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Chapter 3   Power transmission and sizing  101


                     The maximum load speed 300 rev min  1  a non-optimal gearbox with a ratio of 10:1
                     has been selected. The gearbox’s moment of inertia referred to its input shaft is
                                  2
                     4.3   10  4  kg m .
                     The load’s moment of inertia has been determined to 5   10  2  kg m 2
                     The maximum load torque is 6 Nm.

                     Based on this information the minimum motor speed required can be determined
                   including an allowance for voltage fluctuations, in this calculation we use a safety factor of 1.2.
                                   Motor speed ¼ 300   gear ratio   1:2 ¼ 3600 rev min  1

                                             Continuous torque ¼ 0:6Nm
                   Consideration of the motor data provided by a manufacturer indicates that the following
                   motor and drive could meet the requirements.
                     Peak speed: 5000 rev min  1   10% at a terminal voltage of 95 V
                     Continuous stall torque: 1.2 Nm. Peak torque: 8.0 Nm
                     Voltage constant: 0.18 V s rad  1  Current constant: 0.18 Nm A  1
                     Moment of inertia: 1.7  10  4  kg m 2
                     Drive continuous current: 5 A Drive peak torque: 10 A
                     Drive voltage: 95 V
                     The required speed is below the peak motor speed of 5000 rev min  1   10%, and the
                   required torque is below the motor’s continuous torque rating. At the continuous torque
                   requirement, the motor requires 3.33 A, hence the selected drive is suitable.
                     To ensure that the motor-drive combination is acceptable, the acceleration can be
                   determined for the drives peak output of 10 A. At this current the torque generated by the motor
                   is 1.8 Nm, well within the motor rating. Assuming the worst-case situation of the external load
                   being applied during acceleration and using Eq. (2.12):
                                    T peak   T L =n    1:8   0:6                2
                                              ¼                      ¼ 109:0 rad s
                                a L ¼                             2
                                           I L           4  5   10
                                    n I d þ     10 6   10  þ
                                          n 2                 10 2
                   Where I d is the inertia of the motor and gearbox, hence the load will be accelerated to its peak
                   speed of 300 rev min  1  within 0.5 s, which can be considered to be satisfactory. In practice the
                   acceleration rate would be controlled, so that the system, it particular any gearing, would not
                   experience significant shock loads.
                                                                                         nnn

                 3.8.2  Intermittent duty
                 When the acceleration performance is all important, the motor inertia must be consid-
                 ered, and the torque which is necessary to accelerate the total inertia must be determined
                 early in the sizing process. A suitable algorithm is as follows:

                   Using the application requirements and the required speed profile determine the
                   required speeds and acceleration.
                   Estimate the minimum motor torque for the application using Eq. (2.1).
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