# Gearbox service factor and service class explained

Sizing a gearbox (or gearmotor) for an industrialapplication typically begins with determining the appropriate service factor. In simple terms, the service factor is the ratio of the gearbox rated horsepower (or torque) to the application’s required horsepower (or torque). Service factors are defined by the American Gear Manufacturers Association (AGMA), based on the type of gearbox, the expected service duty, and the type of application.

While service factors may seem to be very specific, with thousands of combinations of gearbox types and applications each assigned its own numerical value, the criteria used to determine these values are based not on testing and empirical data, but rather on extensive review and analysis of gearbox manufacturers’ experience.

In general, the horsepower (or torque) rating of a gear tooth is based on the durability of the gear surface — its resistance to pitting — or on its bending fatigue. As the service factor of a gearbox is increased, the relationship between the gear teeth life (based on durability of the gear surface) and load is proportional to the increase in service factor, raised to the 8.78 power. In other words, if the service factor is increased by 30 percent (from 1.0 to 1.30, for example), the gear tooth life will increase 10 times (1.308.78 = 10.01).

To determine the gearbox service factor, start by consulting a set of tables or charts provided by the manufacturer, based on the type of gearing — worm, spiral bevel, helical, and so on. These tables list a wide range of applications ... conveyors, cranes, winders, saws, blowers, and so on, Each is listed with (typically) three levels of service duty the gearbox is expected to see: zero to 3 hours per day; 3 to 10 hours per day; or greater than 10 hours per day.

Each of these application-service duty combinations is assigned a recommended service factor.

Remember the gearbox service factor is much like a safety factor to ensure the gearbox meets the application requirements, taking into account typical operating conditions known to exist for various types of applications.

Once the AGMA-recommended service factor is determined, consider other, non-typical, working conditions that can cause additional stress and wear on the gear teeth, bearings, or lubrication. If any of these conditions exist, increase the service factor accordingly to ensure a sufficient safety margin and life of the gearbox.

Some conditions that may require an increase in the service factor are:

• Elevated temperatures

• Extreme shock loads or vibrations

• Non-uniform loads (cutting versus conveying, for example)

• Cyclic loads (frequent starts and stops)

• High peak versus continuous loads

Once the appropriate gearbox service factor is determined, multiply the service factor by the horsepower (or torque) required for the application, and the result is the output horsepower (or torque) required by the gearbox.

### HOW DOES SERVICE CLASS DIFFER FROM SERVICE FACTOR?

In some cases, manufacturers cite gearbox “service classes” rather than service factors. Service classes are designated as I, II, or III, and are generally translated to numerical service factors of 1.0, 1.4, and 2.0, respectively, to be used in gearbox sizing calculations. It’s common that even if a manufacturer publishes service classes for general application types, they also publish the more specific service factors for specific applications as well.

### WHY DON’T SOME CATALOGS LIST GEARBOX SERVICE FACTORS?

Using service factor to guide the selection of a gearbox is appropriate for applications driven by traditional AC induction motors. But because gearbox output torque, speed, and inertia are much more critical for the proper operation of a servo system, sizing a so-called “servo-rated” gearbox requires a more detailed and exact method. For gearboxes that are used in servo systems, the primary emphasis in the sizing process is on required torque and inertia match.

### CALCULATING THE INERTIA OF A SERVO-DRIVEN SYSTEM

Inertia or more specifically the inertia ratio is one of the most important factors in sizing a servo system. Inertia is defined an object’s resistance to change in velocity and in servo-driven systems, it can be used as a measure of how well the motor is able to control the acceleration and deceleration of the load.

The yardstick by which this is judged is the inertia ratio, which is defined as the inertia of the driven component(s) divided by the inertia of the motor.

J L = inertia of load reflected to motor

J M = inertia of motor

An inertia ratio that is too low means the motor is likely oversized, leading to higher than necessary cost and energy usage. An inertia ratio that is too high means the motor will have a difficult time controlling the load, which results in resonance and causes the system to overshoot its target parameter (position, velocity or torque).

While it seems logical that an inertia ratio of 1:1 would be the goal, it’s not always achievable or cost-effective. Most servo motor manufacturers recommend that the inertia ratio be kept to 10:1 or less, although there are many applications that operate successfully at much higher ratios. The best inertia ratio for an application comes down to the dynamics of the move and the accuracy required.

How to calculate inertia of a drive system: Load inertia includes the inertia of all rotating parts, including the drive (such as a belt and pulley system, screw, or rack and pinion), the load being moved, and the coupling between the load and the motor.

J L = Inertia of load reflected to motor

J D = Inertia of drive (ball screw, belt, rack & pinion)

J E = Inertia of external (moved) load

J C = Inertia of coupling

Manufacturers typically provide the inertia value (or a simple equation to calculate the inertia value) of drive systems, such as ball screws. But if the inertia value isn’t provided, it must be calculated manually. One of the most common ways to do this is to model the drive system as a shape for which the inertia equation is easily defined. Some common examples are solid cylinders and hollow cylinders.

Inertia of solid cylinder (screw or pinion) is:

Inertia of hollow cylinder (pulley) is:

Where m = Mass of cylinder; r = Radius of solid cylinder; r o = Outer radius of hollow cylinder; and r i = Inner radius of hollow cylinder

How to calculate inertia of a load: To determine the inertia of a screwdriven load, the effect of the screw’s lead must be taken into account.

Inertia of load driven by screw is:

m = Mass of load l s = Lead of screw

Note that this equation is based on the screw’s lead (expressed as inches or mm per revolution), not on the pitch (expressed as revolutions per inch or per millimeter).

When a load is driven by a belt and pulley system, the mass of both the load and the belt must be considered ... because both of these components are being driven by the motor.

Rack and pinion systems are treated similarly, with the mass of the pinion being added to the mass of the load.

Inertia of load driven by belt (or rack and pinion):

Where m = mass of load + mass of belt (or pinion)

Also, the radius in this case is the outer radius of the pulley, since this is the axis around which the belt and load are being rotated.

How to reduce inertia ratio: If the inertia ratio is too high, one way to reduce it is to add a gearbox to the system. In this case, the inertia of the load is divided by the square of the gear ratio. Inertia of load with gear reduction:

Where J G = inertia of gearbox and i = Gear ratio

Note that the inertia of the gearbox is added to the system, but its addition is small compared to the reduction provided by the gear ratio, which has an inverse squared effect on the load inertia.