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V012024 • © 2024 Portescap. Specifications subject to change without notice.(6) The total time to move is determined as follows: ∆trun + ∆taccel + ∆tdecel = ∆ttotal 0.37 sec + 0.027 sec + 0.027 sec = 0.42 secThe problem is now solved. The Can Stack stepper motor 42M048C1U is a good first estimate. This motor can be moved slower if more of a safety factor is desired.Example: No ramping acceleration or deceleration control is allowed.Even though no acceleration time is provided, the stepper can lag a maximum of two steps or 180° electrical degrees. If the motor goes from zero steps/sec to f steps/sec the lag time ∆t would be ∆t = 2/fThe torque equation for no acceleration or deceleration is: τJ = JT × f2/2 × 2π/λWhere: JT = rotor inertia (gm2) + load inertia (gm2) = 25 × 10-4 gm2 f = 250 steps/s λ = 48 steps/rev (from 7.5° per step)Example: Friction plus inertia and no acceleration ramping.For this application we are looking for Stepper motor for a continuous duty application. The application requirements are:A tape capstan is to be driven by a stepper motor.Motor operating point: 12 mNm [τf] – frictional load 10 × 10-4 gm2 [JL] – load inertia Continuous operationThe capstan must rotate in 7.5° increments at a rate of 200 steps/s.Since a torque greater than 12 mNm at 200 steps/s is required, consider the Can Stack stepper motor 42M048C1U.The total inertia = motor rotor inertia + load inertia JT = JR + JL JT = 12.5 × 10-4 gm2 + 10 × 10-4 gm2 JT = 22.5 × 10-4 gm2240Engineer’s Appendix