Laguna Hills, Calif.
Edited by
www.baldor.comEllis & Associates CADD/Technical Illustration, ellis-assoc.com/Contacts.phpMachinery’s Handbook, Industrial Press, tinyurl.com/6yevkoo |

There are different ways of selecting electric motors for specific applications. Perhaps the simplest way of approaching motor selection, though, is to ascertain the mechanical or physical requirements of the job and make the electrical requirements match them. For example, if constrained by space or weight requirements, initially select a motor within those parameters. Then, try to use mechanical means (pulleys, gears, gear heads, speed reducers, and so forth) to meet the mechanical requirements.

Designers typically first settle on either an ac or dc motor or gearmotor. Gearmotors are ac or dc motors typically used for higher torque and lower rotational speed. Knowing the torque and speed requirements will help in determining if an ac or dc motor is required.

One of the mechanical limiting factors of electric motors is the bearings. Motors that use bearings will typically last longer than those using bushings. They also typically handle more perpendicular loading to the shaft (radial load), whether horizontally or vertically.

No matter how much torque the motor can generate, it will eventually hit a crossing point where either torque falls off as speed rises or the motor can only maintain a given torque by rotating more slowly. Once these torque versus speed qualities have been established, then you can play with the numbers using the aforementioned accessories.

Let us take a concrete example of a **Baldor Electric Co.** dc motor that generates 11,500 rpm with a 1-in.-pitch-diameter pulley. This configuration will produce a linear speed of 36,128 ipm, or 3,011 fpm, or 602 ips. The pulley size, of course, could be changed to vary the speed or torque. However, some applications might need slower motors with a gearbox. It’s a number game; as the speed requirement rises, the load capability drops, and vice versa.

Consider the example of applying this motor in a conveyor or tangential drive system. Further assume the need to spray 1 fl oz of material over an 18 × 14-in. area using a spray tip which produces 0.050 gallons/min or 0.1067fl-oz/sec at 40 psi.

Motor selection starts by first finding the needed speed (or velocity), and torque. Next comes acceleration, determined by establishing the amount of time required for the move and then solving for shaft speed in rpm.

One determines time in this case by dividing the amount of material to be dispersed by the rate of dispersion, or 1 fl-oz/0.1067 fl-oz/sec = 9.372 sec. To determine linear velocity, divide the length of material by elapsed time, or 18 in./9.372 sec = 1.9206 ips.

In many cases, velocity is the operational requirement which dictates the motor size and/or type. Examples include the speed with which you can transfer a part from one location to another, the rate at which you can fill a container or remove material, or a dispersal rate for sprayed fluid.

To find the rotational velocity in rpm corresponding to this linear velocity, we first convert inches/minute to inches/second and then convert to revolutions. In this example the pulley diameter is 1.003 in. That gives 1.9203ips × 60 sec/min × 1 rev/(1.003 in. × π) = 36.57 rev/min or 0.6rev/sec.

To determine the angular velocity, acceleration and time, we make a simplifying assumption that it takes 1linear in. to reach a constant speed. We next determine the associated arc length for a rotary system, which is 1 in./π = 0.3183 in. The formula for determining arc angle is from the *Machinery’s Handbook*. To use it, we first determine the pulley radius, 1.003/2 = 0.5015. Using the pulley radius and associated arc length, we have an arc angle (57.296 × 0.3183)/0.5015 = 36.3655 decimal degrees, or 0.6347radans. Here, 57.296 is a constant from the *Machinery Handbook*.

To determine final angular velocity, we divide linear velocity by the pulley radius, 1.9206 ips/0.5915 in. = 3.8297rad/sec. To determine final angular acceleration, we use the relationship for acceleration,

a = V^{2}/2θ

where θ = arc angle and V = linear velocity: (3.8297rad/sec^{2} )/(2 × 0.6347) = 11.5540 rad/sec.^{2}

The final angular time or time needed to reach velocity comes from the relationship t2 = 2θ/ω. Solving for t gives √((2×0.6347 rad)/11.554 rad/sec^{2}) = 0.3315 sec.

Of course, the motor must provide more torque if the system needs a higher acceleration rate or a shorter ramp distance. The more torque there is available, the quicker the acceleration to reach the prescribed velocity.

Next comes the calculation of load inertia. In moving real objects and not just theoretical examples, the load on the motor is more than just the load imposed by the object being moved. It also consists of the load comprised of pulleys, belts, couplers, shafts, belt-tension devices, and any other object between the motor and the object being moved. To size a motor properly, you must find the total inertia of all these components as they act on the motor shaft. In this task, it can sometimes be easier to use the actual weight (transformed into mass) of the objects rather than calculating the inertia requirements.

In our example, say the system consists of: load at 96.0oz, two pulleys at 1.0 oz each, and a belt at 0.8 oz. Using the general equation for inertia *I* = *mr*^{2} , where *m* = mass and *r* = the distance to the rotation axis, then the total inertia on the motor, *I* = (96 oz ×( 0.5015 in.)^{2}) + (0.8 oz × (0.5015 in.)^{2}) + ((1 oz × 0.50152 in.) × 2) = 24.8484oz-in.^{2}

Next comes friction considerations. Say in this example you are using a common configuration consisting of two slider rails with four carriage pads carrying the load. Each of the four carriage pads has a coefficient of friction of 0.17. The force due to friction, *F* = μN, where μ = friction coefficient and *N* = force perpendicular to the surface. In this case, *N* = just the mass of the load. So the relationship reduces to *F* = (96 oz × (4 × 0.17) = 65.28 oz. This relationship, in turn, is multiplied by the distance to the rotation axis: 65.28 oz × 0.5015 in. = 32.738 oz-in.

To determine total torque, we first determine torque needed for acceleration. The initial step is to convert total inertia from oz-in.2 to oz-in.-sec2. This is a simple conversion that consists of multiplying total inertia by a factor read from an inertia/torque conversion table, available from a variety of sources: 24.8484 oz-in.^{2} × 0.00259 = 0.0643573 oz-in.-sec^{2}. Next this figure is multiplied by the angular velocity and divided by the time needed to reach that velocity: (0.0643573 oz-in.-sec^{2} × 3.8297 rad/sec)/0.3315 sec = 0.7435 oz-in. Finally, we add the force needed to overcome friction: 0.7435 oz-in. + 32.738 oz-in. = 33.482 oz-in. Thus, most of the torque for acceleration is needed to overcome friction.

The process for determining torque needed for a constant load is similar. The only difference in the equation is that linear velocity, calculated earlier, is used instead of angular velocity, and division is by spray time, also calculated earlier, rather than acceleration time. This gives (0.0643573oz-in.-sec^{2} × 1.9206 ips)/9.372 sec = 0.0132oz-in. To this we once again add the force needed to overcome friction: 0.0132 oz-in. + 32.738 oz-in. = 32.751oz-in. Once again, most of the torque goes into overcoming friction. The total torque is just the sum of the torque needed for acceleration and for handling a constant load: 33.482 + 32.751 = 66.233oz-in.

A point to note is that torque for acceleration will not always be about the same as torque for constant load, as in this case. Do not assume you can just double the torque for constant load and meet the total torque requirement.

**Determining size**

This example didn’t consider deceleration torque. It is not required when solving for maximum torque unless it exceeds the torque needed to accelerate. Another tip: Do not use holding torque to size the motor. Holding torque shows how much the motor will hold at 0 rpm.

Once this analysis leads to a particular motor, the designer should go back and add the motor-rotor inertia to the calculation and recalculate to verify that the total torque required lies well inside the torque-versus-speed curve. If not, the situation calls for the next bigger motor size. As long as the required torque and speed are kept below the motor profile (with a safety factor) all other concerns are not relevant.

Another point to keep in mind: Side load (radial load) and overhang load are established by the motor manufacturer. They must not to be exceeded. Doing so will make the motor fail prematurely. Finally, with the motor installed, it’s best to empirically measure the actual required torque to move the load and find the side load on the motor.

It is common practice to include a safety factor in motor sizing to account for unseen problems. For example, calculations calling for a 66-oz-in. motor might result in using the next size up, a 100-oz-in. motor, to provide a 1.7safety factor. Common safety factors are in the 1.5 to 2.0range.

Empirical measurements can verify calculations. In the above example, a simple fish scale could give a force reading in a pull test to find the amount of force needed to move the load.

One factor that is worth considering is the ratio of load-to-rotor inertia. This entity tends to be important when the motor must accelerate with some precision or stop quickly. It is basically a ratio of how fast a motor will accelerate or decelerate its own mass. This, in turn, bears on the accuracy of the motor shaft position.

Baldor Electric Co. recommends keeping the load-to-rotor-inertia ratio below 5:1. If there is no accuracy requirement other than for starting or stopping the motor, designers need only make the speed and torque requirement fall within the speed-versus-torque chart profile with an allowable safety factor. If the rotor-to-load inertia ratio is too high, the problem will be one of overshooting or undershooting the location of the stop position. The shaft might even oscillate back and forth until settling at the right position.

Thus, the need for precision, or the lack of it, determines whether load-to-rotor inertia must be a significant design parameter. A system with a 1:1 ratio will have optimum precision. A system with a 2:1 ratio or worse will be less so.

As an example, consider the inertia from the earlier example and a motor having 0.00143 oz-in.-sec^{2} rotor inertia. We convert to the same units (using information from widely available tables) to solve for the ratio: 0.00143oz-in.-sec^{2} × 386 ips^{2} = 0.55198 oz-in.^{2} Then 24.8484 oz-in.2/0.55198 oz-in.^{2} = 45. Thus the ratio is 45:1.

If need be, a simple solution for lowering the ratio is either to use a motor with a larger rotor inertia (bigger shaft) or to add a gearhead to match as closely as possible to the load and rotor inertia. Use of a gearhead will reduce output shaft speed at the gearhead and boost torque according to the ratio value. One of the many advantages of gearheads is they can handle a higher radial loading than would be possible by just mounting the device directly to the motor shaft.

Gearboxes offer a significant benefit in that they affect the inertia ratio by a factor of the gearbox ratio squared. Thus to find what gearhead size is needed, we take √(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. Recall that with a gearhead, torque rises and output shaft speed drops with the gear ratio. You can now size the gearhead to a motor by figuring 66 oz-in. × 1.5 (safety factor) = 100 oz-in. of output torque from the gearhead. This gives 100 oz-in./7 = 14oz-in. from the motor through the gearbox and 37 rpm × 7 = 259 rpm from the motor.

In this case, the rpm and torque are more than is required. The controller can fine-tune the shaft speed and torque requirements to reach the final values.