Q: What is the difference between sizing a load for static versus dynamic holding?

A: Torque can informally be thought of as a “rotational force” or “angular force” that causes a change in rotational motion. In static-holding applications, the torque needed for a brake to hold the load is the result of linear force multiplied by a radius.

There are several applications that may need special calculations, but a few examples and guidelines illustrate how to properly size a brake for a broad array of static-holding scenarios.

1. The brake specified should always be at the high-speed end of the anticipated needs if the application has a motor, gearbox, and drive. The brake should also mount at the motor, rather than at the load, which allows use of a smaller brake.

To calculate the necessary torque to hold the load, work backwards from the load. As an example, suppose the gearbox-output torque T is 250 lb-in. and that level is required to drive the load. Then the holding torque *T _{H}* would be:

* T _{H} = (T/G_{R}) K*

where *G _{R}* = gear ratio and

*K*= service factor. The service factor is a way to account for brake wear. The higher the factor, the less the wear. It should be 1.5 to 3 depending on application data. If

*K*is 1.5, then

* T _{H} = *250/10 1.5 = 25 1.5 = 37.5 lb-in

*.*

2. Now consider applications where a motor is driving the load through a ball screw or leadscrew. A ball screw is more efficient, so it will probably need more holding torque than a leadscrew. But the needed holding torque demanded of the brake will vary depending on the type of ball or leadscrew. The determination of the proper brake size may involve efficiency, inertia, speed, and several other factors too numerous to go into here. But the end result should be multiplied by a service factor, usually 1.5. There is a tool at linearmotioneering.com that walks through the calculations for this case.

3. In some applications the brake must mount on an arm to keep a load from falling when power is removed. Simply calculate the torque needed to hold the weight of the arm plus the load (force distance). Again, then multiply by a service factor of 1.5.

4. In applications where you know the motor’s horsepower but little idea of the loads or holding torque, use a general formula to get in the right ballpark of brake selection:

*T _{H}* = 1.25 63,024 (

*(P K*)/ω)

where 1.25 = a factor added to account for holding the load, in addition to stopping; P = motor output power, hp; ω = motor speed, rpm; and *K* = the service factor which can be 1.5 to 3. The 63,024 is a conversion factor to give torque in lb-in. It is the product of 5,252 rpm/lb-ft 12 in./ft.

*— Rocco Dragone *

**Danaher Motion Corp**. Got a question about motion control or mechatronics?

**Ask**Rocco via e-mail at Editorial@ DanaherMotion.com.