The optimal size for a clutch or brake is determined by three things: Required torque, thermal horsepower per engagement, and average required thermal horsepower. Here, we'll discuss the first, learning to calculate torque capacity under static and dynamic conditions.

Dynamic torque

To accurately determine the torque required during acceleration or deceleration, it is necessary to know total inertia, component efficiency, total load torque, and the amounts reflected back to the clutch/brake output shaft. A major consideration is inefficiencies; individual drive components and their power losses during acceleration or deceleration put significant burden on a system — which the clutch or brake must overcome. The total inertial torque is the sum of all the torques of the individual drive components. The dynamic torque is then found by adding total inertial torque and load torque together. With drive efficiency E, load torque T_L, and inertial torque T_i dynamic torque is calculated:

Note that efficiency and load are always factors in a system, but the inertial term only applies when the system is accelerating or decelerating.

Determining inertia

Inertia is the measure of an object's resistance to changes in motion. Rotational inertia is a function of an object's mass and how it's distributed about the rotating axis. The effective radius is where the entire mass of the object is taken to be concentrated. Called the radius of gyration, this is determined by an object's geometry and is designated K. The first step in calculating dynamic torque is to determine the inertia (in the system) that must be accelerated or decelerated. The term used to quantify it is WK². To obtain the WK² in a form that can be used, calculate the inertia of each component in the system that will be cycled — and the inertia reflected from each cycled component back to the clutch and brake. An equivalent WK² can then be attached to the output of the clutch/brake to represent the inertia of all cycled components in the system.

Learning by example

Here we calculate the inertia of several objects involved in a motion system; then we calculate how much of that inertia is seen by the clutch/brake. Assume for the conveyor system on page 26 that:

Acceleration time = 0.4 sec

Deceleration time = 0.13 sec

Cycles per min. = 10

Conveyor efficiency = 0.8

Chain drive efficiency = 0.9

Reducer efficiency = 0.8

Maximum pressure = 60 psig

The procedure to obtain the equivalent or reflected inertia is based on the principle that the total energy in the system is conserved. This means that the reflected inertia of an object is equal to the actual kinetic energy it possesses in the drive system. Since the kinetic energy varies with the square of the speed, the reflected inertia (for the boxes in our example) is the object's actual inertia on the square of the ratio of the operating speed to clutch/brake speed.

Because the weight of each box in our example system is significant, the inertia they reflect to the clutch/brake is also significant. In contrast, the amount of weight the actual conveyor belt sends to the clutch/brake will be small. Nonetheless, the steel pulleys can still affect the overall system.

The sprockets on our system, though attached to the pulleys, will have a different affect on the clutch/brake because of their larger geometry. Using the same general formulas, its calculated weight is 266.72 lbs. and its inertia 0.232 lb-ft². Its inertia is reflected through the chain drive and reducer; its larger diameter reflects more inertia back to the clutch/brake. In contrast, the smaller sprocket on our system contributes significantly less inertia: From the same formulas used to calculate pully and large sprocket values, the small sprocket weighs 66.65 lb with an inertia reflected of 0.06 lb-ft².

After these inertia values are calculated, those of other standard system components can often be obtained from manufacturers. Assuming typical inertia values for reducers, couplings, and clutch/brakes involved, the values for WK² might be 0.17, 0.78, and 0.10 lb-ft² respectively. From these values, the final, total system-reflected inertial torque for our example system is 0.868 + 0.232 + 0.06 + 0.17 + 0.78 + 0.10 = 2.27 lb-ft².