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Dynamic model of robot manipulators

Claudio Melchiorri

Dipartimento di Ingegneria dell’Energia Elettrica e dell’Informazione (DEI)

Universita di Bologna

email: [email protected]

C. Melchiorri (DEI) Dynamic Model 1 / 65

Summary

1 Dynamic modelsIntroductionEuler-Lagrange model


Dynamic models Introduction

Dynamic model of manipulators

Robot Dynamics =⇒ Study of the relation between the applied forces/torquesand 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 thejoint 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 areknown, 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 dynamicmodel 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 ofconnected 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) (andthen of q(t), q(t)), given the vector of generalized forces (torques and/orforces) τ (t) applied to the joints and, in case, the external forces applied tothe 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 toobtain a desired trajectory q(t), q(t), q(t), once the forces applied of theend-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

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 areusually 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

∂qii = 1, . . . , n

being ψi the non-conservative (external or dissipative) generalized forcesperforming work on qi . In robotics, we consider:

τi joint actuator torque[

JTFc

]

iterm due to external (contact) forces

dii qi joint friction torque

Therefore: ψi = τi +[

JTFc

]

i− dii qi .




Since the potential energy does not depend on the velocity, the Euler-Lagrangeequations can be rewritten as

ψi =d

dt

(

∂K

∂qi

)

−∂K

∂qi+∂P

∂qii = 1, . . . , n (1)

This formulation is more convenient since in robotics it is possible to computequite 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 B:

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 as

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 vC and ω, i.e. the trans-lational and rotational velocities of thecenter of mass, are known with respectto 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 byintroducing the rotation matrix R betweeen F ’ and F0

RT v = RT (vC + ω × r) = RT vC + (RTω)× (RT r)

Thereforev′ = 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

vTC vC dm +1

2

∫

B

rTSTSr dm +

∫

B

vTC Sr dm

=1

2

∫

B

vTC vC dm +1

2

∫

B

rTSTSr dm

As a matter of fact, from the definition of CoM (∫

Br dm =

∫

B(pC − p)dm = 0):

∫

B

vTC Sr dm = vTC S

∫

B

r dm = 0




In conclusion

K =1

2

∫

B

vTC vC dm +1

2

∫

B

rTSTSr dm

The first term depends on the linear velocity vC of the center of mass

1

2

∫

B

vTC vC dm =1

2m vTC vC

For the second term, considering that aTb = Tr(a bT ) and the particularstructure of matrix S, we have

1

2

∫

B

rTSTSr dm =1

2

∫

B

Tr(Sr rTST ) dm =1

2Tr(S

∫

B

rrT dm ST )

=1

2Tr(S J ST ) =

1

2ω

T 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 havethe following general expressions:

J =

∫

r2x dm∫

rx ry dm∫

rx rz dm∫

rx ry dm∫

r2y dm∫

ry rz dm∫

rx rz dm∫

ry rz dm∫

r2z dm

I =

∫

(r2y + r2z ) dm −∫

rx ry dm −∫

rx rz dm

−∫

rx ry dm∫

(r2x + r2z ) dm −∫

ry rz dm

−∫

rx rz dm −∫

ry rz dm∫

(r2x + r2y ) dm

The elements of both matrices J ed I depend on vector r, i.e. on the position ofthe generic point of the i-th link with respect to its center of mass, defined in thebase frame.

Since the position of the i-th link depends on the configuration of themanipulator, 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 =

IxxIyy

Izz

=

112m(b2 + c2)

112m(a2 + c2)

112m(a2 + b2)

• Empty cylinder with length h, and external/internal radii a, b

I =

12m(a2 + b2)

112m

(

3(a2 + b2) + h2)

Izz

,

Izz = Iyy

I ′zz = Izz +m(

h2

)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 dueto translation p of the frameof computation:

I = Ic +m(pT p E3×3 − ppT )

Ic body inertia matrix wrt thecenter of massE3×3 (3× 3) identity matrix




In conclusion, the kinetic energy of a rigid body is defined as (Konig Theorem)

K =1

2m vTC vC +

1

2ωT 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 ‖ = vTCvC ), is invariant with

respect to the reference frame used to express the velocity:

vTC vC = (RvC )T (RvC ) = vTC (RTR)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 thefollowing relation:

I′ = R I RT

Then: ωT Iω = ω

′T I′ω′ = (R ω)T (RIRT )(Rω) = ωT (RTR)I(RTR)ω.

Therefore, being (3) invariant with respect to the reference frame, F can be chosen in order tosimplify the computations required for the evaluation of K .




Since the kinetic energy Ki of the generic i-th link is invariant with respect to thereference frame (used to express vC ,ω, I), it is convenient to chose a frameFi 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 thegeometric properties of the link.

On the other hand, normally the rotational velocity ω is defined in the base frameF0 , and therefore it is needed to transform it according to RT

ω, being R therotation matrix between Fi and F0 (known from the kinematic model of themanipulator).




In conclusion, the kinetic energy of a manipulator can be determined when, foreach 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 inwhich it has a constant expression Ii ;

the linear velocity vCi of the center of mass, and the rotational velocity ωi ofthe 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

2miv

TCivCi +

1

2ωTi Ri IiR

Ti ωi

It is now necessary to compute the linear and rotational velocities (vCi and ωi ) asfunctions 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, . . . , qnthrough the Jacobian matrix J. The same methodology can be used to compute thevelocity of a generic point of the manipulator, and in particular the velocity vCi = pCi ofthe center of mass pCi , that results function of the joint velocities q1, . . . , qi only:

pCi = Jiv1q1 + Jiv2q2 + . . .+ Jivi qi = Jiv q

ωi = Jiω1q1 + Jiω2q2 + . . .+ Jiωi qi = Jiωq

where

Jiv =[

Jiv1 . . . Jivi 0 . . . 0]

Jiω =[

Jiω1 . . . Jiωi 0 . . . 0]

with[

JivjJiωj

]

=

[

zj−1 × (pCi − pj−1)zj−1

]

rotational joint[

JivjJiωj

]

=

[

zj−1

0

]

linear joint

being pj−1 the position of the origin of the frame associated to the j-th link.C. Melchiorri (DEI) Dynamic Model 19 / 65



In conclusion, for a n dof manipulator we have:

K =1

2

n∑

i=1

mivTCivCi +

1

2

n∑

i=1

ωTi Ri IiR

Ti ω

=1

2qT

n∑

i=1

[

miJi Tv (q)Jiv (q) + Ji T

ω(q)Ri IiR

Ti J

iω(q)

]

q

=1

2qTM(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 themanipulator configuration q.

Matrix M(q) is called the Inertia Matrix of the manipulator.




Computation of the Potential Energy. For rigid bodies, the only potentialenergy taken into account in the dynamics is due to the gravitational field g. Forthe generic i-th link

Pi =

∫

Li

gTp dm = gT

∫

Li

p dm = gTpCimi

The potential energy does not depend on the joint velocities q, and may beexpressed as a function of the position of the centers of mass. For the wholemanipulator:

P =

n∑

i=1

gTpCimi

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 oneadopted 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 themanipulator. The dynamics is expressed by

ψk =d

dt

(

∂L

∂qk

)

−∂L

∂qkk = 1, . . . , n

The Lagrangian function is

L = K − P =1

2

n∑

i=1

n∑

j=1

Mij qi qj −n∑

i=1

gTpCimi

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

dtqj =

n∑

j=1

Mkj qj +n∑

i=1

n∑

j=1

∂Mkj

∂qiqi qj

Moreover∂L

∂qk=

1

2

n∑

i=1

n∑

j=1

∂Mij

∂qkqi qj −

∂P

∂qkC. Melchiorri (DEI) Dynamic Model 22 / 65



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, andtherefore their computation is relatively simple once the manipulator’s configuration isknown. 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 andconsidering 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 q2j 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 iand j

For the configuration-dependent terms:

gk represents the torque generated on joint k by the gravity force acting on themanipulator in the current configuration.




Recalling that the non-conservative forces ψk are in general composed by

τk joint actuator torque[

JTFc

]

kexternal (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+Dq+ 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 whoseelements 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 thenn∑

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 = 12

[

∂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 matrixsuch 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 followingproperty 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 symmetryof M(q)) that nkj = −njk .

Since matrix N is skew-symmetrix, then

xT N(q, q) x = 0, ∀x




The conditionxT N(q, q) x = 0, ∀x

holds since N(q, q) is skew-symmetric, due to the particular choice of the elements ofmatrix 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

(

qTMq)

= qT (τ −Dq− g(q)− JTF)

The first element is (from the dynamic model Mq = −Cq−Dq − g(q) + τ − JTF):

1

2

d

dt

(

qTMq)

=1

2qTMq+ qTMq =

1

2qTMq+ qT (−Cq−Dq− g(q) + τ − JTF)




Then

1

2qTMq+ qT (−Cq −Dq− g(q) + τ − JTF) = qT (τ −Dq− g(q)− JTF)

from which

1

2qTMq = qTCq =⇒ qT (M− 2C)q = 0

This equation holds ∀q and without any assumption on matrix C(q, q) (it holdsalso if C is not based on the Cristoffel symbols).




In deriving the dynamic model, the actuation system has not been taken intoaccount. 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 thedynamic model derived on the basis of the Euler-Lagrangian formulation.



Example - 1

Dynamic model of a pendulum (one dof manipulator).

Considerθ 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 = 12 I θ

2

Potenzial energy: P = mgL(1− cos θ)

Lagrangian function L: L = 12 I θ

2 −mgL(1− cos θ)



Example - 1

Lagrangian function: L = 12 I θ

2 −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 appliedto 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 throughthe CoM and parallel to z;

ai i-th link length;

aCi distance between joint i and the CoMof the i-th link;

τi torque on joint i;

g gravity force along y;

Pi , Ki potential and kinetic energy of thei-th link.

The dynamic equations will be obtained in two manners:

a) with the “classic” approach, deriving the Lagrangian function (based on the kineticand 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. Thekinetic and potential energies Ki and Pi are:

link 1:

K1 =1

2m1a

2C1θ1

2+

1

2I1θ1

2, P1 = m1gaC1S1

link 2: in this case, the position and velocity of the CoM are

pC2x = a1C1 + aC2C12

pC2y = a1S1 + aC2S12

pC2x = −a1S1θ1 − aC2S12(θ1 + θ2)

pC2y = a1C1θ1 + aC2C12(θ1 + θ2)

then

K2 =1

2m2p

TC2pC2 +

1

2I2(θ1 + θ2)

2, P2 = m2g(a1S1 + aC2S12)

wherepTC2pC2 = a

21 θ1

2+ a

2C2(θ1 + θ2)

2 + 2a1aC2C2(θ21 + θ1θ2)



Example - 2 (classic approach)

Therefore, L = K1 + K2 − P1 − P2 and

τ1 = [m1a2C1 + I1 +m2(a

21 + a2C2 + 2a1aC2C2) + I2]θ1 +

+[m2(a2C2 + a1aC2C2) + I2]θ2 −m2a1aC2S2(2θ1θ2 + θ22) +

+m1gaC1C1 +m2g(a1C1 + aC2C12)

τ2 = [m2(a2C2 + a1aC2C2) + I2]θ1 + (m2a

2C2 + I2)θ2 +

m2a1aC2S2θ21 +m2gaC2C12



Example - 2 (‘robotics’ approach)

The structural properties of the manipulator are exploited for the computation ofthe kinematic and potential energies. For the computation of the velocities of theCoM, one obtains:

J1v =

−aC1S1 0aC1C1 0

0 0

J2v =

−a1S1 − aC2S12 −aC2S12a1C1 + aC2C12 aC2C12

0 0

and

J1ω=

0 00 01 0

J2ω=

0 00 01 1

In this particular case, the frames associated to link 1 and 2 have z axes parallel tothe same axis of F0 , and therefore it is not necessary to consider the rotationmatrices R1,R2 (ω = ωz).




The kinetic energy is computed as

K =1

2qT

[

m1J1 Tv J1v +m2J

2 Tv J2v + J1 T

ωI1J

1ω+ J2 T

ωI2J

2ω

]

q

being

J1 Tω

I1J1ω+ J1 T

ωI2J

2ω=

[

I1 + I2 I2I2 I2

]

The elements of the inertia matrix M(q) are

M11 = m1a2C1 +m2(a

21 + a2C2 + 2a1aC2C2) + I1 + I2

M12 = m2(a2C2 + a1aC2C2) + I2

M22 = m2a2C2 + I2




From M11 = m1a2C1 +m2(a

21 + a

2C2 + 2a1aC2C2) + I1 + I2

M12 = m2(a2C2 + a1aC2C2) + I2

M22 = m2a2C2 + I2

The Christoffel symbols cijk = 12

[

∂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) =

[

hθ2 h(θ1 + θ2)

−hθ1 0

]




Matrix N(q) is

N(q) = M(q)− 2C(q, q)

=

[

2hθ2 hθ2hθ2 0

]

− 2

[

hθ2 h(θ1 + θ2)

−hθ1 0

]

=

[

0 −2hθ1 − hθ22hθ1 + hθ2 0

]

As expected, it results skew-symmetric.




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, from M(q)q+ C(q, q)q+Dq+ g(q) = τ we have

M11θ1 +M12θ2 + c121θ1θ2 + c211θ2θ1 + c221θ22 + g1 = τ1

M21θ1 +M22θ2 + c112θ21 + g2 = τ2

or

[m1a2C1 +m2(a

21 + a2C2 + 2a1aC2C2) + I1 + I2]θ1 + [m2(a

2C2 + a1aC2C2) + I2]θ2

−m2a1aC2S2θ22 − 2m2a1aC2S2θ1θ2

+(m1aC1 +m2a1)gC1 +m2gaC2C12 = τ1

[m2(a2C2 + a1aC2C2) + I2]θ1 + [m2a

2C2 + I2]θ2

+m2a1aC2S2θ21

+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 bounded1

µ2I ≤ M−1(q) ≤

1

µ1I

4 in case of revolute joints, then µ1, µ2 are constant (not function of q) since theelements 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)q1 C(q, q)q is a quadratic function of q2 the generic k-th element of vector c(q, q) = C(q, q)q can also be witten as

ck(q, q) = qTSk(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 onlytranscendental functions (sin(qi ) or cos(qi))

5 in case of prismatic joints, then vb may result a scalar function of q6 for any choice of C(q, q), then matrix N(q, q) = M(q, q)− 2C(q, q) verifies:

qTN(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.

xTN(q, q)x = 0 ∀x




3. Properties of vectors g(q) and Dq

1 the friction term Dq is bounded:

||Dq|| ≤ 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 dependson 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 effectsamong 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+Dq + g(q) = τ

Y(q, q, q)α = τ



Example

Properties of the dynamic model of a2 dof manipulator.

Neglecting friction effects we have:

M(q) =

[

m1a2C1 +m2(a

21 + a2C2 + 2a1aC2C2) + I1 + I2 m2(a

2C2 + a1aC2C2) + I2

m2(a2C2 + a1aC2C2) + I2 m2a

2C2 + I2

]

C(q, q)q = −m2a1aC2S2

[

2θ1θ2 + θ22−θ21

]

, 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 maximumeigenvalues (λmin, λmax) of M(q), ∀q. Computationally, it is easier to define thescalars M1,M2.

The 1–norm of M(q) is alway defined on the basis of the first column:

||M(q)||1 = |m1a2C1+m2(a

21+a2C2+2a1aC2C2)+ I1+ I2|+ |m2(a

2C2+a1aC2C2)+ I2|

which is bounded by

M1 = m1a2C1 +m2(a

21 + 2a2C2) + I1 + 2I2

M2 = m1a2C1 +m2(a

21 + 2a2C2 + 3a1aC2) + I1 + 2I2



Example

2. Bounds on vector C(q, q)q

It results that

||C(q, q)q||1 = |m2a1aC2S2(2θ1θ2 + θ22)|+ |m2a1aC2S2θ21 |

≤ m2a1aC2|2θ1θ2 + θ22 + θ21 |

≤ 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)

=

[

2hθ2 hθ2hθ2 0

]

− 2

[

hθ2 h(θ1 + θ2)

−hθ1 0

]

=

[

0 −2hθ1 − hθ22hθ1 + hθ2 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 intime), then vb, gb are functions of q.



Example

Plots of the eigenvalues of M(q) for the 2 dof planar manipulator

050

100150

200250

300350

0

100

200

300

4000

20

40

60

80

100

120

140

160

180

lm

in = 6.2399

lm

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 obtainedwith:

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‖∞ = 192.5C. Melchiorri (DEI) Dynamic Model 53 / 65


Example

3. Linearity of the dynamic model Y(q, q, q)α = τ

[m1a2C1 +m2(a

21 + a2C2 + 2a1aC2C2) + I1 + I2]θ1 + [m2(a

2C2 + a1a

2C2C2) + I2]θ2

−m2a1aC2S2θ22 − 2m2a1aC2S2θ1θ2

+(m1aC1 +m2a1)gC1 +m2gaC2C12 = τ1

[m2(a2C2 + a1aC2C2) + I2]θ1 + [m2a

2C2 + I2]θ2

+m2a1aC2S2θ21

+m2gaC2C12 = τ2

By inspection, one can define the parameters vector

α = [α1, α2, α3, α4, α5]T

with

α1 = m1aC1, α2 = m1a2C1 + I1, α3 = m2, α4 = m2aC2, α5 = m2a

2C2 + I2



Example

Then

Y(q, q, q) =

[

y11 y12 y13 y14 y150 0 0 y24 y25

]

with

y11 = gC1

y12 = θ1

y13 = a21θ1 + a1gC1

y14 = 2a1C2θ1 + a1C2θ2 − 2a1S2θ1θ2 − a1S2θ21 + gC12

y15 = θ1 + θ2

y24 = a1C2θ1 + a1S2θ21 + 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:

→ τ 0 = Y0α

→ YT0 τ 0 = YT

0 Y0α

→ α = (YT0 Y0)

−1YT0 τ 0 = Y+

0 τ 0

Being Y+0 = (YT

0 Y0)−1YT

0 the (left) pseudo-inverse of Y0.

Note that some of the parameters may result non identifiable!



A Simulink dynamic model

Demux

.

1/sq2

1/sdq2

Demux

Demux2

Mux

Mux2 ddq

dq

Mux

Mux

Mux

Mux3

1/sq1

1/sdq1

MATLABFunctiondir2yn

Clockt

Tempo

Coppie

Disturbo

qPosizioni

dqVelocita`

ddqAccelerazioni

tauTau

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 avertical plane). The dynamic model, described by the Matlab block dir2dyn.m

is obtained with the Euler-Lagrange approach:

M(q)q+ C(q, q)q+Dq+ g(q) = τ + JT (q)Fa



Dynamic model in Simulink

The elements of the inertia matrix M(q) are

M11 = m1a2C1 +m2(a

21 + a2C2 + 2a1aC2C2) + I1 + I2

M12 = m2(a2C2 + a1aC2C2) + I2

M22 = m2a2C2 + I2

Matrix C(q, q) is

C(q, q) =

[

hθ2 h(θ1 + θ2)

−hθ1 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

s1 = sin(q1); s2 = sin(q2); s12 = sin(q1+q2);

c1 = cos(q1); c2 = cos(q2); c12 = cos(q1+q2);

dq12 = dq1 + dq2;C. Melchiorri (DEI) Dynamic Model 59 / 65



%%%%% 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]); % frictionC. Melchiorri (DEI) Dynamic Model 61 / 65



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

1Posizioni q1, q2 (dash)

0 1 2 3 4 5 6 7 8 9 10−10

−5

0

5

10Velocita‘ dq1, dq2 (dash)

0 1 2 3 4 5 6 7 8 9 10−40

−20

0

20

40Accelerazioni ddq1, ddq2 (dash)

0 1 2 3 4 5 6 7 8 9 10−1

−0.5

0

0.5

1Coppie tau1, tau2 (dash)




D = diag[30, 30];

0 1 2 3 4 5 6 7 8 9 10−3

−2

−1

0


0 1 2 3 4 5 6 7 8 9 10−4

−2

0

2


0 1 2 3 4 5 6 7 8 9 10−15

−10

−5

0

5

10


0 1 2 3 4 5 6 7 8 9 10−1

−0.5

0

0.5





D = diag[80, 80];

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 10−3

−2

−1

0


0 1 2 3 4 5 6 7 8 9 10−15

−10

−5

0

5

10


0 1 2 3 4 5 6 7 8 9 10−1

−0.5

0

0.5



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