Control of Robot Manipulators in Joint Space - R. Kelly, V. Santibanez and A. Loria Part 5 pot

Case Study: The Pelican Prototype Robot The purpose of this chapter is twofold: first, to present in detail the model ofthe experimental robot arm of the Robotics lab.. Second, to review

4.4 The Residual Dynamics 103

and with an abuse of notation it may be written as

h(t, ˜q, ˙˜q) = [M(q d)− M(q)] ¨q d + [C(q d , ˙q d)− C(q, ˙q)] ˙q d + g(q d)− g(q).

This function has the characteristic that h(t, 0, 0) = 0 for all t but more

does
˙˜q
, independently of
˜q
.

In order to make this statement formal we need to recall the definition and properties of a continuously differentiable monotonically increasing function: the tangent hyperbolic As a matter of fact, the statement can be shown for a large class of monotonically increasing functions but for clarity of exposition, here we restrict our discussion to

x − e −x

e x + e −x

which is illustrated in Figure 4.1

−1.0

−0.5

0.0

0.5

1.0

tanh(x)

x

Figure 4.1.Graph of tangent hyperbolic: tanh(x) As it is clear from Figure 4.1, tanh(x) is continuous monotonically in-creasing Also, it has continuous derivatives and it satisfies |x| ≥ |tanh(x)| and 1 ≥ |tanh(x)| for all x ∈ IR All these observations are stated formally below Definition 4.1 Vectorial tangent hyperbolic function We define the vectorial tangent hyperbolic function as tanh(x) := ⎡ ⎢ ⎣ tanh(x1)

.

⎤

⎥

104 4 Properties of the Dynamic Model

where x ∈ IR n The first partial derivative of tanh(x) is given by

Property 4.4 Residual dynamics vector h(t, ˜ q, ˙˜q)

¨

1

residual dynamics satisfies



h1
˙˜q
+ kh2
tanh(˜q)
(4.15)

hyper-bolic function introduced in Definition 4.1

Proof According to the definition of the residual dynamics function (4.12),

its norm satisfies

4.4 The Residual Dynamics 105

We wish to upperbound each of the three terms on the right-hand side of

fol-lows that the vector of gravitational torques – considering robots with revolute joints – satisfies the inequalities

g(q d)− g(q d − ˜q)
≤ k g
˜q

g(q d)− g(q d − ˜q)
≤ 2k 

0

0

2k 

k g

˜G

C(G d)− C(G d − ˜G)

2k 

slope k g .

.

Figure 4.2.Belonging region for
C(G d)− C(G d − ˜G)

Regarding the first term on the right-hand side of Inequality (4.16), we

have from Property 4.1 of the inertia matrix M (q), that the following two

inequalities hold:

[M(q d)− M(q d − ˜q)] ¨q d
≤ k M
¨q d
M
˜q
,

[M(q d)− M(q d − ˜q)] ¨q d
≤ 2k 

M
¨q d
M ,

M
¨q d
, which is

Finally it is only left to bound the second term on the right-hand side

of Inequality (4.16) This operation requires the following computations By virtue of Property 4.2 it follows that (see (4.5))



C(q d , ˙q d)− C(q d − ˜q, ˙q d − ˙˜q)q˙d ≤ k

C2
˙q d
2M
˜q

(4.17) Also, observe that the left-hand side of (4.17) also satisfies

106 4 Properties of the Dynamic Model

the terms on the right-hand side also satisfy

4.4 The Residual Dynamics 107

0

0

s2

s1

˜G

s(˜G)

s2

slope s1

.

Figure 4.3. Graph of the function s(˜G)

|s(˜q)| ≤ k h2 tanh(
˜q
) (4.23)

k h2≥ s2

tanh



s2

s1



h2
tanh(˜q)

where we have used the fact that

assumed to satisfy

Thus, Property 4.4 follows

♦♦♦

Residual Dynamics when ˙q d ≡ 0

(4.12) boils down to

108 4 Properties of the Dynamic Model

chap-To close the chapter we summarize, in Table 4.1, the expressions involved

given by Equations (4.21) and (4.22), respectively

Properties 4.1 and 4.2 are proved in

• Spong M., Vidyasagar M., 1989, “Robot dynamics and control”, Wiley,

New York

• Craig J., 1988, “Adaptive control of mechanical manipulators”, Addison

-Wesley, Reading MA

2

˙

M (q) − C(q, ˙q) was established in

• Ortega R and Spong M., 1989, “Adaptive motion control of rigid robots:

A tutorial,” Automatica, Vol 25-6, pp 877–888.

The concept of residual dynamics was introduced in

• Arimoto S., 1995, “Fundamental problems of robot control: Part I: vation in the realm of robot servo–loops”, Robotica, Vol 13, Part 1, pp.

110 4 Properties of the Dynamic Model

“Passivity-and Control Engg Series

Problems 111

2 Is it true that the inertia matrix M (q) is constant if and only if C(q, ˙q) =

0 ? (The matrix C(q, ˙q) is assumed to be defined upon the Christoffel

symbols of the first kind.)

3 Consider the dynamic model (3.33) of robots with (linear) actuators

Sup-pose that there is no friction (i.e f ( ˙q) = 0) Show that

12

4 Consider the equation that characterizes the behavior of a pendulum of

length l and mass m concentrated at the edge and is submitted to the action of gravity g to which is applied a torque τ on the axis of rotation,

Case Study: The Pelican Prototype Robot

The purpose of this chapter is twofold: first, to present in detail the model ofthe experimental robot arm of the Robotics lab from the CICESE ResearchCenter, Mexico Second, to review the topics studied in the previous chaptersand to discuss, through this case study, the topics of direct kinematics andinverse kinematics, which are fundamental in determining robot models.For the Pelican, we derive the full dynamic model of the prototype; in par-ticular, we present the numerical values of all the parameters such as mass,

inertias, lengths to centers of mass, etc This is used throughout the rest of the

book in numerous examples to illustrate the performance of the controllers

that we study We emphasize that all of these examples contain

planar arm with two links connected through revolute joints, i.e it possesses

2 DOF The links are driven by two electrical motors located at the “shoulder”

(base) and at the “elbow” This is a direct-drive mechanism, i.e the axes of

the motors are connected directly to the links without gears or belts

illus-trated in Figure 5.2 The distances from the rotating axes to the centers of

114 5 Case Study: The Pelican Prototype Robot

Figure 5.1.Pelican: experimental robot arm at CICESE, Robotics lab

Figure 5.2.Diagram of the 2-DOF Pelican prototype robot

pass through the respective centers of mass and are parallel to the axis x.

of the first link toward the second link, both being positive counterclockwise

The vector of joint positions q is defined as

q = [q1 q2]T .

5.1 Direct Kinematics 115

The meaning of the diverse constant parameters involved as well as theirnumerical values are summarized in Table 5.1

Table 5.1.Physical parameters of Pelican robot arm

Distance to the center of mass (Link 1) l c1 0.0983 m

Distance to the center of mass (Link 2) l c2 0.0229 m

Inertia rel to center of mass (Link 1) I1 0.1213 kg m2

Inertia rel to center of mass (Link 2) I2 0.0116 kg m2

5.1 Direct Kinematics

The problem of direct kinematics for robot manipulators is formulated as

follows Consider a robot manipulator of n degrees-of-freedom placed on a

fixed surface Define a reference frame also fixed at some point on this face This reference frame is commonly referred to as ‘base reference frame’.The problem of deriving the direct kinematic model of the robot consists inexpressing the position and orientation (when the latter makes sense) of a ref-erence frame fixed to the end of the last link of the robot, referred to the basereference frame in terms of the joint coordinates of the robot The solution tothe so-formulated problem from a mathematical viewpoint, reduces to solving

sur-a geometricsur-al problem which sur-alwsur-ays hsur-as sur-a closed-form solution

Regarding the Pelican robot, we start by defining the reference frame ofbase as a Cartesian coordinated system in two dimensions with its originlocated exactly on the first joint of the robot, as is illustrated in Figure 5.2 The

Cartesian coordinates x and y determine the position of the tip of the second

link with respect to the base reference frame Notice that for the present casestudy of a 2-DOF system, the orientation of the end-effector of the arm makes

no sense One can clearly appreciate that both Cartesian coordinates, x and

116 5 Case Study: The Pelican Prototype Robot

y, depend on the joint coordinates q1 and q2 Precisely it is this correlation

that defines the direct kinematic model,

simply, the Jacobian of the robot Clearly, the following relationship betweenaccelerations also holds,

textbook since we do not use it for control.

5.2 Inverse Kinematics

The inverse kinematic model of robot manipulators is of great importancefrom a practical viewpoint This model allows us to obtain the joint positions

q in terms of the position and orientation of the end-effector of the last link

referred to the base reference frame For the case of the Pelican prototyperobot, the inverse kinematic model has the form

q1

q2 = ϕ

5.2 Inverse Kinematics 117

The derivation of the inverse kinematic model is in general rather complexand, in contrast to the direct kinematics problem, it may have multiple solu-tions or no solution at all! The first case is illustrated in Figure 5.3 Notice

that for the same position (in Cartesian coordinates x, y) of the arm tip there

exist two possible configurations of the links, i.e two possible values for q.

q d2

q d1

y

x

Figure 5.3. Two solutions to the inverse kinematics problem

So we see that even for this relatively simple robot configuration thereexist more than one solution to the inverse kinematics problem

The practical interest of the inverse kinematic model relies on its utility

is more intuitive to specify a task for a robot in end-effector coordinates sothat interest in the inverse kinematics problem increases with the complexity

of the manipulator (number of degrees of freedom)

Thus, let us now make our this discussion more precise by analyticallycomputing the solutions

118 5 Case Study: The Pelican Prototype Robot

The desired joint velocities and accelerations may be obtained via the

differential kinematics1 and its time derivative In doing this one must keep

in mind that the expressions obtained are valid only as long as the robot does

is square and nonsingular These expressions are

configurations Physically, these configurations (for any valid n) represent the

second link being completely extended or bent over the first, as is illustrated

in Figure 5.4 Typically, singular configurations are those in which the effector of the robot is located at the physical boundary of the workspace(that is, the physical space that the end-effector can reach) For instance, thesingular configuration corresponding to being stretched out corresponds to

which is the boundary of the robot’s workspace As for Figure 5.4 the origin

of the coordinates frame constitute another point of this boundary

Having illustrated the inverse kinematics problem through the planar nipulator of Figure 5.2 we stop our study of inverse kinematics since it is

ma-1 For a definition and a detailed treatment of differential kinematics see the book

(Sciavicco, Siciliano 2000) —cf Bibliography at the end of Chapter 1.

5.3 Dynamic Model 119

q d1

x

Figure 5.4. “Bent-over” singular configuration

beyond the scope of this text However, we stress that what we have seen inthe previous paragraphs extends in general

In summary, we can say that if the control is based on the Cartesian dinates of the end-effector, when designing the desired task for a manipulator’send-effector one must take special care that the configurations for the latter

coor-do not yield singular configurations Concerning the controllers studied in thistextbook, the reader should not worry about singular configurations since theJacobian is not used at all: the reference trajectories are given in joint coor-dinates and we measure joint coordinates This is what is called “control injoint space”

Thus, we leave the topic of kinematics to pass to the stage of modeling

that is more relevant for control, from the viewpoint of this textbook, i.e.

dynamics

5.3 Dynamic Model

In this section we derive the Lagrangian equations for the CICESE type shown in Figure 5.1 and then we present in detail, useful bounds onthe matrices of inertia, centrifugal and Coriolis forces, and on the vector ofgravitational torques Certainly, the model that we derive here applies to anyplanar manipulator following the same convention of coordinates as for ourprototype

proto-5.3.1 Lagrangian Equations

Consider the 2-DOF robot manipulator shown in Figure 5.2 As we havelearned from Chapter 3, to derive the Lagrangian dynamics we start by writing

120 5 Case Study: The Pelican Prototype Robot

it may be decomposed into the sum of the two parts:

• the product of half the mass times the square of the speed of the center of

mass; plus

• the product of half its moment of inertia (referred to the center of mass)

times the square of its angular velocity (referred to the center of mass)

us now develop in more detail, the corresponding mathematical expressions

To that end, we first observe that the coordinates of the center of mass of link

1, expressed on the plane x–y, are

On the other hand, the coordinates of the center of mass of link 2, expressed

on the plane x–y are

5.3 Dynamic Model 121

Similarly, the potential energy may be decomposed as the sum of the

potential energy is zero at y = 0, we obtain

122 5 Case Study: The Pelican Prototype Robot

Thus, the dynamic equations of the robot (5.3)-(5.4) constitute a set of

is, of the form (3.1)

5.3 Dynamic Model 123

5.3.2 Model in Compact Form

For control purposes, it is more practical to rewrite the Lagrangian dynamicmodel of the robot, that is, Equations (5.3) and (5.4), in the compact form

We emphasize that the appropriate state variables to describe the dynamic

terms of these state variables, the dynamic model of the robot may be writtenas

Properties of the Dynamic Model

We present now the derivation of certain bounds on the inertia matrix, the trix of centrifugal and Coriolis forces and the vector of gravitational torques.The bounds that we derive are fundamental to properly tune the gains of thecontrollers studied in the succeeding chapters We emphasize that, as stud-ied in Chapter 4, some bounds exist for any manipulator with only revoluterigid joints Here, we show how they can be computed for CICESE’s Pelicanprototype illustrated in Figure 5.2

ma-124 5 Case Study: The Pelican Prototype Robot

M11(q)M22(q) − M21 (q)2

we only need to compute the determinant of M (q), that is,

1l c22m22[1− cos2(q

2)]

Notice that only the last term depends on q and is positive or zero Hence,

Inequality (5.5) constitutes an important property for control purposes

then this matrix is positive semidefinite (resp positive definite) See Horn R A.,

Johnson C R., 1985, Matrix analysis, p 473.

3 See also Remark 2.1 on page 25

126 5 Case Study: The Pelican Prototype Robot

Consider again the vector of centrifugal and Coriolis forces C(q, ˙q) ˙q written

is more intuitive to specify a task for a robot in end-effector coordinates sothat interest in the inverse kinematics problem increases... point of this boundary

Having illustrated the inverse kinematics problem through the planar nipulator of Figure 5. 2 we stop our study of inverse kinematics since it is

ma-1