Robot Dynamics – Part I: The Euler-Lagrange Approach

Robot Dynamics I – PPGEE/UFMG 1 / 98

Robot Dynamics – Part I:The Euler-Lagrange Approach

Leonardo A. B. Torres

August, 2020.

Foundations of Lagrangian Mechanics

Foundations ofLagrangianMechanics

Introduction

Introduction

Introduction

IntroductionVirtualDisplacements

VirtualDisplacements

Virtual Work

Generalized ForcesGeneralized Forces:External orNon-conservativeTorques

Generalized Forces:External orNon-conservativeTorques



The Total VirtualWork done byConstraint Forces isZero


Introduction


Introduction

Introduction

Introduction



Virtual Work







There are very good textbooks that can be consulted toaid in understanding what we are going to discuss. Thefollowing two volumes were the most influential in thepreparation of these notes:

■ (Spong et al., 2005): Our basic source. Our primaryobjective is to explain in detail some of the derivationsin Chapter 7.

■ (Lanczos, 1986): This is an excelent book thatpresents mechanics in a very elegant way.

Introduction


Introduction

Introduction

Introduction



Virtual Work







To understand the Euler-Lagrange approach to Mechanics,consider the robotic mechanism depicted below:

q2

x2

ag

q1

x1

r1

O

X2

X1

Introduction


Introduction

Introduction

Introduction



Virtual Work








q2

x2

ag

q1

x1

r1

O

X2

X1

It is very important to define the Inertial Reference Frame!

Introduction


Introduction

Introduction

Introduction



Virtual Work








q2

x2

ag

q1

x1

r1

O

X2

X1

A

B

C

~rA

~rB

Introduction


Introduction

Introduction

Introduction



Virtual Work







Initially, just to simplify matters, lets consider that eachpart of the robot can be viewed as equivalent point-masselements m1 and m2:

O

X2

X1

~rA

⊕m1

~rB

⊕m2

Introduction


Introduction

Introduction

Introduction



Virtual Work







Notice that by considering the Robot Configuration Spaceinstead of the Robot Workspace we are automaticallyincorporating the constraints that determine how therelative positions of the point-masses m1 and m2 shouldevolve with time:

~rA = FAK (~q) ⇒ ~rA ≡ ~rA(~q),

~rB = FBK (~q) ⇒ ~rB ≡ ~rB(~q);

where FAK (·) : R

2 → R2 and FB

K (·) : R2 → R

2 are theForward Kinematics maps associated with the centers ofmass 1 and 2, respectively, with ~q = [q1 q2]

⊤.

Virtual Displacements


Introduction

Introduction

Introduction



Virtual Work







■ Any variation of the point-mass elements positions inthe Workspace that do not violate the mechanicalconstraints, i.e. they are consistent with themovements the mechanism can perform, are calledvirtual displacements:

δ~rA =∂~rA

∂q1δq1 +

∂~rA

∂q2δq2;

δ~rB =∂~rB

∂q1δq1 +

∂~rB

∂q2δq2.

They are virtual because they don’t need tocorrespond to real displacements, i.e. δ~rA and δ~rB areconsidered just as possible/kinematically consistentinfinitesimal movements, not actual displacements.



Introduction

Introduction

Introduction



Virtual Work







■ Any variation of the point-mass elements positions inthe Workspace that do not violate the mechanicalconstraints, i.e. they are consistent with themovements the mechanism can perform, are calledvirtual displacements:

δ~rA =∂~rA

∂q1δq1 +

∂~rA

∂q2δq2;

δ~rB =∂~rB

∂q1δq1 +

∂~rB

∂q2δq2.

Notice that although δ~rA depends on δ~rB (due to theunderlying mechanical constraints), δq1 and δq2 areindependent variables!



Introduction

Introduction

Introduction



Virtual Work







Suppose the corresponding Jacobian matrices associatedwith the Forward Kinematic maps can be written as:

JA =

[∂~rA

∂q1|

∂~rA

∂q2

]

;

JB =

[∂~rB

∂q1|

∂~rB

∂q2

]

such that

δ~rA = JAδ~q;δ~rB = JBδ~q;

(1)

with δ~q = [δq1 δq2]⊤.

Virtual Work


Introduction

Introduction

Introduction



Virtual Work







■ The work done on a mass-point element, consideringa virtual displacement and the force applied to it, isgiven by:

δWA = ~F⊤A δ~rA,

δWB = ~F⊤B δ~rB.

The Total Virtual Work is δW = ~F⊤A δ~rA + ~F⊤

B δ~rB.In this case, one has that:

δW = ~F⊤A [JAδ~q] + ~F⊤

B [JBδ~q] ,

=[

~F⊤A JA + ~F⊤

B JB

]

︸︷︷︸

~F⊤

δ~q.

Generalized Forces


Introduction

Introduction

Introduction



Virtual Work







From the previous discussion, we can define:

■ Generalized Force:

~F = J⊤A~FA + J⊤

B~FB (2)

Notice that the unit of a Generalized Force doesn’t haveto be N (Newtons), but the product between aGeneralized Force and a Virtual DisplacementδW = ~F⊤δ~q is necessarily energy in J (Joules).

The power of this concept lies on the fact that δ~q can bechosen arbitrarily, and the corresponding displacements ofthe point-mass elements will automatically be consistentwith the underlying mechanical constraints.

Generalized Forces: External or Non-conservativeTorques


Introduction

Introduction

Introduction



Virtual Work







If there are not only external forces applied to themechanical system, but also external torques, one has that

δW = ~F⊤

A δ~rA + ~F⊤

B δ~rB + · · ·

+ ~T⊤1 δ~θ1 + ~T⊤

2 δ~θ2 + · · · ,

with ~FA, ~FB, . . . the external forces; ~Ti, i = 1, 2, . . . , nq,the resultant external torque applied to the Robot’s parti; δ~rA, δ~rB, . . . the virtual displacements of the points ofapplication of each external forces; and δ~θi = ~ωiδt thevirtual angular displacements for each part of the Robot.



Introduction

Introduction

Introduction



Virtual Work







A virtual angular displacement δ~θi can be understood asan allowed small rotation, i.e. it is a small variation in theorientation of the part i, that is consistent with themechanical constraints of the Robot.In this case

~ωi =d~θi

dt, i = 1, 2, . . . , nq,

and one expects that

~θi ≡ ~θi(~q) ⇒ δ~θi =

(

∂~θi

∂~q

)

δ~q =

(∂~ωi

∂~q

)

δ~q,

where the fact that ∂~θi∂~q = ∂~ωi

∂~q(see slide 92) was used.



Introduction

Introduction

Introduction



Virtual Work







Therefore,

~τ⊤δ~q =[

~F⊤

A JA + ~F⊤

B JB + · · ·

+~T⊤1

(∂~ω1

∂~q

)

+ ~T⊤2

(∂~ω2

∂~q

)

+ · · ·

]

δ~q.

Consequently

~τ =J⊤

A~FA + J⊤

B~FB + · · ·

+

(∂~ω1

∂~q

)⊤

~T1 +

(∂~ω2

∂~q

)⊤

~T2 + · · ·

A similar expression can be found in Siciliano and Khatib(2008), Section 3.3.5, and in a more involved derivationin Kane and Levinson (2005).



Introduction

Introduction

Introduction



Virtual Work







Some important observations:

■ It doesn’t matter the specific point of application of aTorque vector, but only the body in which it is applied(and the corresponding angular velocity of the body).

■ When representing external pure torques (couples)applied by actuators that are rigidly attached to therobotic mechanism (e.g. an electrical motor in one ofthe robot’s joints), the corresponding ReactionTorque, which is applied to the robot’s part wherethe actuator is anchored, has to be accounted for, inthe last expression, as an external contribution too.

The Total Virtual Work done by Constraint Forces isZero


Introduction

Introduction

Introduction



Virtual Work







Suppose that the resultant forces applied to each materialpoint of the Robot can be decomposed in the followingway:

~FA = ~F externalA + ~F constr

A ;

~FB = ~F externalB + ~F constr

B .

In this case the Generalized Force can be broken up into:

~F = ~Fexternal + ~Fconstr,

where

~Fexternal = J⊤

A~F externalA + J⊤

B~F externalB ,

~Fconstr = J⊤

A~F constrA + J⊤

B~F constrB ,



Introduction

Introduction

Introduction



Virtual Work







A very important hypothesis we will adopt is:

(

~Fconstr)⊤

δ~q = 0,

i.e. the net contribution (the sum) of Generalized Forcesassociated with the constraints to the Total Virtual Workis zero.



Introduction

Introduction

Introduction



Virtual Work







In practice this means that the constraint forces areorthogonal to the Virtual Displacements or, when theconstraint forces are parallel to them, they form anaction-reaction pair that acts in opposite directions withcontributions to the total Virtual Work that cancel eachother out.

Since the Euler-Lagrange approach is based on Work andEnergy, the constraint forces will not show up explicitly inthe equations because of this assumption.



Introduction

Introduction

Introduction



Virtual Work







To try to show that the previous assumption of zeroVirtual Work done by contraint forces makes sense,consider the following holonomic constraint in Robot’sWorkspace:

h(~rA, ~rB) = 0.

And also assume that there is no external force applied tothe underlying mechanism. Despite of this, the aboveconstraint has to be fulfilled at all times.



Introduction

Introduction

Introduction



Virtual Work







For example, consider the fact that the points A, B andC (center of rotation) in the Robotic Mechanism we havebeen considering so far should always be on the same lineconnecting them, i.e.

(~rA − ~rB)⊤(~rB − ~pC) = ‖~rA − ~rB‖‖~rB − ~pC‖,

where ~rA = [xA yA]⊤ and ~rB = [xB yB]

⊤ are variablequantities, and ~pC = [c1 c2]

⊤ is a constant/fixed columnvector.

q2

x2

ag

q1

x1

r1



Introduction

Introduction

Introduction



Virtual Work







In this case, since the holonomic constraint is satisfied ∀t,we necessarily have that:

δh = 0 ⇒∂h

∂~rAδ~rA +

∂h

∂~rBδ~rB = 0, (3)

where ∂h∂~rA

is the row vector that corresponds to thegradient of the function h:

∂h

∂~rA=

[∂h

∂xA

∂h

∂yA

]

.

On the other hand, from the previous hypothesis we knowthat:

( ~F constrA )⊤δ~rA + ( ~F constr

B )⊤δ~rB = 0. (4)



Introduction

Introduction

Introduction



Virtual Work







By comparing expressions (3) and (4) we can see that ifthe constraint forces are such that:

~F constrA = λA

(∂h∂~rA

)⊤

,

~F constrB = λB

(∂h∂~rB

)⊤

,

with λA, λB ∈ R, this would make each constraint forceindeed orthogonal to the surface defined by theholonomic constraint h(~rA, ~rB) = 0.



Introduction

Introduction

Introduction



Virtual Work







If this is not the case, expression (4):

( ~F constrA )⊤δ~rA + ( ~F constr

B )⊤δ~rB = 0,

⇒[

( ~F constrA )⊤JA + ( ~F constr

B )⊤JB

]

︸︷︷︸

Generalized Constraint Force ( ~Fconstraints)⊤

δ~q = 0

would also be verfied if ~F constrA and ~F constr

B comprise an

action-reaction pair of forces, i.e. ~F constrA = −~F constr

B ,which are applied to the same point, i.e. A = B, or suchthat they are applied to points that “move together”when there are changes in the configuration variables inthe sense that JA(~q) = JB(~q).

The Principle of Virtual Work


Introduction

Introduction

Introduction



Virtual Work







■ As we know from Newton’s second law, if theresultant force acting on the system is zero, thesystem will be in equilibrium, meaning that there willbe no accelerations with respect to an inertialreference frame.

■ In this case

~F =∑

i

~Fi =∑

i

~F externali + ~F constr

i = 0.

■ Therefore,

( ~F )⊤δ~r = ( ~F)⊤δ~q = 0,

=

(∑

i

~Fexternali + ~Fconstr

i

)⊤

δ~q = 0.

The Principle of Virtual Work


Introduction

Introduction

Introduction



Virtual Work







In this case, since

(

~Fconstr)⊤

δ~q = 0,

one has that

■ The Total Virtual Work done by all External Forcesand Torques is zero if, and only if, the mechanicalsystem is in equilibrium, i.e. accelerations are null:

δW = ( ~Fexternal)⊤δ~q = 0 ⇔ Equilibrium.

This principle can be extended to the case when the sumof external forces is nonzero by introducing the concept ofInertia Forces.

Inertia Forces and Effective Forces


Introduction

Introduction

Introduction



Virtual Work







■ Inertia Forces: the time derivative of the linearmomentum associated with the mass elements:

~FIA = −d

dt(m1~vA) ,

~FIB = −d

dt(m2~vB) .

■ Effective Forces: the sum of the applied ExternalForces and the Inertia Forces:

~FeA = ~FA + ~FIA,

~FeB = ~FB + ~FIB.

D’Alembert Principle


Introduction

Introduction

Introduction



Virtual Work







Generalizing the Principle of Virtual Work we have that

■ D’Alembert Principle: the mechanical system wouldbe in equilibrium if to the External Forces we addedthe Inertia Forces, i.e. the Total Virtual Work done bythe Effective Forces is Zero:

δW = 0 ⇒[

~FA − ddt (m1~vA)

]⊤

δ~rA+[

~FB − ddt (m2~vB)

]⊤

δ~rB = 0.

(5)

Notice that in the above expression the sum is zero, andthis does not imply that each term should be null sincethe Virtual Displacements δ~rA and δ~rB are notindependent variations.

The Euler-Lagrange Equations


The Euler-LagrangeEquations

Total Virtual Work –Analysis

Total Virtual Workdone by ExternalForcesTotal Virtual Workdone by InertiaForcesTotal Virtual Workdone by InertiaForcesTotal Virtual Workdone by InertiaForcesTotal Virtual Workdone by InertiaForcesTotal Virtual Workdone by InertiaForcesThe Euler-LagrangeEquation

The Euler-LagrangeEquation


Rigid Body KineticEnergy


Total Virtual Work – Analysis









From D’Alembert Principle:

δW = 0 ⇔

(

~F⊤A δ~rA + ~F⊤

B δ~rB

)

︸︷︷︸

δWext

+

−

[d

dt(m1~vA)

⊤ δ~rA +d

dt(m2~vB)

⊤ δ~rB

]

︸︷︷︸

δWInercia

= 0. (6)

And each additive term above can be analyzed separately.

Total Virtual Work done by External Forces









As we have seen before, the Virtual Work done byExternal Forces can be rewritten depending on theGeneralized Forces and on the changes of the Mechanismconfiguration, such that:

δWext = ~F⊤A (JAδ~q) + ~F⊤

B (JBδ~q) ,

=[

~F⊤A JA + ~F⊤

B JB

]

︸︷︷︸

~F⊤

δ~q.

δWext = ~F⊤δ~q (7)

Total Virtual Work done by Inertia Forces









The Virtual Work done by the Inertia Forces is:

−δWInertia =d

dt(m1~vA)

⊤ δ~rA︸︷︷︸

1st term

+d

dt(m2~vB)

⊤ δ~rB︸︷︷︸

2nd term

. (8)

The first term above can be expanded to become

d

dt(m1~vA)

⊤ δ~rA =

d

dt(m1~vA)

⊤

[∂~rA

∂q1δq1 +

∂~rA

∂q2δq2

]

=

[d

dt(m1~vA)

⊤ ∂~rA

∂q1

]

δq1 +

[d

dt(m1~vA)

⊤ ∂~rA

∂q2

]

δq2.










The first term in the last expression can also be rewrittenas:

[d

dt(m1~vA)

⊤ ∂~rA

∂q1

]

︸︷︷︸

u′.v

=

d

dt

(

m1~v⊤A

∂~rA

∂q1

)

︸︷︷︸

(u.v)′

−m1~v⊤A

d

dt

(∂~rA

∂q1

)

,

︸︷︷︸

u.v′

from which we can do the following substitutions (see theproofs for these facts at the end of this presentation):

∂~rA

∂q1=

∂~vA

∂q1e

d

dt

(∂~rA

∂q1

)

=∂~vA

∂q1.










Using the previous substitutions one can define theso-called “Kinetic Energy of Mass 1”:

KA =1

2m1~v

⊤A~vA (9)

from which we can see that:

∂KA

∂q1= m1~v

⊤A

∂~vA

∂q1,

∂KA

∂q1= m1~v

⊤A

∂~vA

∂q1.

[d

dt(m1~vA)

⊤ ∂~rA

∂q1

]

=

[d

dt

(∂KA

∂q1

)

−∂KA

∂q1

]










Therefore it is possible to write, from (8), that:

[d

dt(m1~vA)

⊤ ∂~rA

∂q1

]

δq1 =

[d

dt

(∂KA

∂q1

)

−∂KA

∂q1

]

δq1;

[d

dt(m1~vA)

⊤ ∂~rA

∂q2

]

δq2 =

[d

dt

(∂KA

∂q2

)

−∂KA

∂q2

]

δq2;

[d

dt(m2~vB)

⊤ ∂~rB

∂q1

]

δq1 =

[d

dt

(∂KB

∂q1

)

−∂KB

∂q1

]

δq1;

[d

dt(m2~vB)

⊤ ∂~rB

∂q2

]

δq2 =

[d

dt

(∂KB

∂q2

)

−∂KB

∂q2

]

δq2;

where the “Kinetic Energy of Mass 2” is given by:

KB =1

2m2~v

⊤B~vB (10)










By summing up all the individual contributions from eachmass element of the Robot, the Total Kinetic Energy ofthe mechanism will be

K = KA +KB ,

and, from (8), one has that

−δWInertia =

[d

dt

(∂K

∂~q

)

−∂K

∂~q

]⊤

δ~q (11)

Notice that we only consider the Total Kinetic Energy ofthe Robot. We haven’t had to include the so-called“internal reaction forces” between parts of the mechanicalsystem, or the constraint forces that guarantee kinematicconsistency for the Robot movement.

The Euler-Lagrange Equation









From (6), (7), (8) and (11), and using the Principle ofD’Alembert, we have that:

δWext = −δWInertia,

~F⊤δ~q =

[d

dt

(∂K

∂~q

)

−∂K

∂~q

]⊤

δ~q,

And therefore:

~F =

[d

dt

(∂K

∂~q

)

−∂K

∂~q

]

, (12)

because δ~q is an arbitrary variation!This is one expression of the famous Euler-LagrangeEquation.










However, it is common pratice to separate the GeneralizedForces in two vector quantities:

~F = ~τ +

[

−∂V

∂~q

]

with ~τ the vector of Non-conservative ExternalGeneralized Forces, and V (~q) a Potential Functionassociated with the Conservative External GeneralizedForces, e.g. the weight force resulting from the earthgravitational field. Moreover, it is usual to define theso-called Lagrangian Function:

L(

~q, ~q)

= K(

~q, ~q)

− V (~q) .










From the previous definitions, and considering (12), wecan finally write down the most common form of theEuler-Lagrange Equation:

~τ =

[d

dt

(∂L

∂~q

)

−∂L

∂~q

]

(13)

The importance of this equation is that it represents thetime evolution of the mechanical system configurationvector ~q, only from the knowledge of the differencebetween the Total Kinetic Energy and the Total PotentialEnergy of the Robot, without explicitly taking intoaccount neither the constraint forces that make the Robotmovement kinematically consistent, nor the internalreaction forces between parts of the Robot.

Rigid Body Kinetic Energy




Rigid Bodies TotalKinetic Energy


Rigid Bodies:Translation andRotationRigid Bodies:Translation andRotationRigid Bodies: somedefinitionsRigid Bodies: somedefinitionsRigid Bodies: somedefinitionsThe Rigid BodyKinetic Energy

The Rigid BodyKinetic Energy


Angular Velocity andSkew-symmetricMatricesAngular Velocity and


Rigid Bodies Total Kinetic Energy











If we consider that each part of the Robot is comprised bya large number of point-mass elements, the Robot KineticEnergy will be given by:

K = K1︸︷︷︸

Part 1

+ K2︸︷︷︸

Part 2

+ . . .+ Kp︸︷︷︸

Part p

+ . . . ,

where

Kp =1

2

np∑

i=1

mi~v⊤i ~vi.

with ~vi the velocity vectors of each point-mass elementmi from body p, represented in theInertial Reference Frame previously defined.

Rigid Bodies Total Kinetic Energy











Indeed, continuing with the example we have investigatedso far, we have 2 rigid bodies and not two point-masselements.

yx

y

xy

xθ1

C

Parte 1

Parte 2

We will see that the Kinetic Energy of each rigid bodycan be decomposed in two portions: one associated withthe translation of the rigid body center of mass itself, andanother one related to the rotation of the rigid bodyaround its center of mass.

Rigid Bodies: Translation and Rotation











The velocity of a point-mass element i can be split in twoterms:

~vi = ~vcm + ~vi/cm (14)

with ~vcm the rigid body center of mass velocity vector,and ~vi/cm the point-mass element velocity i relative tothis center of mass.By definition, ~vi/cm has to be associated with a rotationsince mass element i of a rigid body cannot reduce orincrease its distance to the center of mass, i.e.:

~vi/cm = ~ω × ~ri/cm (15)

where ~ω is the rigid body angular velocity vector, and~ri/cm the mas element i position relative to the c.m.

Rigid Bodies: Translation and Rotation











In our example, we can see below the relationshipbetween vectors ~ri = ~rcm + ~ri/cm:

y

x

yx

C

~ri

mi ~ri/cm

~rcm

Notice that all the vectors depicted above could berepresented in the Inertial/World Reference Frame, orwith respect to a Reference Frame attached to a part ofthe Robot.

Rigid Bodies: some definitions











It is important to keep in mind the following definitions:

■ Center of Mass Position:

~rcm =1

M

np∑

i=1

mi~ri. (16)

■ Center of Mass Velocity:

~vcm =d~rcm

dt=

1

M

np∑

i=1

mi~vi. (17)

M is the rigid body Total Mass.












From (16) we have that:

np∑

i=1

mi~ri/cm = 0

because

np∑

i=1

mi~ri/cm =

np∑

i=1

(mi~ri −mi~rcm) =

np∑

i=1

mi~ri

︸︷︷︸

M~rcm

−

( np∑

i=1

mi

)

~rcm = 0,












Similarly, from equation (17) one has that:

np∑

i=1

mi~vi/cm = 0

because

np∑

i=1

mi~vi/cm =

np∑

i=1

(mi~vi −mi~vcm) =

k∑

i=1

mi~vi

︸︷︷︸

M~vcm

−

( np∑

i=1

mi

)

~vcm = 0,

The Rigid Body Kinetic Energy











From the previous definitions, we can write the followingexpression for the Total Kinetic Energy associated withRigid Body1 p:

Kp =1

2

np∑

i=1

mi

(~vcm + ~vi/cm

)⊤ (~vcm + ~vi/cm

),

=1

2

np∑

i=1

mi~v⊤cm~vcm + ~v⊤cm

( np∑

i=1

mi~vi/cm

)

+

1

2

np∑

i=1

mi~v⊤

i/cm~vi/cm

1So far, considered to be comprised by countable np point-mass

elements.












Kp =1

2

np∑

i=1

mi

(~vcm + ~vi/cm

)⊤ (~vcm + ~vi/cm

),

=1

2

np∑

i=1

mi~v⊤cm~vcm + ~v⊤cm

✟✟✟✟✟✟✟✟✯0( np∑

i=1

mi~vi/cm

)

+

1

2

np∑

i=1

mi~v⊤

i/cm~vi/cm












Kp =1

2

( np∑

i=1

mi

)

~v⊤cm~vcm +1

2

np∑

i=1

mi~v⊤

i/cm~vi/cm,

=1

2Mp~v

⊤cm~vcm

︸︷︷︸

Translation

+1

2

np∑

i=1

mi~v⊤

i/cm~vi/cm

︸︷︷︸

Rotation

.

As we will see, the component of the Kinetic Energyassociated with the rotation of the rigid body can berewritten to better reveal its relation with the rigid bodyangular velocity vector ~ω.

Angular Velocity and Skew-symmetric Matrices











The mass element mi position relative to the c.m. ~ri/cmcan be represented with respect to the Inertial ReferenceFrame ~r Inertial

i/cm , or with respect to a Reference Frame

attached to the rigid body ~rBodyi/cm . We know that these

representations are related by a Rotation Matrix, suchthat

~r Inertiali/cm = R~r

Bodyi/cm ,

with R ∈ SO(3), that is, RR⊤ = I, and det{R} = +1.Computing the time derivative of the above expression,one has that

~v Inertiali/cm = R~r

Bodyi/cm +R

✟✟✟✟✟✯0(

~rBodyi/cm

)

= R(

R⊤~r Inertiali/cm

)

.












From the previous expression, it follows that

RR⊤~r Inertiali/cm = ~ω Inertial × ~r Inertial

i/cm = S(~ω)~r Inertiali/cm ,

i.e. the vector cross-product is equivalent to a matrixmultiplication using the special matrix S(~ω):

S(~ω) = RR⊤ (18)

Matrix S(~ω) is Skew-symmetric, i.e.

RR⊤ = I ⇒d

dt

(

RR⊤

)

= 0,

RR⊤

︸︷︷︸

S(~ω)

+RR⊤

︸︷︷︸

S⊤(~ω)

= 0 ⇒ S(~ω) = −S⊤(~ω) .












From the skew-symmetry of S(~ω) a series of propertiescan be derived for its elements sij :

sij = −sji, ∀i, j;

sii = −sii ⇒ sii = 0, ∀i,

and, therefore, matrix S(~ω) can be rewritten as

S(~ω) =

0 −ωz ωy

ωz 0 −ωx

−ωy ωx 0

(19)

Thus, from the knowledge of R one can find out thecomponents ωx, ωy and ωz of the angular velocity vector~ω, since S(~ω) = RR⊤.

Skew-symmetric Matrices and the Cross-Product











S(·) can also be seen as an operator, in the sense that

~ω × ~r =

ωx

ωy

ωz

×

rxryrz

=

ωyrz − ωzryωzrx − ωxrzωxry − ωyrx

,

= S(~ω)

rxryrz

,

=

0 −ωz ωy

ωz 0 −ωx

−ωy ωx 0

rxryrz

.

And, therefore, any vector cross-product ~a×~b can bewritten as S(~a)~b, with S(~a) = −S⊤(~a).

Skew-symmetric Matrices and the Cross-Product











To consider S(~a) as an operator, whose result depends onthe components of ~a, is very interesting because thisallow us to obtain the following expression

~ω × ~r = −~r × ~ω,

= −S(~r)~ω.

And, therefore,

~v⊤i/cm~vi/cm =(~ω × ~ri/cm

)⊤ (~ω × ~ri/cm

),

=[−~ri/cm × ~ω

]⊤ [−~ri/cm × ~ω

],

=[−S

(~ri/cm

)~ω]⊤ [

−S(~ri/cm

)~ω],

= ~ω⊤

[

S(~ri/cm

)⊤S(~ri/cm

)]

~ω.

Rotational Kinetic Energy











The former expression makes it possible to compute theKinetic Energy portion associated with the RotationalKinetic Energy of a given part p of the Robot, assumingthat this part is comprised by np point-mass elements,such that:

Krotationp =

1

2

np∑

i=1

mi~ω⊤

[

S(~ri/cm

)⊤S(~ri/cm

)]

~ω,

=1

2

np∑

i=1

mi~ω⊤

[

−S(~ri/cm

)2]

~ω,

=1

2~ω⊤

{ np∑

i=1

mi

[

−S(~ri/cm

)2]}

~ω,

=1

2~ω⊤

Is~ω.

where Is is a 3× 3 real matrix called Inertia Tensorassociated with the specific rigid body part of the Robot.

Inertia Tensor











To compute Is considering that each Robot’s part p is arigid body formed by an infinite number of point-masselements whose volumes tend to zero, the formersummations must be replaced by volume integrals:

Is =

∫

Vp

[

−S(~ri/cm

)2]

ρp(~ri/cm

)dV

︸︷︷︸

≈ dmi

, (20)

Is =

Ixx −Ixy −Ixz ,

−Ixy Iyy −Iyz,

−Ixz −Iyz Izz,

,

with ρp(~ri/cm

)the mass density function evaluated at

~ri/cm; Ixx, Iyy and Izz are called moments of inertiaaround x, y and z axes, respectively; and Ixy, Ixz and Iyzare called products of inertia.

Inertia Tensor – Computation











By expanding expression (20), and considering that~ri/cm = [x y z]⊤, we have that:

−S(~ri/cm

)2= −

0 −z y

z 0 −x

−y x 0

2

−S(~ri/cm

)2=

(y2 + z2) −xy −xz

−xy (x2 + z2) −yz

−xz −yz (x2 + y2)

,

and, therefore, the moments of inertia will be given byIxx =

∫

vol(y2 + z2)ρpdV , Iyy =

∫

vol(x2 + z2)ρpdV ,

Izz =∫

vol(x2 + y2)ρpdV ; and the products of inertia will

be given by Ixy =∫

vol(xy)ρpdV , Iyz =∫

vol(yz)ρpdV ,Ixz =

∫

vol(xz)ρpdV .

Inertia Tensor – Reference Frames











Notice that, in the previous development, the componentsof ~ri/cm = [x y z]⊤ for each point-mass element wererepresented with respect to the Inertial Reference Frame.This makes very involved the calculation of thesummations or volume integrals defined previously, sincex, y and z are changing as long as the rigid body p ismoving, and we end up with a time varying Inertia Tensor.However, it is easy to see that:

~r Inertiali/cm = R~r

Bodyi/cm ,

for some R ∈ SO(3).












Notice that it is always true that:

R(

~a×~b)

= (R~a)×(

R~b)

,

R(

S(~a)~b)

= [S(R~a)][

R~b]

.

From the last expression, it follows that

RS(~a) = S(R~a)R,⇒ RS(~a)R⊤ = S(R~a)

for vectors ~a, and arbitrary rotation matrices R.












On the other hand, the computation of the Inertia Tensordepends on the following development:

−S(

~r Inertiali/cm

)2= −

[

S(

~r Inertiali/cm

)

S(

~r Inertiali/cm

)]

,

= −[

S(

R~rBodyi/cm

)

S(

R~rBodyi/cm

)]

,

and using the previous property, we can write that

S(

R~rBodyi/cm

)

S(

R~rBodyi/cm

)

= RS(

~rBodyi/cm

)

✟✟✟✯I

R⊤RS(

~rBodyi/cm

)

R⊤,

and, therefore,

−S(

~r Inertiali/cm

)2= R

[

−S(

~rBodyi/cm

)2]

R⊤












From the previous expression, one has that

Is =

∫

Vp

[

−S(

~r Inertiali/cm

)2]

ρpdV,

=

∫

Vp

R

[

−S(

~rBodyi/cm

)2]

R⊤ρpdV,

= R

(∫

Vp

[

−S(

~rBodyi/cm

)2]

ρpdV

)

︸︷︷︸

Ib

R⊤.

Therefore, the relation between a time-varying Is and theconstant Ib computed with respect to the body fixedReference Frame is:

Is = R Ib R⊤.

Rotational Kinetic Energy – Final Expressions











From the previous developments, the Rotational KineticEnergy associated with a Robot’s part p is given by

Krotationalp =

1

2

(

~ω Inertial)⊤

Is

(

~ω Inertial)

,

=1

2

(

R~ω Body)⊤ (

RIbR⊤

)(

R~ω Body)

,

Krotationalp =

1

2

(

~ω Body)⊤

Ib

(

~ω Body)

.

Rigid Body Kinetic Energy – Summary











The Total Kinetic Energy of a Robot’s part p is given by:

Kp =1

2Mp

(

~v Inertialcm

)⊤ (

~v Inertialcm

)

︸︷︷︸

Translation

+

1

2

(

~ω Body)⊤

Ibp

(

~ω Body)

︸︷︷︸

Rotation

(21)

with Mp ∈ R+ the mass of the part p and Ibp the

corresponding Inertia Tensor (a positive definitesymmetric matrix).

Center of Mass Velocity as a Function of ~q and ~q











To use the Euler-Lagrange equation (13) we need toexpress the Kinetic Energy as a function of ~q and ~q.

To do this, notice that:

~v Inertialcm =

d

dt

[

~r Inertialcm (~q)

]

,

and, therefore,

~v Inertialcm = Jcm(~q)~q, (22)

where Jcm(~q) is the Jacobian matrix ascociated with theForward Kinematics that gives ~r Inertial

cm from theknowledge of ~q.

Angular Velocity as a Function of ~q and ~q











As we have seen in previous classes of this course, we canget an expression for the angular velocity ~ω Body

associated to a specific Robot’s part depending on ~q and~q. Remember that, from (19), the angular velocity isgiven by

S(

~ω Inertial)

= RR⊤.

where R ∈ SO(3) is the Rotation Matrix in the expression~r Inertial = R~r Body. On the other hand, we know that

S(

~ω Inertial)

= S(

R~ω Body)

= RS(

~ω Body)

R⊤;

and, therefore,

S(

~ω Body)

= R⊤R (23)












On the other hand, notice that:

R (~q) =[~rx (~q) | ~ry (~q) | ~rz (~q)

];

R⊤ (~q) =

~r⊤x (~q)

~r⊤y (~q)

~r⊤z (~q)

;

R(

~q, ~q)

=[

∂~rx∂~q ~q

∂~ry∂~q ~q

∂~rz∂~q ~q

]

,

where ∂~rx∂~q ,

∂~ry∂~q e ∂~rx

∂~q are Jacobian matrices associatedwith each column of R(~q).












From (23), and considering that:

S(

~ω Body)

=

0 −ωz ωy

ωz 0 −ωx

−ωy ωx 0

,

one has that, for example,

ωx

(

~q, ~q)

= −~r⊤y

(∂~rz

∂~q

)

~q,

ωy

(

~q, ~q)

= ~r⊤x

(∂~rz

∂~q

)

~q,

ωz

(

~q, ~q)

= −~r⊤x

(∂~ry

∂~q

)

~q.












Finally, it is possible to write that

~ω Body ≡ W (~q) ~q (24)

with

W (~q) =

−~r⊤y

(∂~rz∂~q

)

~r⊤x

(∂~rz∂~q

)

−~r⊤x

(∂~ry∂~q

)

.

Rigid Bodies Kinetic Energy











Using expressions (22) and (24), in (21), one has that

Kp =1

2Mp~q

⊤ [Jcm(~q)]⊤ [Jcm(~q)] ~q

︸︷︷︸

Translation

+

1

2~q⊤ [W (~q)]⊤ Ibp [W (~q)] ~q,︸︷︷︸

Rotation

Hence the Total Kinetic Energy associated with a Robot’spart p is a quadratic function of the so-called generalizedvelocity ~q, i.e.:

Kp =1

2~q⊤Mp(~q) ~q

Robot Total Kinetic Energy











From the previous expressions, we have that:

K =

n∑

p=1

Kp =

n∑

p=1

{1

2~q⊤Mp(~q) ~q

}

.

with

Mp = [Jp,cm(~q)]⊤Mp [Jp,cm(~q)] + [Wp(~q)]

⊤Ibp [Wp(~q)] ,

where Jp,cm(~q) and Wp(~q) are the matrices definedpreviously that depend on which Robot’s part p that it isbeing considered.

Robot Total Kinetic Energy











This means that, for any Robot in the world, its TotalKinetic Energy, resulting from summing up all thecontributions from each one of its parts, is indeed aquadratic function of the generalized velocity ~q:

K =1

2~q⊤M(~q) ~q (25)

where

M(~q) = M1(~q) +M2(~q) + . . .+Mn(~q).

is the so-called Mass Matrix.

The Mass Matrix











The Mass Matrix M(~q) has some very interestingproperties: it is symmetric and positive definite.Therefore, M−1(~q) always exists, because the eigenvaluesof M(~q) are real and positive.

From (25), the Mass Matrix can also be defined from theknowledge of the Total Kinetic Energy K, such that itselements Mij(~q) are given by:

Mij(~q) =∂2K

∂qi∂qj(26)

Robot Total Potential Energy











To each Robot’s part p we can also associated a PotentialFunction Vp(~q), whose gradient determines the directionof the conservative force that acts on that part:~τ conservativep = −

∂Vp

∂~q .

The Robot Total Potential Energy can be written as

V (~q) =n∑

p=1

Vp(~q) (27)

Obs.: usually the Potential Energy is associated with thegravity that manifests itself as weight vector forcesapplied at the center of mass of each part p of the Robot.

The Robot Lagrangian Function











From (25) and (27), the Robot Lagrangian function isgiven by

L(~q, ~q) = K(~q, ~q)− V (~q),

L(~q, ~q) =1

2~q⊤M(~q) ~q − V (~q).

Finally, the Robot Euler-Lagrange equation is

~τ =d

dt

(∂L

∂~q

)

−∂L

∂~q.

Robot Canonical Equations




Robot CanonicalEquations

Canonical Equations

Canonical Equations

Canonical Equations– DetailsCanonical Equations– DetailsCanonical Equations– DetailsCanonical Equations– DetailsCanonical Equations– DetailsCanonical Equations– DetailsCanonical Equations– DetailsCanonical Equations– DetailsCanonical Equations– DetailsThe RobotCanonical Equation

The RobotCanonical Equations


Canonical Equations





Canonical Equations

Canonical Equations




From the Robot Euler-Lagrange equation, notice that:

d

dt

(∂L

∂~q

)

=d

dt

(∂K

∂~q

)

,

∂L

∂~q=

∂

∂~q

[

K(~q, ~q)− V (~q)]

.

And since K = 12 ~q

⊤M(~q)~q, one has that2

∂K

∂~q= M(~q)~q,

2See Proof 3 at the end of these lecture notes.

Canonical Equations





Canonical Equations

Canonical Equations




Therefore, we can write that:

d

dt

(∂K

∂~q

)

=d

dt

(

M(~q)~q)

= M(~q)~q +dM(~q)

dt︸︷︷︸

?

~q,

and

∂

∂~q

[

K(~q, ~q)− V (~q)]

=1

2

∂[

~q⊤M(~q)~q]

∂~q− g(~q),

where g(~q) = ∂V (~q)∂~q .

In the previous equations, the expressions dM(~q)dt and

12

∂[~q⊤M(~q)~q]∂~q need to be further explained.

Canonical Equations – Details





Canonical Equations

Canonical Equations




The Euler-Lagrange equations is a vector equation that isequivalent to the following set of scalar equations:

τ1 =d

dt

(∂L

∂q1

)

−∂L

∂q1,

τ2 =d

dt

(∂L

∂q2

)

−∂L

∂q2, (28)

...

τnq =d

dt

(∂L

∂qnq

)

−∂L

∂qnq

,

with τ1, τ2, . . . , τnq the components of the generalizedforce ~τ .






Canonical Equations

Canonical Equations




Therefore, to really understand the expression dM(~q)dt we

need to remember that dM(~q)dt ~q should give us a column

vector, whose k-th element will be given by

{dM(~q)

dt~q

}

k

=

nq∑

j=1

{dM

dt

}

kj

qj ,

=

nq∑

j=1

{dMkj

dt

}

qj ,

=

nq∑

j=1

{ nq∑

i=1

∂Mkj

∂qiqi

}

qj . (29)






Canonical Equations

Canonical Equations




Similarly, the k-th element in the column vector∂K∂~q = 1

2

∂[~q⊤M(~q)~q]∂~q will be

1

2~q⊤

∂M

∂qk~q =

1

2~q⊤

∑nq

j=1

∂M1j

∂qkqj ,

∑nq

j=1

∂M2j

∂qkqj ,

...

∑nq

j=1

∂Mnqj

∂qkqj ,

,

=1

2

nq∑

i=1

nq∑

j=1

∂Mij

∂qkqj

qi. (30)






Canonical Equations

Canonical Equations




Combining expressions (29) and (30) above; exchangingthe indexes i and j in (30), and noticing that Mij = Mji,one has that one of the terms in the k-th Euler-Lagrangeequation can be expanded in the following way:

{dM

dt~q −

∂K

∂~q

}

k

=

nq∑

j=1

{ nq∑

i=1

(∂Mkj

∂qi−

1

2

∂Mji

∂qk

)

qi

}

qj ,

and this last result looks like the multiplication of a rowvector Γk

⊤ by the column vector ~q, with

Γk⊤ =

[Γk1 Γk2 . . .Γk nq

].






Canonical Equations

Canonical Equations




From the previous developments, we have that

dM

dt~q −

∂K

∂~q= C(~q, ~q)~q,

with the Coriolis and Centrifugal Matrix given by

C(~q, ~q) =

Γ11 Γ12 . . . Γ1nq

Γ21 Γ22 . . . Γ2nq

......

. . ....

Γnq 1 Γnq 2 . . . Γnq nq

where

Γkj =

nq∑

i=1

(∂Mkj

∂qi−

1

2

∂Mji

∂qk

)

qi (31)






Canonical Equations

Canonical Equations




A related development usually found in Robotics consistsin rewriting expression (29) such as:

nq∑

j=1

{ nq∑

i=1

∂Mkj

∂qiqi

}

qj =

nq∑

j=1

{ nq∑

i=1

1

2

∂Mkj

∂qiqi

}

qj +

nq∑

j=1

{ nq∑

i=1

1

2

∂Mkj

∂qiqi

}

qj ,

=

nq∑

j=1

{nq∑

i=1

1

2

∂Mkj

∂qiqi

}

qj +

nq∑

i=1

nq∑

j=1

1

2

∂Mki

∂qjqj

qi,

where index j was exchanged with index i in the lastsummation.






Canonical Equations

Canonical Equations




By changing the order of the summations in the lastexpression:

nq∑

j=1

{ nq∑

i=1

∂Mkj

∂qiqi

}

qj =

nq∑

j=1

{ nq∑

i=1

1

2

∂Mkj

∂qiqi

}

qj +

nq∑

j=1

{ nq∑

i=1

1

2

∂Mki

∂qjqi

}

qj ,

Therefore,

nq∑

j=1

{nq∑

i=1

∂Mkj

∂qiqi

}

qj =1

2

nq∑

j=1

{nq∑

i=1

(∂Mkj

∂qi+

∂Mki

∂qj

)

qi

}

qj

(32)






Canonical Equations

Canonical Equations




Again by exchanging the indexess i and j in (30), andusing (32), one has that:

{dM

dt~q −

∂K

∂~q

}

k

=

nq∑

j=1

{nq∑

i=1

1

2

(∂Mkj

∂qi+

∂Mki

∂qj−

∂Mij

∂qk

)

qi

}

qj ,

from which we can define the so-called ChristoffelSymbols of the first kind:

cijk =1

2

(∂Mkj

∂qi+

∂Mki

∂qj−

∂Mij

∂qk

)

(33)

Note that, with fixed k, one hast that cijk = cjik.






Canonical Equations

Canonical Equations




In this case the Coriolis and Centrifugal Matrix elementsdefined in (31) can be rewritten as

Γkj =

nq∑

i=1

cijk(~q)qi (34)

and each scalar k-th Euler-Lagrange equation will begiven by

nq∑

j=1

Mkj qj +

nq∑

j=1

Γkj

(

~q, ~q)

qj + gk(~q) = τk,

where gk(~q) is the k-th element of the vector

g(~q) = ∂V (~q)∂~q ; and τk is the k-th element of the

generalized force vector ~τ .

The Robot Canonical Equation





Canonical Equations

Canonical Equations




Finally, the so-called Canonical Equation or NormalForm Equation of the Robot can be compactly written as

M (~q ) ~q + C(

~q, ~q)

~q + g (~q ) = ~τ , (35)

with M(~q) ∈ Rnq × R

nq a real symmetric positive definitematrix called mass matrix, or inertia matrix ;C(~q, ~q) ∈ R

nq × Rnq is the Coriolis and Centrifugal terms

matrix obtained from M(~q); g(~q) ∈ Rnq is the total

potential energy gradient; and ~τ ∈ Rnq is the vector of

non-conservative generalized forces (which includesexternal and friction forces and torques).

The Robot Canonical Equations – Observations





Canonical Equations

Canonical Equations




■ The importance of (35) is not in the systematic waywe have used to derive it, but actually in the fact thatwe can see more clearly what are the properties ofeach term in this equation that will be true for everyrobot in the world.

■ With respect to this last point, notice that very oftenit is much easier/simpler, regarding the algebraicdevelopment of the final Robot Canonical Equation,to depart from (28) and to apply the Euler-Lagrangeapproach instead of writing explicitly the elements Γkj

in C(~q, ~q) following some tedious recipe.

The Robot Canonical Equations – InterestingProperties





Canonical Equations

Canonical Equations




From the previous developments, it is possible to provethat

■ M(~q) is a square positive definite real matrix.Therefore, it will always exist the inverse matrixM−1(~q). Moreover, if the configuration ~q is alwaysbounded, one has that:

0 < λmin(~q)I ≤ M(~q) ≤ λmax(~q)I;

with λmin the smallest real and positive eigenvalue,and λmax the greatest real and positive eigenvalue ofM(~q). (every M(~q) eigenvalue is real and positive).






Canonical Equations

Canonical Equations




■ The matrix M − 2C is skew symmetric, i.e.

[

M − 2C]

⊤ = −[

M − 2C]

,

and, therefore, ~q⊤[

M − 2C]

~q = 0, since

~q⊤[

M − 2C]

~q = a ∈ R

a = a⊤,

a = ~q⊤[

M − 2C]⊤

~q,

a = −~q⊤[

M − 2C]

~q,

a = −a ⇒ a = 0.






Canonical Equations

Canonical Equations




Linearity with respect to the constitutive parameters:

■ The Robot Canonical Equation is linear with respectto parameters formed by combining masses, inertiamoments and products, and geometric lengths, suchthat:

M(~q)~q + C(~q, ~q)~q + g(~q) = ~τ ,

Y (~q, ~q, ~q)~Θ = ~τ ,

with Y (~q, ~q, ~q) ∈ Rl × R

l a regression matrix(sometimes difficult to be derived); and ~Θ ∈ R

l is avector of parameters that depend on the Robot massdistribution and geometry.

Mathematical Proofs





Mathematical Proofs

Proof 1

Proof 2

Proof 3

Proof 3 (cont.)

Proof 4

Proof 4 (cont.)

Bibliography


Proof 1





Mathematical Proofs

Proof 1

Proof 2

Proof 3

Proof 3 (cont.)

Proof 4

Proof 4 (cont.)

Bibliography


∂~rA

∂q1=

∂~vA

∂q1. or

∂~θi

∂q1=

∂~ωi

∂q1

~vA =d~rA

dt,

=∂~rA

∂q1q1 +

∂~rA

∂q2q2,

∴

∂~vA

∂q1=

∂~rA

∂q1

Similarly,

~ωi =d~θi

dt=

∂~θi

∂q1q1 +

∂~θi

∂q2q2 ∴

∂~ωi

∂q1=

∂~θi

∂q1.

Proof 2





Mathematical Proofs

Proof 1

Proof 2

Proof 3

Proof 3 (cont.)

Proof 4

Proof 4 (cont.)

Bibliography


d

dt

(∂~rA

∂q1

)

=∂~vA

∂q1

d

dt

(∂~rA

∂q1

)

=∂

∂q1

[∂~rA

∂q1

]

q1 +∂

∂q2

[∂~rA

∂q1

]

q2,

=∂2~rA

∂q21q1 +

∂2~rA

∂q2∂q1q2,

=∂

∂q1

[∂~rA

∂q1q1 +

∂~rA

∂q2q2

]

,

=∂~vA

∂q1.

Proof 3





Mathematical Proofs

Proof 1

Proof 2

Proof 3

Proof 3 (cont.)

Proof 4

Proof 4 (cont.)

Bibliography


∂K

∂~q= M(~q)~q

General case:∂(~x⊤A~x)

∂~x = 2A~x for any symmetric matrix

A whose elements do not depend on the vector ~x. Noticethat

F (x1, x2, . . . , xn) = ~x⊤A~x =n∑

i=1

n∑

j=1

aijxj

︸︷︷︸

fi(x1,x2,...,xn)

xi,

=n∑

i=1

fixi.

Proof 3 (cont.)





Mathematical Proofs

Proof 1

Proof 2

Proof 3

Proof 3 (cont.)

Proof 4

Proof 4 (cont.)

Bibliography


If we consider ∂F∂~x as a column vector with elements ∂F

∂xk,

k = 1, 2, . . . , n, we can write that:

∂F

∂xk=

(n∑

i=1

∂fi

∂xkxi

)

+ fk,

=

(n∑

i=1

aikxi

)

+n∑

j=1

akjxj .

By replacing index i by the index variable j in the firstsummation, and relying on the symmetry of A, i.e.ajk = akj , one has that

∂F

∂xk= 2

n∑

j=1

akjxj ,= k-th element in the muliplication 2A~x.

Proof 4





Mathematical Proofs

Proof 1

Proof 2

Proof 3

Proof 3 (cont.)

Proof 4

Proof 4 (cont.)

Bibliography


If N = M − 2C, with M ≡ M(~q) and C ≡ C(~q, ~q), then

N = −N⊤.

By using (33) in (34), and noticing that the element kj inmatrix M can be written as

{

M}

kj=

nq∑

i=1

∂Mkj

∂qiqi,

it is possible to write that every element Nkj in N will begiven by

Nkj =

nq∑

i=1

{

��∂Mkj

∂qi−

(

��∂Mkj

∂qi+

∂Mki

∂qj−

∂Mij

∂qk

)}

qi,

Proof 4 (cont.)





Mathematical Proofs

Proof 1

Proof 2

Proof 3

Proof 3 (cont.)

Proof 4

Proof 4 (cont.)

Bibliography


Therefore, every element Nkj in N can be written as

Nkj =

nq∑

i=1

{∂Mki

∂qj−

∂Mij

∂qk

}

qi,

and we reach the conclusion that Nkj = −Njk, i.e.N = −N⊤.

Bibliography





Mathematical Proofs

Proof 1

Proof 2

Proof 3

Proof 3 (cont.)

Proof 4

Proof 4 (cont.)

Bibliography


Kane, T. R. and Levinson, D. A. (2005). Dynamics,Theory and Applications. The Internet-First UniversityPress. Freely available from Cornell University.

Lanczos, C. (1986). The Variational Principles ofMechanics. Dover Publications, fourth edition.

Siciliano, B. and Khatib, O., editors (2008). Handbookof Robotics. Springer-Verlag Berlin Heidelberg, firstedition.

Spong, M. W., Hutchinson, S., and Vidyasagar, M.(2005). Robot Modeling and Control. John Wiley &Sons, Inc., first edition.

Robot Dynamics – Part I: The Euler-Lagrange Approach

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Transcript of Robot Dynamics – Part I: The Euler-Lagrange Approach