Edward Cowern
Baldor Electric
Fort Smith, Ark.

Edited by Miles Budimir

Motors in hydraulic applications typically have largely   varying load requirements. It's possible to reduce motor size by analyzing   the duty cycle and computing RMS horsepower loading.

Motors in hydraulic applications typically have largely varying load requirements. It's possible to reduce motor size by analyzing the duty cycle and computing RMS horsepower loading.


The graph shows the widely varying horsepower demands   of a typical hydraulic application. The 12.5-hp peak demand does not necessarily   mean that a 12.5-hp motor must be used. The RMS horsepower loading technique   usually shows that a smaller motor can be used.

The graph shows the widely varying horsepower demands of a typical hydraulic application. The 12.5-hp peak demand does not necessarily mean that a 12.5-hp motor must be used. The RMS horsepower loading technique usually shows that a smaller motor can be used.


Many applications involving hydraulics and hydraulically-driven machines have load requirements that fluctuate greatly. In some cases, the peak loads last for relatively short periods during the normal cycle of the machine. It might seem initially that a motor should be sized to handle the worst part of the load cycle. For example, it would be natural to utilize a 20-hp motor for a case needing 18 hp for a period of time.

There's a more practical approach, however. It takes advantage of a motor's ability to handle substantial overloads as long as they are relatively short compared to the total cycle time.

A technique for calculating whether or not the motor can handle a particular cycling application is called the RMS (root mean squared) horsepower loading method. The calculations required are relatively simple.

The RMS calculations take into account that heat buildup within the motor is greater at a 50% overload than at normal operating conditions. Thus, the weighted average horsepower is what is significant. The RMS calculations determine the weighted average horsepower.

Besides reducing the size and cost of a motor for a particular application, RMS loading also helps improve the overall efficiency and power factor on a duty cycle-type load. For example, when an oversized motor is operated on a light load, the efficiency is generally fairly low. So working the motor harder (with a higher average horsepower) will generally result in improved overall efficiency and reduced operating cost.

The following equation calculates the RMS loading:

where Prms = rms horsepower, tn = time in seconds of the nth part of the cycle, and Pn = horsepower demand for the nth portion of the cycle.

The easiest way to approach this type of calculation is to use the information from the graph to make several columns as shown in the calculation table.

In this case, the total cycle time is 28 sec and the summation of horsepower squared multiplied by time for the individual steps in the cycle is 1,134.4. Substituting this into the equation yields:

At first glance, it appears that a 7.5-hp motor could handle the loading required by this duty cycle. But there's one further check needed to see if the motor has enough pullout or breakdown torque to handle the worst portion of the duty cycle without stalling. The motor data sheet usually has such information.

Designers should use an additional safety factor because the pullout torque of the motor varies with the square of the applied voltage. Thus for a motor running at 90% of rated voltage, the amount of pullout torque available is only 0.92 or approximately 80% of the value at full-rated voltage. This is why it's never safe to use the full value of the pullout torque to see if the motor can handle the overload. In practice, it is wise to stay below 80% of the rated pullout for a determination of adequacy.

As an example, suppose that a 7.5-hp, open dripproof motor has a breakdown torque of 88.2 lb-ft and a full-load operating torque of 22.3 lb-ft. Dividing breakdown torque by the full-load operating torque and multiplying by 100% gives an actual pullout torque of 395%. Utilizing 80% of this value, the available, safe pullout torque would be 316%.

For the duty cycle shown, the required pullout torque percentage can be determined by the ratio of maximum horsepower to rated horsepower:

Because the available pullout torque at 90% of rated voltage is 316%, this 7.5-hp motor would be more than adequate for this application.

The previous equation and example serves for applications where the duty cycle repeats itself continuously without interruption. When a duty cycle involves a period of shut-off time, the following equation applies:

where ts = number of seconds the motor is stopped, C = 3 for open dripproof motors, and C = 2 for totally enclosed motors.

This equation is essentially the same as the previous one. The modification reflects the fact that during the nonoperating time when the motor is at a standstill, it also loses its ability to self-cool.

The total amount of time for which RMS loading can be adequately calculated depends somewhat on the size of the motor. But in general, it would be safe to use this method for duty cycles totaling less than five minutes from start to finish. Designers confronted with times stretching beyond five minutes should talk to the motor manufacturer for a more detailed analysis.