Jacobians – Velocity Transformation
Craig Carignan
Glen Henshaw
October 5, 2021
Abstract
Jacobian (Ref Sec. 5.7; Exercises 5.17, SS 3.1, SS 3.2)
0.1
Jacobians
The “Jacobian” is actually a “Jacobian matrix” which relates the differentials
of one coordinate set to another:
y1
=
f1 (x1 , x2 )
(1)
y2
=
f2 (x1 , x2 )
(2)
=
f (x)
(3)
or
y
Taking partial derivatives of both sides, we get
δy =
∂f
=
∂x
"
or
ẏ =
∂f
δx
∂x
∂f1
∂x1
∂f2
∂x1
∂f1
∂x2
∂f2
∂x2
#
∂f
ẋ ≡ J ẋ
∂x
(4)
Note that J ≡ J(x) if f (x) is nonlinear.
In robotics, the two coordinate sets are typically the joint angles and the
end effector pose (position and orientation):
i i
V
Jtran
=
q̇
(5)
i
i
ω
Jrot
|{z}
| {z } | {z } joint
tool
velocity
[M ]
Jacobian
J
[M ×N ]
1
rates
[N ]
where
q̇ i =
θ̇i
ḋi
revolute
prismatic
So the Jacobian J transforms vectors of joint rates to vectors of end effector
rates:
q̇
0.2
ẋ
EXAMPLE: Polar robot translation Jacobian
y
6
5
4
r
3
2
1
θ
0
0
1
2
3
4
5
6 x
Let’s first write the forward kinematics for this arm. From inspection,
or
x
= r cos θ
y
= r sin θ
r cos θ
x
=
y
r sin θ
| {z } | {z }
p
f (r,θ)
or
p = f (q)
2
where
p
=
[x y]t
q
=
[r θ]t
∂x
∂x
∂θ
∂y
∂θ
then
∂r
∂y
∂r
ṗ = Jtran (q) ⇒ Jtran (q)
=
cos θ
sin θ
−r sin θ
r cos θ
(6)
This is the “direct differentiation” method. Works okay for arms with few
degrees of freedom. Completely intractable for most arms.
Okay, what about rotation?
Ω = Ṙ(q)R(q)T
where
−ωz
0
ωx
0
Ω , ωz
−ωy
and
ωy
−ωx
0
ωx
ω = ωy .
ωx
Ṙ =
N
X
∂R(q)
j=1
∂qj
q̇j
can be rewritten as
ω = Jrot q̇
See Exercise 5.16. and the RPR Wrist Jacobian handout.
This is the “Direct Differentiation” method for the rotation Jacobian, and
it is very tedious.
0.3
Cross–product method
There is an alternative method called the “cross–product method” which is more
computationally efficient and derives from the velocity propagation method. It’s
based on the insight that each element of the Jacobian describes the instantaneous motion of the end effector along some direction in terms of the motion of
each joint. Graphically:
3
where the circle is the set of points swept out by the end effector and the
solid arrow shows the current velocity of the end effector, which of course is
normal to the circle. Finding the direction of this vector is, of course, a simple
matter of crossing the unit vector passing through the link’s center of rotation
(out of the plane of the page) with the vector from the center of rotation to the
end effector.
Formally, let
i
ẑj , i (j ẑj )
i.e. the third column of ij R = ij R j zj . Then,
i
Jrot =
i
ẑ1 ,i ẑ2 , · · · ,i ẑN Λ
where
Λ , diag(λi ) with λi =
1
0
revolute
prismatic
and
i
h
i
ẑ1 × i ( 1 pN +1 ), i ẑ2 × i ( 2 pN +1 ), i ẑ3 × i ( 3 pN +1 ) Λ
+ (IN ×N − Λ) i ẑ1 , i ẑ2 , · · · , i ẑN
Jtrans =
i
4
y
6
5
4
3
2
1
0
0
0.4
1
2
3
4
6 x
5
Changing the frame of the Jacobian
Given B J, find A J.
By definition,
B
Vp
B
ω
=B J(q)q̇
Velocity is a free vector, so:
A
Vp
A
ω
0.5
=
=
A B
BR Vp
A B
BR ω
A
⇒
Vp
A
ω
=
A
BR
03×3
03×3
A
BR
B
J(q)q̇)
Example: 3–link planar manipulator
A: Find i Jrot ⇒ 01 R 12 R
c1 −s1 0
c12 −s12 0
c123 −s123
0
s1 c1 0 , 02 R = s12 c12 0 , 03 R = s123 c123
1R =
0
0
1
0
0
1
0
0
0
0 0 0
0
0
Jrot = 0 0 0 = 2 Jrot ⇒ 0 ω 3 = 2 ω 3 =
1 1 1
θ̇1 + θ̇2 + θ̇3
0.5.1
0.5.2
B: Find i Jtrans
(1) Use differentiation method
l1 c1 + l2 c12 + l3 c123
0
p4 = l1 s1 + l2 s12 + l3 s123
0
5
0
0
1
Factoring,
0
−l1 s1 − l2 s12 − l3 s123
jtrans =
= l1 c1 + l2 c12 + l3 c123
∂θ
0
∂ 0 p4
c12
0
Jtran = 20 R 0 J = −s12
0
or
s12
c12
0
−l2 s12 − l3 s123
l2 c12 + l3 c123
0
0
−l1 s1 − l2 s12 − l3 s123
0 l1 c1 + l2 c12 + l3 c123
1
0
l1 s2 − l3 s3
= l1 c2 + l2 + l3 c3
0
−l3 s3
l2 + l3 c3
0
−l3 s123
l3 c123
0
−l2 s12 − l3 s123
l2 c12 + l3 c123
0
−l3 s3
l3 c3
0
Notice that 2 Jtran is much simpler than 0 Jtran .
(2) Use “Cross Product” method
0
so
Jtran =
0
ẑ2 × 0 ( 2 p4 ) 0 ẑ3 × 0 ( 3 p4 )
0
0
ẑ1 = 0 ẑ2 = 0 ẑ3 = 0
1
l3
3
p4 = 0
0
l2
2
p3 = 0
0
l1
1
p2 = 0
0
ẑ1 × 0 ( 1 p4 )
0
l2 c12 + l3 c123
0 2
( p4 ) = 0 ( 2 p3 ) + 0 ( 3 p4 ) = l2 s12 + l3 s123
0
l1 c1 + l2 c12 + l3 c123
0 1
( p4 ) = 0 ( 1 p2 ) + 0 ( 2 p4 ) = l1 s1 + l2 s12 + l3 s123
0
Note that these two equations can be obtained recursively. Performing the cross
product yields the same result for 0 Jtrans .
To compute 2 Jtrans :
2
Jtrans =
2
ẑ1 × 2 ( 1 p4 )
2
ẑ2 × 2 ( 2 p4 )
6
2
ẑ3 × 2 ( 3 p4 )
−l3 s123
l3 c123
0
In our example, 2 ẑ1 = 2 ẑ2 =2 ẑ3 = [0 0 1]T , so starting with the last term:
l3 c3
2
ẑ3 × 2 ( 3 p4 ) = l3 s3
0
then
l2
l3 c3
2
ẑ2 × 2 ( 2 p4 ) = 2 ( 2 p3 ) + 2 ( 3 p4 ) = 0 + l3 s3
0
0
and finally
c2
2
ẑ1 × 2 ( 1 p4 ) = −s2
0
s2
c2
0
0
l1
l2 + l3 c3
l2 + l2 c2 + l3 c3
= −l2 s2 + l3 s3
0 0 + l3 s3
1
0
0
0
Performing the cross product yields the same results for 2 Jtrans . Notice how
much easier it is to compute using this method!
7
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