Fir 05 dynamics

Dynamic model of robot manipulators
Claudio Melchiorri
Dipartimento di Elettronica, Informatica e Sistemistica (DEIS)
Universit`a di Bologna
email: claudio.melchiorri@unibo.it
C. Melchiorri (DEIS) Dynamic Model 1 / 65

Summary
1 Dynamic models
Introduction
Euler-Lagrange model

Dynamic models Introduction
Dynamic model of manipulators
Robot Dynamics =⇒ Study of the relation between the applied forces/torques
and the resulting motion of an industrial manipulator.
Similarly to kinematics, also for the dynamics it is possible to deﬁne two “models”:
Direct model: once the forces/torques applied to the joints, as well as the
joint positions and velocities are known, compute the joint accelerations
¨q = f (q, ˙q, τ)
and then
˙q = ¨q dt, q = ˙q dt
Inverse model: once the joint accelerations, velocities and positions are
known, compute the corresponding forces/torques
τ = f −1
(¨q, ˙q, q) = g(¨q, ˙q, q)

Normally, a manipulator is composed by an open kinematic chain, and its dynamic
model is aﬀected by several “drawbacks”:
low rigidity (elasticity in the structure and in the joints)
potentially unknown parameters (dimensions, inertia, mass,. . . )
dynamic coupling among links.
Other non linear eﬀects are usually introduced by the actuation system:
friction
dead zones
. . .
In any case, in the derivation of the dynamic model, an ideal case of a series of
connected rigid bodies is made.

Two problems may be defined for the study of the dynamic model:
Direct dynamic model: computation of the time evolution of ¨q(t) (and
then of q(t), ˙q(t)), given the vector of generalized forces (torques and/or
forces) bτ(t) applied to the joints and, in case, the external forces applied to
the end-effector, and the initial conditions q(t = t0), ˙q(t = t0).
τ(t) =⇒ ¨q(t) ( ˙q(t), q(t) )
Inverse dynamic problem: computation of the vector τ(t) necessary to
obtain a desired trajectory ¨q(t), ˙q(t), q(t), once the forces applied of the
end-effector are known.
¨q(t), ˙q(t), q(t) =⇒ τ(t)

There are several reasons for studying the dynamics of a manipulator:
• simulation: test desired motions without resorting to real experimentation
• analysis and synthesis of suitable control algorithms
• analysis of the structural properties of the manipulator since the design phase.
Two approaches for the definition of the dynamic model:
EULER-LAGRANGE approach.
First approach to be developed. The dynamic model obtained in this manner is simpler and
more intuitive, and also more suitable to understand the effects of changes in the
mechanical parameters. The links are considered altogether, and the model is obtained
analytically. Drawbacks: the model is obtained starting from the kinetic and potential
energies (non intuitive); the model is not computationally efficient.
NEWTON-EULER approach.
Based on a computationally efficient recursive technique that exploits the serial structure of
an industrial manipulator. On the other hand, the mathematical model is not expressed in
closed form.
Obviously, the two techniques are equivalent (provide the same results).

Dynamic models Euler-Lagrange model
Generalized variables, or Lagrange coordinates:
Independent variables used to describe the position of rigid bodies in the space.
For the same physical system, more choices for the Lagrangian coordinates are
usually possible.
In robotics =⇒ joint variables q1, q2, . . . qn.

From physics, we know that it is possible to deﬁne:
1 The Kinetic Energy function K(q, ˙q)
2 The Potential Energy function P(q)
and therefore the Lagrangian function L(q, ˙q) = K(q, ˙q) − P(q)
The Euler-Lagrange equations are
ψi =
d
dt
∂L
∂ ˙qi
−
∂L
∂qi
i = 1, . . . , n
being ψi the non-conservative (external or dissipative) generalized forces
performing work on qi . In robotics, we consider:
τi joint actuator torque
JT
Fc i
term due to external (contact) forces
dii ˙qi joint friction torque
Therefore: ψi = τi + JT
Fc i
− dii ˙qi .

Since the potential energy does not depend on the velocity, the Euler-Lagrange
equations can be rewritten as
ψi =
d
dt
∂K
∂ ˙qi
−
∂K
∂qi
+
∂P
∂qi
i = 1, . . . , n (1)
This formulation is more convenient since in robotics it is possible to compute
quite easily the terms K and P from the geometric properties of the manipulator.
Then, by applying (1), the dynamic model is obtained.
Note that
K =
n
i=1
Ki P =
n
i=1
Pi

Computation of the Kinetic Energy. For a rigid body:
The mass can be computed by
m =
B
ρ(x, y, z) dx dy dz
where ρ(x, y, z) is the mass density (assumed constant): ρ(x, y, z) = ρ.
The center of mass (CoM) is
pC =
1
m B
p(x, y, z)ρ dx dy dz =
1
m B
p(x, y, z) dm
The kinetic energy results
K =
1
2 B
vT
(x, y, z)v(x, y, z)ρ dx dy dz
=
1
2 B
vT
(x, y, z)v(x, y, z) dm

Let assume that the translational vC and
rotational ω velocities of the center of
mass are known with respect to an inertial
frame F0 .
The velocity of a generic point p′
of the body is
v = vC + ω × (p′
− pC ) = vC + ω × r (2)
The velocity expressed in a frame F’ ﬁxed to the rigid body may be computed by
introducing the rotation matrix R betweeen F’ and F0
RT
v = RT
(vC + ω × r) = RT
vC + (RT
ω) × (RT
r)
Therefore
v′
= v′
C + ω′
× r′

Since the product ω × r in (2) can be expressed as S(ω) r, we have:
K =
1
2 B
vT
(x, y, z)v(x, y, z) dm
=
1
2 B
(vC + Sr)T
(vC + Sr) dm
=
1
2 B
vT
C vC dm +
1
2 B
rT
ST
Sr dm +
B
vT
C Sr dm
=
1
2 B
vT
C vC dm +
1
2 B
rT
ST
Sr dm
As a matter of fact, from the deﬁnition of CoM ( B r dm = B (pC − p)dm = 0):
B
vT
C Sr dm = vT
C S
B
r dm = 0

In conclusion
K =
1
2 B
vT
C vC dm +
1
2 B
rT
ST
Sr dm
The ﬁrst term depends on the linear velocity vC of the center of mass
1
2 B
vT
C vC dm =
1
2
m vT
C vC
For the second term, considering that aT
b = Tr(a bT
) and the particular
structure of matrix S, we have
1
2 B
rT
ST
Sr dm =
1
2 B
Tr(Sr rT
ST
) dm =
1
2
Tr(S
B
rrT
dm ST
)
=
1
2
Tr(S J ST
) =
1
2
ωT
I ω I : body inertia matrix
Also this term depends on the velocity of the center of mass (in this case ω).

Matrix J, Euler matrix, and matrix I, body inertia matrix, are symmetric, and have
the following general expressions:
J =


r2
x dm rx ry dm rx rz dm
rx ry dm r2
y dm ry rz dm
rx rz dm ry rz dm r2
z dm


I =


(r2
y + r2
z ) dm − rx ry dm − rx rz dm
− rx ry dm (r2
x + r2
z ) dm − ry rz dm
− rx rz dm − ry rz dm (r2
x + r2
y ) dm


The elements of both matrices J ed I depend on vector r, i.e. on the position of
the generic point of the i-th link with respect to its center of mass, deﬁned in the
base frame.
Since the position of the i-th link depends on the conﬁguration of the
manipulator, matrices J and I are in general functions of the joint variables q!

Examples of body inertia matrices
Homogeneous bodies of mass m, with axes of symmetry.
• Parallelepiped with sides a (length/height), b, c (base)
I =


Ixx
Iyy
Izz

 =


1
12
m(b2 + c2)
1
12
m(a2 + c2)
1
12
m(a2 + b2)


• Empty cylinder with length h, and external/internal radii a, b
I =


1
2
m(a2 + b2)
1
12
m 3(a2 + b2) + h2
Izz

 ,
Izz = Iyy
I′
zz = Izz + m h
2
2
single axis translation theorem
x0
y0
z0
a
b
c
x0
y0
z0z′
0
a
b
h
h/2
Steiner theorem:
changes of body inertia due
to translation p of the frame
of computation:
I = Ic + m(pT p E3×3 − ppT )
Ic body inertia matrix wrt the
center of mass
E3×3 (3 × 3) identity matrix

In conclusion, the kinetic energy of a rigid body is deﬁned as (Konig Theorem)
K =
1
2
m vT
C vC +
1
2
ωT
Iω (3)
Both terms depend only on the velocity of the rigid body.
The ﬁrst term, being related to the magnitude of a vector ( vC = vT
C vC ), is invariant with
respect to the reference frame used to express the velocity:
vT
C vC = (RvC )T
(RvC ) = vT
C (RT
R)vC , ∀ R
This property holds also for the second term: the product ωT Iω is invariant with respect to the
reference frame. As a matter of fact, the body inertia matrix is transformed according to the
following relation:
I′
= R I RT
Then: ωT Iω = ω′T
I′ω′ = (R ω)T (RIRT )(Rω) = ωT (RT R)I(RT R)ω.
Therefore, being (3) invariant with respect to the reference frame, F can be chosen in order to
simplify the computations required for the evaluation of K.

Since the kinetic energy Ki of the generic i-th link is invariant with respect to the
reference frame (used to express vC , ω, I), it is convenient to chose a frame
Fi ﬁxed to the link itself, with origin in the center of mass.
In this manner matrix I is constant and simple to be computed on the basis of the
geometric properties of the link.
On the other hand, normally the rotational velocity ω is deﬁned in the base frame
F0 , and therefore it is needed to transform it according to RT
ω, being R the
rotation matrix between Fi and F0 (known from the kinematic model of the
manipulator).

In conclusion, the kinetic energy of a manipulator can be determined when, for
each link, the following quantities are known:
the link mass mi ;
the inertia matrix Ii , computed wrt a frame Fi ﬁxed to the center of mass in
which it has a constant expression ˜Ii ;
the linear velocity vCi of the center of mass, and the rotational velocity ωi of
the link (both expressed in F0 );
the rotation matrix Ri between the frame ﬁxed to the link and F0 .
The kinetic energy Ki of the i-th link has the form:
Ki =
1
2
mi vT
Ci vCi +
1
2
ωT
i Ri
˜Ii RT
i ωi
It is now necessary to compute the linear and rotational velocities (vCi and ωi ) as
functions of the Lagrangian coordinates (i.e. the joint variables q).

The end-eﬀector velocity may be computed as a function of the joint velocities ˙q1, . . . , ˙qn
through the Jacobian matrix J. The same methodology can be used to compute the
velocity of a generic point of the manipulator, and in particular the velocity vCi = ˙pCi of
the center of mass pCi , that results function of the joint velocities ˙q1, . . . , ˙qi only:
˙pCi = Ji
v1 ˙q1 + Ji
v2 ˙q2 + . . . + Ji
vi ˙qi = Ji
v ˙q
ωi = Ji
ω1 ˙q1 + Ji
ω2 ˙q2 + . . . + Ji
ωi ˙qi = Ji
ω ˙q
where
Ji
v = Ji
v1 . . . Ji
vi 0 . . . 0
Ji
ω = Ji
ω1 . . . Ji
ωi 0 . . . 0
with
Ji
vj
Ji
ωj
=
zj−1 × (pCi − pj−1)
zj−1
rotational joint
Ji
vj
Ji
ωj
=
zj−1
0
linear joint
being pj−1 the position of the origin of the frame associated to the j-th link.

In conclusion, for a n dof manipulator we have:
K =
1
2
n
i=1
mi vT
Ci vCi +
1
2
n
i=1
ωT
i Ri
˜Ii RT
i ω
=
1
2
˙qT
n
i=1
mi Ji T
v (q)Ji
v (q) + Ji T
ω (q)Ri
˜Ii RT
i Ji
ω(q) ˙q
=
1
2
˙qT
M(q)˙q
=
1
2
n
i=1
n
j=1
Mij (q)˙qi ˙qj
where M(q) is a n × n, symmetric and positive deﬁnite matrix, function of the
manipulator conﬁguration q.
Matrix M(q) is called the Inertia Matrix of the manipulator.

Modello di Eulero-Lagrange
Computation of the Potential Energy. For rigid bodies, the only potential
energy taken into account in the dynamics is due to the gravitational ﬁeld g. For
the generic i-th link
Pi =
Li
gT
p dm = gT
Li
p dm = gT
pCi mi
The potential energy does not depend on the joint velocities ˙q, and may be
expressed as a function of the position of the centers of mass. For the whole
manipulator:
P =
n
i=1
gT
pCi mi
In case of ﬂexible link, one should consider also terms due to elastic forces.
K e P are computed (once mi and ˜I are known) with a procedure similar to the one
adopted for the forward kinematic model:
• K → matrices Ji
e Ri , • P → pCi position of the centers of mass

Once K and P are known, it is possible to compute the dynamic model of the
manipulator. The dynamics is expressed by
ψk =
d
dt
∂L
∂ ˙qk
−
∂L
∂qk
k = 1, . . . , n
The Lagrangian function is
L = K − P =
1
2
n
i=1
n
j=1
Mij ˙qi ˙qj −
n
i=1
gT
pCi mi
Then
∂L
∂ ˙qk
=
∂K
∂ ˙qk
=
n
j=1
Mkj ˙qj
and
d
dt
∂L
∂ ˙qk
=
n
j=1
Mkj ¨qj +
n
j=1
d Mkj
dt
˙qj =
n
j=1
Mkj ¨qj +
n
i=1
n
j=1
∂Mkj
∂qi
˙qi ˙qj
Moreover
∂L
∂qk
=
1
2
n
i=1
n
j=1
∂Mij
∂qk
˙qi ˙qj −
∂P
∂qk

The Lagrangian equations have the following formulation
n
j=1
Mkj ¨qj +
n
i=1
n
j=1
∂Mkj
∂qi
−
1
2
∂Mij
∂qk
˙qi ˙qj +
∂P
∂qk
= ψk k = 1, . . . , n
By deﬁning the term hkji (q) as
hkji (q) =
∂Mkj (q)
∂qi
−
1
2
∂Mij (q)
∂qk
and gk (q) as
gk (q) =
∂P(q)
∂qk
the following equations are ﬁnally obtained
n
j=1
Mkj (q)¨qj +
n
i=1
n
j=1
hkji (q)˙qi ˙qj + gk(q) = ψk k = 1, . . . , n

The elements Mkj (q), hijk (q), gk (q) are function of the joint position only, and
therefore their computation is relatively simple once the manipulator’s configuration is
known. They have the following physical meaning:
For the acceleration terms:
Mkk is the moment of inertia about the k-th joint axis, in a given configuration and
considering blocked all the other joints
Mkj is the inertia coupling, accounting the effect of acceleration of joint j on joint k
For the quadratic velocity terms:
hkjj ˙q2
j represents the centrifugal effect induced on joint k by the velocity of joint j
(notice that hkkk = ∂Mkk
∂qk
= 0)
hkji ˙qi ˙qj represents the Coriolis effect induced on joint k by the velocities of joints i
and j
For the configuration-dependent terms:
gk represents the torque generated on joint k by the gravity force acting on the
manipulator in the current configuration.

Recalling that the non-conservative forces ψk are in general composed by
τk joint actuator torque
JT
Fc k
external (contact) forces
dkk ˙qk joint friction torque
the Lagrangian equations
n
j=1
Mkj (q)¨qj +
n
i=1
n
j=1
hkji (q)˙qi ˙qj + gk(q) = ψk k = 1, . . . , n
can be written in matrix form as
M(q)¨q + C(q, ˙q)˙q + D˙q + g(q) = τ + JT
(q)Fc
This matrix equation is known as the dynamic model of the manipulator.

Euler-Lagrange model - Some considerations
The product C(q, ˙q)˙q = n
i=1
n
j=1 hkji (q)˙qi ˙qj is a (n × 1) vector v whose
elements are quadratic functions of the joint velocities ˙qj .
The k-th element vk of this vector is:
vk =
n
j=1
Ckj ˙qj
where the elements Ckj are computed as
Ckj =
n
i=1
cijk ˙qi
with
cijk =
1
2
∂Mkj
∂qi
+
∂Mki
∂qj
−
∂Mij
∂qk
Christoﬀel Symbols (4)

The elements of matrix C(q, ˙q) are computed as follows. From
n
i=1
n
j=1
hkji ˙qi ˙qj =
n
i=1
n
j=1
∂Mkj
∂qi
−
1
2
∂Mij
∂qk
˙qi ˙qj
by exchanging the sum (i, j) and exploiting the symmetry one obtains
n
i=1
n
j=1
∂Mkj
∂qi
=
1
2
n
i=1
n
j=1
∂Mkj
∂qi
+
∂Mki
∂qj
and then
n
i=1
n
j=1
∂Mkj
∂qi
−
1
2
∂Mij
∂qk
=
n
i=1
n
j=1
1
2
∂Mkj
∂qi
+
∂Mki
∂qj
−
∂Mij
∂qk
=
n
i=1
n
j=1
cijk
where cijk = 1
2
∂Mkj
∂qi
+ ∂Mki
∂qj
−
∂Mij
∂qk
are the so-called Christoﬀel Symbols.
Since matrix M(q) is symmetric, for a given k then cijk = cjik .
The elements of matrix C(q, ˙q)are then computed as
[C(q, ˙q)]k,j =
n
i=1
cijk ˙qi (5)

This is not the only possible expression for matrix C(q, ˙q). In general, any matrix
such that
n
j=1
cij ˙qj =
n
j=1
n
k=1
hijk ˙qk ˙qj
can be considered. The choice (4) is preferred since in this case the following
property is verified.
Property. Matrix N(q, ˙q), defined as
N(q, ˙q) = ˙M(q, ˙q) − 2C(q, ˙q) (6)
in which the elements of C(q, ˙q) are defined as
cijk =
1
2
∂Mkj
∂qi
+
∂Mki
∂qj
−
∂Mij
∂qk
[C(q, ˙q)]k,j =
n
i=1
cijk ˙qi
results skew-symmetric, i.e. nkj = −njk , nkk = 0.

In fact, by considering the generic element nkj , one obtains
nkj =
d Mkj
dt
− 2[C]kj
=
n
i=1
∂Mkj
∂qi
− (
∂Mkj
∂qi
+
∂Mki
∂qj
−
∂Mij
∂qk
) ˙qi
=
n
i=1
∂Mij
∂qk
−
∂Mki
∂qj
˙qi
from which it follows (if indices k and j are exchanged, because of the symmetry
of M(q)) that nkj = −njk .
Since matrix N is skew-symmetrix, then
xT
N(q, ˙q) x = 0, ∀x

The condition
xT
N(q, ˙q) x = 0, ∀x
holds since N(q, ˙q) is skew-symmetric, due to the particular choice of the elements of
matrix C(q, ˙q). On the other hand, the condition
˙qT
N(q, ˙q) ˙q = 0
holds for any choice of matrix C(q, ˙q) (from the energy conservation principle).
The evolution over time of the kinetic energy K must be equal to the work generated by
the forces acting at joints:
d K
dt
=
1
2
d
dt
˙qT
M˙q = ˙qT
(τ − D˙q − g(q) − JT
F)
The ﬁrst element is (from the dynamic model M¨q = −C˙q − D˙q − g(q) + τ − JT
F):
1
2
d
dt
˙qT
M˙q =
1
2
˙qT ˙M˙q + ˙qT
M¨q =
1
2
˙qT
M˙q + ˙qT
(−C˙q − D˙q − g(q) + τ − JT
F)

Then
1
2
˙qT
M˙q + ˙qT
(−C˙q − D˙q − g(q) + τ − JT
F) = ˙qT
(τ − D˙q − g(q) − JT
F)
from which
1
2
˙qT
M˙q = ˙qT
C˙q =⇒ ˙qT
( ˙M − 2C)˙q = 0
This equation holds ∀˙q and without any assumption on matrix C(˙q, q) (it holds
also if C is not based on the Cristoﬀel symbols).

In deriving the dynamic model, the actuation system has not been taken into
account. This is normally composed by:
motors
reduction gears
trasmission system.
The actuation system has several effects on the dynamics:
if motors are installed on the links, then masses and inertia are changed
it introduces its own dynamics (electromechanical, pneumatic, hydraulic, ...)
that may be non negligible (e.g. in case of lightweight manipulators)
it introduces additional nonlinear effects such as backslash, friction, elasticity,
...
Notice that these effects could be considered by introducing suitable terms in the
dynamic model derived on the basis of the Euler-Lagrangian formulation.

Example - 1
Dynamic model of a pendulum (one dof manipulator).
Consider
θ joint variable,
τ joint torque,
m mass,
L distance between center of mass and joint,
d viscous friction coeﬃcient,
I inertia seen at the rotation axis.
Kinetic energy: K = 1
2 I ˙θ2
Potenzial energy: P = mgL(1 − cos θ)
Lagrangian function L: L = 1
2 I ˙θ2
− mgL(1 − cos θ)

Example - 1
Lagrangian function: L = 1
2 I ˙θ2
− mgL(1 − cos θ)
from which
∂L
∂ ˙θ
= I ˙θ,
d
dt
∂L
∂ ˙θ
= I ¨θ,
∂L
∂θ
= −mgL sin θ
The generalized Lagrangian force in this case must account for the torque applied
to the joint and for the friction eﬀect:
ψ = τ − d ˙θ
From the general expression
ψ =
d
dt
∂L
∂ ˙θ
−
∂L
∂θ
we have the following second order diﬀerential equation
I ¨θ + d ˙θ + mgL sin θ = τ

Example - 2
Dynamic model of a 2 dof manipulator. Consider:
θi i-th joint variable;
mi i-th link mass ;
˜Ii i-th link inertia, about an axis through
the CoM and parallel to z;
ai i-th link length;
aCi distance between joint i and the CoM
of the i-th link;
τi torque on joint i;
g gravity force along y;
Pi , Ki potential and kinetic energy of the
i-th link.
The dynamic equations will be obtained in two manners:
a) with the “classic” approach, deriving the Lagrangian function (based on the kinetic
and potential energy K, P)
b) exploiting the particular structure of a manipulator (Jacobian, ...).

Example - 2 (classic approach)
We chose as generalized coordinates the joint variables q1 = θ1; q2 = θ2. The
kinetic and potential energies Ki and Pi are:
link 1:
K1 =
1
2
m1a2
C1
˙θ1
2
+
1
2
˜I1
˙θ1
2
, P1 = m1gaC1S1
link 2: in this case, the CoM position and velocity are



pC2x = a1C1 + aC2C12
pC2y = a1S1 + aC2S12



˙pC2x = −a1S1
˙θ1 − aC2S12( ˙θ1 + ˙θ2)
˙pC2y = a1C1
˙θ1 + aC2C12( ˙θ1 + ˙θ2)
then
K2 =
1
2
m2 ˙pT
C2 ˙pC2 +
1
2
˜I2( ˙θ1 + ˙θ2)2
, P2 = m2g(a1S1 + aC2S12)
where
˙pT
C2 ˙pC2 = a2
1
˙θ1
2
+ a2
C2( ˙θ1 + ˙θ2)2
+ 2a1aC2C2( ˙θ2
1 + ˙θ1
˙θ2)

Example - 2 (classic approach)
Therefore, L = K1 + K2 − P1 − P2 and
τ1 = [m1a2
C1 + ˜I1 + m2(a2
1 + a2
C2 + 2a1aC2C2) + ˜I2]¨θ1 +
+[m2(a2
C2 + a1aC2C2) + ˜I2]¨θ2 − m2a1aC2S2(2 ˙θ1
˙θ2 + ˙θ2
2) +
+m1gaC1C1 + m2g(a1C1 + aC2C12)
τ2 = [m2(a2
C2 + a1aC2C2) + ˜I2] ¨θ1 + (m2a2
C2 + ˜I2)¨θ2 +
m2a1aC2S2
˙θ2
1 + m2gaC2C12

Example - 2 (‘robotics’ approach)
The structural properties of the manipulator are exploited for the computation of
the kinematic and potential energies. For the computation of the CoM velocities,
one obtains:
J1
v =


−aC1S1 0
aC1C1 0
0 0

 J2
v =


−a1S1 − aC2S12 −aC2S12
a1C1 + aC2C12 aC2C12
0 0


and
J1
ω =


0 0
0 0
1 0

 J2
ω =


0 0
0 0
1 1


In this particular case, the frames associated to link 1 and 2 have z axes parallel to
the the same axis of F0 , and therefore it is not necessary to introduce the
rotation matrices R1, R2 (ω = ωz ).

The kinetic energy is computed as
K =
1
2
˙qT
m1J1 T
v J1
v + m2J2 T
v J2
v + J1 T
ω
˜I1J1
ω + J2 T
ω
˜I2J2
ω ˙q
being
J1 T
ω
˜I1J1
ω + J1 T
ω
˜I2J2
ω =
˜I1 + ˜I2
˜I2
˜I2
˜I2
The elements of the inertia matrix M(q) are
M11 = m1a2
C1 + m2(a2
1 + a2
C2 + 2a1aC2C2) + ˜I1 + ˜I2
M12 = m2(a2
C2 + a1aC2C2) + ˜I2
M22 = m2a2
C2 + ˜I2

From M11 = m1a2
C1 + m2(a2
1 + a2
C2 + 2a1aC2C2) + ˜I1 + ˜I2
M12 = m2(a2
C2 + a1a2
C2C2) + ˜I2
M22 = m2a2
C2 + ˜I2
The Christoﬀel symbols cijk = 1
2
∂Mkj
∂qi
+ ∂Mki
∂qj
−
∂Mij
∂qk
are
c111 =
1
2
∂M11
∂q1
= 0
c121 = c211 =
1
2
∂M11
∂q2
= −m2a1aC2S2 = h
c221 =
∂M12
∂q2
−
1
2
∂M22
∂q1
= h
c112 =
∂M21
∂q1
−
1
2
∂M11
∂q2
= −h
c122 = c212 =
∂M22
∂q1
= 0
c222 =
∂M22
∂q2
= 0
=⇒ Matrix C(q, ˙q) is
C(q, ˙q) =
h ˙θ2 h( ˙θ1 + ˙θ2)
−h ˙θ1 0

Matrix N(q) is
N(q) = ˙M(q) − 2C(q, ˙q)
=
2h ˙θ2 h ˙θ2
h ˙θ2 0
− 2
h ˙θ2 h( ˙θ1 + ˙θ2)
−h ˙θ1 0
=
0 −2h ˙θ1 − h ˙θ2
2h ˙θ1 + h ˙θ2 0

For the potential energy, we have:
P1 = m1gaC1S1
P2 = m2g(a1S1 + aC2S12)
Then
P = P1 + P2 = (m1aC1 + m2a1)gS1 + m2gaC2S12
g1 =
∂P
∂θ1
= (m1aC1 + m2a1)gC1 + m2gaC2C12
g2 =
∂P
∂θ2
= m2gaC2C12

Summarizing, we have
M11
¨θ1 + M12
¨θ2 + c121
˙θ1
˙θ2 + c211
˙θ2
˙θ1 + c221
˙θ2
2 + g1 = τ1
M21
¨θ1 + M22
¨θ2 + c112
˙θ2
1 + g2 = τ2
or
[m1a2
C1 + m2(a2
1 + a2
C2 + 2a1aC2C2) + ˜I1 + ˜I2]¨θ1 + [m2(a2
C2 + a1a2
C2C2) + ˜I2]¨θ2
−m2a1aC2S2
˙θ2
2 − 2m2a1aC2S2
˙θ1
˙θ2
+(m1aC1 + m2a1)gC1 + m2gaC2C12 = τ1
[m2(a2
C2 + a1aC2C2) + ˜I2]¨θ1 + [m2a2
C2 + ˜I2]¨θ2
+m2a1aC2S2
˙θ2
1
+m2gaC2C12 = τ2
=⇒ Same result!

Properties of the Euler-Lagrangian dynamic model
The Euler-Lagrange dynamic model is characterized by some structural properties,
concerning in particular:
1 The inertia matrix M(q);
2 The vector c(q, ˙q) = C(q, ˙q)q;
3 The vectors g(q) and D ˙q;
4 Linearity with respect to the geometric/mechanical parameters.

1. Properties of matrix M(q)
1 M(q) ∈ IRn×n
is symmetric, positive definite and depends on the manipulator
configuration q
2 M(q) is upper and lower bounded:
µ1I ≤ M(q) ≤ µ2I
that is
xT
(M(q) − µ1I)x ≥ 0 xT
(µ2I − M(q))x ≥ 0
3 also M−1
(q) is bounded
1
µ2
I ≤ M−1
(q) ≤
1
µ1
I
4 in case of revolute joints, then µ1, µ2 are constant (not function of q) since the
elements of M(q) are functions of sin(qi ) or cos(qi )
5 in case of prismatic joints, µ1, µ2 may result scalar functions of q
6 since M(q) is bounded, then
M1 ≤ ||M(q)|| ≤ M2
for some properly defined norm (1, 2, p, ∞)

2. Properties of vector c(q, ˙q) = C(q, ˙q)˙q
1 C(q, ˙q)˙q is a quadratic function of ˙q
2 the generic k-th element of vector c(q, ˙q) = C(q, ˙q)˙q can also be witten as
ck (q, ˙q) = ˙qT
Sk (q)˙q
Sk(q) =
1
2
∂mk
∂q
+
∂mk
∂q
T
−
∂M
∂qk
mk = k-th col. of M
3 it results that
||C(q, ˙q)˙q|| ≤ vb||˙q||2
4 in case of rotative joints, then vb is constant (not function of q) since we have only
transcendental functions (sin(qi ) or cos(qi ))
5 in case of prismatic joints, then vb may result a scalar function of q
6 for any choice of C(q, ˙q), then matrix N(q, ˙q) = ˙M(q, ˙q) − 2C(q, ˙q) veriﬁes:
˙qT
N(q, ˙q)˙q = 0
7 with a proper choice of the elements of C(q, ˙q) (Christoﬀel symbols), matrix
N(q, ˙q) = ˙M(q, ˙q) − 2C(q, ˙q) is skew-symmetric, i.e.
xT
N(q, ˙q)x = 0 ∀x

3. Properties of vectors g(q) and D˙q
1 the friction term D˙q is bounded:
||D˙q|| ≤ dmax ||˙q||
2 the gravity term g(q) is bounded
||g(q)|| ≤ gb(q)
3 in case of revolute joints, gb is constant (does not depend on q) since qi depends
on sinusoidal functions (sin(qi ) or cos(qi ))
4 in case of prismatic joints, then gb may result function of q.

4. Linearity properties (in the geometrical/mechanical parameters)
The dynamic model of a manipulator:
in general is a non linear function of q, ˙q, ¨q and with dynamic coupling eﬀects
among the joints,
is a linear function of the geometrical/mechanical parameters of the links (i.e.
masses, inertia, friction, ...)
M(q)¨q + C(q, ˙q)˙q + D˙q + g(q) = τ
Y(q, ˙q, ¨q)α = τ

Example
Properties of the dynamic model of a
2 dof manipulator.
Neglecting friction eﬀects we have:
M(q) =
m1a2
C1 + m2(a2
1 + a2
C2 + 2a1aC2C2) + ˜I1 + ˜I2 m2(a2
C2 + a1aC2C2) + ˜I2
m2(a2
C2 + a1aC2C2) + ˜I2 m2a2
C2 + ˜I2
C(q, ˙q)˙q = −m2a1aC2S2
2 ˙θ1
˙θ2 + ˙θ2
2
− ˙θ2
1
, g(q)=
(m1aC1 + m2a1)gC1 + m2gaC2C12
m2gaC2C12
Consider (for the sake of simplicity) the 1-norm || · ||1, and θ1, θ2 ∈ [−π/2, π/2].

Example
1. Bounds on the inertia matrix.
The scalar quantities µ1, µ2 can be defined as the minimum and maximum
eigenvalues (λmin, λmax ) of M(q), ∀q. Computationally, it is easier to define the
scalars M1, M2.
The norm–1 of M(q) is alway defined on the basis of the first column:
||M(q)||1 = |m1a2
C1 +m2(a2
1 +a2
C2 +2a1aC2C2)+˜I1 +˜I2|+|m2(a2
C2 +a1aC2C2)+˜I2|
which is bounded by
M1 = m1a2
C1 + m2(a2
1 + 2a2
C2) + ˜I1 + 2˜I2
M2 = m1a2
C1 + m2(a2
1 + 2a2
C2 + 3a1aC2) + ˜I1 + 2˜I2

Example
2. Bounds on vector C(q, ˙q)˙q
It results that
||C(q, ˙q)˙q||1 = |m2a1aC2S2(2 ˙θ1
˙θ2 + ˙θ2
2)| + |m2a1aC2S2
˙θ2
1|
≤ m2a1aC2|2 ˙θ1
˙θ2 + ˙θ2
2 + ˙θ2
1|
≤ m2a1aC2(| ˙θ1| + | ˙θ2|)2
= vb||˙q||2
Moreover N(q, ˙q) = ˙M(q, ˙q) − 2C(q, ˙q), (h = −m2a1aC2S2):
N(q) = ˙M(q) − 2C(q, ˙q)
=
2h ˙θ2 h ˙θ2
h ˙θ2 0
− 2
h ˙θ2 h( ˙θ1 + ˙θ2)
−h ˙θ1 0
=
0 −2h ˙θ1 − h ˙θ2
2h ˙θ1 + h ˙θ2 0

Example
3. Bounds on the gravity vector g(q)
||g(q)||1 = |(m1aC1 + m2a1)gC1 + m2gaC2C12| + |m2gaC2C12|
≤ (m1aC1 + m2a1)g + 2m2gaC2
= gb
Notice that if one of the joints is prismatic (and therefore ai , aCi may vary in
time), then vb, gb are functions of q.

Example
Plots of the eigenvalues of M(q) for the 2 dof planar manipulator
0
50
100
150
200
250
300
350
0
100
200
300
400
0
20
40
60
80
100
120
140
160
180
l
m
in = 6.2399
l
m
ax = 161.2601
q1 [deg]
Andamento autovalori minimo e massimo di M(q)
q2 [deg]
λmin = 6.2399
λmax = 161.2601
In this case, the eigenvalues depend only on q2! These results have been obtained
with:
m1 = m2 = 50kg; I1 = I2 = 10; a1 = a2 = 1; ac1 = ac2 = 0.5
One obtains
M1 = 117, 5, M2 = 192.5 e M 1 = 192.5, M 2 = 161.26, M ∞ = 192.5

Example
3. Linearity of the dynamic model Y(q, ˙q, ¨q)α = τ
[m1a2
C1 + m2(a2
1 + a2
C2 + 2a1aC2C2) + ˜I1 + ˜I2]¨θ1 + [m2(a2
C2 + a1a2
C2C2) + ˜I2]¨θ2
−m2a1aC2S2
˙θ2
2 − 2m2a1aC2S2
˙θ1
˙θ2
+(m1aC1 + m2a1)gC1 + m2gaC2C12 = τ1
[m2(a2
C2 + a1aC2C2) + ˜I2]¨θ1 + [m2a2
C2 + ˜I2]¨θ2
+m2a1aC2S2
˙θ2
1
+m2gaC2C12 = τ2
By inspection, one can deﬁne the parameters vector
α = [α1, α2, α3, α4, α5]T
with
α1 = m1aC1, α2 = m1a2
C1 + ˜I1, α3 = m2, α4 = m2aC2, α5 = m2a2
C2 + ˜I2

Example
Then
Y(q, ˙q, ¨q) =
y11 y12 y13 y14 y15
0 0 0 y24 y25
with
y11 = gC1
y12 = ¨θ1
y13 = a2
1
¨θ1 + a1gC1
y14 = 2a1C2
¨θ1 + a1C2
¨θ2 − 2a1S2
˙θ1
˙θ2 − a1S2
˙θ2
1 + gC12
y15 = ¨θ1 + ¨θ2
y24 = a1C2
¨θ1 + a1S2
˙θ2
1 + gC12
y25 = ¨θ1 + ¨θ2
The elements yij depend on q, ˙q, ¨q, g and on a1.

Example
Identiﬁcation of the dynamic parameters
From:
Y(q, ˙q, ¨q)α = τ
by measuring along a proper trajectory the quantities q, ˙q, ¨q, τ one obtains:
→ τ0 = Y0α
→ YT
0 τ0 = YT
0 Y0α
→ α = (YT
0 Y0)−1
YT
0 τ0 = Y+
0 τ0
Being Y+
0 = (YT
0 Y0)−1
YT
0 the (left) pseudo-inverse of Y0.
Note that some of the parameters may result non identiﬁable!

A Simulink dynamic model
Demux
.
1/s
q2
1/s
dq2
Demux
Demux2
Mux
Mux2 ddq
dq
Mux
Mux
Mux
Mux3
1/s
q1
1/s
dq1
MATLAB
Function
dir2yn
Clock
t
Tempo
Coppie
Disturbo
q
Posizioni
dq
Velocita`
ddq
Accelerazioni
tau
Tau
Mux
Mux1
q
tau
sc_d2dof.m
Inizializzazione con
esed2dof.m
The model takes into account the dynamics of a planar 2 dof arm, with gravity (arm in a
vertical plane). The dynamic model, described by the Matlab block dir2dyn.m
is obtained with the Euler-Lagrange:
M(q)¨q + C(q, ˙q)˙q + D˙q + g(q) = τ + JT
(q)Fa

Dynamic model in Simulink
The elements of the inertia matrix M(q) are
M11 = m1a2
C1 + m2(a2
1 + a2
C2 + 2a1aC2C2) + ˜I1 + ˜I2
M12 = m2(a2
C2 + a1aC2C2) + ˜I2
M22 = m2a2
C2 + ˜I2
Matrix C(q, ˙q) is
C(q, ˙q) =
h ˙θ2 h( ˙θ1 + ˙θ2)
−h ˙θ1 0
h = −m2a1aC2S2
The terms in the gravity vector are:
g1 = (m1aC1 + m2a1)gC1 + m2gaC2C12
g2 = m2gaC2C12

function ddq = dir2dyn(u);
% function ddq = dir2dyn(q,dq,tau,tauint);
% Dynamic model of a planar 2 dof manipulator
% Input:
% dq -->> joint velocity vector
% q -->> joint position vector
% tau -->> joint torques
% tauint -->> interaction forces
% Output:
% ddq -->> joint acceleration vector
global I1z I2z m1 m2 L1 L2 Lg1 Lg2 g D
q1 = u(1); q2 = u(2);
dq = u(3:4); dq1 = u(3); dq2 = u(4);
tau = u(5:6); tauint = u(7:8);
% Kinematic functions
ßs1 = sin(q1); s2 = sin(q2); s12 = sin(q1+q2);
c1 = cos(q1); c2 = cos(q2); c12 = cos(q1+q2);

%%%%% Elements of the Inertia Matrix M
M11 = I1z+I2z+Lg1^2*m1+m2*(L1^2+Lg2^2+2*L1*Lg2*c2);
M12 = I2z+m2*(Lg2^2+L1*Lg2*c2);
M22 = I2z+Lg2^2*m2;
M = [M11, M12; M12, M22];
%%%%% Coriolis and centrifugal elements
C11 = -(L1*dq2*s2*(Lg2*m2)); C12 = -(L1*dq12*s2*(Lg2*m2));
C21 = m2*L1*Lg2*s2*dq1; C22 = 0;
C = [C11 C12; C21 C22];
%%%%% Gravity :
g1 = m1*Lg1*c1+m2*(Lg2*c12 + L1*c1); g2 = m2*Lg2*c12;
G = g*[g1; g2];
%%%%% Computation of acceleration
ddq = inv(M) * (tau + tauint - C*dq - D*dq - G);

% Simulation of the dynamic model of a planar 2 dof manipulator
% Simulink scheme: sc_d2dof.m
% Definition and initialization of global variables
% Run the simulation RK45
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%% Robots’ Coefficients %%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
clear I1z I2z m1 m2 L1 L2 Lg1 Lg2 g D
global I1z I2z m1 m2 L1 L2 Lg1 Lg2 g D
I1z = 10; I2z = 10;
m1 = 50.0; m2 = 50.0;
L1 = 1; L2 = 1;
Lg1=L1/2; Lg2=L2/2;
g = 9.81;
D = 0.0 * eye(2); % friction
% D = diag([30,30]); % friction
% D = diag([80,80]); % friction

COPPIA = 0; % joint torques
DISTURBO = 0; % joint torques due to interaction with environment
%%%% Iniziatialization and run
TI = 0; TF = 10;
ERR = 1e-3;
TMIN = 0.002; TMAX = 10*TMIN;
OPTIONS = [ERR,TMIN,TMAX,0,0,2];
X0 = [0 0 0 0];
[ti,xi,yi]=rk45(’sc_d2dof’,TF,X0,OPTIONS);

D = diag[0, 0];
0 1 2 3 4 5 6 7 8 9 10
−4
−3
−2
−1
0
1
Posizioni q1, q2 (dash)
0 1 2 3 4 5 6 7 8 9 10
−10
−5
0
5
10
Velocita‘ dq1, dq2 (dash)
0 1 2 3 4 5 6 7 8 9 10
−40
−20
0
20
40
Accelerazioni ddq1, ddq2 (dash)
0 1 2 3 4 5 6 7 8 9 10
−1
−0.5
0
0.5
1
Coppie tau1, tau2 (dash)

D = diag[30, 30];
0 1 2 3 4 5 6 7 8 9 10
−3
−2
−1
0
1
0 1 2 3 4 5 6 7 8 9 10
−4
−2
0
2
4
0 1 2 3 4 5 6 7 8 9 10
−15
−10
−5
0
5
10
15
0 1 2 3 4 5 6 7 8 9 10
−1
−0.5
0
0.5
1

D = diag[80, 80];
0 1 2 3 4 5 6 7 8 9 10
−3
−2
−1
0
1
2
0 1 2 3 4 5 6 7 8 9 10
−3
−2
−1
0
1
0 1 2 3 4 5 6 7 8 9 10
−15
−10
−5
0
5
10
15
0 1 2 3 4 5 6 7 8 9 10
−1
−0.5
0
0.5
1

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