Kinematic modelofrobotmanipulators (1)

Kinematic Model of Robot Manipulators
Claudio Melchiorri
Dipartimento di Ingegneria dell’Energia Elettrica e dell’Informazione (DEI)
Universit`a di Bologna
email: claudio.melchiorri@unibo.it
C. Melchiorri (DEIS) Kinematic Model 1 / 164

Summary
1 Kinematic Model
2 Direct Kinematic Model
3 Inverse Kinematic Model
4 Diﬀerential Kinematics
5 Statics - Singularities - Inverse diﬀerential kinematics
6 Inverse kinematics algorithms
7 Measures of performance

Kinematic Model Introduction
Kinematic Model
In robotics, there are two main ‘kinematic’ problems:
1. Forward (direct) Kinematic Problem: once the joint position, velocity,
acceleration are known, compute the corresponding variables of the end-eﬀector in
a given reference frame (e.g. a Cartesian frame).
=⇒ Forward kinematic model:
a function f deﬁned between the joint space IRn
and the work space IRm
:
x = f(q) x ∈ IRm
, q ∈ IRn

Kinematic Model
2. Inverse Kinematic Problem: computation of the relevant variables
(positions, velocities, accelerations) from the work space to the joint space.
=⇒ Inverse Kinematic Model:
function g = f−1
from IRm
to IRn
:
q = g(x) = f−1
(x) q ∈ IRn
, x ∈ IRm
Some common (somehow arbitrary) definitions must be adopted ⇒ for the
same manipulator, different (although equivalent) kinematic models can be
defined.

Kinematic Model – Example: a 2 dof planar robot
Forward kinematic model:
x = l1 cos θ1 + l2 cos(θ1 + θ2)
y = l1 sin θ1 + l2 sin(θ1 + θ2)
φ = θ1 + θ2
An easy problem...
Inverse kinematic model:
cos θ2 =
x2
0 + y2
0 − l2
1 − l2
2
2l1l2
, sin θ2 = ± (1 − cos2 θ2)
θ2 = atan2(sin θ2, cos θ2)
k1 = l1 + l2 cos θ2, k2 = l2 sin θ2
sin θ1 =
y0k1 − x0k2
k2
1
+ k2
2
, cos θ1 =
y0 − k1 sin θ1
k2
θ1 = atan2(sin θ1, cos θ1)
• The solution is not so simple.
• Two possible solutions (sign of sin θ2).

Kinematic Model
Homogeneous Transformations are used for the deﬁnition of the kinematic model.
A robotic manipulator is a mechanism composed by a chain of rigid bodies, the
links, connected by joints.
A reference frame is associated to each link, and homogeneous transformations
are used to describe their relative position/orientation.

Kinematic Model
A convention for the description of robots.
Each link is numbered from 0 to n, in order to be univocally identiﬁed in the
kinematic chain: L0, L1, . . . , Ln.
=⇒ Conventionally, L0 is the “base” link, and Ln is the ﬁnal (distal) link.
Each joint is numbered, from 1 to n, starting from the base joint: J1, J2, . . . , Jn.
=⇒ According to this convention, joint Ji connects link Li−1 to link Li .
A manipulator with n + 1 links has n joints.

Kinematic Model
The motion of the joints changes the end-effector position/orientation in the work
space.
The position and the orientation of the end-effector result to be a (non linear)
function of the n joint variables q1, q2, ..., qn, i.e.
p = f(q1, q2, ..., qn) = f(q)
q = [q1 q2 . . . qn]T
is defined in the joint space IRn
,
p is defined in the work space IRm
.
Usually, p is composed by:
some position components (e.g. x, y, z, wrt a Cartesian reference frame)
some orientation components (e.g. Euler or RPY angles).

Kinematic Model Denavit-Hartenberg Parameters
Kinematic Model
Need of deﬁning a systematic and possibly unique method for the deﬁnition of the
kinematic model of a robot manipulator:
DENAVIT-HARTENBERG NOTATION
A reference frame is assigned to each link, and homogeneous transformations
matrices are used to describe the relative position/orientation of these frames.
The reference frames are assigned according to a particular convention, and
therefore the number of parameters needed to describe the pose of each link, and
consequently of the robot, is minimized.

Denavit-Hartenberg Parameters
Problem:
How to assign frames to the links in order to minimize the number of parameters?
Generally speaking, 6 parameters are necessary to describe the position and the
orientation of a rigid body in the 3D space (a rigid body has 6 dof), and therefore
6 parameters are required to describe Fb in Fa .
Under some hypotheses, only 4 parameters are required: the Denavit-Hartenberg
Parameters.
Given two reference frames F0 and F6 in the 3D space, 4 cases are possible:

Most general case: Skew Axes.
PROBLEM: Find a sequence of elementary homogeneous transformations relating
two generic reference frames F0 e F6 , with skew axes z0 and z6.
SOLUTION: Infinite solutions are possible.
It is desirable to define A sequence so that the kinematic model is defined
univocally and using the minimum number of parameters.

Denavit-Hartenberg Parameters - Procedure
A common normal n exists among two skew z axes. Let us deﬁne:
d the distance between the origin of F0 and the intersection point of z0 with n
a the distance between z0 and z6 along n
Apply the following sequence of translations/rotations:
1 Translate the origin of F0 along z0 for the quantity d: the frame F1 is obtained
2 Rotate (ccw) F1 about z1 by the angle θ until x1 is aligned with n:F2 is obtained
3 Translate F2 along x2 (= n) for a: F3 is obtained, with origin on the z6 axis
4 Rotate (ccw) F3 about x3 by α, so that z3 is aligned with z6: F4 is obtained
5 Translate F4 along z4 for the quantity b until F6 : the frame F5 is obtained
6 Rotate F5 about z5 by the angle φ: F6 is reached

Denavit-Hartenberg Parameters - Procedure

Six cyclic transformations have been employed to move from F0 to F6 : 3
translations and 3 rotations.
There is a translation-rotation pattern:
0
T6 = Tras(z0, d)Rot(z1, θ)Tras(x2, a)Rot(x3, α)Tras(z4, b)Rot(z5, φ) (1)
The ﬁrst 4 transformations are of particular interest: 2 couples of translations and
rotations about two axes (note that z0 = z1 and x2 = x3):
0
H4 = Tras(z0, d)Rot(z1, θ)Tras(x2, a)Rot(x3, α)
=




Cθ −SθCα SθSα aCθ
Sθ CθCα −CθSα aSθ
0 Sα Cα d
0 0 0 1





Matrix 0
T6 can be expressed in terms of H matrices by adding to (1)
a null translation along x6, obtaining the frame F7
a null rotation about x7, obtaining the frame F8
Therefore we have
0
T8 = Tras(z0, d)Rot(z1, θ)Tras(x2, a)Rot(x3, α)
Tras(z4, b)Rot(z5, φ)Tras(x6, 0)Rot(x7, 0)
that is expressed by cyclic transformations.

If another frame F12 is given, it is possible to move from F6 to F12 by means of a
sequence similar to (1). Then, the transformation from F0 to F12 is
0
T12 = 0
H4 Tras(z4, b)Rot(z5, φ)Tras(z6, d′
)Rot(z7, θ′
)Tras(x8, a′
)Rot(x9, α′
)
Tras(z10, b′
)Rot(z11, φ′
)Tras(x12, 0)Rot(x13, 0)
Since a translation and a rotation about the same axis may commute, i.e.
Rot(z5, φ)Tras(z6, d′
) = Tras(z6, d′
)Rot(z5, φ)
we have that
Tras(z4, b)Rot(z5, φ)Tras(z6, d′
)Rot(z7, θ′
) = Tras(z4, b)Tras(z6, d′
)Rot(z5, φ)Rot(z7, θ′
)
= Tras(z4, b + d′
)Rot(z5, φ + θ′
)

In conclusion, the transformation between F0 and F12 is expressed by two DH
transformations expressed by H matrices:
the ﬁrst one with parameters d, θ, a, α,
the second one with parameters (b + d′
), (φ + θ′
), a′
, α′
(and a third one with parameters b′
, φ′
, 0, 0).

In general, only frames F0 and F4 are of interest, and not the intermediate ones
(F1 -F3 ). Therefore, F4 will be indicated from now as F1 . The transformation
0
H4 is then indicated as 0
H1.
0
H1 = Tras(z0, d)Rot(z1, θ)Tras(x2, a)Rot(x3, α)
=




Cθ −SθCα SθSα aCθ
Sθ CθCα −CθSα aSθ
0 Sα Cα d
0 0 0 1




The frames associated to each link are used only for the deﬁnition of the
kinematic model of the robot: usually their position/orientation may be freely
assigned and do not depend by other constraints.
Therefore, these frames are assigned in order to minimize the number of
parameters required for the deﬁnition of the kinematic model.

As a matter of fact, if F0 and F6 are two frames associated to two consecutive
links, and the position and orientation of F6 are not constrained by other
considerations, it is possible to choose F4 as the frame of the second link (NOT
F6 ), reducing in this manner to 4 the number of parameters: b and φ are not
necessary.
Then, the transformation between two consecutive links is 0
H4.

In conclusion:
Although in general 6 parameters are necessary to specify the relative position and
orientation of two frames F0 and F1 , only 4 parameters are suﬃcient (d, θ, a, α)
by assuming that:
1 The axis x1 intersects z0
2 The axis x1 is perpendicular to z0
These parameters are known as the Denavit-Hartenberg Parameters.

Consider now a generic manipula-
tor. Li−1, Li : consecutive links
Ji ed Ji+1 i relative joints
The motion axis of Ji deﬁnes the direction of
zi−1 (frame Fi−1 ) associated to the proximal
link
zi (Fi ) is aligned with the motion axis of the
following joint
The origin of Fi is at the intersection of zi with
the common normal ai between zi−1 and zi
If a common normal does not exist (ai = 0),
the origin of Fi is placed on zi−1
If the two axes intersect, the origin is placed at
the intersection
If the two axes coincide, also the origins of
Fi−1 and Fi coincide
xi (Fi ) is directed along the common normal
yi is chosen in order to obtain a proper frame.

Conclusion: the position and the orientation of two consecutive frames, and
therefore of the related links, may be deﬁned by the four Denavit-Hartenberg
parameters:
ai = length of the common normal between the axes of two consecutive joints
αi = ccw angle between zi−1 the axis of joint i, and zi , axis of joint i + 1
di = distance between the origin oi−1 of Fi−1 and the point pi ,
θi = ccw angle between the axis xi−1 and the common normal ˆpi oi about
zi−1.

The parameters ai and αi are constant and depend only on the link geometry:
ai is the link length
αi is the link twist angle
between the joints’ axes.
Considering the two other parameters, depending on the joint type one is
constant and the other one may change in time:
prismatic joint: di is the joint variable and θi is constant;
rotational joint: θi is the joint variable and di is constant.

The homogeneous transformation matrix relating the frames Fi−1 and Fi is
i−1
Hi = Trasl(zi−1, di ) Rot(zi−1, θi ) Trasl(xi , ai ) Rot(xi , αi )
=




1 0 0 0
0 1 0 0
0 0 1 di
0 0 0 1








Cθi −Sθi 0 0
Sθi Cθi 0 0
0 0 1 0
0 0 0 1








1 0 0 ai
0 1 0 0
0 0 1 0
0 0 0 1








1 0 0 0
0 Cαi −Sαi 0
0 Sαi Cαi 0
0 0 0 1




=




Cθi −Sθi Cαi Sθi Sαi ai Cθi
Sθi Cθi Cαi −Cθi Sαi ai Sθi
0 Sαi Cαi di
0 0 0 1




known as Canonical Transformation.
In literature, matrix i−1
Hi is also indicated as i−1
Ai .

Each matrix i−1
Hi is a function of the i-th joint variable, di or θi depending on the
joint type. For notational ease, the joint variable is generically indicated as qi , i.e.:
qi = di for prismatic joints
qi = θi for rotational joints
Therefore: i−1
Hi = i−1
Hi (qi ).
In case of a manipulator with n joints, the relationship between frame F0 and
frame Fn is:
0
Tn = 0
H1(q1) 1
H2(q2)...n−1
Hn(qn)
This equation expresses the position and orientation of the last link wrt the base
frame, once the joint variables q1, q2, . . . , qn are known.
This equation is the kinematic model of the manipulator.

Reference Configuration of a Canonical Transformation
A generic homogenous transformation 0
Tn may be expressed as a function of n
canonical transformations
0
Tn =
n
i=1
i−1
Hi
If all the rotational joint variables are null, i.e. θi = 0, and all the prismatic joints
variables are at the minimum value, i.e. dj = min(dj ) (with θj = 0), the so-called
Reference Configuration for 0
Tn is obtained.
Note that for prismatic joints the value θj may be imposed by the manipulator
structure (and be not null). Also in these cases, it is arbitrarily considered null. A
similar consideration holds also for rotational joints (θi = 0):
The reference configuration may be non physically reachable by the
manipulator.

Kinematic Model
In the reference conﬁguration, the matrices i−1
Hi are:
i−1
Hi = i−1
Hi |θi =0 =



1 0 0 ai
0 Cαi −Sαi 0
0 Sαi Cαi di
0 0 0 1


 rotational joints
i−1
Hi = i−1
Hi |θi =0; di =min(di ) =



1 0 0 ai
0 Cαi −Sαi 0
0 Sαi Cαi min(di )
0 0 0 1


 prismatic joints
The rotational part of these matrices indicates a rotation about the xi axis.
Therefore, by composing all the i−1
Hi , the xi axes results only translated (their
orientation does not change).
In this conﬁguration, all the xi axes have the same direction (they are aligned).

Direct Kinematic Model Procedure for assigning frames
Direct Kinematic Model
Claudio Melchiorri

Kinematic Model
A procedure to assign frames to the links of a manipulator
Need of common conventions, in order to define univocally the kinematic equations.
First step: definition of the base frame F0 . In this case it is usually posible to consider
not only the kinematic configuration of the manipulator but also other considerations,
related e.g. to the work space. However, according to the DH convention, usually F0 is
chosen so that z0 coincides with the motion axis of J1.
F0 = ?
Fn = ?
Also Fn is assigned considering not only the robot’s kinematics, since a motion axis for
the last link does not exist. =⇒ F0 and Fn are arbitrarily chosen!

Kinematic Model
The Denavit-Hartenberg convention does not define univocally the frames
associated to the links. As a matter of fact, the frames may be assigned with
some arbitrariness in the following cases:
1 F0 : only the direction of z0 may be univocally defined, while in general the
origin o0 and the orientation of x0 and y0 are not assigned;
2 Fn : only xn is constrained to be perpendicular to zn−1 (i.e. to Jn). Since the
joint n + 1 does not exist, it is not possible to define the other elements;
3 parallel consecutive axes: it is not defined univocally the common normal line;
4 intersecting consecutive axes: the direction of xi is not defined;
5 prismatic joint: only zi is defined.
In these cases, it is possible to exploit the arbitrariness in order to simplify the
kinematic model, for example by posing the origin of consecutive frames in the
same point, or aligning axes of consecutive frames, and so on.

Procedure for assigning the frames
The frames are assigned to the links with the following procedure:
1. The joints and links are numbered (joints from 1 to n; links from 0 to n).
Links Li−1 and Li are adjacent (proximal and distal, respectively), connected
by the joint Ji (whose variable is qi );
2. Deﬁnition of the axes zi , i = 0, . . . , n − 1, aligned with the joint motion
directions (rotation/translation);
(N.B. zi is the motion direction of joint Ji+1: z0 → J1; z1 → J2; . . .)
3. Deﬁnition of F0 , with origin in any point of z0, and axes x0, y0 ‘properly’
chosen;

Steps 4 - 6 are repeated for i = 1, . . . , n − 1
4. Definition of Fi . Three cases are possible:
a) the axes of joints Ji and Ji+1 have a common normal:
the origin of Fi is placed at the intersection point of zi with the common
normal between zi−1 and zi
b) the axes of Ji e Ji+1 intersect:
the origin of Fi is placed at the intersection point between zi−1 and zi
c) the axes of joints Ji and Ji+1 are coincident or parallel:
if Ji is rotational, the origin of Fi is chosen so that di = 0
if Ji is prismatic, the origin of Fi is placed at a joint limit
5. Definition of xi along the common normal between zi−1 and zi (if exists)
with positive direction from Ji to Ji+1; if zi−1 intersects zi , then the following
joints are considered;
6. Definition of yi in order to obtain a proper frame.

Finally:
7. Define on coincident with on−1;
8. Define xn perpendicular to zn−1;
9. If Jn is rotational, choose zn parallel to zn−1;
If Jn is prismatic, it is possible to choose zn freely;
10. Define yn in order to obtain a proper frame.
Note that:
The position of on and its orientation zn are arbitrary
In this manner, the frame Fn is different wrt the frame of the end-effector t
T
(with axes n, s, a). Therefore, it is in general necessary to define a constant
homogeneous transform matrix to take into account this difference.

Once the n frames Fi (i = 1, . . . , n) are deﬁned, the corresponding 4n DH
parameters di , ai , αi , θi can be easily determined, and therefore also the
matrices i−1
Hi can be computed. The kinematic model is then obtained.
Then:
a) Deﬁne the DH Parameters Table
b) Compute the homogeneous transformation matrices i−1
Hi , i = 1, . . . , n
c) Compute the direct kinematic function
0
Tn = 0
H1
1
H2 . . . n−1
Hn

Direct Kinematic Model Examples
Example
Let us consider the following 3 dof manipulator:

Example
Step 1: Assign numbers to joints and links

Example
Step 2: Choice of the zi axes (joint rotation/translation motion axes)

Example
Step 3: Choice of F0

Example
Steps 4 - m: Deﬁnition of F1 ... Fn

Example
Finally (optional): choice of the tool frame

Example
Let’s consider a 2 dof planar manipulator:
Denavit-Hartenberg parameters
d θ a α
L1 0 θ1 a1 0o
L2 0 θ2 a2 0o
The i−1
Hi matrices result:
0
H1 =




C1 −S1 0 a1C1
S1 C1 0 a1S1
0 0 1 0
0 0 0 1




1
H2 =




C2 −S2 0 a2C2
S2 C2 0 a2S2
0 0 1 0
0 0 0 1





Example
Then
0
T2 = 0
H1
1
H2 =
n s a p
0 0 0 1
=




C12 −S12 0 a1C1 + a2C12
S12 C12 0 a1S1 + a2S12
0 0 1 0
0 0 0 1




The vectors n, s, a express the orientation of the manipulator (rotation about z0),
while p deﬁnes the end eﬀector position (plane x0 − y0).

Example
zi−1 axes aligned with the motion direc-
tion of Ji
Note that:
di = 0: distances among common
normal lines
ai : distances among the joint axes
Ji
αi : angle between zi−1 and zi
about xi
With the DH convention, the origin of F2 is coincident with F1 . In this case, one
obtains:
0
T2 =



C12 −S12 0 a1C1
S12 C12 0 a1S1
0 0 1 0
0 0 0 1


 Then 2
Tt =



1 0 0 a2
0 1 0 0
0 0 1 0
0 0 0 1




Example: 3 dof anthropomorphic manipulator
Table of the Denavit-Hartenberg parameters
d θ a α
L1 0 θ1 0 90o
L2 0 θ2 a2 0o
L3 0 θ3 a3 0o
Matrices i−1
Hi
0
H1=




C1 0 S1 0
S1 0 −C1 0
0 1 0 0
0 0 0 1



 ,1
H2=




C2 −S2 0 a2C2
S2 C2 0 a2S2
0 0 1 0
0 0 0 1



 ,2
H3=




C3 −S3 0 a3C3
S3 C3 0 a3S3
0 0 1 0
0 0 0 1





Kinematic model:
0
T3 =




C1C23 −C1S23 S1 C1(a2C2 + a3C23)
S1C23 −S1S23 −C1 S1(a2C2 + a3C23)
S23 C23 0 a2S2 + a3S23
0 0 0 1




The orientation of z3 depends only on the
ﬁrst joint J1; pz does not depend on θ1.

Check if the model is correct
With θ1 = θ2 = θ3 = 0o
0
T3 =



1 0 0 a2 + a3
0 0 −1 0
0 1 0 0
0 0 0 1


 x0
z0
y0
F0
a2 a3
y3
z3
F3
With θ1 = θ2 = θ3 = 90o
0
T3 =



0 0 1 0
−1 0 0 −a3
0 −1 0 a2
0 0 0 1



x0
z0
y0
F0
a2
a3
z3
y3
x3 F3

Another choice for the last frame could be
In this case, the last frame does not re-
spect the DH convention:
=⇒ x3 does not intersect z2!
There are two possible manners to obtain the kinematic model.

There are two possible manners to obtain the kinematic model:
1) Use the previous model and add a constant rotation, in this case
T =




0 0 1 0
1 0 0 0
0 1 0 0
0 0 0 1




and then
0
T3,new = 0
T3T =




−C1S23 S1 C1C23 C1(a2C2 + a3C23)
−S1C23 −C1 S1C23 S1(a2C2 + a3C23)
C23 0 S23 a2S2 + a3S23
0 0 0 1




The unit vector s depends only on the ﬁrst joint. The position pz does not depend
on θ1.

2) Use the DH convention by adding suitable (ﬁctitious) i−1
Hi matrices.
In this case, it is necessary to add a rotation of π/2 about z and a rotation of π/2
about x and therefore the ‘DH’ angles θ = α = π/2.
x2
y2
z2
J3
L3
x3
y3
z3
y′
3
x′
3
z′
3
z′′
3
x′′
3
y′′
3

The new DH parameters table (the joint angle θ3 and the new angle θ are deﬁned
about the same axis and then it is possible to simply add them together):
d θ a α
L1 0 θ1 0 90o
L2 0 θ2 a2 0o
L3 0 θ3 + 90o
a3 90o
The i−1
Hi matrices are
0
H1=




C1 0 S1 0
S1 0 −C1 0
0 1 0 0
0 0 0 1



 ,1
H2=




C2 −S2 0 a2C2
S2 C2 0 a2S2
0 0 1 0
0 0 0 1



 ,2
H3=




−S3 0 C3 a3C3
C3 0 S3 a3S3
0 1 0 0
0 0 0 1





The kinematic model results:
0
T3,new =




−C1S23 S1 C1C23 C1(a2C2 + a3C23)
−S1C23 −C1 S1C23 S1(a2C2 + a3C23)
C23 0 S23 a2S2 + a3S23
0 0 0 1




The unit vector s depends only on the ﬁrst joint. The position pz does not depend
on θ1.

Example: 3 dof spherical manipulator
Denavit-Hartenberg parameters:
d θ a α
L1 0 θ1 0 −90o
L2 d2 θ2 0 90o
L3 d3 0 0 0o
The Denavit-Hartenberg matrices are:
0
H1 =




C1 0 −S1 0
S1 0 C1 0
0 −1 0 0
0 0 0 1



 , 1
H2 =




C2 0 S2 0
S2 0 −C2 0
0 1 0 d2
0 0 0 1



 , 2
H3 =




1 0 0 0
0 1 0 0
0 0 1 d3
0 0 0 1





The kinematic model results:
0
T3 =
n s a p
0 0 0 1
=




C1C2 −S1 C1S2 −d2S1 + d3C1S2
C2S1 C1 S1S2 d2C1 + d3S1S2
−S2 0 C2 d3C2
0 0 0 1




The third joint J3 does not aﬀect the orientation, s depends only on J1.

If θ1 = θ2 = 0o
, d3 = d
0
T3 =



1 0 0 0
0 1 0 d2
0 0 1 d
0 0 0 1


 y0
z0
x0
F0
d2
d
y3
z3
x3
F3
If θ1 = θ2 = 90o
, d3 = d
0
T3 =



0 −1 0 −d2
0 0 1 d
−1 0 0 0
0 0 0 1


 y0
z0
x0
F0
−d2
d
z3
x3
y3
F3

Example: 3 dof spherical wrist
Denavit-Hartenberg parameters
d θ a α
L4 0 θ4 0 −90o
L5 0 θ5 0 90o
L6 d6 θ6 0 0o
Note the numbers starting from 4...
Then
3
H4 =




C4 0 −S4 0
S4 0 C4 0
0 −1 0 0
0 0 0 1



 ,4
H5 =




C5 0 S5 0
S5 0 −C5 0
0 1 0 0
0 0 0 1



 ,5
H6 =




C6 −S6 0 0
S6 C6 0 0
0 0 1 d6
0 0 0 1





The kinematic model is:
3
T6 =




C4C5C6 − S4S6 −S4C6 − C4C5S6 C4S5 d6C4S5
S4C5C6 + C4S6 C4C6 − S4C5S6 S4S5 d6S4S5
−S5C6 S5S6 C5 d6C5
0 0 0 1




In this case, the rotation matrix has the same expression as an Euler ZYZ rotation
matrix.
REuler (φ, θ, ψ) = Rot(z0, φ)Rot(y1, θ)Rot(z2, ψ)
=


CφCθCψ − SφSψ −CφCθSψ − SφCψ CφSθ
SφCθCψ + CφSψ −SφCθSψ + CφCψ SφSθ
−SθCψ SθSψ Cθ


It means that the manipulator’s joints θ4, θ5 and θ6 are equivalent to the Euler
ZYZ angles.

If θ1 = θ2 = θ3 = 0o
3
T6 =




1 0 0 0
0 1 0 0
0 0 1 d6
0 0 0 1




If θ1 = θ2 = θ3 = 90o
3
T6 =




−1 0 0 0
0 0 1 d6
0 1 0 0
0 0 0 1





Example: Stanford manipulator
By composing the 3 dof spherical manipu-
lator with the spherical wrist, the so-called
“Stanford manipulator” is obtained, a 6
dof robot.
Since the frames have been deﬁned in a
consistent manner, the kinematic model is
simply obtained by multiplying the matri-
ces 0
T3 of the arm and 3
T6 of the wrist.
Then
0
T6 = 0
T3
3
T6 =
n s a p
0 0 0 1

Example: Stanford manipulator
where
n =


−S1(S4C5C6 + C4S6) + C1(C2(C4C5C6 − S4S6) − S2S5C6)
C1(S4C5C6 + C4S6) + S1(−S2S5C6 + C2(C4C5C6 − S4S6))
−C2S5C6 − S2(C4C5C6 − S4S6)


o =


−S1(C4C6 − S4C5S6) + C1(−C2(S4C6 + C4C5S6) + S2S5S6)
C1(C4C6 − S4C5S6) + S1(S2S5S6 − C2(S4C6 + C4C5S6))
C2S5S6 + S2(S4C6 + C4C5S6)


a =


−S1S4S5 + C1(S2C5 + C2C4S5)
C1S4S5 + S1(S2C5 + C2C4S5)
C2C5 − S2C4S5


p =


−d2S1 + d3C1S2 + d6(C1S2C5 + C1C2C4S5 − S1S4S5)
d2C1 + d3S1S2 + d6(S1S2C5 + S1C2C4S5 + C1S4S5)
d3C2 + d6(C2C5 − S2C4S5)



Example: PUMA 260
Joint variables θi are defined about the
zi−1 axes; a2 is the distance between z1
and z2 (in this case parallel), d3 is the
offset between the origins of F2 and F3 ,
and d4 is the offset between the origins of
F3 and F4 . Frames F4 and F5 coincides.
The αi angles are either 0o
or ±90o
.
d θ a α
L1 0 θ1 0 90o
L2 0 θ2 a2 0o
L3 −d3 θ3 0 90o
L4 d4 θ4 0 −90o
L5 0 θ5 0 90o
L6 d6 θ6 0 0o

Example: PUMA 260
Canonical transformation matrices:
0
H1=




C1 0 S1 0
S1 0 −C1 0
0 1 0 0
0 0 0 1



 ,1
H2=




C2 −S2 0 a2C2
S2 C2 0 a2S2
0 0 1 0
0 0 0 1



 ,2
H3=




C3 0 S3 0
S3 0 −C3 0
0 1 0 −d3
0 0 0 1




3
H4=




C4 0 −S4 0
S4 0 C4 0
0 −1 0 d4
0 0 0 1



 ,4
H5=




C5 0 S5 0
S5 0 −C5 0
0 1 0 0
0 0 0 1



 ,5
H6=




C6 −S6 0 0
S6 C6 0 0
0 0 1 d6
0 0 0 1





Example: PUMA 260
0
T3 =




C1C2C3 − C1S2S3 S1 C1C3S2 + C1C2S3 a2C1C2 − d3S1
C2C3S1 − S1S2S3 −C1 C3S1S2 + C2S1S3 C1d3 + a2C2S1
C3S2 + C2S3 0 −(C2C3) + S2S3 a2S2
0 0 0 1




3
T6 =




C4C5C6 − S4S6 −(C6S4) − C4C5S6 C4S5 C4d6S5
C5C6S4 + C4S6 C4C6 − C5S4S6 S4S5 d6S4S5
−(C6S5) S5S6 C5 d4 + C5d6
0 0 0 1




0
T6 =0
T6 = 0
T3
3
T6 =
n s a p
0 0 0 1

Example: PUMA 260
n =
S1(C5C6S4 + C4S6) + C1(C2(−(C6S3S5) + C3(C4C5C6 − S4S6)) − S2(C3C6S5 + S3(C4C5C6 − S4S6)))
−(C1(C5C6S4 + C4S6)) + S1(C2(−(C6S3S5) + C3(C4C5C6 − S4S6)) − S2(C3C6S5 + S3(C4C5C6 − S4S6)))
S2(−(C6S3S5) + C3(C4C5C6 − S4S6)) + C2(C3C6S5 + S3(C4C5C6 − S4S6))
s=
S1(C4C6 − C5S4S6)+C1(C2(S3S5S6 + C3(−(C6S4) − C4C5S6)) − S2(−(C3S5S6) + S3(−(C6S4) − C4C5S6)))
−(C1(C4C6 − C5S4S6))+S1(C2(S3S5S6 + C3(−(C6S4) − C4C5S6))−S2(−(C3S5S6)+S3(−(C6S4) − C4C5S6)))
S2(S3S5S6 + C3(−(C6S4) − C4C5S6)) + C2(−(C3S5S6)+S3(−(C6S4) − C4C5S6))
a =
S1S4S5 + C1(C2(C5S3 + C3C4S5) − S2(−(C3C5) + C4S3S5))
−(C1S4S5) + S1(C2(C5S3 + C3C4S5) − S2(−(C3C5) + C4S3S5))
S2(C5S3 + C3C4S5) + C2(−(C3C5) + C4S3S5)
p =
S1(−d3 + d6S4S5) + C1(a2C2 + C2((d4 + C5d6)S3 + C3C4d6S5) − S2(−(C3(d4 + C5d6)) + C4d6S3S5))
−(C1(−d3 + d6S4S5)) + S1(a2C2 + C2((d4 + C5d6)S3 + C3C4d6S5) − S2(−(C3(d4 + C5d6)) + C4d6S3S5))
a2S2 + S2((d4 + C5d6)S3 + C3C4d6S5) + C2(−(C3(d4 + C5d6)) + C4d6S3S5)

Example: planar 4 dof manipulator (redundant)
DH parameters
d θ a α
L1 0 θ1 a1 0o
L2 0 θ2 a2 0o
L3 0 θ3 a3 0o
L4 0 θ4 a4 0o
All the i−1
Hi matrices have the same
structure
i−1
Hi =




Ci −Si 0 ai Ci
Si Ci 0 ai Si
0 0 1 0
0 0 0 1





Example: planar 4 dof manipulator (redundant)
Then
0
T4 = 0
H1
1
H2
2
H3
3
H4 =
n s a p
0 0 0 1
=




C1234 −S1234 0 a1C1 + a2C12 + a3C123 + a4C1234
S1234 C1234 0 a1S1 + a2S12 + a3S123 + a4S1234
0 0 1 0
0 0 0 1




The vectors n, s, a define the end-effector orientation (rotation about z), while p
defines its position (on the x − y plane, pz = 0).
=⇒ The procedure can be applied also to redundant manipulators.

Inverse Kinematic Model
Claudio Melchiorri

Inverse Kinematic Model Introduction
Direct Kinematic Model:
The direct kinematic model consists in a function f(q) mapping the joint
position variables q ∈ IRn
to the position/orientation of the end effector.
The definition of f(q) is conceptually simple, and a general approach for its
computation has been defined.

Inverse Kinematic Model:
The inverse kinematics consists in finding a function g(x) mapping the
position/orientation of the end-effector to the corresponding joint variables q:
the problem is not simple!
A general approach for the solution of this problem does not exist
On the other hand, for the most common kinematic structures, a scheme for
obtaining the solution has been found. Unfortunately
The solution is not unique. In general we have:
No solution (e.g. starting with a position x not in the workspace);
A finite set of solutions (one or more);
Infinite solutions.
We seek for closed form solutions, and not based on numerical techniques:
The analytic solution is more efficient from the computational point of view;
If the solutions are known analytically, it is possible to select one of them on
the basis of proper criteria.

In order to obtain a closed form solution to the inverse kinematic problem, two
approaches are possible:
An algebraic approach, i.e. elaborations of the kinematic equations until a
suitable set of (simple) equations is obtained for the solution
A geometric approach based, when possible, on geometrical considerations,
dependent on the kinematic structure of the manipulator and that may help
in the solution.

Inverse Kinematic Model Algebraic Approach
Algebraic Approach
For a 6 dof manipulator, the kinematic model is described by the equation
0
T6 = 0
H1(q1) 1
H2(q2) . . . 5
H6(q6)
equivalent to 12 equations in the 6 unknowns qi , i = 1, . . . , 6.
Example: spherical manipulator (only 3 dof)
T=



0.5868 −0.6428 0.4394 −0.4231
0.5265 0.7660 0.3687 0.9504
−0.5736 0.0000 0.8192 0.4096
0 0 0 1


=



C1C2 −S1 C1S2 −d2S1 + d3C1S2
C2S1 C1 S1S2 d2C1 + d3S1S2
−S2 0 C2 d3C2
0 0 0 1



Since both the numerical values of 0
T6 and the structure of the i−1
Hi matrices
are known, by suitable pre- / post-multiplications it is possible to obtain
[ 0
H1(q1) . . .i−1
Hi (qi )]−1 0
T6 [ j
Hj+1(qj+1) . . .5
H6(q6)]−1
= i
Hi+1(qi+1) . . .j−1
Hj (qj ), i < j
obtaining 12 new equations for each couple (i, j), i < j.
By selecting the most simple equations among all those obtained, it might be
possible to obtain a solution to the problem.

Inverse Kinematic Model Geometric Approach
Geometric Approach
General considerations that may help in ﬁnding solutions with algebraic techniques
do not exist, being these strictly related to the mathematical expression of the
direct kinematic model. On the other hand, it is often possible to exploit
considerations related to the geometrical structure of the manipulator.
PIEPER APPROACH
Many industrial manipulators have a kinematically decoupled structure, for
which it is possible to decompose the problem in two (simpler) sub-problems:
1 Inverse kinematics for the position (x, y, z) → q1, q2, q3
2 Inverse kinematics for the orientation R → q4, q5, q6.

Geometric Approach
PIEPER APPROACH: Given a 6 dof manipulator, suﬃcient condition to ﬁnd a
closed form solution for the IK problem is that the kinematic structure presents:
three consecutive rotational joints with axes intersecting in a single point
or
three consecutive rotational joints with parallel axes.

Geometric Approach
In many 6 dof industrial manipulators, the ﬁrst 3 dof are usually devoted to
position the wrist, that has 3 additional dof give the correct orientation to the
end-eﬀector.
In these cases, it is quite simple to decompose the IK problem in the two
sub-problems (position and orientation).

Geometric Approach
In case of a manipulator with a spherical wrist, a natural choice is to decompose
the problem in
1 IK for the point pp (center of the spherical wrist)
2 solution of the orientation IK problem
Therefore:
1 The point pp is computed since 0
T6 is known (submatrices R and p):
pp = p − d6a
pp depends only on the joint variables q1, q2, q3;
2 The IK problem is solved for q1, q2, q3;
3 The rotation matrix 0
R3 related to the ﬁrst three joints is computed;
4 The matrix 3
R6 = 0
RT
3 R is computed;
5 The IK problem for the rotational part is solved (Euler).

Inverse Kinematic Model Examples
Solution of the spherical manipulator
Direct kinematic model:
0
T3 =
n s a p
0 0 0 1
=




C1C2 −S1 C1S2 −d2S1 + d3C1S2
C2S1 C1 S1S2 d2C1 + d3S1S2
−S2 0 C2 d3C2
0 0 0 1




If the whole matrix 0
T3 is known, it is possible to compute:



θ1 = atan2 (−sx , sy )
θ2 = atan2 (−nz , az )
d3 = pz / cosθ2
=⇒
Unfortunately, according to
the Pieper approach only p
is known!

We known only the position vector p.
From 0
T3 = 0
H1
1
H2
2
H3 we have
(0
H1)−1 0
T3 =




C1 S1 0 0
0 0 −1 0
−S1 C1 0 0
0 0 0 1








nx sx ax px
ny sy ay py
nz sz az pz
0 0 0 1



=




C2 0 S2 d3S2
S2 0 −C2 −d3C2
0 1 0 d2
0 0 0 1




= 1
H2
2
H3

By equating the position vectors,
1
pp =


px C1 + py S1
−pz
−pxS1 + py C1

 =


d3S2
−d3C2
d2


The vector 1
pp depends only on θ2 and d3! Let’s deﬁne a = tan θ1/2, then
C1 =
1 − a2
1 + a2
S1 =
2a
1 + a2
By substitution in the last element of 1
pp
(d2 + py )a2
+ 2px a + d2 − py = 0 =⇒ a =
−px ± p2
x + p2
y − d2
2
d2 + py
Two possible solutions! ((p2
x + p2
y − d2
2 ) > 0??)
Then
θ1 = 2 atan2 (−px ± p2
x + p2
y − d2
2 , d2 + py )

Vector 1
pp is defined as
1
pp =


px C1 + py S1
−pz
−pxS1 + py C1

 =


d3S2
−d3C2
d2


From the first two elements
pxC1 + py S1
−pz
=
d3S2
−d3C2
from which
θ2 = atan2 (px C1 + py S1, pz )
Finally, if the first two elements are squared and added together
d3 = (px C1 + py S1)2 + p2
z

Note that two possible solutions exist considering the position of the end-eﬀector
(wrist) only. If also the orientation is considered, the solution (if exists) is unique.
We have seen that the relation (p2
x + p2
y − d2
2 ) > 0 must hold:

Numerical example: Given a spherical manipulator with d2 = 0.8 m in the pose
θ1 = 20o
, θ2 = 30o
, d3 = 0.5 m
we have
0
T3 =




0.8138 −0.342 0.4698 −0.0387
0.2962 0.9397 0.171 0.8373
−0.5 0 0.866 0.433
0 0 0 1




• The solution based on the whole matrix 0
T3 is: θ1 = 20o
, θ2 = 30o
, d3 = 0.5.
• The solution based on the vector p gives:
a) θ1 = 20o
, θ2 = 30o
, d3 = 0.5 (with the “+” sign).
b) θ1 = −14.7o
, θ2 = −30o
, d3 = 0.5 (with the “-” sign).

• The solution based on the vector p gives:
a) θ1 = 20o
, θ2 = 30o
, d3 = 0.5 (with the “+” sign).
b) θ1 = −14.7o
, θ2 = −30o
, d3 = 0.5 (with the “-” sign).

Solution of the 3 dof anthropomorphic manipulator
From the kinematic structure, one obtains
θ1 = atan2 (py , px )
Concerning θ2 and θ3, note that the arm
moves in a plane deﬁned by θ1 only.
One obtains
C3 =
p2
x + p2
y + p2
z − a2
2 − a2
3
2a2a3
S3 = ± 1 − C2
3
θ3 = atan2 (S3, C3)
Moreover, by geometrical arguments, it is possible to state that:
θ2 = atan2 (pz , p2
x + p2
y ) − atan2 (a3S3, a2 + a3C3)

Solution of the 3 dof anthropomorphic manipulator
Note that also the solution is valid
θ1 = π + atan2 (py , px ), θ′
2 = π − θ2
Then, FOUR possible solutions exist:
shoulder right - elbow up; shoulder right - elbow down;
shoulder left - elbow up; shoulder left - elbow down;
Same position, but diﬀerent orientation!
Note that the conditions px = 0, py = 0 must hold (singular conﬁguration).

Solution of the spherical wrist
Let us consider the matrix
3
R6 =


nx sx ax
ny sy ay
nz sz az


From the direct kinematic equations one obtains
3
R6 =


C4C5C6 − S4S6 −S4C6 − C4C5S6 C4S5
S4C5C6 + C4S6 C4C6 − S4C5S6 S4S5
−S5C6 S5S6 C5



3
R6 =
C4C5C6 − S4S6 −S4C6 − C4C5S6 C4S5
S4C5C6 + C4S6 C4C6 − S4C5S6 S4S5
−S5C6 S5S6 C5
The solution is then computed as (ZYZ Euler angles):
θ5 ∈ [0, π]:
θ4 = atan2 (ay , ax )
θ5 = atan2 ( a2
x + a2
y , az )
θ6 = atan2 (sz , −nz )
θ5 ∈ [−π, 0]:
θ4 = atan2 (−ay , −ax )
θ5 = atan2 (− a2
x + a2
y , az )
θ6 = atan2 (−sz , nz )

Numerical example: Let us consider a spherical wrist in the pose
θ4 = 10o
θ5 = 20o
, θ6 = 30o
Then
3
R6 =
0.7146 −0.6131 0.3368
0.6337 0.7713 0.0594
−0.2962 0.1710 0.9397
Therefore, if
• θ5 ∈ [0, π] θ4 = 10o
θ5 = 20o
, θ6 = 30o
• θ5 ∈ [−π, 0] θ4 = −170o
θ5 = −20o
, θ6 = −30o
Note that the PUMA has an anthropomorphic structure (4 solutions) and a
spherical wrist (2 solutions):
=⇒ 8 diﬀerent conﬁgurations are possible!

Diﬀerential Kinematics
Diﬀerential Kinematics
Claudio Melchiorri

Differential Kinematics Introduction
Differential Kinematics: the Jacobian matrix
In robotics it is of interest to define, besides the
mapping between the joint and workspace posi-
tion/orientation (i.e. the kinematic equations),
also
The relationship between the joints and
end-effector velocities:
v
ω
⇐⇒ ˙q
The relationship between the force applied
on the environment by the manipulator and
the corresponding joint torques
f
n
⇐⇒ τ
These two relationships are based on a linear operator, a matrix J, called the
Jacobian of the manipulator.

Differential Kinematics Introduction
The Jacobian matrix
In robotics, the Jacobian is used for several purposes:
To define the relationship between joint and workspace velocities
To define the relationship between forces/torques between the spaces
To study the singular configurations
To define numerical procedures for the solution of the IK problem
To study the manipulability properties.

Diﬀerential Kinematics The Jacobian Matrix
Velocity domain
The translational and rotational velocities are considered separately.
Let us consider two frames F0 (base frame) and F1 (integral with the rigid body).
The translational velocity of point p of the rigid body, with respect to F0 , is
deﬁned as the derivative (with respect to time) of p, denoted as ˙p:
˙p =
d p
dt

Velocity domain
With respect to the rotational velocity, two different definitions are possible:
1 A triple γ ∈ IR3
giving the orientation of F1 with respect to F0 (Euler,
RPY,... angles) is adopted, and its derivative is used to define the rotational
velocity ˙γ
˙γ =
dγ
dt
2 An angular velocity vector ω is defined, giving the rotational velocity of a
third frame F2 with origin coincident with F0 and axes parallel to F1 .
The velocity vector ω is placed in the ori-
gin, and its direction coincides with the
instantaneous rotation axis of the rigid
body.

Analytical and Geometrical expressions of the Jacobian
The two descriptions lead to diﬀerent results concerning the expression of the
Jacobian matrix, in particular in the part relative to the rotational velocity.
One obtains respectively the:
Analytic Jacobian JA
Geometric Jacobian JG
Two problems:
1) Integration of the rotational velocity
˙γdt =⇒ γ: orientation of the rigid body
ωdt =⇒ ??

Example: Let’s consider a rigid body and the following rotational velocities
Case a)
ω = [π/2, 0, 0]T
0 ≤ t ≤ 1
ω = [0, π/2, 0]T
1 < t ≤ 2
Case b)
ω = [0, π/2, 0]T
0 ≤ t ≤ 1
ω = [π/2, 0, 0]T
1 < t ≤ 2
By integrating the velocities in the two cases, one obtains:
2
0
ωdt = [π/2, π/2, 0]T

Case a)
ω = [π/2, 0, 0]T
0 ≤ t ≤ 1
ω = [0, π/2, 0]T
1 < t ≤ 2
Case b)
ω = [0, π/2, 0]T
0 ≤ t ≤ 1
ω = [π/2, 0, 0]T
1 < t ≤ 2
2
0
ωdt = [π/2, π/2, 0]T
On the other hand, the rotation matrices in the two cases are:
Ra =


0 1 0
0 0 −1
−1 0 0

 Rb =


0 0 1
1 0 0
0 1 0


The integration of ω does not have a clear physical interpretation.

2) On the other hand:
ω represents the velocity components about the three axes of F0 ,
the elements of ˙γ are deﬁned with respect to frame that: a) is not Cartesian
(its axes are not orthogonal each other); b) varies in time according to γ.

The two expressions of the Jacobian matrix physically deﬁne the same
phenomenon (velocity of the manipulator), and therefore a relationship between
them must exist. For example, if the Euler angles φ, θ, ψ are used for the triple γ,
it is possible to show that
ω =


0 − sinφ cos φ sin θ
0 cos φ sin φ sin θ
1 0 cos θ

 ˙γ = T(γ) ˙γ
Note that matrix T(γ) is singular when
sin θ = 0.
In this case, some rotational velocities
may be expressed by ω and not by ˙γ, e.g.
ω = [cos φ, sin φ, 0]T
.
These cases are called representation singu-
larities of γ.

Deﬁnition of matrix T(γ):
˙φ : [ωx , ωy , ωz ]T
= ˙φ
0
0
1
ωz = ˙φ
˙θ : [ωx , ωy , ωz ]T
= ˙θ
−Sφ
Cφ
0
ωx = −Sφ
˙θ
ωy = Cφ
˙θ
˙ψ : [ωx , ωy , ωz ]T
= ˙ψ
−CφSθ
SφSθ
Cθ



ωz = Cθ
˙ψ
ωx = SθCφ
˙ψ
ωy = SθSφ
˙ψ

If sin θ = 0, then the components perpendicular to z of the velocity expressed by ˙γ
are linearly dependent (ω2
x + ω2
y = ˙θ2
), while physically this constraint may not
exist!
From:
ω =


0 − sin φ cos φ sin θ
0 cos φ sin φ sin θ
1 0 cos θ

 ˙γ
one obtains:


0 −Sφ 0
0 Cφ 0
1 0 1




˙φ
˙θ
˙ψ

 =⇒


−Sφ
˙θ
Cφ
˙θ
˙φ + ˙ψ

 =


ωx
ωy
ωz

 =⇒
ω2
x + ω2
y = ˙θ2
ωz = ˙φ + ˙ψ

In general, given a triple of angles γ, a transformation matrix T(γ) exists such that
ω = T(γ) ˙γ
Once matrix T(γ) is known, it is possible to relate the analytical and geometrical
expressions of the Jacobian matrix:
v
ω
=
I 0
0 T(γ)
˙p
˙γ
Then
JG =
I 0
0 T(γ)
JA = TA(γ)JA

Differential Kinematics Analytical Jacobian
Analitycal Jacobian
The analytical expression of the Jacobian is obtained by differentiating a vector
x = f(q) ∈ IR6
, that defines the position and orientation (according to some
convention) of the manipulator in F0 .
By differentiating f(q), one obtains
dx =
∂f(q)
∂q
dq
= J(q)dq
where the m × n matrix
J(q) =
∂f(q)
∂q
=


∂f1
∂q1
∂f1
∂q2
. . . ∂f1
∂qn
. . .
∂fm
∂q1
∂fm
∂q2
. . . ∂fm
∂qn

 J(q) ∈ IRm×n
is called Jacobian matrix or JACOBIAN of the manipulator.

Analitycal Jacobian
If the inﬁnitesimal period of time dt is considered, on obtains
d x
dt
= J(q)
d q
dt
that is
˙x =
v
˙γ
= J(q) ˙q
relating the velocity vector ˙x expressed in F0 and the joint velocity vector ˙q.
The elements Ji,j of the Jacobian are non linear functions of q(t): therefore
these elements change in time according to the value of the joint variables.
The Jacobian dimensions depend on the number n of joints and on the
dimension m of the considered operative space (J(q) ∈ IRm×n
).

Example: 2 dof manipulator
d θ a α
L1 0 θ1 a1 0o
L2 0 θ2 a2 0o
The end-effector position is
px = a1C1 + a2C12
py = a1S1 + a2S12
pz = 0
If γ is composed by the Euler angles φ, θ, ψ defined about axes z0, y1, z2, and
considering that the z axes of the base frame and of the end effector are parallel,
the total rotation is equivalent to a single rotation about z0 and therefore:


φ
θ
ψ

 =


θ1 + θ2
0
0



Euler angles:


φ
θ
ψ

 =


θ1 + θ2
0
0


By diﬀerentiation of the expressions of p and γ one obtains
˙p
˙γ
=








−a1S1 − a2S12 −a2S12
a1C1 + a2C12 a2C12
0 0
1 1
0 0
0 0








˙q
= J(q) ˙q

Differential Kinematics Geometric Jacobian
Geometric Expression of the Jacobian
The geometric expression of the Jacobian is obtained considering the rotational
velocity vector ω.
Each column of the Jacobian matrix defines the effect of the i-th joint on the
end-effector velocity. It is divided in two terms.
The first term considers the effect of ˙qi on the linear velocity v, while the second
one on the rotational velocity ω, i.e.
v
ω
= J ˙q =⇒ J =
Jv1 Jv2 . . . Jvn
Jω1 Jω2 . . . Jωn
Therefore
v = Jv1 ˙q1 + Jv2 ˙q2 + . . . + Jvn ˙qn
ω = Jω1 ˙q1 + Jω2 ˙q2 + . . . + Jωn ˙qn
The analytic and geometric Jacobian differ for the rotational part.
In order to obtain the geometric Jacobian, a general method based on the
geometrical structure of the manipulator is adopted.

Derivative of a Rotation Matrix
Let’s consider a rotation matrix R = R(t) and R(t)RT
(t) = I .
Let’s derive the equation: R(t)RT
(t) = I =⇒ ˙R(t)RT
(t) + R(t) ˙RT
(t) = 0
A 3 × 3 (skew-symmetric) matrix S(t) is obtained
S(t) = ˙R(t)RT
(t)
As a matter of fact
S(t) + ST
(t) = 0 =⇒ S =


0 −ωz ωy
ωz 0 −ωx
−ωy ωx 0


Then
˙R(t) = S(t)R(t)
This means that the derivative of a rotation matrix is expressed as a function of
the matrix itself.

Physical interpretation:
Matrix S(t) is expressed as a function of a vector ω(t) = [ωx , ωy , ωz]T
representing the angular velocity of R(t).
Consider a constant vector p′
and the vector p(t) = R(t)p′
. The time derivative
of p(t) is
˙p(t) = ˙R(t)p′
which can be written as
˙p(t) = S(t)R(t)p′
= ω × (R(t) p′
)
This last results, i.e. ˙p(t) = ω × (R(t) p′
), is well known from the classic
mechanics of rigid bodies.

Moreover
R S(ω) RT
= S(R ω)
i.e. the matrix form of S(ω) in a frame rotated by R is the same as the
skew-symmetric matrix S(R ω) of the vector ω rotated by R.
Consider two frames F and F’, which diﬀer by the rotation R (ω′
= R ω). Then
S(ω′
) operates on vectors in F’ and S(ω) on vectors in F.
Consider a vector v′
a in F’ and assume that some operations must be performed
on that vector in F (then using S). It is necessary to:
1 Transform the vectors from F’ to F (by RT
)
2 Use S(ω)
3 Transform back the result to F’ (by R)
That is: v′
b = R S(ω) RT
v′
a
equivalent to the transformation using S(ω′
): v′
b = S(ω′
) v′
a

Example
Consider the elementary rotation about z
Rot(z, θ) =


cos θ − sin θ 0
sin θ cos θ 0
0 0 1


If θ is a function of time
S(t)=


− ˙θ sin θ − ˙θ cos θ 0
˙θ cos θ − ˙θ sin θ 0
0 0 0




cos θ sin θ 0
− sin θ cos θ 0
0 0 1

=


0 − ˙θ 0
˙θ 0 0
0 0 0

 = S(ω(t))
Then
ω =


0
0
˙θ

 i.e. a rotational velocity about z.

Geometric Jacobian
The end-effector velocity is a linear composition of the joint velocities
v = Jv1 ˙q1 + Jv2 ˙q2 + . . . + Jvn ˙qn
ω = Jω1 ˙q1 + Jω2 ˙q2 + . . . + Jωn ˙qn
Each column of the Jacobian
matrix defines the effect of the
i-th joint on the end-effector ve-
locity.

Geometric Jacobian
It is possible to show (see Appendix) that the i-th column of the Jacobian can be
computed as
Jvi
Jωi
=
0
zi−1 × (0
pn − 0
pi−1)
0
zi−1
revolute joint
Jvi
Jωi
=
0
zi−1
0
prismatic joint
where 0
zi−1 and 0
ri−1,n = 0
pn − 0
pi−1 depend on the joint variables
q1, q2, ..., qn. In particular:
0
pn is the end-effector position, defined in the first three elements of the last
column of 0
Tn = 0
H1(q1) . . . n−1
Hn(qn);
0
pi−1 is the position of Fi−1, defined in the first three elements of the last
column of 0
Ti−1 = 0
H1(q1) . . . i−2
Hi−1(qi−1);
0
zi−1 is the third column of 0
Ri−1:
0
Ri−1 = 0
R1(q1) 1
R2(q2) . . . i−2
Ri−1(qi−1)

Diﬀerential Kinematics Examples
The Jacobian is computed as
J =
z0 × (p2 − p0) z1 × (p2 − p1)
z0 z1
The origins of the frames are
p0 =


0
0
0

 p1 =


a1C1
a1S1
0

 p2 =


a1C1 + a2C12
a1S1 + a2S12
0


and the rotational axes are
z0 = z1 =


0
0
1



Then
z0 × (p2 − p0) = −


0 0 a1S1 + a2S12
0 0 −a1C1 − a2C12
−a1S1 − a2S12 a1C1 + a2C12 0




0
0
1


=


−a1S1 − a2S12
a1C1 + a2C12
0


z1 × (p2 − p1) = −


0 0 a2S12
0 0 −a2C12
−a2S12 a2C12 0




0
0
1

 =


−a2S12
a2C12
0


In conclusion:
J(q) =








−a1S1 − a2S12 −a2S12
a1C1 + a2C12 a2C12
0 0
0 0
0 0
1 1









Jacobian:
J(q) =








−a1S1 − a2S12 −a2S12
a1C1 + a2C12 a2C12
0 0
0 0
0 0
1 1








If the orientation is not of interest, only the ﬁrst two rows may be considered
J(q) =
−a1S1 − a2S12 −a2S12
a1C1 + a2C12 a2C12

Example: 3dof anthropomorphic manipulator
The canonical transformation matrices are
0
H1 =




C1 0 S1 0
S1 0 −C1 0
0 1 0 0
0 0 0 1




1
H2 =




C2 −S2 0 a2C2
S2 C2 0 a2S2
0 0 1 0
0 0 0 1




2
H3 =




C3 −S3 0 a3C3
S3 C3 0 a3S3
0 0 1 0
0 0 0 1




and the kinematic model
0
T3 =




C1C23 −C1S23 S1 C1(a2C2 + a3C23)
S1C23 −S1S23 −C1 S1(a2C2 + a3C23)
S23 C23 0 a2S2 + a3S23
0 0 0 1





The Jacobian results
J =
z0 × (p3 − p0) z1 × (p3 − p1) z2 × (p3 − p2)
z0 z1 z2
where
p0 = p1 =


0
0
0

 p2 =


a2C1C2
a2S1S2
a2S2

 p3 =


C1(a2C2 + a3C23)
S1(a2C2 + a3C23)
a2S2 + a3S23


The rotational axes are
z0 =


0
0
1

 z1 = z2 =


S1
−C1
0



Therefore
J =








−S1(a2C2 + a3C23) −C1(a2S2 + a3S23) −a3C1S23
C1(a2C2 + a3C23) −S1(a2S2 + a3S23) −a3S1S23
0 a2C2 + a3C23 a3C23
0 S1 S1
0 −C1 −C1
1 0 0








- Only three rows are linearly independent (3 dof).
- Note that it is not possible to achieve all the rotational velocities ω in IR3
.
- Moreover S1ωy = −C1ωx (ωx = S1
˙θ2 + S1
˙θ3, ωy = −C1
˙θ2 − C1
˙θ3, ).
By considering the linear velocity only, one obtains:
J =


−S1(a2C2 + a3C23) −C1(a2S2 + a3S23) −a3C1S23
C1(a2C2 + a3C23) −S1(a2S2 + a3S23) −a3S1S23
0 a2C2 + a3C23 a3C23



Note that:
˙θ1 does not affect vz (nor ωx , ωy )
ωz depends only by ˙θ1
S1ωy = −C1ωx : ωx and ωy are not independent
the first three rows may also be obtained by derivation of 0
p3
In the “linear velocity” case (i.e. the first three rows only)
det(J) = −a2a3S3(a2C2 + a3C23)
Therefore det(J) = 0 in two cases:
S3 = 0 =⇒ θ3 =
0
π
(a2C2 + a3C23) = 0 i.e. when the wrist is on the z axis (px = py = 0):
shoulder singularity

Example 3 dof spherical manipulator
Canonical transformation matrices
0
H1 =




C1 0 −S1 0
S1 0 C1 0
0 −1 0 0
0 0 0 1



 ,1
H2 =




C2 0 S2 0
S2 0 −C2 0
0 1 0 d2
0 0 0 1




2
H3 =




1 0 0 0
0 1 0 0
0 0 1 d3
0 0 0 1




Kinematic model:
0
T3 =




C1C2 −S1 C1S2 −d2S1 + d3C1S2
C2S1 C1 S1S2 d2C1 + d3S1S2
−S2 0 C2 C2d3
0 0 0 1





The Jacobian is
J =
z0 × (p3 − p0) z1 × (p3 − p1) z2
z0 z1 0
with
z0 =


0
0
1

 z1 =


−S1
C1
0

 z2 =


C1S2
S1S2
C2


and
p0 = p1 =


0
0
0

 p2 =


−d2S1
d2C1
0

 p3 =


−d2S1 + d3C1S2
d2C1 + d3S1S2
C2d3



Then
J =








−d2C1 − d3S1S2 d3C1C2 C1S2
−d2S1 + d3C1S2 d3S1C2 S1S2
0 −d3S2 C2
0 −S1 0
0 C1 0
1 0 0








Note that:
˙q3 does not aﬀect ω;
ωz depends only on ˙q1;
S1ωy = −C1ωx.

J =








−d6S4S5 d6C4C5 0
d6C4S5 d6C5S4 0
0 −d6S5 0
0 −S4 C4S5
0 C4 S4S5
1 0 C5








By choosing d6 = 0, i.e. the origin of F6 is in the intersection point of the three
joint axes, then
J =






0 0 0
0 0 0
0 0 0
0 −S4 C4S5
0 C4 S4S5
1 0 C5






With this expression, however, the linear
velocity of the end-eﬀector is not com-
puted.
det(J) = 0 =⇒ S5 = 0, i.e. θ5 = 0, π.
In this case it is not possible to determine
individually ˙θ4 and ˙θ6.

Example: PUMA 260
Only revolute joints are present:
J =
z0 × (p6 − p0) z1 × (p6 − p1) z2 × (p6 − p2)
z0 z1 z2
z3 × (p6 − p3) z4 × (p6 − p4) z5 × (p6 − p5)
z3 z4 z5

Example: PUMA 260
If d6 = 0:
J =






−d3C1 − S1(a2C2 + d4S23) C1(d4C23 − a2S2) d4C1C23
−d3S1 + C1(a2C2 + d4S23) S1(d4C23 − a2S2) d4S1C23
0 a2C2 + d4S23 d4S23
0 S1 S1
0 −C1 −C1
1 0 0
0 0 0
0 0 0
0 0 0
C1S23 S1C4 − C1C23S4 C1S23C5 + C1C23C4S5 + S1S4S5
S1S23 −C1C4 − S1C23S4 S1S23C5 + S1C23C4S5 − C1S4S5
−C23 −S23S4 −C23C5 + S23C4S5







Statics - Singularities - Inverse diﬀerential kinematics
Statics - Singularities - Inverse diﬀerential kinematics
Claudio Melchiorri

Statics - Singularities - Inverse diﬀerential kinematics Statics
Forces
The relation ˙x = J˙q maps velocities deﬁned in the joint space to corresponding
velocities in the operational space. On this basis, exploiting the virtual work
principle, a similar mapping can be established considering the force domain.
Since the work, computed as the product between the applied force and the
displacement, is invariant with respect to the frame used to compute it, one
obtains:
wT
dx = τT
dq
being w = [fT
nT
]T
a 6 × 1 vector composed by the linear forces f and torques n
applied to the manipulator, and τ the n × 1 vector collecting the forces/torques
applied to the joints.

Statics - Singularities - Inverse diﬀerential kinematics Statics
Forces
By recalling that
dx = J(q)dq
one obtains
τ = JT
(q)w
deﬁning the relationship between the joint torque vector τ and the force vector w,
applied to the manipulator.
J(q) =⇒ mapping between ˙q and [vT
, ωT
]T
JT
(q) =⇒ mapping between [fT
, nT
]T
and τ
˙q (τ)
J(q)
JT (q)
˙x (w)

Statics - Singularities - Inverse diﬀerential kinematics Transformation of reference frame for the Jacobian
Transformation of reference frame for the Jacobian
Let’s consider two frames Fa and
Fb attached to a rigid body (“robot”).
If a ˙x is the end-eﬀector velocity in Fa , then
a
˙x = a
J ˙q
It is known that it is possible to transform the same velocity in another frame
Fb as
b
˙x = b
Ga
a
˙x
where b
Ga is the transformation matrix between the two frames:
b
Ga =
b
Ra − b
Ra
a
Pab
0 b
Ra

Statics - Singularities - Inverse diﬀerential kinematics Transformation of reference frame for the Jacobian
Transformation of reference frame for the Jacobian
Then
b
˙x = b
J ˙q = b
Ga
a
˙x = b
Ga
a
J ˙q
and therefore the transformation for the Jacobian between the two frames Fa and
Fb is given by
b
J = b
Ga
a
J
Similar considerations hold in the force domain (where the transpose Jacobian is
used).
Note that if the two frames are not rigidly attached to the robot, then the
Jacobian transformation between them is deﬁned only by the rotation matrix b
Ra:
b
J =
b
Ra 0
0 b
Ra
a
J

Statics - Singularities - Inverse differential kinematics Kinematic singularities
Singular configurations
The Jacobian is a 6 × n matrix mapping the IRn
joint velocity space to the IR6
operational velocity space:
˙x = J(q)˙q
From an infinitesimal point of view, this relationship is expressed as
dx = J(q)dq
that can be interpreted as a relationship between infinitesimal displacements in
IRn
and IR6
.
Singular configurations or Singulariities
In general, rank(J(q)) = min (6, n). On the other hand, since the elements of J
are function of the joints, some robot configurations exist such that the Jacobian
looses rank. These configurations are denoted as kinematic singularities.
In these configurations, there are “directions” (vectors ˙x) in IR6
without any
correspondent “direction” (˙q) in IRn
: these directions cannot be actuated and the
robot ”looses” a motion capability.

Singular configurations
The singular configurations are important for several reasons:
1 They represents configurations in which the motion capabilities of the robot are
reduced: it is not possible to impose arbitrary motions of the end-effector
2 In the proximity of a singularity, small velocities in the operational space may
generate large (infinite) velocities in the joint space
3 They correspond to configurations that have not a well defined solution to the
inverse kinematic problem: either no solution or infinite solutions.
There are two types of singularities:
1 Boundary singularities, that correspond to points on the border of the workspace,
i.e. when the robot is either fully extended or retracted. These singularities may be
easily avoided by not driving the manipulator to the border of its workspace
2 Internal singularities, that occur inside the reachable workspace and are generally
caused by the alignment of two or more axes of motion, or else by the attainment
of particular end-effector configurations. These singularities constitute a serious
problem, as they can be encountered anywhere in the reachable workspace for a
planned path in the operational space.

Example
Consider the Jacobian J(q)
J =
1 1
0 0
Then rank(J) = 1 and
dx = dq1 + dq2
dy = 0
For any value of dq1, dq2, then dy = 0.
Any vector dx whose second element is not null represents an instantaneous
motion that cannot be achieved.

Example
Jacobian:
J =
1 1
0 0
In the force domain
τ = JT
f =
1 0
1 0
fx
fy
then
τ1 = fx
τ2 = fx
then, fy does not aﬀect the joint torques, and any torque vector such that
τ1 = −τ2 does not generate any force on the environment.

Example
Consider the 2 dof planar manipulator:
J(q) =
−a1S1 − a2S12 −a2S12
a1C1 + a2C12 a2C12
If θ1 = θ2 = 0, then
J(q) =
0 0
a1 + a2 a2
=
0 0
2 1
if a1 = a2 = 1
Therefore
dx = 0
dy = 2 dq1 + dq2
Moreover
JT
(q) =
0 2
0 1
=⇒
τ1 = 2 fy
τ2 = fy
If τ2 = −2 τ1 → fx = fy = 0.

Example
J(q) =
−a1S1 − a2S12 −a2S12
a1C1 + a2C12 a2C12
Consider the velocity vector ˙x = [−1, −1]T
. By computing ˙q = J−1
(q)˙x:
If θ1 = 0o
, θ2 = 1o
then ˙q =
−58.9
115.59 (rad/sec) =
−3374.7
6622.8 (o
/sec)
If θ1 = 0o
, θ2 = 10o
then ˙q =
−6.67
12.43 (rad/sec) =
−382.16
712.18 (o
/sec)

Example
J(q) =
−a1S1 − a2S12 −a2S12
a1C1 + a2C12 a2C12
Plot of ˙q = J(q)−1
˙x
with ˙x =
−1
−1 and
θ1 = 0, θ2 ∈ [0.0005, 0.01] rad
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01
−3000
−2000
−1000
0
1000
2000
3000
4000
5000
q2 [rad]
dq1,dq2[rad/sec]
Velocita‘ di giunto (dq2 dash)

Singularity decoupling
In case of complex structures, the analysis of the kinematic singularities via the
Jacobian determinant det(J) may result quite difficult.
In case of manipulators with spherical wrist, by similarity with the inverse
kinematics, it is possible to decompose the study of the singular configurations
into two sub-problems:
arm singularities (e.g. the first three joints)
wrist singularities
If J ∈ IR6×n
then
J =
J11 J12
J21 J22
where, since the last three joints are of the revolute type:
J12 = [z3 × (p6 − p3), z4 × (p6 − p4), z5 × (p6 − p5)] J22 = [z3, z4, z5]

Singularity decoupling
Singularities depend on the mechanical structure, not on the frames chosen to
describe kinematics. Therefore, it is convenient to choose the origin of the
end-eﬀector frame at the intersection of the wrist axes, where also the last frames
are placed. Then:
J12 = [0, 0, 0] =⇒ J =
J11 0
J21 J22
In this manner, J is a block lower-triangular matrix, and
det(J) = det(J11)det(J22)
The singularities are then decoupled, since
det(J11) = 0 gives the arm singularities, while
det(J22) = 0 gives the wrist singularities.

Statics - Singularities - Inverse differential kinematics Inverse Differential Kinematics
Inverse Differential Kinematics
The ‘direct’ relationship between joint and operational space velocities:
˙x = J(q)˙q
is defined by the m × n Jacobian matrix J.
Inverse problem: ˙x =⇒ ˙q
A solution of the linear system ˙x = J(q)˙q is seeked.
In case m = n, the inverse of the Jacobian matrix is employed:
˙q = J−1
(q)˙x
Otherwise, it is necessary to use its (Moore-Penrose) pseudo-inverse
˙q = J+
(q)˙x
where:
J+
= JT
(JJT
)−1
if m < n (right pseudo-inverse : JJ+
= I)
J+
= (JT
J)−1
JT
if m > n (left pseudo-inverse: J+
J = I)

If m = n there are two cases:
J
a) m < n
J
b) m > n
• a): JJ+
= Im
J JT
(JJT
)−1
= Im =⇒ J+
= JT
(JJT
)−1
• b): J+
J = In
(JT
J)−1
JT
J = In =⇒ J+
= (JJT
)−1
JT

Solution of ˙x = J˙q if m = n, det(J) = 0
J(q)
J−1
(q)
The equation ˙x = J˙q (as well as ˙q = J−1 ˙x) has an unique solution:
∀ ˙xo → ∃ ! ˙qo = J−1
˙xo such that ˙xo = J˙qo = J J−1
˙xo

Solution of ˙x = J˙q if m < n
N(J)
IRn
0
J(q)
IRm
• rank(J) = min(m, n) = m → R(J) = IRm
• ∀ ˙x ∃ ˙q such that ˙x = J˙q (multiple solutions!)
• ˙q = J+
˙x ∃ N(J) such that ∀ ˙q ∈ N(J) → ˙x = J ˙q = 0 dim(N(J)) = n − m
→ ˙q = J+
˙x + qN → ˙x = J (J+
˙x + ˙qN ) = ˙x, ∀˙qN ∈ N(J)
→ ˙q = J+
˙x + (I − J+
J)y general expression of the solution (I − J+
J) base of N(J)
• ˙q = J+
˙x has minimum norm

Solution oﬁ ˙x = J˙q if m > n
IRn
R(J)
IRm
• rank(J) = min(m, n) = n
• ∀ ˙q ∃ ! ˙x such that ˙x = J˙q ∈ R(J)
• ∀ ˙x ∈ R(J) ∃ ! ˙q such that ˙x = J˙q (˙q = J−1
˙x)
• If ˙x ∈ R(J) → ∃ ˙q such that ˙x = J˙q
• If ˙x0 ∈ R(J) → ∃ ˙q0 = J+
˙x0 → ˙x = J˙q0 = JJ+
˙x0 = ˙x0 (JJ+
= I)
• ˙x − ˙x0 is minimum

Solution of ˙x = J˙q in general if m = n
N(J)
IRn
R(J)
IRm
• rank(J) = p < min(m, n)
• The solution given by the pseudo-inverse ˙qs = J+
˙x is such that (˙xs = J˙qs ):
˙xs − ˙x the norm of the error is minimum
˙qs the norm of the solution is minimum

Statics - Singularities - Inverse differential kinematics Accelerations
Accelerations
If accelerations are of interest, by differentiating ˙x = J˙q one obtains
¨x = J(q)¨q +
d J(q)
dt
˙q
If an acceleration vector ¨x is given in the operational space, the corresponding
vector ¨q is computed as a solution of
b = J(q)¨q
being b = ¨x − d J(q)
dt ˙q. Then
¨q = J−1
b if the inverse of the Jacobian exists
¨q = J+
b otherwise
If the Jacobian is not square (e.g. in case of redundant manipulators (m < n) or with
less than 6 dof), an exact solution of ˙x = J˙q (b = J¨q) exists iff ˙x ∈ R(J) (b ∈ R(J)).
A vector a is in R(A) iff
rank([A, a]) = rank[A]

Inverse kinematics algorithms
Claudio Melchiorri

Inverse kinematics algorithms Introduction
The Jacobian matrix can be used also for the solution of the inverse kinematic
problem.
If a desired trajectory is known in terms of the velocity v(kT) = vk, a simple
approach is to consider
qk+1 = qk + J−1
(qk )vk T
equivalent to a numerical integration over time of the velocity. This operation has
two major drawbacks aﬀecting the computation of the exact solution:
numerical drifts
initialization problems
These problems may be avoided by implementing a feedback scheme accounting
for the operational space errors e = xd − x. Then
˙e = ˙xd − ˙x = ˙xd − Ja(q)˙q (2)
The vector ˙q must be chosen so that the error e converges to zero.

Inverse kinematics algorithms Jacobian (pseudo)-inverse
Scheme 1: Jacobian (pseudo)-inverse
Assuming that Ja is invertible, then ˙q = J−1
a (q) (˙xd + Ke)
By substituting this expression in (2) one obtains
˙e + Ke = 0
representing, if K is positive deﬁnite, an asymptotically stable system. Note that
the convergence velocity depends on K.
K J−1
a (q)
f(·)
❥❥ ✲ ✲
✛
✲✲✲✲
✻
❄
−
xd
˙xd
e ˙q q
x

Inverse kinematics algorithms Jacobian transpose
Scheme 2: Jacobian transpose
This scheme is based on the Lyapunov approach. A Lyapunov function must be
found guaranteeing the convergence to zero of the error e.
Let’s assume
V (e) =
1
2
eT
K e
with K symmetric and positive deﬁnite. In this manner
V (e) > 0, ∀e = 0, V (0) = 0
By diﬀerentiating V (e) one obtains
˙V = eT
K˙e = eT
K˙xd − eT
K˙x
= eT
K˙xd − eT
KJa(q)˙q

Scheme 2: Jacobian transpose
From
˙V = eT
K˙xd − eT
KJa(q)˙q
By choosing
˙q = JT
a (q)Ke
one obtains
˙V = eT
K˙xd − eT
KJa(q)JT
a (q)Ke
If ˙xd = 0 then ˙V < 0 and the system is asymptotically stable if Ja is full rank.
If Ja is not full rank, then the condition ˙q = 0 may be obtained also with e = 0
(Ke ∈ null(JT
a )).
K JT
a (q)
f(·)
❥ ✲ ✲
✛
✲✲
✻−
xd e ˙q q
x
✲

Example
Consider the non linear function
z = f(q) =
x
y
=
q3
1 + sin(q1q2)
q1q3
2 + sin(q2
1 + 2q2)
The Jacobian is
J(q) =
3q2
1 + q2cos(q1q2) q1cos(q1q2)
q3
2 + 2q1cos(q2
1 + 2q2) 3q1q2
2 + 2cos(q2
1 + 2q2)
Assuming z0 = [0.1, 0.2]T
, ﬁnd q0 = f−1
(z0).

Example
With null initial conditions (q1 = q2 = 0), the following results are computed with
the two algorithms (Ts = 0.001 s).
K = 100
(pseudo-) Inverse J−1
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
−0.6
−0.4
−0.2
0
0.2
Algoritmo J^−1; valori x, y (dash)
time [s]
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
−1.5
−1
−0.5
0
Valori q1, q2 (dash)
time [s]
Transpose JT
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
−0.6
−0.4
−0.2
0
0.2
Algoritmo J^T; valori x, y (dash)
time [s]
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2
−1.5
−1
−0.5
0
time [s]

Example
K = 1000
(pseudo-) Inverse J−1
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02
−0.5
0
0.5
Algoritmo J^−1; valori x, y (dash)
time [s]
0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018 0.02
−1.5
−1
−0.5
0
time [s]
Transpose JT
Unstable

Measures of performance
Measures of performance
Claudio Melchiorri

Measures of performance Introduction
Manipulator performance
Deﬁnition of criteria in order to compute the performance, in both the
velocity and force domains, achievable by a manipulator.
“Attitude” of a manipulator in a given conﬁguration to perform a given task.
Considered “criteria”:
Manipulability ellipsoids;
Manipulability measures;
Polytopes.

Measures of performance Manipulability ellipsoids
Manipulability ellipsoids
The manipulability ellipsoids consider the mechanical gain of the manipulator, in terms
of both velocity and force.
VELOCITY Consider a unit sphere in the joint velocity space
˙qT
˙q = 1
This sphere may be considered as a “cost” (energy entering in the system); it is of
interest to compute the corresponding entity in the operational space (i.e. the achieved
performance). From ˙x = J˙q we have
˙xT
(JJT
)+
˙x = 1
This is a quadratic form representing, in the operational space IRm
, an ellipsoid.
As known from geometry, this ellipsoid has shape and orientation deﬁned by the kernel of
the quadratic form, i.e. by (JJT
)+
:
Direction of the principal axes: deﬁned by the eigenvectos ui of JJT
;
Length of the principal axes: given by the singular values of J, σi = λi (JJT ) .
As a matter of fact, from the singular value decomposition it follows that:
J = USVT
JJT
= US2
UT
(JJT
)+
= US−2
UT

FORCE
Consider the mapping in the force/torque domain
τT
τ = 1
from which (τ = JT
w):
wT
JJT
w = 1
This equation deﬁnes an ellipsoid in the operational force space IRm
.
Similarly to the velocity case, we have that:
The principal axes of the ellipsoid are directed along the eigenvectors ui di JJT
;
Their length are proportional to the inverse of the singular values σi di J.
This is an important and remarkable result, that conﬁrms the duality of the velocity and
force spaces:
The directions along which good performance are obtained in the velocity domain are
directions along which poor performance are obtained in the force domain, and viceversa

Some considerations:
Actuation needs a “large” gain; it is better along directions corresponding to
large singular values;
Control: needs a “small” gain; it is better along directions whit smaller
singular values (better sensitivity).
=⇒ The “optimal direction” for velocity (force) actuation is also the “optimal
direction” for controlling the force (velocity)

Example: 2 dof planar manipulator
Velocity ellipsoids:
-3 -2 -1 0 1 2 3
-3
-2
-1
0
1
2
3
Force ellipsoids:
-3 -2 -1 0 1 2 3
-3
-2
-1
0
1
2
3

Other manipulability measures
1 Manipulability measure
w(q) = det(JJT )
proportional to the volume of the velocity manipulability ellipsoid. If the
manipulator is not redundant, then
w(q) = |det(J)|
2 Another criterion is
w(q) =
σmin
σmax
i.e. the ratio between the minimum and maximum eigenvalues (the inverse of the
conditioning number of J). The more this fraction is “close” to one, the more the
ellipsoid is close to a sphere, and all the “directions” give more or less the same
performance (no “worst case” directions).
3 It is also possible to consider
w(q) = σmin
giving the “minimum” achievable performance (along any direction in IRm
).

Measures of performance Velocity and force polytopes
Velocity and force polytopes
The actuation system has physical limits both in the velocity and force/torque
domain (maximum values):
˙qi,min ≤ ˙qi ≤ ˙qi,max
τi,min ≤ τi ≤ τi,max
These bounds geometrically represent polytopes (e.g. volumes delimited by planar
surfaces) in the joint velocity and force spaces IRn
: Pq, Pτ .
The Jacobian matrix allows to transform these polytopes from the joint to the
operational space. Since the mapping is linear (expressed through the matrix
operator J(q)), one obtains volumes in the operational space still delimited by
planar surfaces: the polytopes Pv , Pw in IRm
.

Velocity and force polytopes
Directions along which the maximum performance are obtained in the operational
space are those corresponding to one of the vertices of Pv , Pw .
These directions may be computed with techniques and algorithms from the
operational research ﬁeld, and are less “elegant” with respect those employed to
compute the manipulability ellipsoids.
As general rule, however, we can observe that the maximum performance in the
operational space are achieved in the vertices of the polytopes Pq, Pτ .

Example: 2 dof planar manipulator
Velocity polytopes:
-3 -2 -1 0 1 2 3
-3
-2
-1
0
1
2
3
Force polytopes:
-3 -2 -1 0 1 2 3
-3
-2
-1
0
1
2
3

Example
Velocity ellipsoid and polytope
-0.5 0 0.5 1 1.5 2 2.5
-1
-0.5
0
0.5
1
1.5
2
2.5
Force ellipsoid and polytope
-0.5 0 0.5 1 1.5 2 2.5
-1
-0.5
0
0.5
1
1.5
2
2.5

Example
Velocity and force ellipsoids and polytopes
-0.5 0 0.5 1 1.5 2 2.5
-1
-0.5
0
0.5
1
1.5
2
2.5

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