# How to choose service factor of a gear unit

Tuesday - 29/09/2020 11:14
The effect of the driven machine on the gear unit is taken into account with sufficient accuracy using the service factor fB. The service factor is determined according to the daily operating time and the starting frequency.
Three load classifications are taken into consideration depending on the inertia acceleration factor. You can read off the service factor applicable to your application in the graphic below. The product of the ascertained service factor and the required output torque must be less than the maximum permitted output torque. Service Factor for the gearbox reducer

 Load Characteristic Operating Time / Day (Hour) ≦ 2 hrs 2~10 hrs 10~24 hrs Uniform Load 1.00 1.3 1.6 Medium Load 1.3 1.6 1.8 Heavy Shock Load 1.6 1.8 2.0

(I) Uniform, permitted inertia acceleration factor ≤ 0.2
(II) Moderate shock load, permitted inertia acceleration factor ≤ 3
(III) Severe shock load, permitted inertia acceleration factor ≤ 10

Inertia acceleration factor

The inertia acceleration factor is calculated as follows:

Inertia acceleration factor “All external mass moments of inertia” are the mass moments of inertia of the driven machine and the gear unit, scaled down
to the motor speed. The calculation for scaling down to the motor speed is performed using the following formula: JX = Reflected inertia on the motor shaft
J = Inertia referenced to the output shaft of the gear unit
n = Output speed of the gear unit
nM = Motor speed

“Inertia at the motor end” is the mass moment of inertia of the motor and, if fitted, the brake and the flywheel fan (Z fan).

Service factor: fB

The method for determining the maximum permitted continuous torque Tamax and using this to derive the service factor
fB = Tamax / Ta is not defined in a standard and varies greatly from manufacturer to manufacturer.

Example
Inertia acceleration factor 2.5 (load classification II), 14 hours/day operating time and 300 cycles/hour produce a service
factor fB = 1.51 