 0
11
11
J d C C d S C
J d C C S
J d C
= + +
= +
= This is the ARM determinate,it would be zero whenever
Cos(2) = 0 (90 or 270)U
niversitiKualaLumpurMalaysiaFrance
Institute