- Stepper Motors
When determining the torque requirements of a Stepper Motor application, the effects of Inertia are often over looked. Many stepper applications are low-performance systems that accelerate and run at a low velocity, then decelerate to a stop.
These systems typically do not require careful inertia calculations to size correctly since most of the torque seen by the drive is friction and load torque that does not vary significantly with changes in inertia. In this case, the torque required is:
where F = force in ounces
r = radius in inches.
Torque (T) is equal to force (F) times the radius at which that force acts (r). The units that are easiest relating to the torque produced by stepper are ounces and inches, so T (oz·in) = F (oz) · r(in) so the units are simple.
When designing systems that are higher in performance, the torque resulting from the acceleration of system inertia may far exceed the torque from friction and load. In these systems, you must add the torque resulting from acceleration to the torque resulting from friction and load.
Inertia calculations seem intimidating to many people, but there are many online aids that can help. The basic formula is very similar to one that almost everyone knows: F=MA which says that force is equal to mass times acceleration. The one that calculates torque due to inertia has the form: T=I α which says that torque required (T) is equal to moment of inertia (I) times the angular acceleration (α). The units are far from simple, so here the ones easiest to use with oz · in: T (oz·in) = I (oz· in· sec2 ) · α (radians).
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