Fundamentals of robotic mechanical systems_ theory, methods, and algorithms-Springer (2003)

340 8. Kinematics of Complex Robotic Mechanical Systems FIGURE 8.13. A planar parallel manipulator. FIGURE 8.14. A spherical parallel manipulator. TLFeBOOK

8.3 Kinematics of Parallel Manipulators 341 Now we write the second vector equation of eq.(8.97a) using the foregoing definitions, which yields (θ˙1)J fJ × (bJ + b4)eJ + (θ˙2)J gJ × (bJ + b4)eJ + b˙J eJ = c˙ J where b4 is the same for all legs, since all have identical architectures. Now we can eliminate (θ˙2)J from the foregoing equation by dot-multiplying its two sides by gJ , thereby producing (θ˙1)J gJ × fJ · (bJ + b4)eJ + +gJT (eJ eJT )c˙ J = gT c˙ J where an obvious exchange of the cross and the dot in the above equation has taken place, and expression (8.101a) for b˙J has been recalled. Now it is a simple matter to solve for (θ˙1)J from the above equation, which yields (θ˙1)J = − gJT (1 − eJ eTJ )c˙ J ∆J with ∆J defined as ∆J ≡ (bJ + b4)eJ × fJ · gJ (8.104) Moreover, we can obtain the above expression for (θ˙1)J in terms of the platform twist by recalling eq.(8.98), which is reproduced below in a more suitable form for quick reference: c˙ J = CJ t where t is the twist of the platform, the 3 × 6 matrix CJ being defined as CJ ≡ [ RJ 1 ] in which RJ is the cross-product matrix of rJ and 1 is the 3 × 3 identity matrix. Therefore, the expression sought for (θ˙1)J takes the form (θ˙1)J = −1 gJT (1 − eJ eJT )CJ t, J = I, II, . . . , V I (8.105a) ∆J A similar procedure can be followed to find (θ˙2)J , the final result being (θ˙2)J = 1 fJT (1 − eJ eJT )CJ t, J = I, II, . . . , V I (8.105b) ∆J thereby completing the calculations required to obtain the rates of all actu- ated joints. Note that the unit vectors involved in those calculations, eJ , fJ , and gJ , are computed from the leg inverse kinematics, as discussed above. TLFeBOOK

342 8. Kinematics of Complex Robotic Mechanical Systems Planar and Spherical Manipulators The velocity analysis of the planar and spherical parallel manipulators of Figs. 8.13 and 8.14 are outlined below: Using the results of Subsection 4.8.2, the velocity relations of the Jth leg of the planar manipulator take the form JJ θ˙ J = t, J = I, II, III (8.106) where JJ is the Jacobian matrix of this leg, as given by eq.(4.105), while θ˙ J is the 3-dimensional vector of joint rates of this leg, i.e., JJ ≡ 1 1 1 ,  θ˙J1  J = I, II, III ErJ 1 ErJ 2 ErJ 3 θ˙ J ≡  θ˙J2  , θ˙J 3 For purposes of kinematic velocity control, however, we are interested only in the first joint rate of each leg; i.e., all we need to determine in order to produce a desired twist of the end-effector is not all of the foregoing nine joint rates, but only θ˙I1, θ˙II1, and θ˙III1. Thus, we want to eliminate from eq.(8.106) the unactuated joint rates θ˙J2 and θ˙J3, which can be readily done if we multiply both sides of the said equation by a 3-dimensional vector nJ perpendicular to the second and the third columns of JJ . This vector can be most easily determined as the cross product of those two columns, namely, as n ≡ jJ2 × jJ3 = −rTJ 2 ErJ 3 rJ2 − rJ3 Upon multiplication of both sides of eq.(8.106) by nJT , we obtain −rTJ2ErJ3 + (rJ2 − rJ3)T ErJ1 θ˙J1 = −(rTJ2ErJ3)ω + (rJ2 − rJ3)T c˙ (8.107) and hence, we can solve directly for θ˙J1 from the foregoing equation, thereby deriving θ˙J 1 = −(rTJ2ErJ3)ω + (rJ2 − rJ3)T c˙ (8.108a) −rTJ2ErJ3 + (rJ2 − rJ3)T ErJ1 Note that eq.(8.107) can be written in the form jJ θ˙J1 = kJT t, J = I, II, III (8.108b) with jJ and kJ defined, for J = I, II, III, as jJ ≡ (rJ2 − rJ3)T ErJ1 − rJT2ErJ3, kJ ≡ [ rJT2ErJ3 (rJ2 − rJ3)T ]T If we further define θ˙ ≡ [ θ˙I1 θ˙II2 θ˙III3 ]T TLFeBOOK