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84 Electric Drives and Electromechanical Systems


                         Table 3.1  Approximate efficiencies for a range of lead and
                         ball screws.
                         System type                                       Efficiency
                         Ball screw                                        0.95
                         Lead screw                                        0.90
                         Rolled-ball lead screw                            0.80
                         ACME threaded lead screw                          0.40




             Energisation of the windings will cause the lead screw to move a defined distance, which
             is typically in the range 0.025e0.1 mm depending on the step angle and the lead of the
             lead screw. For a motor with a step angle of q radians, fitted to a lead screw of lead L, the
             incremental linear step, S, is given by,
                                                      qL
                                                   S ¼                                   (3.19)
                                                      2p



             nnn
               Example 3.2
               Determine the speed and torque requirements for the following lead screw application:
                 The length ðL s Þ of a lead screw is 1 m, its radius ðRÞ is 20 mm and is manufactured from
                                    3                       1
                  steel r ¼ 7850 kg m  . The lead ðLÞ is 6 mm rev . The efficiency ε of the lead screw is
                  0.85.
                 The total linear mass (M L ) to be moved is 150 kg. The coefficient of friction (m) between
                  the mass and its slipway is 0.5. A 50 N linear force (F L ) is being applied to the mass.
                 The maximum speed of the load (V L ) is 6 m min  1  and the system is required to reach
                  this speed in a time (t)1 s.
                  The mass of the lead screw and its inertia are calculated first:
                                                        M s R 2
                                     2                     s           3     2
                              M s ¼ rpR L s ¼ 9:86 kg and I s ¼  ¼ 1:97   10  kg m
                                     s
                                                         2
               The total inertia can be calculated by adding the reflected inertia from the load to the lead
               screw’s inertia:
                                                    2
                                                 L             3     2

                                     I tot ¼ I s þ M L  ¼ 2:11   10  kg m
                                                2p
               The torque required to drive the load against the external and frictional forces is given by
                                              LF L  LM L gm
                                         T ext ¼  þ      ¼ 0:75 Nm
                                              2p     2p
               The maximum input speed and acceleration required is given by,
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