Oriental Motor U.S.A. Corp.
Proper motor sizing is important in these days of cost consciousness. Oversized motors are luxuries that are not cost effective. But a motor that's too small simply won't move the load. Fortunately, motor sizing is a simple arithmetic exercise if the parameters are known.
Ultimately, we are looking for a torque and speed value. Once these are determined, it's a simple matter to look at a torque/speed curve to find the right motor. Often the speed value is known. Torque, though, is a bit more difficult to calculate.
Torque is actually made up of two components; a constant torque and an acceleration torque. Constant torques are forces that always act on the motor. They usually stem from friction and gravity, but there can be other sources. An example might be a spring exerting a force on the load, and hence on the motor. Constant torques are usually easy to calculate or measure with a torque wrench on the motor shaft. The motor must overcome constant torque when running.
Acceleration torque is due to the load inertia and is present only during acceleration or deceleration. When accelerating up to speed, the motor must have enough acceleration torque to overcome the inertia. When decelerating, the motor must have enough braking torque to also overcome the inertia. More often than not, the total required torque value will be a combination of constant torque and acceleration torque. But while we can often measure constant torque, it is dangerous and difficult to do the same for acceleration torque. Instead, we often must calculate it. Fortunately, that is not too difficult.
Almost all loads can be modeled as simple components such as rods, discs, and cylinders. Equations for calculating inertias of simple shapes are available in physics text books or other reference materials. For a known inertia, we can determine the acceleration torque by
where TA = acceleration torque (ozin.), VF = final velocity (rad/sec), VO = initial velocity (rad/sec), tA = acceleration time (sec), and JL = load inertia (oz-in.-sec2).
Different units can be used in the above equation with the proper conversion factors. This equation converts everything up front, making calculations easier.
As an example, suppose a load has an inertia of 0.026 oz-in.-sec2. We want to accelerate from 500 to 2,500 rpm within 0.8 sec. The constant torque is 15 oz-in. How much torque is required and, just as important, what is the required power? Converting the rpm values into angular velocities yields 52.3 rad/sec and 261.6 rad/sec. Therefore, the acceleration torque is
With a safety factor of 30%, the required torque at 2,500 rpm is
(6.8 + 15) 1.3 = 28.4 oz-in.
The required power can be determined by
where P = required power (W), T = torque (oz-in.), and N = speed (rpm).
In this example, the motor must be rated for at least (28.4) (2,500)/1,340 = 53 W.
One final consideration is inertia matching. Brushless dc motors in closed-loop speed-control applications employ velocity feedback. As such, loop gains must be preset to control the response of the motor to deviations from the set speed. A high gain means the motor will respond quickly with overshoot, whereas a low gain means the motor will respond more slowly with little or no overshoot. Regardless of gain setting, the motor will stay stable only if certain inertia limitations are met. These limitations are usually in manufacturer literature. Exceeding them leads to instability.