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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 importantly, the residual dynamics h(t, ˜ q, ˙ ˜ q) has the virtue of not growing faster than ˙ ˜ q and ˜ q. Moreover it may grow arbitrarily fast only when so 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 tanh(x)= e x − e −x e x + e −x which is illustrated in Figure 4.1. −4 −3 −2 −101234 −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(x 1 ) . . . tanh(x n ) ⎤ ⎥ ⎦ (4.13) 104 4 Properties of the Dynamic Model where x ∈ IR n . The first partial derivative of tanh(x) is given by ∂tanh ∂x (x)=:Sech 2 (x)=diag{sech 2 (x i )} (4.14) where sech(x i ):= 1 e x i − e −x i . The vectorial tangent hyperbolic function satisfies the following properties. For any x , ˙ x ∈ IR n •tanh(x)≤α 1 x •tanh(x)≤α 2 •tanh(x) 2 ≤ α 3 tanh(x) T x • Sech 2 (x) ˙ x ≤ α 4 ˙ x where α 1 , ···,α 4 > 0. With tanh(x) defined as in (4.13), the constants α 1 = 1,α 2 = √ n, α 3 =1,α 4 =1. Property 4.4. Residual dynamics vector h(t, ˜ q, ˙ ˜ q) The vector of residual dynamics h(t, ˜ q, ˙ ˜ q)ofn × 1 depends on the position errors ˜ q, velocity errors ˙ ˜ q, and on the desired joint motion —q d , ˙ q d , and ¨ q d — that is supposed to be bounded. In this respect, we denote by ˙ q d M and ¨ q d M the supreme values over the norms of the desired velocity and acceleration. In addition, h(t, ˜ q, ˙ ˜ q) has the following property: 1. There exist constants k h1 ,k h2 ≥ 0 such that the norm of the residual dynamics satisfies h(t, ˜ q, ˙ ˜ q) ≤ k h1 ˙ ˜ q + k h2 tanh( ˜ q) (4.15) for all ˜ q, ˙ ˜ q ∈ IR n , where tanh( ˜ q) is the vectorial tangent hyper- bolic function introduced in Definition 4.1. Proof. According to the definition of the residual dynamics function (4.12), its norm satisfies h(t, ˜ q, ˙ ˜ q) ≤[M(q d ) − M(q d − ˜ q)] ¨ q d + C(q d , ˙ q d ) − C(q d − ˜ q, ˙ q d − ˙ ˜ q) ˙ q d + g(q d ) − g(q d − ˜ q) . (4.16) 4.4 The Residual Dynamics 105 We wish to upperbound each of the three terms on the right-hand side of this inequality. We start with g(q d ) − g(q d − ˜ q). From Property 4.3 it 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 for all q d , ˜ q ∈ IR n and where we have used g(q)≤k for the second in- equality. This may be illustrated as in Figure 4.2 where g(q d ) − g(q d − ˜ q) is in the dotted region, for all q d , ˜ q ∈ IR n . 0 0 2k k g ˜ ( d ) − ( d − ˜ ) 2k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . slope k g . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 4.2. Belonging region for ( d ) − ( d − ˜ ) 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 , where for the second inequality, we used that M(x) ¨ q d ≤k M ¨ q d , which is valid for all x ∈ IR n . 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 C 1 ˙ q d M ˙ ˜ q + k C 2 ˙ q d 2 M ˜ q. (4.17) Also, observe that the left-hand side of (4.17) also satisfies 106 4 Properties of the Dynamic Model C(q d , ˙ q d ) − C(q d − ˜ q, ˙ q d − ˙ ˜ q) ˙ q d ≤C(q d , ˙ q d ) ˙ q d + C(q d − ˜ q, ˙ q d − ˙ ˜ q) ˙ q d (4.18) but in view of the fact that C(q, x)y≤k C 1 xy for all q, x, y ∈ IR n , the terms on the right-hand side also satisfy C(q d , ˙ q d ) ˙ q d ≤k C 1 ˙ q d 2 M and, C(q d − ˜ q, ˙ q d − ˙ ˜ q) ˙ q d ≤ k C 1 ˙ q d M ˙ q d − ˙ ˜ q ≤ k C 1 ˙ q d 2 M + k C 1 ˙ q d M ˙ ˜ q . Using the latter in (4.18) we obtain C(q d , ˙ q d ) − C(q d − ˜ q, ˙ q d − ˙ ˜ q) ˙ q d ≤ 2k C 1 ˙ q d 2 M + k C 1 ˙ q d M ˙ ˜ q . (4.19) Hence, to bound the norm of the residual dynamics (4.16) we use (4.17) and (4.19), as well as the previous bounds on the first and third terms. This yields that h(t, ˜ q, ˙ ˜ q) also satisfies h(t, ˜ q, ˙ ˜ q) ≤ k C 1 ˙ q d M ˙ ˜ q + k g + k M ¨ q d M + k C 2 ˙ q d 2 M ˜ q, and h(t, ˜ q, ˙ ˜ q) ≤ k C 1 ˙ q d M ˙ ˜ q +2 k + k M ¨ q d M + k C 1 ˙ q d 2 M for all ˜ q ∈ IR n . In other terms, h(t, ˜ q, ˙ ˜ q) ≤ k C 1 ˙ q d M ˙ ˜ q + s( ˜ q) (4.20) where the scalar function s( ˜ q) is given by s( ˜ q)= s 1 ˜ q if ˜ q <s 2 /s 1 s 2 if ˜ q≥s 2 /s 1 with s 1 = k g + k M ¨ q d M + k C 2 ˙ q d 2 M , (4.21) and s 2 =2 k + k M ¨ q d M + k C 1 ˙ q d 2 M . (4.22) 4.4 The Residual Dynamics 107 0 0 s 2 s 1 ˜ s( ˜ ) s 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . slope s 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 4.3. Graph of the function s( ˜ ) The plot of s( ˜ q) is shown in Figure 4.3. It is clear that s( ˜ q)maybe upperbounded by the tangent hyperbolic function of ˜ q that is, |s( ˜ q)|≤k h2 tanh( ˜ q) (4.23) where k h2 is any number that satisfies k h2 ≥ s 2 tanh s 2 s 1 . (4.24) Thus, we conclude that h(t, ˜ q, ˙ ˜ q) in (4.20) satisfies h(t, ˜ q, ˙ ˜ q) ≤ k h1 ˙ ˜ q + k h2 tanh( ˜ q) where we have used the fact that tanh( ˜ q) ≤tanh( ˜ q), where tanh( ˜ q) is the vectorial tangent hyperbolic function (4.13) and k h1 is assumed to satisfy k C 1 ˙ q d M ≤ k h1 . (4.25) Thus, Property 4.4 follows. ♦♦♦ Residual Dynamics when ˙q d ≡ 0 In the situation when ˙ q d ≡ 0, and therefore ¨ q d ≡ 0, the residual dynamics (4.12) boils down to 108 4 Properties of the Dynamic Model h(t, ˜ q, ˙ ˜ q)=g(q d ) − g(q d − ˜ q), = g(q d ) − g(q). Notice also that s 1 and s 2 in (4.21) and (4.22) respectively, become s 1 = k g , s 2 =2k . With this information and what we know about k h1 from (4.25) and about k h2 from (4.24), we conclude that these constants k h1 =0, k h2 ≥ 2k tanh 2k k g . From this last inequality one can show that k h2 satisfies k h2 ≥ k g . Thus, we finally conclude from (4.20) and (4.23) that h(t, ˜ q, ˙ ˜ q) = g(q d ) − g(q d − ˜ q)≤k h2 tanh( ˜ q), ≤ k h2 ⎡ ⎢ ⎣ tanh(˜q 1 ) . . . tanh(˜q n ) ⎤ ⎥ ⎦ , for all q d , ˜ q ∈ IR n . 4.5 Conclusions Properties 4.1, 4.2, 4.3 and 4.4 are exhaustively used in the succeeding chap- ters in the stability analysis of the control schemes that we present. In particu- lar, Property 4.1 is used to construct non-negative functions and occasionally, Lyapunov functions to study stability and convergence properties for equilib- ria in robot control systems. To close the chapter we summarize, in Table 4.1, the expressions involved in the computation of the main constants introduced, where s 1 and s 2 are given by Equations (4.21) and (4.22), respectively. Bibliography 109 Table 4.1. Bounds on the matrices involved in the Lagrangian model Bound Definition β n (max i,j,q |M ij (q)|) k M n 2 max i,j,k,q ∂M ij (q) ∂q k k C 1 n 2 max i,j,k,q C k ij (q) k C 2 n 3 max i,j,k,l,q ∂C k ij (q) ∂q l k g n max i,j,q ∂g i (q) ∂q j k h1 k c1 ˙ d M k h2 s 2 tanh s 2 s 1 Bibliography 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. The property of skew-symmetry of 1 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: Inno- vation in the realm of robot servo–loops”, Robotica, Vol. 13, Part 1, pp. 19–27. • Arimoto S., 1995, “Fundamental problems of robot control: Part II: A nonlinear circuit theory towards an understanding of dexterous motions, Robotica, Vol. 13, Part 2, pp. 111–122. 110 4 Properties of the Dynamic Model One version of the proof of Property 4.4 on the residual dynamics h(t, ˜ q, ˙ ˜ q) is presented in • Santib´a˜nez V., Kelly R., 2001, “PD control with feedforward compensation for robot manipulators: Analysis and experimentation”, Robotica, Vol. 19, pp. 11–19. For other properties of robot manipulators not mentioned here and rele- vant to control, see • Ortega R., Lor´ıa A., Nicklasson P. J., Sira-Ram´ırez H., 1998, “Passivity- based control of Euler-Lagrange Systems Mechanical, Electrical and Elec- tromechanical Applications”, Springer-Verlag: London, Communications and Control Engg. Series. Problems 1. Consider the simplified Cartesian mechanical device of Figure 4.4. x 0 y 0 z 0 q 1 q 2 x 0 y 0 z 0 q 1 q 2 m 1 m 2 Figure 4.4. Problem 1 a) Obtain the dynamic model using Lagrange’s equations. Specifically, determine M(q), C(q, ˙ q) and g(q). b) Verify that the matrix 1 2 ˙ M(q) − C(q, ˙ q) is skew-symmetric. c) Express the dynamic model in terms of the state vector [q 1 q 2 ˙q 1 ˙q 2 ] T . Under which conditions on the external τ 1 and τ 2 do there exist equilibria? 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 1 2 ˙ M (q) − C(q, ˙ q)= 1 2 ˙ M(q) − C(q, ˙ q) . 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, ml 2 ¨q + mgl sin(q)=τ where q is the angular position of the pendulum with respect to the ver- tical. Show that there exists a constant β such that T 0 τ(s)˙q(s) ds ≥ β, ∀ T ∈ IR + . Hint: Using Property 4.3, show that for any T ≥ 0, T 0 ˙q(s)sin(q(s)) ds ≥−K with K ≥ 0. 5 Case Study: The Pelican Prototype Robot The purpose of this chapter is twofold: first, to present in detail the model of the experimental robot arm of the Robotics lab. from the CICESE Research Center, Mexico. Second, to review the topics studied in the previous chapters and to discuss, through this case study, the topics of direct kinematics and inverse 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 experimen- tation results. Thus, the chapter is organized in the following sections: • direct kinematics; • inverse kinematics; • dynamic model; • properties of the dynamic model; • reference trajectories. For analytical purposes, further on, we refer to Figure 5.2, which represents the prototype schematically. As is obvious from this figure, the prototype is a 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. The manipulator arm consists of two rigid links of lengths l 1 and l 2 , masses m 1 and m 2 respectively. The robot moves about on the plane x–y as is illus- trated in Figure 5.2. The distances from the rotating axes to the centers of mass are denoted by l c1 and l c2 for links 1 and 2, respectively. Finally, I 1 and [...]... Length of Link 1 l1 0.26 m Length of Link 2 l2 0.26 m Distance to the center of mass (Link 1) lc1 0.0983 m Distance to the center of mass (Link 2) lc2 0.0229 m Mass of Link 1 m1 6 .52 25 kg Mass of Link 2 m2 2.0 458 kg 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 Gravity acceleration g 9.81 m/s2 5. 1 Direct Kinematics The problem of direct kinematics... derivatives of the Jacobian and thereby obtains expressions for the velocities in Cartesian coordinates, is called differential kinematics This topic is not studied in more detail in this textbook since we do not use it for control 5. 2 Inverse Kinematics The inverse kinematic model of robot manipulators is of great importance from a practical viewpoint This model allows us to obtain the joint positions q in. .. not yield singular configurations Concerning the controllers studied in this textbook, the reader should not worry about singular configurations since the Jacobian is not used at all: the reference trajectories are given in joint coordinates and we measure joint coordinates This is what is called control in joint space Thus, we leave the topic of kinematics to pass to the stage of modeling that is... 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 prototype robot, the inverse kinematic model has the form q1 q2 = ϕ−1 (x, y) 5. 2 Inverse Kinematics 117 where ϕ−1 : Θ → IR2 and Θ ⊆ IR2 The derivation of the inverse kinematic model is in general rather complex and, in contrast to the direct kinematics problem,... interest of the inverse kinematic model relies on its utility T to define desired joint positions q d = [qd1 qd2 ] from specified desired positions xd and yd for the robot s end-effector Indeed, note that physically, it is more intuitive to specify a task for a robot in end-effector coordinates so that interest in the inverse kinematics problem increases with the complexity of the manipulator (number of degrees... of base as a Cartesian coordinated system in two dimensions with its origin located 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 case study of a 2-DOF system, the orientation of the end-effector of the arm makes no sense... 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 manipulator of Figure 5. 2 we stop our study of inverse kinematics since it is 1 For a definition and a detailed treatment of differential kinematics see the book (Sciavicco, Siciliano 2000) —cf Bibliography at the end of Chapter... makes sense) of a reference frame fixed to the end of the last link of the robot, referred to the base reference frame in terms of the joint coordinates of the robot The solution to the so-formulated problem from a mathematical viewpoint, reduces to solving a geometrical problem which always has a closed-form solution Regarding the Pelican robot, we start by defining the reference frame of base as a.. .114 5 Case Study: The Pelican Prototype Robot Figure 5. 1 Pelican: experimental robot arm at CICESE, Robotics lab y g Link 1 l1 lc1 x I1 m1 Link 2 m2 q1 I2 q2 lc2 l2 Figure 5. 2 Diagram of the 2-DOF Pelican prototype robot I2 denote the moments of inertia of the links with respect to the axes that pass through the respective centers of mass and are parallel to the axis x The degrees of freedom... vertical position, and q2 , which is measured relative to the extension 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 1 15 The meaning of the diverse constant parameters involved as well as their numerical values are summarized in Table 5. 1 Table 5. 1 Physical parameters of Pelican robot arm Description . Ortega R. , Lor´ a A., Nicklasson P. J., Sira-Ram´ırez H., 1998, “Passivity- based control of Euler-Lagrange Systems Mechanical, Electrical and Elec- tromechanical Applications”, Springer-Verlag:. concept of residual dynamics was introduced in • Arimoto S., 19 95, “Fundamental problems of robot control: Part I: Inno- vation in the realm of robot servo–loops”, Robotica, Vol. 13, Part 1, pp. 19–27. •. diverse constant parameters involved as well as their numerical values are summarized in Table 5. 1. Table 5. 1. Physical parameters of Pelican robot arm Description Notation Value Units Length of
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