How to calculate which electric motor you need: Practical guide with examples and useful tips

Engineers are constantly called upon to select electric motors for industrial applications that can perform the required task and have the lowest possible lifetime cost. There are many ways to make this selection based on price, performance, manufacturer and even colour. However, one of the simplest and most effective methods is to calculate the mechanical and physical requirements of the application, size the electrical requirements to match those requirements and then choose a motor that meets them.

For example, if you have an application that has limited space or needs to be lightweight, you should start with a motor that meets those parameters. If necessary, you can use mechanical means such as pulleys, gears, speed reducers, etc. to meet the mechanical requirements.

Typically, designers like you first choose between an AC or DC motor, or a geared motor. Geared motors are AC or DC motors that are typically used to obtain higher torque and lower rotational speed. Calculating your torque and speed requirements will help you determine whether you need an AC or DC motor.

One of the limiting mechanical factors of electric motors is the bearings. Motors that use bearings tend to have a longer service life than those that use bushings. They also tend to carry a higher perpendicular load on the shaft (radial load), either horizontally or vertically.

No matter how much torque the engine can generate, eventually the torque decreases as the speed increases, or the engine can only maintain the given torque by reducing its rotation. Once these torque versus speed qualities are established, engineers can adjust the numbers using the mechanical means mentioned above.

Example of electric motor dimensioning

Let's examine an example of a direct current (DC) motor rotating at 11,500 rpm with a 1-inch pitch diameter pulley. This generates a linear speed of 36,128 inches per minute (equivalent to 3,011 feet per minute or 602 inches per second). Changing the pulley size would alter the speed or torque. However, there are applications that require slower engines with a gearbox. It is one of those immutable characteristics; as speed increases, load capacity decreases.

Assume that the DC motor will be used to power a conveyor or tangential drive system. Add the complication that in the application, a spray nozzle will need to send one fluid ounce of material over an 18 × 14 inch area using a nozzle that ejects 0.050 gallons per minute or 0.1067 ounces per second at 40 psi.

To select the right motor, first find the required speed and torque. Then, the acceleration is found by setting the time required for the movement and solving for the shaft speed in rpm.

To obtain the allotted time, divide the amount of material dispersed by the dispersion rate:

1 fluid ounce / 0.1067 fluid ounces per second = 9.372 seconds.

To convert that into linear velocity, divide the length of the material by the elapsed time:

18 inches / 9.372 seconds = 1.9206 inches per second.

To find the rotational speed in rpm corresponding to 1.9206 inches per second of linear speed, convert inches per minute to inches per second and then convert that to revolutions. In this example, the pulley diameter of 1.003 inches leads to:

1.9203 inches per second × 60 seconds per minute × 1 revolution / (1.003 inches × π) = 36.57 revolutions per minute or 0.6 revolutions per second.

To determine the final angular velocity, acceleration and time, assume that the motor reaches a constant speed after the equivalent of 1 linear inch of travel. The associated arc length for a rotating system becomes 1 inch / π = 0.3183 inches.

The formula for determining the angle of arc can be found in the Machinery Manual. To use it, calculate the radius of the pulley:

1.003 / 2 = 0.5015.

Using the radius of the pulley and the associated arc length, we obtain an arc angle: (57.296 × 0.3183) / 0.5015 = 36.3655 decimal degrees, or 0.634 radians (57.296 is a constant from the Machinery's Handbook).

To find the final angular acceleration, use the equation for acceleration: a = V^2 / (2θ), where θ is the arc angle and V is the linear velocity. Substituting the values of the variables: (3.8297 rad/s^2) / (2 × 0.6347) = 11.5540 rad/s^2.

The final angular time or the time needed to reach the velocity comes from the relation: t^2 = (2θ) / ω. Solving for t, we obtain:

√((2 × 0.6347 rad) / 11.554 rad/s^2) = 0.3315 s.

Of course, the engine must provide more torque to achieve a higher acceleration rate or a shorter ramp distance. The more torque available, the greater the acceleration to achieve the required speed.

How to calculate the load inertia for engine sizing

Next, calculate the load inertia. When moving objects, the load on the motor is not only the load of the moving object. It also includes the loads of the pulleys, belts, couplings and any other devices or mechanisms between the motor and the moving object. To size the motor, find the total inertia of all these components acting on the motor shaft. Sometimes it is easier to do this using actual weights (converted to masses) of the devices rather than calculating the inertia requirements.

Calculate Friction

In this example, two sliding rails with four carriage pads carry the load. Each pad has a coefficient of friction of 0.17. The force due to friction is calculated using F = μN, where μ is the coefficient of friction and N is the force perpendicular to the surface. In this case, N is the mass of the load. The equation then becomes:

F = (96 oz × (4 × 0.17) = 65.28 oz.

This force, in turn, is multiplied by the distance to the axis of rotation:

65.28 ounces × 0.5015 inches = 32.738 ounce-inches.

To get the total torque, determine the torque required for acceleration. The first step is to convert the total inertia from ounces-inches^2 to ounces-inches-sec^2. This is a simple conversion that consists of multiplying the total inertia by a factor read from an inertia/torque conversion table, available from various sources:

24.8484 ounce-inches^2 × 0.00259 = 0.0643573 ounce-inches-sec^2.

This value is multiplied by the angular velocity and divided by the time required to reach that velocity:

(0.0643573 oz-inches-sec^2 × 3.8297 rad/sec)/0.3315 sec = 0.7435 oz-inches.

Finally, the force necessary to overcome the friction is added:

0.7435 ounce-inches + 32.738 ounce-inches = 33.482 ounce-inches.

Size

This example does not consider deceleration torque. It is not necessary to include it when calculating the maximum torque unless it exceeds the torque required for acceleration. Another tip: do not use holding torque for sizing motors. Holding torque is the maximum torque applied without causing shaft rotation when the motor is not energised.

Once this analysis leads to a particular motor, the designer must go back and add the rotor inertia of the motor to the calculation and recalculate to verify that the total torque required is well within the torque versus speed curve. If this is not the case, a larger motor will be required. As long as the required torque and speed are kept below the motor profile (with a safety margin), all other concerns are irrelevant.

Other factors to take into account when calculating electric motors

Note that the loads are set by the engine manufacturer. They should not be exceeded unless the goal is to cause the motor to fail prematurely. Finally, after installing the motor, empirically measure the actual torque required to move the load and find the lateral load on the motor. These empirical measurements can verify the calculations. In this example, a simple fishing scale could provide designers with a force reading in a pull test to determine the amount of force required to move the load.

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It is common and prudent to include a safety factor in the sizing of the electric motor to account for unanticipated problems. For example, if calculations call for a 66 oz-in motor, then it is advisable to choose the next larger size, a 100 oz-in motor, to provide a safety factor of 1.7. Common factors of safety range from 1.5 to 2.0.

Another factor worth considering is the ratio of load inertia to rotor inertia. This ratio is important if the engine is to accelerate with any precision or stop quickly. Basically, it is a ratio of how fast the engine accelerates or decelerates its own mass. This, in turn, affects the accuracy of the motor shaft position.

Some motor manufacturers recommend keeping the ratio of load inertia to rotor inertia below 5:1. If there are no precision requirements beyond starting or stopping the motor, engineers need only ensure that the speed and torque requirements are within the speed versus torque profile of the motor and add a safety factor. If the rotor-to-load inertia ratio is too high, the motor will overshoot or undershoot the stall position. Or the shaft could swing back and forth before settling into the proper position.

Therefore, the need for accuracy, or lack thereof, determines whether the ratio of load inertia to rotor inertia is a significant design parameter. Systems with a ratio of 1:1 will have the best accuracy. Those with a ratio of 2:1 or worse will be less so.

If necessary, a simple way to reduce the ratio is to use a motor with higher rotor inertia (larger shaft) or to add a gearbox to match the load and rotor inertia as closely as possible. The use of a gearbox reduces the output shaft speed and increases the torque according to the ratio value. An advantage of gearboxes is that they handle higher radial loads than would be possible by mounting the device on the motor shaft.

Gearboxes also affect the inertia ratio by a factor of the square of the gear ratio. To determine what size gearbox is needed, we solve:

e √(24.8484 oz-in.^2) / (0.55198 oz-in.^2) = 6.7

This indicates a gear ratio of 6.7:1, rounded to 7:1. Remember that with gearboxes, torque increases and output shaft speed decreases with gear ratio. Now size the gearbox for a motor by calculating 66 oz-in. × 1.5 (safety factor) = 100 oz-in. of gearbox output torque. This gives 100 oz-in. / 7 = 14 oz-in. of engine through the gearbox and 37 rpm × 7 = 259 engine rpm.

In this case, the speed and torque exceed the requirements. The controller can fine tune the spindle speed and torque requirements to achieve the final values.

In this article, we have discussed the basic data necessary for the selection and dimensioning of electric motors for various industrial applications. Here are the final tips for calculating and dimensioning electric motors:

1. Engineers must select engines that can perform the required task and have a low operating cost over their lifetime.
2. An effective way to select a motor is to determine the mechanical and physical requirements of the application, establish the corresponding electrical requirements, and then find a motor that meets them.
3. The choice between alternating current (AC) and direct current (DC) motors depends on the torque and speed requirements of the application.
4. Bearings are preferable to bushings in terms of durability and load carrying capacity.
5. The size of the pulley and the use of gears can adjust the speed and torque of the motor to suit the mechanical requirements.
6. Load inertia and how to calculate it was also discussed, as well as the importance of considering the safety factor in engine sizing.
7. The ratio of load inertia to rotor inertia can affect the accuracy of the motor in terms of acceleration and deceleration.
8. In the case of heavy loads or additional torque requirements, a gearbox can be used to adapt the motor to the application.

Correct selection and dimensioning of an electric motor involves considering various factors such as torque and speed requirements, load and rotor inertia, and the possible need for gearboxes. By taking these aspects into account, engineers can ensure optimum performance and long motor life in industrial applications.