Page 59 - Sensing, Intelligence, Motion : How Robots and Humans Move in an Unstructured World
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34    A QUICK SKETCH OF MAJOR ISSUES IN ROBOTICS

                                   link i
                                center of mass
                                                 r * i


                                                             origin of
                                                p * i
                       origin of                             link i + 1,
                        link i,                               joint i
                      joint i − 1

                                            r i     p i
                                   p i − 1




                                            arm base

                    Figure 2.5  Balance of forces and torques acting on a single link.


           the accelerations ¨ r 1 and ¨ r 2 of the links’ centers of mass by Newton’s sec-
           ond law,

                                        f 1 = m 1 ¨ r 1
                                                                          (2.9)
                                        f 2 = m 2 ¨ r 2

           From these equations, accelerations ¨ r i of the centers of mass can be derived.

                Let ω i be the angular velocity vector of the center of mass of link i.
                Let ˙ω i be the corresponding angular acceleration.
                Let I i be the inertia matrix of link i.


              Then torques are related to angular velocities and accelerations by Euler’s
           equations,


                                   n 1 = I 1 ˙ω 1 + ω 1 × I 1 ω 1
                                                                         (2.10)
                                   n 2 = I 2 ˙ω 2 + ω 2 × I 2 ω 2
           For our planar two-link manipulator shown in Figure 2.1, the torque is normal
           to the arm’s plane. Rotary inertia through the centers of mass of links 1 and 2
           are [7]

                                                    2
                                          2
                                  I 1 = m 1 l /12 + m 1 R /4
                                         1                               (2.11)
                                          2
                                                    2
                                  I 2 = m 2 l /12 + m 2 R /4
                                         2
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