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Selecting linear actuators, made easy

Oct. 1, 2000
Finding the right electric linear actuator for an application usually requires knowing many application details and making several calculations. For the time-strapped engineer, however, there are ways to narrow the selection to one or two safe approximations first

For various reasons, a user may not be able to execute all the steps necessary to find the best and most economical electric linear actuator from many choices. There are some steps that must be made, though, for a preliminary selection. Once the user has that preliminary selection, an application engineer from the actuator supplier or distributor can perform all the steps needed for final selection.

For a replacement actuator, there are the additional steps needed to review the application as if it were new. New actuator features and capabilities may resolve some problems from the original installation. Or, the original actuator may have been incorrectly applied, and without a review, those possible mistakes may be perpetuated.

“Whether the user is an electrical engineer or a mechanical engineer, the electric linear actuator uses principles from both engineering disciplines,” says Mr. Roy Schreffler, design engineer, SKF Specialty Products Co., Bethlehem, Pa. “Thus, power, defined in watts, is usually the first requirement found. To get mechanical power out of an electric linear actuator, it is necessary to put electrical power in. Mechanical power-out is usually the easier requirement to define because all that is needed is the force, or load that will be moved, and how fast it must move.”

If the parameters are in metric or SI units, multiply the force (in Newtons) times the speed (in meters/sec) to obtain watts. (To convert pounds to Newtons, multiply pounds by 4.448; to convert inches to meters, multiply inches by 0.0254.)

Mechanical power out (Po ):
Po = F x v
= force, N
= Velocity, m/s

But electrical power-in, duty cycle, and actuator efficiency also affect the preliminary selection. This information is usually found through performance graphs and charts from suppliers’ catalogs. Every supplier presents this information differently, but typically there are graphs for force vs. speed and force vs. current draw at some voltage. Such data are frequently presented in two graphs or combined in one. In others, the current draw is in tabular form. In addition, factors will be given based on a duty-cycle curve or in tabular form.

“For a preliminary actuator selection using a graph that shows lines for force (F), speed (V), and current draw (I), (Figure 1), the engineer should find the point on the force axis that represents the load (F) to be moved,” continues Mr. Schreffler. “Then, find where that point intersects the required speed on the speed axis (V). For this example, the actuator uses a standard 24 Vdc motor and offers three gear ratio choices, A, B, and C.”

“Assume an application requires an actuator to move a load of 3,000 N, point a on the graph in Figure 1 for gear ratio A. The intersecting point on the speed axis for this load is approximately 10 mm/sec. Thus, as long as the application requires the actuator to move the load at 10 mm/sec or less, this actuator is a good preliminary selection. The mechanical power-out available with this actuator is 3,000 N times 0.010 m/sec, or 30 W.”

To find the electrical power-in parameter, find the 3,000 N point on the force (F) axis and look across at the corresponding I line intersection point (labeled b) on the current-draw X axis. This point shows 7 A as the current draw. The formula for electrical power-in (Pi)is:

Pi = E x I E = voltage, V I = current, A

Multiply the two numbers to find an electrical power-in requirement of 168 W.


The next step is to establish the dutycycle factor. (Some manufacturers may refer to this as a de-rating factor). “Depending on this factor, the preliminary actuator selection may not meet application requirements,” says Mr. Schreffler.

The duty-cycle tells how often an actuator will operate in an application and the amount of time between operations. Because the power lost due to inefficiency goes as heat, the actuator component with the lowest allowable temperature — usually the motor — sets the duty-cycle limit for the complete actuator. Although, there are some heat losses from friction in the gearbox and by ballscrew and acme-screw drive systems.

To calculate duty cycle, for example, assume an actuator runs for 10 sec cumulative, up and down, then doesn’t run for another 40 sec. The duty cycle is 10/(40 + 10), or 20%.

Based on the earlier example, a preliminary actuator was selected for a power-out requirement of 30 W. Assume the customer now indicates that the actuator will operate 60% of the time. The application engineer will probably refer to a graph similar to Figure 2, which may or may not be in a product catalog. To show how dramatically duty cycle can impact actuator selection, at this duty cycle factor of 60%, the actuator would have to be de-rated by 80%. This means the customer really needs an actuator that is rated for 150 W, five times the power-out of the original selection! The customer has several choices, select a more powerful actuator, reduce the duty cycle, reduce the load, or reduce the speed of the application. (These options may not be practical). If duty cycle is increased, the load or the speed must be reduced. Conversely, if load or the speed decreases, duty cycle can increase.

“If the actuator is used on a machine or production device, duty cycle is easy to calculate,” says Mr. Schreffler. “In other applications where the actuator will be used infrequently or unpredictably, estimate the worst case scenario.”

“Operating on the edge of the manufacturer’s power curves runs the risk of the actuator running hot,” Mr. Schreffler continues. “In some applications, if the duty cycle is 10% or less, the actuator can run to the limit of its power curves.”


It is important to know electric linear actuator efficiency, but this parameter is usually not given in manufacturers’ catalogs. Knowing the efficiency provides an idea of how hot the actuator may get in operation. If that actuator uses a ball screw, it will also indicate whether holding brakes are needed on the system. And, if the system uses batteries as the power supply, the efficiency can also help determine how long the batteries may last.

“If the application requires that the actuator not move after electricity shuts off,” says Mr. Schreffler, “and there is no holding brake on the motor, the actuator must have an efficiency of about 27% or less. This sounds horrible, but if it were 100% efficient without a brake, the actuator would move (backdrive) on its own even after it was positioned electrically.”

You can calculate the efficiency from performance curves. The efficiency is found by dividing mechanical power-out by electrical power-in. From the first example of 30 W mechanical power-out and 168 W electrical power-in, that actuator has an efficiency of approximately 18%.

Actuator life

One last thing to consider before final selection — how long the actuator will last in the application.

Most actuators cannot be easily repaired. Some components, such as the motor or screw portion, are replaceable. Many catalogs indicate acme or ball screw life at a certain load, or include mathematical formulas to calculate life based on application parameters. It’s a good idea to have screw life and motor life about equal. If the actuator will use a dc brush motor, brush life will affect actuator life too.

In addition, when replacing an existing actuator, be sure the application engineer or supplier has all the necessary information. Use existing documentation such as sales brochures and CAD and assembly drawings. Or supply a unit that the actuator goes into. This can speed turnaround in matching the proper actuator.

Information for this article was provided by SKF Specialty Products Co.

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