There are a great many applications especially in hydraulics and hydraulically-driven machines that have greatly fluctuating load requirements. In some cases, the peak loads last for relatively short periods during the normal cycle of the machine. At first glance, it might seem that a motor would have to be sized to handle the worst part of the load cycle. For example, if a cycle included a period of time where 18 HP is required, then the natural approach would be to utilize a 20 HP motor. A more practical approach to these types of “duty cycle loads” takes advantage of an electric motor’s ability to handle substantial overload conditions as long as the period of overload is relatively short compared to the total time involved in the cycle.

The method of calculating whether or not the motor will be suitable for a particular cycling application is called the RMS (root mean squared) horsepower loading method. The calculations required to properly size a motor for this type of application are relatively simple and are presented in this paper.

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

In addition to reducing the size and cost of a motor for a particular application, RMS loading also offers the advantage of being able to improve the overall efficiency and power factor on a duty cycle type of 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.

In order to use the RMS method of horsepower determination, the duty cycle has to be spelled out in detail as shown in the following example.

Step Horsepower Duration (seconds)
1 3 3
2 7.5 10
3 2.5 12
4 12.5 3
Repeats continuously.

In order to determine the RMS loading for the previous cycle, we can use the formula:

The easiest way to approach this type of calculation is to make several columns as shown below and fill in the details underneath.

Step Horsepower HP2 Duration (seconds) HP2x Time
1 3.0 9.0 3 27.0
2 7.5 56.3 10 563.0
3 2.5 6.3 12 75.6
4 12.5 156.3 3 468.8
TOTAL 28 1134.4

In this case, the total time of the cycle is 28 seconds and the summation of horsepower squared times time for the individual steps in the cycle is 1134.4. when inserted into the equation, the RMS horsepower comes out to be:

At first glance, it appears that a 7-1/2 HP motor would be adequate to handle the loading required by this duty cycle. One further check has to be made and that is to determine if the motor has adequate pullout torque (breakdown torque) to handle the worst portion of the duty cycle without stalling. In this case, you would have to refer to the manufacturer’s data for the motor and determine the percent of pullout torque that is available.

An additional safety factor should be used because the pullout torque of the motor varies with the applied voltage. In fact, the pullout torque varies in relation to applied voltage squared. Thus, when the motor is running on 90% of rated voltage the amount of pullout torque available is only .9 x .9 or approximately 80% of the value that it has at full rated voltage. For this reason, it is never safe to use the full value of the pullout torque to determine if the overload can be handled. As a rule of good practice, it is wise not to use more than 80% of the rated pullout for a determination of adequacy.

In this case, referring to the Baldor CD ROM, we would find that a 7-1/2 HP, open drip proof motor with a catalog number M3311T, has a breakdown torque of 88.2 ft. lbs. and a full load operating torque of 22.3 ft. lbs. Thus, the actual pullout torque is 395% and utilizing 80% of this value, we would find that 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 as follows:

% Pullout torque required = 12.5 (Max. HP Point)/7.5 (Selected HP) x 100 = 167%

Since the available pullout torque at 90% of rated voltage is 316%, this 7-1/2 HP motor would be more than adequate to handle this application.

The previous formula and example can be used for applications where the duty cycle repeats itself continuously, without interruption. When a duty cycle involves a period of shut-off time, a different formula is used. That formula is shown below.

This formula is the same as the previous one but it is modified to reflect the fact that during the non-operating (motor is at standstill) time, it also loses its capability of cooling itself.

The total amount of time for which RMS loading can be adequately calculated would depend somewhat on the size of the motor but, in general, it would be safe to utilize this method for duty cycles that total less than 5 minutes from start to finish (of one complete cycle). If the total time is beyond 5 minutes, then the application should be referred to the motor manufacturer for more detailed analysis.


RMS horsepower loading is a very practical way to reduce motor horsepower requirements on cycling loads. With reduced motor horsepower also come a reduction in physical size and a reduction in initial cost, along with somewhat improved efficiency and reduced operating costs. If the selection procedure is handled carefully, you can expect to get very good performance and reliability from the completed unit.

On servomotors and other adjustable speed applications, similar calculations are frequently made. In these cases armature amperes or required torques are substituted in place of horsepower. The resulting RMS amperes or RMS torque requirement is then compared to the motor’s continuous and peak ratings to determine adequacy.

If you should have any questions regarding this method of sizing, please feel free to give us a call.

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