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.18) supposes that thelinks are rigid, that is, they do not present any torsion or any other defor-mation phenomena.On the other hand, we also considered that the jointsbetween each pair of links are stiff and frictionless.The incorporation of thesephenomena in the dynamic model of robots is presented in this and the fol-lowing section.Friction effects in mechanical systems are phenomena that depend on mul-tiple factors such as the nature of the materials in contact, lubrication of the 76 3 Robot Dynamicslatter, temperature, etc.For this reason, typically only approximate models offriction forces and torques are available.Yet, it is accepted that these forcesand torques depend on the relative velocity between the bodies in contact.Thus, we distinguish two families of friction models: the static models, inwhich the friction force or torque depends on the instantaneous relative ve-locity between bodies and, dynamic models, which depend on the past valuesof the relative velocity.ÙThus, in the static models, friction is modeled by a vector f(q) " IRn thatÙ Ùdepends only on the joint velocity q.Friction effects are local, that is, f(q)may be written as¡# ¤#f1(Ùq1)¢# ¥#¢# ¥#f2(Ùq2)Ùf(q) =¢# ¥#.¢# ¥#.¢# ¥#.£# ¦#.fn(Ùqn)An important feature of friction forces is that they dissipate energy, thatis,Ù Ù ÙqT f(q) > 0 " q = 0 " IRn.A classical static friction model is one that combines the so-called viscousÙand Coulomb friction phenomena.This model establishes that the vector f(q)is given byÙ Ù Ùf(q) =Fm1q + Fm2 sign(q) (3.24)where Fm1 and Fm2 are n×n diagonal positive definite matrices.The elementsof the diagonal of Fm1 correspond to the viscous friction parameters while theelements of Fm2 correspond to the Coulomb friction parameters.Furthermore,in the model given by (3.24)¡# ¤#sign(q1)Ù¢# ¥#sign(q2)ÙÙsign(q) =¢#.¥#¢# ¥#.£# ¦#.sign(qn)Ùand sign(x) is the sign function , given by1 if x>0sign(x) =-1 if x0 is the friction coefficient" k >0 is the stiffness coefficient of the spring" g is the acceleration of gravity" Ä is the applied force" q is the vertical position of the mass m with respect to origin of theplane x y.Write the model in the form ‹ = f(t, x) where x =[q q]T.Ùa) What restrictions must be imposed on Ä so that there exist equilibria?b) Is it possible to determine Ä so that the only equilibrium is the origin,x = 0 " IR2 ? Problems 91ymkqxfFigure 3.14.Problem 2z0 z0m2zo2q2 q2l2l1q1 m1 q1yo2 y0x0 y0 x0Figure 3.15.Problems 3 and 43.Consider the mechanical arm shown in Figure 3.15.Assume that the potential energy U(q1, q2) is zero when q1 = q2 = 0.Determine the vector of gravitational torques g(q),g1(q1, q2)g(q) =.g2(q1, q2)4.Consider again this mechanical device but with its simplified descriptionas depicted in Figure 3.15.a) Obtain the direct kinematics model of the device, i.e.determine therelationsy02 = f1(q1, q2)z02 = f2(q1, q2). 92 3 Robot Dynamicsb) The analytical Jacobian J(q) of a robot is the matrix¡# ¤#" " "Õ1(q) Õ1(q) · · · Õ1(q)¢# ¥#"q1 "q2 "qn¢# ¥#¢# ¥#" " "¢# ¥#Õ2(q) Õ2(q) · · · Õ2(q)¢# ¥#""qn¥#J(q) = Õ(q) =¢# "q1 "q2¢# ¥#"q.¢# ¥#.¢# ¥#¢# ¥#£# ¦#" " "Õm(q) Õm(q) · · · Õm(q)"q1 "q2 "qnwhere Õ(q) is the relation in the direct kinematics model (x = Õ(q)),n is the dimension of q and m is the dimension of x.Determine theJacobian.5.Consider the 2-DOF robot shown in Figure 3.16, for which the meaningof the constants and variables involved is as follows:z´q2 m2l1lc1 Ä2I1, m1q1Ä1yxFigure 3.16.Problem 5" m1, m2 are the masses of links 1 and 2 respectively;" I1 is the moment of inertia of link 1 with respect to the axis parallelto the axis x which passes through its center of mass; the moment ofinertia of the second link is supposed negligible;" l1 is the length of link 1;" lc1 is the distance to the center of mass of link 1 taken from its rotationaxis;" q1 is the angular position of link 1 measured with respect to the hori-zontal (taken positive counterclockwise); Problems 93" q2 is the linear position of the center of mass of link 2 measured fromthe edge of link 1;" ´ is negligible (´ = 0).Determine the dynamic model and write it in the form ‹ = f(t, x) wherex =[q q]T.Ù6.Consider the 2-DOF robot depicted in Figure 3.17.Such a robot has atransmission composed of a set of bar linkage at its second joint.Assumethat the mass of the lever of length l4 associated with actuator 2 is negli-gible.zl4l1link 1l1xactuator 1q1lc3lc1I3I1, m1 ym3l4lc2 actuator 2q2I2, m2link 1Figure 3.17.Problem 6Determine the dynamic model.Specifically, obtain the inertia matrixÙM(q) and the centrifugal and Coriolis matrix C(q, q).Hint: See the robot presented in Example 3.3.Both robots happen to bemechanically equivalent when taking m3 = I3 = ´ =0. 4Properties of the Dynamic ModelIn this chapter we present some simple but fundamental properties of thedynamic model for n-DOF robots given by Equation (3.18), i.e.¨ Ù ÙM(q)q + C(q, q)q + g(q) =Ä.(4.1)In spite of the complexity of the dynamic Equation (4.1), which describesthe behavior of robot manipulators, this equation and the terms which consti-tute it have properties that are interesting in themselves for control purposes.Besides, such properties are of particular importance in the study of controlsystems for robot manipulators.Only properties that are relevant to controldesign and stability analysis via Lyapunov s direct method (see Section 2.3.4in Chapter 2) are presented.The reader is invited to see the references at theend of the chapter to prove further.These properties, which we use extensively in the sequel, may be classifiedas follows:" properties of the inertia matrix M(q);Ù" properties of the centrifugal and Coriolis forces matrix C(q, q);" properties of the gravitational forces and torques vector g(q);" properties of the residual dynamics.Each of these items is treated independently and constitute the material ofthis chapter.Some of the proofs of the properties that are established belowmay be consulted in the references which are listed at the end of the chapterand others are developed in Appendix C.4.1 The Inertia MatrixThe inertia matrix M(q) plays an important role both in the robot s dynamicmodel as well as in control design.The properties of the inertia matrix which is 96 4 Properties of the Dynamic Model1Ù Ùclosely related to the kinetic energy function K = qTM(q)q, are exhaustively2used in control design for robots.Among such properties we underline thefollowing.Property 4.1.Inertia matrix M(q)The inertia matrix M(q) is symmetric positive definite and has dimensionn × n.Its elements are functions only of q.The inertia matrix M(q) satisfiesthe following properties.There exists a real positive number ± such thatM(q) e" ±I " q " IRn1.where I denotes the identity matrix of dimension n×n.The matrixM(q)-1 exists and is positive definite.For robots having only revolute joints there exists a constant ² >0such that»Max{M(q)} d"² " q " IRn.One way of computing ² is2.² e" n max |Mij(q)|i,j,qwhere Mij(q) stands for the ijth element of the matrix M(q).For robots having only revolute joints there exists a constant kM >0 such thatM(x)z - M(y)z d"kM x - y z (4.2)3.for all vectors x, y, z " IRn.One simple way to determine kM isas follows"Mij(q)kM e" n2 max.(4.3)i,j,k,q "qkFor robots having only revolute joints there exists a number kM >0 such that4.M(x)y d"kM yfor all x, y " IRn.The reader interested in the proof of Inequality (4.2) is invited to seeAppendix C. 4 [ Pobierz caÅ‚ość w formacie PDF ]
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.18) supposes that thelinks are rigid, that is, they do not present any torsion or any other defor-mation phenomena.On the other hand, we also considered that the jointsbetween each pair of links are stiff and frictionless.The incorporation of thesephenomena in the dynamic model of robots is presented in this and the fol-lowing section.Friction effects in mechanical systems are phenomena that depend on mul-tiple factors such as the nature of the materials in contact, lubrication of the 76 3 Robot Dynamicslatter, temperature, etc.For this reason, typically only approximate models offriction forces and torques are available.Yet, it is accepted that these forcesand torques depend on the relative velocity between the bodies in contact.Thus, we distinguish two families of friction models: the static models, inwhich the friction force or torque depends on the instantaneous relative ve-locity between bodies and, dynamic models, which depend on the past valuesof the relative velocity.ÙThus, in the static models, friction is modeled by a vector f(q) " IRn thatÙ Ùdepends only on the joint velocity q.Friction effects are local, that is, f(q)may be written as¡# ¤#f1(Ùq1)¢# ¥#¢# ¥#f2(Ùq2)Ùf(q) =¢# ¥#.¢# ¥#.¢# ¥#.£# ¦#.fn(Ùqn)An important feature of friction forces is that they dissipate energy, thatis,Ù Ù ÙqT f(q) > 0 " q = 0 " IRn.A classical static friction model is one that combines the so-called viscousÙand Coulomb friction phenomena.This model establishes that the vector f(q)is given byÙ Ù Ùf(q) =Fm1q + Fm2 sign(q) (3.24)where Fm1 and Fm2 are n×n diagonal positive definite matrices.The elementsof the diagonal of Fm1 correspond to the viscous friction parameters while theelements of Fm2 correspond to the Coulomb friction parameters.Furthermore,in the model given by (3.24)¡# ¤#sign(q1)Ù¢# ¥#sign(q2)ÙÙsign(q) =¢#.¥#¢# ¥#.£# ¦#.sign(qn)Ùand sign(x) is the sign function , given by1 if x>0sign(x) =-1 if x0 is the friction coefficient" k >0 is the stiffness coefficient of the spring" g is the acceleration of gravity" Ä is the applied force" q is the vertical position of the mass m with respect to origin of theplane x y.Write the model in the form ‹ = f(t, x) where x =[q q]T.Ùa) What restrictions must be imposed on Ä so that there exist equilibria?b) Is it possible to determine Ä so that the only equilibrium is the origin,x = 0 " IR2 ? Problems 91ymkqxfFigure 3.14.Problem 2z0 z0m2zo2q2 q2l2l1q1 m1 q1yo2 y0x0 y0 x0Figure 3.15.Problems 3 and 43.Consider the mechanical arm shown in Figure 3.15.Assume that the potential energy U(q1, q2) is zero when q1 = q2 = 0.Determine the vector of gravitational torques g(q),g1(q1, q2)g(q) =.g2(q1, q2)4.Consider again this mechanical device but with its simplified descriptionas depicted in Figure 3.15.a) Obtain the direct kinematics model of the device, i.e.determine therelationsy02 = f1(q1, q2)z02 = f2(q1, q2). 92 3 Robot Dynamicsb) The analytical Jacobian J(q) of a robot is the matrix¡# ¤#" " "Õ1(q) Õ1(q) · · · Õ1(q)¢# ¥#"q1 "q2 "qn¢# ¥#¢# ¥#" " "¢# ¥#Õ2(q) Õ2(q) · · · Õ2(q)¢# ¥#""qn¥#J(q) = Õ(q) =¢# "q1 "q2¢# ¥#"q.¢# ¥#.¢# ¥#¢# ¥#£# ¦#" " "Õm(q) Õm(q) · · · Õm(q)"q1 "q2 "qnwhere Õ(q) is the relation in the direct kinematics model (x = Õ(q)),n is the dimension of q and m is the dimension of x.Determine theJacobian.5.Consider the 2-DOF robot shown in Figure 3.16, for which the meaningof the constants and variables involved is as follows:z´q2 m2l1lc1 Ä2I1, m1q1Ä1yxFigure 3.16.Problem 5" m1, m2 are the masses of links 1 and 2 respectively;" I1 is the moment of inertia of link 1 with respect to the axis parallelto the axis x which passes through its center of mass; the moment ofinertia of the second link is supposed negligible;" l1 is the length of link 1;" lc1 is the distance to the center of mass of link 1 taken from its rotationaxis;" q1 is the angular position of link 1 measured with respect to the hori-zontal (taken positive counterclockwise); Problems 93" q2 is the linear position of the center of mass of link 2 measured fromthe edge of link 1;" ´ is negligible (´ = 0).Determine the dynamic model and write it in the form ‹ = f(t, x) wherex =[q q]T.Ù6.Consider the 2-DOF robot depicted in Figure 3.17.Such a robot has atransmission composed of a set of bar linkage at its second joint.Assumethat the mass of the lever of length l4 associated with actuator 2 is negli-gible.zl4l1link 1l1xactuator 1q1lc3lc1I3I1, m1 ym3l4lc2 actuator 2q2I2, m2link 1Figure 3.17.Problem 6Determine the dynamic model.Specifically, obtain the inertia matrixÙM(q) and the centrifugal and Coriolis matrix C(q, q).Hint: See the robot presented in Example 3.3.Both robots happen to bemechanically equivalent when taking m3 = I3 = ´ =0. 4Properties of the Dynamic ModelIn this chapter we present some simple but fundamental properties of thedynamic model for n-DOF robots given by Equation (3.18), i.e.¨ Ù ÙM(q)q + C(q, q)q + g(q) =Ä.(4.1)In spite of the complexity of the dynamic Equation (4.1), which describesthe behavior of robot manipulators, this equation and the terms which consti-tute it have properties that are interesting in themselves for control purposes.Besides, such properties are of particular importance in the study of controlsystems for robot manipulators.Only properties that are relevant to controldesign and stability analysis via Lyapunov s direct method (see Section 2.3.4in Chapter 2) are presented.The reader is invited to see the references at theend of the chapter to prove further.These properties, which we use extensively in the sequel, may be classifiedas follows:" properties of the inertia matrix M(q);Ù" properties of the centrifugal and Coriolis forces matrix C(q, q);" properties of the gravitational forces and torques vector g(q);" properties of the residual dynamics.Each of these items is treated independently and constitute the material ofthis chapter.Some of the proofs of the properties that are established belowmay be consulted in the references which are listed at the end of the chapterand others are developed in Appendix C.4.1 The Inertia MatrixThe inertia matrix M(q) plays an important role both in the robot s dynamicmodel as well as in control design.The properties of the inertia matrix which is 96 4 Properties of the Dynamic Model1Ù Ùclosely related to the kinetic energy function K = qTM(q)q, are exhaustively2used in control design for robots.Among such properties we underline thefollowing.Property 4.1.Inertia matrix M(q)The inertia matrix M(q) is symmetric positive definite and has dimensionn × n.Its elements are functions only of q.The inertia matrix M(q) satisfiesthe following properties.There exists a real positive number ± such thatM(q) e" ±I " q " IRn1.where I denotes the identity matrix of dimension n×n.The matrixM(q)-1 exists and is positive definite.For robots having only revolute joints there exists a constant ² >0such that»Max{M(q)} d"² " q " IRn.One way of computing ² is2.² e" n max |Mij(q)|i,j,qwhere Mij(q) stands for the ijth element of the matrix M(q).For robots having only revolute joints there exists a constant kM >0 such thatM(x)z - M(y)z d"kM x - y z (4.2)3.for all vectors x, y, z " IRn.One simple way to determine kM isas follows"Mij(q)kM e" n2 max.(4.3)i,j,k,q "qkFor robots having only revolute joints there exists a number kM >0 such that4.M(x)y d"kM yfor all x, y " IRn.The reader interested in the proof of Inequality (4.2) is invited to seeAppendix C. 4 [ Pobierz caÅ‚ość w formacie PDF ]