Chapter 5 Jacobians: Velocities and Static Forces
Angular velocity of a rotating frame {B} in terms of {A}
B  A  B
A
(  to denote the magnitude of  )
(5.6)
Linear velocity of Q in frame {B} which is moving relative to {A}
VQ  AVBOrg  BAR BVQ
A
(5.7)
VQ
K
Q
ΩB
{B}
{A}
Rotational velocity of frame {B} with respect to {A}, AΩB, applied to AQB,
VQ  A B  A Q
A
, a vector cross product
(5.10)
ΔQ = Qt sinθ ωΔt
A
ΩB
A
Qt sinθ
VQ = AQB sinθ AΩB
Qt’
Qt
θ
With QB moving at velocity BVQ in {B} and frame {B} moving at AVBOrg with respect to {A},
the linear and rotational velocity of Q in moving and turning frame {B} with respect to {A},
VQ  AVBOrg  BAR BVQ  A B BAR BQ
A
(5.13)
Property of orthonormal rotation matrix for velocity analysis
Taking a derivative of RR  I n
T
R RT  RR T  R RT  ( R RT )T  0n
So, R RT  R R 1 is a skew-symmetric matrix (in the form of S + S-1 = 0).
(5.16)
Velocity of vector P due to rotating frame
VP  BAR BP
A
(5.22)
VP  BAR BAR 1 A P
A
VP  BAS AP ,
(5.24)
A
S is a skew-symmetric matrix (S+S-1=0)
Skew-symmetric matrices and vector cross product
 0

If S    z
  y

 x 
 Px 
y 



 
  x  ,  =  y  , and P=  Py  and then
  z 
 Pz 
0 
 z
0
x
   z Py   y Pz 


SP    P =   z Px   x Pz 
  y Px   x Py 


(5.27)
Then, from (5.24)
VP  A B AP
A
(5.28)
From (2.80) and sin( x)  x, cos( x)  1 for a small value of x according to the Taylor expansion,
 1

RK ( )   k z 
 k y 

 k z 
1
k x 
k y  

 k x  
1 
(5.33)
Taking the derivative w.r.t. θ
 0
 k z k y 


R   k z
0
 k x
 k y k x
0 

(5.35)’
Transposing R(t) to the LHS, and recognizing the RHS as a skew symmetric matrix,
 0
 k z k y   0

 
0
 k x  =   z
R R 1  R RT   k z
 k y k x
0    y

where    x
y
z
  k 
T
x
 z
0
x

y 

 x 
0 
T
k y k z    Kˆ
Ω = angular velocity vector,
K = instantaneous axis of rotation
(5.36)
(5.37)
Velocity Propagation for revolute joints – More equations!
The angular velocity of link i+1 rotating about its Z axis, projected onto link i which also rotates,
i
i 1
i 1  i i  i 1iRi 1 i 1Zˆ i 1  i i  i 1iR
Multiplying both sides by
i 1
i
i 1
i
0

T
0 i 1
(5.44)
R
Rii1 i1i1 ii1Rii  i1 i1Zˆi1
(5.45)
Linear velocity of frame {i+1}, dropping the last term as the frame origin is constant,
i1 i i  ii i Pi1
i
Multiplying both sides by
(5.46)
i 1
i
R
i1  ii1R( i i1  ii i Pi1 )
i 1
(5.47)
For prismatic joint (no rotational component in the frame, but a translational one.)
i 1 i i1R ii
i 1
i1 ii1R(i i ii i Pi1 )  di 1 i1Zˆi1
i 1
Jacobians
Given a system of six non-linear equations for six X’s and six Y’s.
Y = F(X)
Y 
F
X
X
Y  J ( X )X
Y  J ( X ) X
A Jacobian in robotics is a matrix of partial derivatives that maps a joint velocity vector
0

     

1
2
n

T
into a velocity vector of the end-effector:
   x  y  z  x  y  z 
T
0
 0J ()
0
0
 0J ()1 0
(5.48)
 f1
 
 1
 f 2
0
J ()   1
......
 f
 6
 1
f1
 2
f 2
 2
......
f 6
 2
f1 
 6 

f 2 
......
for six axis robots
 6 
...... ...... 
f 6 
......

 6 
......
Changing a Jacobian reference frame in {B} to frame {A}:
 A   BA R
A   
   0
0   B   BA R

A  B  
R



B 
 0
 AR
Thus, A J ()  B
0
0 B

J ()
A 
R
B 
(5.70)
0 B
 J ()
R
(5.71)
A
B

Find J ()  




 and 0 J ()  




3

 from (5.55) and (5.57)


Singularities
  J ()1
(5.72)
If a matrix is singular, its determinant is zero, and so its inverse cannot be found. As such, under
a singular condition, the end effector velocity cannot be translated into the joint angular velocity.
Types of singularities:
1) Work space boundary singularity – Occurs when the robot arm is fully stretched out with
the end effector reaching the outer boundary.
2) Work space interior singularity – Occurs when two links line up to fold with the end
effector inside the work space boundary.
Static Forces in Manipulator
n  P f
n  P f sin 
------ a cross vector product
i
ni = torque exerted on link i by link i-1, express in {i}
i
fi = force exerted on link i by link i-1, express in {i}
θ i = angle between ifi and iηi
i
Pi+1 = displacement of link i+1, viewed in {i}
{i+1}
ni+1
{i}
i
Pi-1
fi+1
ni
fi+1
Equilibrium (counter balancing) of Force & Moment at a single link - Propagation Equations:
i
f i  i f i 1  i 1i R i 1f i1
(5.80)
i
ni i ni1  iPi1i f i1 i1i R i1ni1  iPi1i f i1
(5.81)
Joint torque at equilibrium –revolute joints = (Moment vector) (Joint axis vector)
 i  i niT zˆi
i
(5.82)
Joint torque at equilibrium –prismatic joints = (Force vector) (Joint axis vector)
 i if iT i zˆi
(5.83)
The partial derivatives of (5.82) constitute a Jacobian. See Ex. 5.7.
Development of Jacobian for converting Force into Torque
Work done in Cartesian space = Work done in Joint space
F  X    
(6 x 1 vectors)
(5.91)
Rewriting in notation for matrix multiplication
F T X   T 
(5.94)
Since X  J by definition,
F T J   T 
(5.95)
Since    ,
FT J   T
Transposing the two sides,
( F T J )T  ( T )T
  JTF
A Jacobian transpose maps the gripper force into equivalent joint torques.
(5.96)
Force and Velocity Transformation in the tool frame
{A}=Revolute, {B}=Fixed, per Fig. 5.13
 
F 
(6x1) velocity vector:     , and (6x1) force/moment vector: F   
 
N 
For the Velocity Transformation, starting from (5.45) and (5.47)
B  ABR A A  B B Zˆ B
(5.45)
B  ABR( AB  AA APB )
(5.47)
B
B
Setting B
 0 …….(why?) and
 BB   AB R  ABR APBOrg   A A 
B   
 A 
B
AR
 B   0
  A 
(5.101)
 AB R A A  ABR( APBOrg  A  A )


B A
A R A


Note that
a)
b)
B
A
A
R( AAAPB ) ABR( APB AA )  ABR APB AA
 i

PBOrg   A   Px
 x

A
j
Py
y
k   i ( Py  z  Pz  y )   0
 
 
Px    j ( Pz  x  Px  z )    Pz
 z  k ( Px  y  Py  x )  Py
 Pz
0
Px
Py   x 

 Px   y 
0    z 
 A A   BA R B B  BAR( ABR APBOrg  A  A )  BA R B B  APBOrg BA R BB )
A   


A B
A B
B R B
B R B
 A  
 

 AR
 B
 0
A
PBOrg BAR  BB 
 B 
A
BR
  B 
(5.102)
The Force-Moment transformation is derived from (5.80) and (5.81):
i
f i if i1 i1iR i1f i1
(5.80)
i
ni i ni1  iPi1i f i1 i1i R i1ni1  iPi1i f i1
(5.81)
 A FA   BA R
A   A
A
 N A   PBOrg  B R
(5.105)
0   B FB 

A  B
 NB 
B R
The relationship between Velocity transformation and Force-Moment transformation:
T ( ABT )T
A
B f
(5.107)
Exercise 5.3
θ3
L2
θ1
L1
θ2
Jacobian derived from the velocity propagation from Base to Tip
Example 5.3: - RRR robot with frozen joint 3.
i ,i i ,i Pi 1 are 3x1 vectors. The vector cross products in (5.47) ii i Pi 1 are shown below.
i
i j k   0 
1
11P2   0 0 1   l11 
l1 0 0   0 
k  
0
i j




2
2 2 P3   0 0 1  2   l2 (1  1 )
l2 0

0  
0


l1s21


3
3
2
2
2
3  2 R (  2  2  P3 )  l1c21  l2 (1  1 )


0


0
3
c12  s12 0
R   s12 c12 0
 0
0 1
 l1s11  l2 s12 (1  1 )


0
3  23R( 2 2  22 2 P3 )   l1c11  l2 c12 (1  1 ) 


0


Note that the components of  3 may also be determined geometrically.
0
Jacobian derived from Static Force propagation from Tip to Base
Jacobian derived from direct Differentiation of the kinematic equations
By observation of the geometric link-frame diagram, the kinematic equations are:
(5.55)
(5.56)
(5.57)
0
P4Org
 4 Px   L1c1  L2c1c2  L3c1c23 
 
  4 Py    L1s1  L2 s1c2  L3 s1c23 
 4 Pz  

L2 s2  L3 s23
 
Taking partial derivatives to arrive at a Jacobian,
 Px
 
 1
P
0
J ( )   y
 1
 P
 z
 1
Px
 2
Py
 2
Pz
 2
Px 
 3 

Py 
 3 
Pz 

 3 
0
4
Once J ( ) is found, J ( ) can be found from:
4
J ( )04R 0J ( ) ,
where 0 R  4 R is readily calculated from the rotational matrixes.
4
0 T