Fir 03 rbody

Ridig Body Motion – Homogeneous Transformations
Claudio Melchiorri
Dipartimento di Elettronica, Informatica e Sistemistica (DEIS)
Universit`a di Bologna
email: claudio.melchiorri@unibo.it
C. Melchiorri (DEIS) Ridig Body Motion – Homogeneous Transformations 1 / 80

Summary
1 Ridig Body Motion
Rotations
Translations
2 Homogeneous transformations

Ridig Body Motion
Ridig Body Motion
Claudio Melchiorri

Ridig Body Motion
Rigid body motions - Homogeneous transformations
Description of the manipulator kinematic properties:
Description of the geometric characteristics of the robot’s motion (position,
velocity, acceleration), without considering the forces applied to it
The solution of the kinematic problem is based on:
deﬁnition of a reference frame associated to each link of the manipulator
a procedure for the computation of the relative motion (position, velocity,
acceleration) of these frames due to joints’ movements.
It is necessary to introduce some conventions to describe the position/orientation
of rigid bodies and their motion in the space:
kinematic properties of a rigid body and how to describe them
homogenous transformations
description of position and velocity (force) vectors in diﬀerent reference
frames.

Ridig Body Motion
Rigid body and its representation
A manipulator is composed by a series of rigid
bodies, the links, connected by joints that allow a
relative motion.
RIGID BODY: idealization of a solid body of ﬁ-
nite size in which deformation is neglected: the
distance between any two given points of a rigid
body remains constant in time regardless of exter-
nal forces exerted on it.
||p(t) − q(t)|| = d(p(t), q(t)) = cost

Ridig Body Motion
Some assumptions:
The 3D operational space is represented by the vector space IR3
,
In the 3D space, are deﬁned the inner product
uT
v =
n
i=1
ui vi = ||u||||v|| cos θ u, v ∈ IRn
and the Euclidean norm
||u|| = uT
u =
n
i=1
u2
i u ∈ IRn
We often use Cartesian (right-handed) reference frames, with homogenous
dimensions along the axes
The base frame is an inertial frame.

Ridig Body Motion
General deﬁnition of norm:
u = uT
W u
being W a matrix:
symmetric
positive deﬁnite
Often, W is a diagonal matrix.
Since W = VT
V, then:
u = uT
W u = (uT
VT
)(Vu) = xT
x

Ridig Body Motion
In IR3
, a rigid body has 6 degrees of freedom (dof):
3 for the position, x, y, z;
3 for the orientation, α, β, γ.
In general, a rigid body in IRn
has
n dof for the position
n(n − 1)/2 dof for the orientation

Ridig Body Motion
Roto-translation motion: most general motion of a rigid body in the space,
composed by a rotation about an axis (instantaneous axis of rotation) and a
translation along the same axis.
Problem of describing the instantaneous position/orientation of a rigid body with
respect to a ﬁxed base frame
Let 0
o1 be the origin of the frame F1 ﬁxed to the
rigid body, expressed in F0 . Each point of the
body has coordinates
1
p = [1
px
1
py
1
pz ]T
constant with respect to F1 .
Since the body moves, the same point has coor-
dinates 0
p, expressed in F0 , variable in time
0
p = [0
px
0
py
0
pz ]T

Ridig Body Motion
First problem: if point p is known in
F1 , compute the equivalent represen-
tation in F0 .
1
p =⇒ 0
p
The problem is solved by using the Homogeneous Transformation Matrix 0
T1:
0
T1 =
0
R1
0
o1
0 0 0 1
=




nx sx ax ox
ny sy ay oy
nz sz az oz
0 0 0 1




deﬁning the transformation (roto-translation) between F1 and F0 .

Ridig Body Motion
The problem is decomposed into two parts:
1 F0 and F1 share the same origin, and have a diﬀerent orientation in space
2 F0 and F1 have parallel axes but a diﬀerent origin (translation).

Ridig Body Motion
Rotations
Claudio Melchiorri

Ridig Body Motion Rotations
Rotations
Consider two reference frames F0 and F1 with the same origin, i.e. o1 ≡ o0.
Given a vector 0
v in F0 , its components vx , vy , vz are the orthogonal projections
of 0
v on the coordinate axes.
vx = 0
vT
i = ||0
v|| cos α1
vy = 0
vT
j = ||0
v|| cosα2
vz = 0
vT
k = ||0
v|| cos α3
i, j, k: unit vectors deﬁning the directions of x0, y0, z0

Rotations
If 0
v indicates an axis of F1 , e.g. 0
i1, then 0
i1 = [0
ix
0
ij
0
iz]T
, where
0
ix = 0
iT
1 i = cos α1
0
iy = 0
iT
1 j = cos α2
0
iz = 0
iT
1 k = cos α3
This is a well known result:
0
i1 = [cos α1, cos α2, cos α3]T
the components of a unit vector with respect to a reference frame are its direction
cosines.
A similar results holds for the other directions 0
j1 and 0
k1.

Rotations
Once the direction cosines of the three axes of F1 with respect to F0 are known,
the matrix R may be deﬁned:
R =




0
iT
1 i 0
jT
1 i 0
kT
1 i
0
iT
1 j 0
jT
1 j 0
kT
1 j
0
iT
1 k 0
jT
1 k 0
kT
1 k




0
i1, 0
j1, 0
k1: axes of F1 expressed in F0
i, j, k: axes of F0 .

Rotations
EXAMPLE
By projecting the unit vectors i1, j1, k1 on i0, j0, k0, the components of the
principal axes of F1 in F0 are obtained:



i1 = [0, 1/
√
2, −1/
√
2]T
j1 = [0, 1/
√
2, 1/
√
2]T
k1 = [1, 0, 0]T

Rotations
The rotation matrix between F0 and F1 is obtained from these three vectors:
R =


0 0 1
1√
2
1√
2
0
− 1√
2
1√
2
0


In general
x1 y1 z1
x0
y0
z0


r11 r12 r13
r21 r22 r23
r31 r32 r33



Rotations
Usually, in robotics the symbols n, s, a, are used to indicate the axes x1, y1, z1,
then
0
R1 = [0
n 0
s 0
a] =


nx sx ax
ny sy ay
nz sz az


defining the relative orientation between F0 and F1 .
Symbols n, s, a refer to a frame fixed on the end-
effector (e.g. gripper) with
z axis (a) along the approach direction
y axis (s) in the sliding plane of the fingers
x axis (n) in the normal direction with respect to
y, z.

Rotations
Rotation matrix:
R = [n s a] =


nx sx ax
ny sy ay
nz sz az


From the conditions:
nT
a = sT
a = sT
n = 0
||n|| = ||s|| = ||a|| = 1
it follows that R is an orthonormal matrix, i.e.
R RT
= RT
R = I3 I3: 3 × 3 identity matrix
A rotation matrix is always invertible. By pre-multiplying by R−1
we have
R−1
= RT
i.e. 0
R−1
1 = 1
R0 = 0
RT
1

Elementary rotations
Consider two frames F0 and F1 with coincident origins.
Rotations of θ about the x0, y0, and z0 axes

In the ﬁrst case, F1 is obtained with a rotation of an angle θ about the x0 axis of
F0 .
From
R =


0
iT
1 i 0
jT
1 i 0
kT
1 i
0
iT
1 j 0
jT
1 j 0
kT
1 j
0
iT
1 k 0
jT
1 k 0
kT
1 k


we have
0
R1 = Rot(x, θ) =


1 0 0
0 cos θ − sin θ
0 sin θ cos θ



Similarly, considering rotations about y0 and z0:
0
R1 = Rot(y, θ) =


cos θ 0 sin θ
0 1 0
− sinθ 0 cos θ


0
R1 = Rot(z, θ) =


cos θ − sin θ 0
sin θ cos θ 0
0 0 1


Rotation matrices R relate diﬀerent reference frames.

Rotations
Another interpretation for rotation matrices.
Let us consider a rotation of point 0
p1 = [7, 3, 2]T
by 90o
about z0.
The matrix expressing the rotation is
R1 = Rot(z, 90o
) =


cos 90o
− sin 90o
0
sin 90o
cos 90o
0
0 0 1


=


0 −1 0
1 0 0
0 0 1


Therefore:
0
p2 =


−3
7
2

 =


0 −1 0
1 0 0
0 0 1




7
3
2



Rotations
Consider now a second rotation of 90o
about y0:
0
p3 =


2
7
3

 = R2
0
p2 =


0 0 1
0 1 0
−1 0 0




−3
7
2



Rotations
By combining the two rotations one obtains
R = R2R1 =


0 0 1
0 1 0
−1 0 0




0 −1 0
1 0 0
0 0 1

 =


0 0 1
1 0 0
0 1 0


from which
0
p3 =


2
7
3

 = R 0
p1 =


0 0 1
1 0 0
0 1 0




7
3
2


Rotation matrices “rotates” vectors
with respect to a ﬁxed reference frame.

Axis/angle rotations
Rotation θ about a generic unit vector w = [wx wy wz]T
.
The rotation of the angle θ about w is equivalent
to the following procedure:
Aligne w with z0
Rotate by θ about w ≡ z0
Restore w in its original position.
Each rotation is performed with respect to F0 , then:
Rot(w, θ) = Rot(z0, α)Rot(y0, β)Rot(z0, θ)Rot(y0, −β)Rot(z0, −α)

Axis/angle rotations
Moreover, since ||w|| = 1, we have:
sin α =
wy
w2
x + w2
y
cos α =
wx
w2
x + w2
y
sin β = w2
x + w2
y cos β = wz
The matrix R representing the rotation is therefore given by
R(w, θ) =


wx wxVθ + Cθ wy wx Vθ − wz Sθ wz wxVθ + wy Sθ
wx wy Vθ + wz Sθ wy wy Vθ + Cθ wz wy Vθ − wxSθ
wx wz Vθ − wy Sθ wy wz Vθ + wxSθ wz wz Vθ + Cθ


being Cθ = cos θ, Sθ = sin θ, e Vθ = vers θ = 1 − cos θ.

Proprieties of rotations
1. Not all the orthogonal matrices (RT
R = I) for which the following conditions
nT
a = sT
a = sT
n = 0
||n|| = ||s|| = ||a|| = 1
are satisﬁed represent rotations. For example, matrix
S =


1 0 0
0 −1 0
0 0 1


does not represent a rotation, but rather a “specular” transformation.
It is not possible, starting from F0 , to ob-
tain frame F1 with a rotation. F1 may be
obtained only by means of a specular reﬂec-
tion.
This is not physically feasible for a rigid
body.

If matrix R represents a rigid body rotation, then
det(R) = 1
Because of their properties, the rotation matrices in IR3
belong to a “special set”,
the Special Orthogonal group of order 3, i.e. So(3).
More in general, the set of n × n matrices R satisfying the two conditions
RRT
= RT
R = I
det(R) = +1
is called So(n): Special Orthogonal group in IRn
=⇒ Special: det(R) = +1
=⇒ Orthogonal: RRT
= RT
R = I
So(n) = {R ∈ IRn×n
: RRT
= RT
R = I, det(R) = +1}

2. The equations
0
R1 = [ 0
n 0
s 0
a ] =


nx sx ax
ny sy ay
nz sz az

 0
R−1
1 = 1
R0 = 0
RT
1
allow to consider the relative rotation of two frames, and to transform in
F0 vectors deﬁned in F1 . The expression of 1
p in F0 is given by:
0
p = 0
R1
1
p =


nx sx ax
ny sy ay
nz sz az

 1
p

The composition of more rotations is expressed by a simple matrix multiplication:
Given n + 1 reference frames F0 , ..., Fn with coincident origins and relative
orientation expressed by i−1
Ri , i = 1, . . . n, and given the vector n
p in Fn , then
0
p = 0
R1
1
R2...n−1
Rn
n
p
A note about computational complexity:
0
p = (0
R1
1
R2
2
R3) 3
p = 0
R3
3
p →
63 products
42 summations
0
p = (0
R1 (1
R2 (2
R3
3
p)
2p
)
1p
) →
27 products
18 summations
3. From 0
p = 0
R1
1
p it follows that a rotation applied to vector 1
p is a linear
function: 0
p = r(1
p) = 0
R1
1
p. Given two vectors p, q and two scalar
quantities a, b, we have
r(ap + bq) = ar(p) + br(q)

4. Rotations do not change the amplitude of a vector:
Ra = a
As a matter of fact:
Ra = aT
RT
Ra = aT
a = a
5. The inner product, and then the angle between two vectors, is invariant with
respect to rotations:
aT
b = (Ra)T
(Rb)
As a matter of fact:
(Ra)T
(Rb) = aT
RT
Rb = aT
b = a b cosθ
6. Since R is an orthogonal matrix, the following property holds
R(a × b) = Ra × Rb

7. In general, the product of rotation matrices does not commute:
RaRb = RbRa
Except the trivial case of the identity matrix (i.e. when R = I3), rotations
commute only if the rotation axis is the same!
Consider the two rotations by a 90o
angle about the x0 and y0 axes:
Ra = Rot(x, 900
) =


1 0 0
0 0 −1
0 1 0

 Rb = Rot(y, 90o
) =


0 0 1
0 1 0
−1 0 0



1)
2)
Case 1) Rb followed by Ra:
R = RbRa =


0 1 0
0 0 −1
−1 0 0


Case 2) Ra followed by Rb:
R = RaRb =


0 0 1
1 0 0
0 1 0



8. It may be of interest to define a sequence of rotations with respect to F0 , and
not with respect to the current frame Fi as assumed until now.
Consider two rotations R1 = Rot(y, φ) and R2 = Rot(z, θ)
about the axes y0 and z0 of F0 .
What is the result of applying first R1 and then R2?
Consider the vector 0
p in F0 . After the first rotation R1, the new expression of the
vector (still wrt F0 ) is
0
p1 = R1
0
p
Since also the second rotation is about an axis of F0 , we have
0
p2 = R2
0
p1 = R2 R1
0
p
More in general, given n consecutive rotations Ri , i = 1, . . . , n defined with respect to
the same reference frame F0 , then
0
pn = Rn Rn−1 . . . R1
0
p

Then, there are two diﬀerent possibilities to deﬁne a sequence of consecutive
rotations:
1 If each rotation is expressed wrt the current frame Fn , Fn−1, . . . , F0 , then
the equivalent rotation matrix 0
Rn is obtained by post-multiplication of the
matrices i−1
Ri .
0
p = 0
R1
1
R2 . . . n−1
Rn
n
p
2 If matrices Ri , i = 1, . . . , n describe rotations about an axis of the base frame
F0 , the equivalent matrix is obtained by pre-multiplication of the matrices.
0
pn = Rn Rn−1 . . . R1
0
p

Interpretations of a rotation matrix
In summary, a rotation matrix 0
R1 has three equivalent interpretations:
1. 0
R1 describes the mutual orientation of two
reference frames F0 and F1 ; the columns of
0
R1 are the direction cosines of the axes of
F1 expressed in F0
2. 0
R1 deﬁnes the coordinate transformation
between the coordinates of a point
expressed in F0 and in F1 (common origin)
3. 0
R1 rotates a vector 0
va to 0
vb in a given
reference frame F0

Representations of rotations
A rotation is described with a 3 × 3 matrix with 9
elements:
R =


nx sx ax
ny sy ay
nz sz az


On the other hand, a rigid body in IR3
has 3 rotational dof
→
Three parameters should be sufficient
to describe its orientation
A 3 × 3 matrix, although computationally efficient, is redundant. Among the 9
elements of R one can define the following relations:
nT
a = sT
a = sT
n = 0
||n|| = ||s|| = ||a|| = 1
Note that it is sufficient to know 6 elements of R to define completely the matrix.
If only 5 (or less) elements are known, R cannot be determined univocally.

Representations of rotations
Theoretically, only 3 parameters are suﬃcient to describe the orientation of a rigid
body in the 3D space.
There are representations based on 3 parameters only (minimal representations),
more “compact” than rotation matrices, although computationally less convenient.
Among these representations, we have:
Euler angles: three consecutive rotations about axes z, y′
, z′′
Roll, Pitch and Yaw angles: three consecutive rotations about axes z0, y0, x0
Axis/Angle representation: a unitary rotation axis r and the angle θ

Euler angles
Euler angles (φ, θ, ψ) represents three rotations, applied sequentially about the
axes z0, y1, z2 of the current frame F0 , F1 , F2 .
Consider a base frame F0 . By applying the three rotations we have
- A frame F1 obtained with the rotation φ about z0
- A frame F2 obtained from F1 with the rotation θ about y1
- A frame F3 obtaine from F2 with the rotation ψ about z2

Euler angles
By composing the three rotations, the total rotation from F0 to F3 is
0
R3 = 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



Euler angles
Rotation matrix corresponding to the Euler angles:
REuler (φ, θ, ψ) =


CφCθCψ − SφSψ −CφCθSψ − SφCψ CφSθ
SφCθCψ + CφSψ −SφCθSψ + CφCψ SφSθ
−SθCψ SθSψ Cθ


Inverse problem: compute the Euler angles corresponding to a given rotation
matrix R:
R =


r11 r12 r13
r21 r22 r23
r31 r32 r33

 → (φ, θ, ψ) ?

Atan2 function
arctan(y
x ) = arctan(−y
−x ), gives results in two quadrants (in [−π/2, +π/2])
atan2 is the arctangent with output values in the four quadrants:
two input arguments
gives values in [−π, +π]
undefined only for (0, 0)
uses the sign of both arguments to define the output quadrant
based on arctan function with output values in [−π/2, +π/2]
available in main languages (C++, Matlab, . . . )
atan2 (y, x) =



arctan(y/x) x > 0
π + arctan(y/x) y ≥ 0, x < 0
−π + arctan(y/x) y < 0, x < 0
π/2 y > 0, x = 0
−π/2 y < 0, x = 0
undefined y = 0, x = 0

Euler angles
Two cases are possible:
1 r2
13 + r2
23 = 0 → sin θ = 0. By assuming 0 < θ < π (sin θ > 0), one obtains:



φ = atan2 (r23, r13);
θ = atan2 ( r2
13 + r2
23, r33);
ψ = atan2 (r32, −r31)
or, with −π < θ < 0, (sin θ < 0):



φ = atan2 (−r23, −r13);
θ = atan2 (− r2
13 + r2
23, r33);
ψ = atan2 (−r32, r31)
Two possible sets of solutions depending on the sign of sin θ.

Euler angles
2. r2
13 + r2
23 = 0 (θ = 0, π and cos θ = ±1). By choosing θ = 0 (cos θ = 1)
one obtains
θ = 0
φ + ψ = atan2 (r21, r11) = atan2 (−r12, r11);
On the other hand, if θ = π (cos θ = −1)
θ = π
φ − ψ = atan2 (−r21, −r11) = atan2 (−r12, −r11);
In both cases, inﬁnite solutions are obtained (only the sum or diﬀerence of φ
and θ is known).
Being θ = 0, π, the rotations by the angles φ and ψ occur about parallel (the
same) axes, i.e. the z axis.

Roll, Pitch, Yaw
Consider three consecutive rotations about the axes of the base frame F0 :
A rotation ψ about x0, (yaw),
A rotation θ about y0, (pitch)
A rotation φ about z0, (roll).

Roll, Pitch, Yaw
By properly composing the three rotations:
0
R3 = RRPY (φ, θ, ψ) = Rot(z0, φ)Rot(y0, θ)Rot(x0, ψ)
=


CφCθ −SφCψ + CφSθSψ SφSψ + CφSθCψ
SφCθ CφCψ + SφSθSψ −CφSψ + SφSθCψ
−Sθ CθSψ CθCψ



Roll, Pitch, Yaw
Rotation matrix corresponding to the RPY angles:
RRPY (φ, θ, ψ) =


CφCθ −SφCψ + CφSθSψ SφSψ + CφSθCψ
SφCθ CφCψ + SφSθSψ −CφSψ + SφSθCψ
−Sθ CθSψ CθCψ


Inverse problem: compute the RPY angles corresponding to a given rotation
matrix R:
R =


r11 r12 r13
r21 r22 r23
r31 r32 r33

 → (φ, θ, ψ) (?)

Roll, Pitch, Yaw
Two cases are possible:
1 r2
11 + r2
21 = 0 → cos θ = 0. By choosing θ ∈ [−π/2, π/2], one obtains:



φ = atan2 (r21, r11);
θ = atan2 (−r31, r2
32 + r2
33);
ψ = atan2 (r32, r33);
Otherwise, if θ ∈ [π/2, 3π/2]:



φ = atan2 (−r21, −r11);
θ = atan2 (−r31, − r2
32 + r2
33);
ψ = atan2 (−r32, −r33);

Roll, Pitch, Yaw
2. r2
11 + r2
21 = 0 → cos θ = 0: θ = ±π/2 and inﬁnite solutions are possible
(sum or diﬀerence of ψ and φ).
It may be convenient to (arbitrarily) assign a value (e.g. ±90o
) to one of the
two angles (φ or ψ) a ±90o
and then compute the remaining one:
θ = ±π/2;
φ − ψ = atan2 (r23, r13) = atan2 (−r12, r22);

Angle/Axis representation
It is possible to describe any rotation in 3D by means of the rotation angle θ and
the corresponding rotation axis w
R =


w2
x (1 − Cθ) + Cθ wx wy (1 − Cθ) − wz Sθ wx wz (1 − Cθ) + wy Sθ
wx wy (1 − Cθ) + wz Sθ w2
y (1 − Cθ) + Cθ wy wz (1 − Cθ) − wx Sθ
wx wz (1 − Cθ) − wy Sθ wy wz (1 − Cθ) + wx Sθ w2
z (1 − Cθ) + Cθ


4 parameters: wx, wy , wz, θ
with the condition: w2
x + w2
y + w2
z = 1
=⇒ 3 dof

Angle/Axis representation
Inverse problem: compute the axis w and the angle θ corresponding to a given
rotation matrix R:
θ = acos
r11 + r22 + r33 − 1
2
w =
1
2 sin θ


r32 − r23
r13 − r31
r21 − r12


The trace of a rotation matrix depends only on the (cosine of) rotation angle.
This representation suﬀers of some drawbacks:
It is not unique: Rot(w, θ) = Rot(−w, −θ)
(we can arbitrarily assume 0 ≤ θ ≤ π)
If θ = 0 then Rot(w, 0) = I3 and w is indeﬁnite
There are numerical problems if θ ≈ 0: in this case sin θ ≈ 0 and problems
may arise in computing w

Example
Compute the rotation matrix corresponding to the RPY angles 0o
, 45o
, 90o
, i.e.:
1 A rotation of 90o
about x0, (yaw)
2 A rotation of 45o
about y0, (pitch)
3 A rotation of 0o
about z0, (roll)
RPY rotations: roll = 0o , pitch = 45o , yaw = 90o
One obtains:
R =


1√
2
1√
2
0
0 0 −1
− 1√
2
1√
2
0



Example
R =


1√
2
1√
2
0
0 0 −1
− 1√
2
1√
2
0


RPY rotations: roll = 0o , pitch = 45o , yaw = 90o
• The Euler angles corresponding to this rotation matrix are
φ = −90o
; θ = 90o
; ψ = 45o
• Considering the Angle/Axis representation, one obtains:
rotation of the angle θ = 98.42o
about the axis w = [0.863, 0.357, −0.357]T

Translations
Claudio Melchiorri

Ridig Body Motion Translations
Translations
Rotations between two coordinate frames can be expressed in matrix form:
0
p = R 1
p
This is not possible for translations! → It is not possible to deﬁne a 3 × 3 matrix
P so that a translation can be expressed as a matrix/vector multiplication
0
p = P 1
p =⇒ not possible!!
A translation of a vector 0
p by 0
o corresponds to a
vectorial summation
0
q = 0
p + 0
o
Then
0
qx = 0
px + 0
ox
0
qy = 0
py + 0
oy
0
qz = 0
pz + 0
oz

Translations
Then, a translation is expressed as a function t(p) = p + o
In general: t(ap + bq) = ap + bq + o = at(p) + bt(q)
Then, translations are not linear transformations!
The most general transformation between two coordinate frames cannot be
represented by a 3 × 3 matrix.
The composition of a rotation and a
translation is obtained from
0
q = 0
p + 0
o
by considering vector p deﬁned in F1 , ro-
tated and translated with respect to F0 .

Translations
Since vectors can be added only if they are deﬁned with respect to the same
coordinate system:
0
p = 0
R1
1
p + 0
o1
being 0
o1 the translation from F0 to F1 and 0
R1 the mutual rotation.

Ridig Body Motion
Homogeneous transformations
Claudio Melchiorri

It is of interest to put in matrix form the equation
0
p = 0
R1
1
p + 0
o1
since, in case of successive transformations, one could obtain expressions similar to
0
p = 0
R1
1
R2...n−1
Rn
n
p
For this purpose, it is possible to add to matrix R the vector 0
o as fourth column;
in this manner a 3 × 4 matrix is obtained
M = [n s a 0
o1] = [R 0
o1]
A square matrix is obtained by adding to M the row [0 0 0 1].

The Homogeneous Transformation Matrix 0
T1 is obtained
0
T1 =
0
R1
0
o1
0 0 0 1
=




nx sx ax ox
ny sy ay oy
nz sz az oz
0 0 0 1




This matrix represents the transformation (rotation-translation) between F0 and
F1 .

By defining the homogeneous 4-dimensional (in IR4
) vector:
[pT
1]T
= [px py pz 1]T
one obtains
0
p
1
= 0
T1
1
p
1
=
0
R1
1
p + 0
o1
1
The subspace IR3
defined by the first three components represents the
transformation 0
p = 0
R1
1
p + 0
o1 on vectors in IR3
, and physically represents
the rigid body motion from F1 to F0 ; the fourth component is not affected by
the matrix multiplication.

Given the homogeneous transformation matrices 0
T1, from F1 to F0 , and 1
T2,
from F2 to F1 , the composition
0
T2 = 0
T1
1
T2
can be applied to vectors [ 2
pT
1]T
deﬁned in F2 , and the result is
0
T1(1
T2
2
p
1
) = 0
T1
1
R2
2
p + 1
o2
1
=
0
R1
1
R2
2
p + 0
R1
1
o2 + 0
o
1

This is equivalent to the product of the homogeneous matrix
0
T2 =
0
R1
1
R2
0
R1
1
o2 + 0
o
0 0 0 1
= 0
T1
1
T2
with the vector [ 2
pT
1]T
.
In general we have
0
Tn = 0
T1
1
T2 . . . n−1
Tn

Homogeneous transformation of coordinates
The coordinate transformation bewteen two reference frames F0 and F1 may be
expressed by a 4 × 4 matrix 0
T1:
0
p = 0
T1
1
p
Of particular interest are the elementary transformations, i.e. simple rotations or
translations along the coordinate axes.
All the coordinate transformations may be obtained by combinations of these
elementary transformations.

Elementary rotations and translations
The homogeneous transformation matrices corresponding to rotations of an angle
θ about the axes x, y, z of F0 are:
Rot(x, θ)=



1 0 0 0
0 Cθ −Sθ 0
0 Sθ Cθ 0
0 0 0 1


 , Rot(y, θ)=



Cθ 0 Sθ 0
0 1 0 0
−Sθ 0 Cθ 0
0 0 0 1


 , Rot(z, θ)=



Cθ −Sθ 0 0
Sθ Cθ 0 0
0 0 1 0
0 0 0 1



being Cθ = cos θ, Sθ = sin θ.
The homogeneous transformation T corresponding to the translation of vector
v = [vx vy vz ]T
is
T = Trasl(v) =




1 0 0 vx
0 1 0 vy
0 0 1 vz
0 0 0 1





Example
The frame F1 , with respect to F0 , is translated of
1 along x0 and of 3 along y0 , moreover, it is rotated
by 30o
about z0.
The transformation matrix from F1 to F0 is
0
T1 =




0.866 −0.500 0 1
0.500 0.866 0 3
0 0 1 0
0 0 0 1





Example
Consider a point deﬁned in F1 by
1
p =


2
1
0


Its coordinates in F0 are
0
p = 0
T1
1
p =




0.866 −0.500 0 1
0.500 0.866 0 3
0 0 1 0
0 0 0 1








2
1
0
1



 =




2.232
4.866
0
1





Example
Consider the point in F0
0
pa =


2
1
0


Let us apply to 0
p a translation of 1 along x0, of 3 along y0; then rotate the vector
by 30o
about z0. The result is obtained by multiplying vector 0
p by the matrix
T =




0.866 −0.500 0 1
0.500 0.866 0 3
0 0 1 0
0 0 0 1





Example
Then
0
pb = T 0
pa = T =




0.866 −0.500 0 1
0.500 0.866 0 3
0 0 1 0
0 0 0 1








2
1
0
1



 =




2.232
4.866
0
1




The same numerical result is obtained, although the physical interpretation is
diﬀerent.

Interpretations of a homogeneous transformation matrix
Similarly to rotation matrices, also an homogeneous transformation matrix 0
T1
0
T1 =
0
R1
0
v
0 0 0 1
has three possible physical interpretations:
1 Description of F1 in F0 : in particular 0
v represents the origin of F1 with
respect to F0 , and the elements of 0
R1 give the direction of the axes of F1
2 Coordinate transformation of vectors between F1 and F0 , 1
p → 0
p;
3 Translates and rotates a generic vector 0
pa to 0
pb in a given reference frame
F0

Inverse transformation
Once the position/orientation of F1 with respect to F0 are known, deﬁned by the
homogeneous transformation matrix 0
T1, it is simple to compute the inverse
transformation 1
T0 = (0
T1)−1
, deﬁning the position/orientation of F0 with
respect to F1 .
From
0
T1
1
T0 = I4
it follows that
1
T0 =
0
RT
1 −0
RT
1 v
0 0 0 1
=




nx ny nz −vT
n
sx sy sz −vT
s
ax ay az −vT
a
0 0 0 1





Inverse transformation
As a matter of fact, if
1
T0 =
M x
0 0 0 1
then
R v
0 0 0 1
M x
0 0 0 1
=




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1




and therefore
I3 = RM
0 = Rx + v
=⇒
M = RT
x = −RT
v

Example
Given
T =




0 0 1 2
1 0 0 1
0 1 0 0
0 0 0 1




compute its inverse transformation.
Solution:
T−1
= RT
−RT
v
0 0 0 1
=



nx ny nz −vT
n
sx sy sz −vT
s
ax ay az −vT
a
0 0 0 1


 =



0 1 0 −1
0 0 1 0
1 0 0 −2
0 0 0 1




The term homogeneous derives from projective geometry.
Equations describing lines and planes in projective geometry in IR3
are
homogeneous in the four variables x1, x2, x3, x4 in IR4
.
In IR3
these equations, affine transformations, are non homogeneous in
x1, x2, x3, (lines or planes not passing through the origin present a constant
term - non function of x1, x2, x3 - in their expression).
In general, computations of an affine transformations in IRn−1
may be
expressed as homogeneous linear transformations in IRn
:
0
p = 0
R1
1
p + 0
o1 affine transformation
0
p′
=
0
p
1
=
0
R1
0
o1
0 1
1
p
1
homogeneous transformation

The most general expression for an homogeneous transformation is
T =
D3×3 p3×1
f1×3 s
=
Deformation Translation
Perspective Scale
Note the terms D, f1×3 and s.
These quantities, in robotics, are alway assumed as:
A rotation matrix (D = R),
The null vector [0 0 0] (f1×3 = [0 0 0]),
The unit gain (s = 1)
In other cases (e.g. computer graphics), these quantities are used to obtain
deformations, perspective distortions, change of scaling factors, and so on (in
general: eﬀects non applicable to rigid bodies).

Example
Consider the transformation matrix
T =




a 0 0 0
0 b 0 0
0 0 c 0
0 0 0 1




This transformation applies a diﬀerent “gain” along the three reference axes
x, y, z. A deformation of the body is obtained.
Point p = [1, 1, 1]T
is transformed to
p1 = T p =




a 0 0 0
0 b 0 0
0 0 c 0
0 0 0 1








1
1
1
1



 =




a
b
c
1




With this matrix, a cube is transformed in a parallelepiped.

Example
Similarly, the transformation
T =




1 0 0 0
0 1 0 0
0 0 1 0
0 −1
f
0 1




performs a perspective transformation along y. The coordinates x, y, z of a point
p are transformed in
x′
=
x
1 − y/f
y′
=
y
1 − y/f
z′
=
z
1 − y/f
0 10 20 30 40 50 60 70 80
0
10
20
30
40
50
60
70
80
x
y
z
f = −20

Equations with homogeneous transformations
Usually, in robotics it is necessary to specify the position/orientation of an object
with respect to different reference frames (e.g. wrt to the end-effector, to the base
frame, to other machines/tools, . . . ).
The following equation must be verified if the robot has to grasp the object:
B B
T6E = OG

Equations with homogeneous transformations
B B
T6E = OG
Usually, matrices B, O, G, E are known (and constant). Therefore, the equation
can be solved in terms of B
T6
B
T6 = B−1
OGE−1
The robot conﬁguration is then obtained.
Otherwise, the object position O (if not known) can be computed as
O = B B
T6EG−1

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