A. MECHANICALLY

Braking systems use devices to gain a mechanical advantage. The most common device for this purpose is leverage. Look at this simple lever system;

A lever is placed on a pivot called the fulcrum. As the distance from A to C is four feet, and from C to B is one foot, the ratio is four to one (4:1). Power has been multiplied by the leverage principle. If a 100 lb. downward force is applied at point A, then the upward force at point B is 400 lbs.

This is the result of the mechanical advantage of leverage.

Compare the Points A, C, B to the previous lever diagram.

B. USE OF AIR

Force can also be multiplied by the use of air to gain a further mechanical advantage. Everyone has felt the power of air on a windy day. Air can be compressed (squeezed) into a much smaller space than that amount of air normally would occupy. For instance, air is compressed in tires to support the weight of a vehicle. The smaller the space into which air is squeezed, the greater the air’s resistance will be to being squeezed. This resistance creates pressure, which is used to gain mechanical advantage.

If a constant supply of compressed air were directed through a pipe that was one inch square, and if a one inch square plug were placed in the pipe, the compressed air would push against the plug. Holding a scale against the plug would register how many pounds of force were being exerted by the air against the plug.

If the scale registered 10 pounds, for example, then it could be said the force was 10 pounds on the one square inch surface of the plug. This would be 10 pounds per square inch (P.S.I.).

The more the air in the supply tank has been compressed, the greater the force that would be exerted on the face of the plug.

C. LEVERAGE & AIR PRESSURE

In actual operation, pipes are round and plugs are diaphragms of flexible material acting against push rods. If compressed air of 120 pounds per square inch (P.S.I.) acts on a diaphragm of 30 square inches, 3,600 lbs. of force is produced (120 x 30). Apply this force to a push rod to move a six-inch slack adjuster operating a cam and the total force equals 21,600 inch pounds torque (3,600 x 6), or 1,800 foot pounds torque (21,600 ÷ 12). It requires 25 – 30 foot pounds of torque to tighten the wheel on a car. This comparison illustrates the power obtained from using mechanical leverage and air pressure combined.