The efficiency of a gear or belt-driven system is calculated by dividing useful output power by input power and multiplying by :
Efficiency is always below in a real mechanical system because some energy is dissipated through friction, belt slippage, vibration, sound, deformation, and heat.
Efficiency compares power transfer, not simply rotational speeds. Define the transmission boundary: input is the power delivered to the driving gear or pulley, while useful output is the power available at the driven shaft. Both values must describe the same operating period and use watts.
Calculating Input and Output Power
For a rotating shaft, calculate mechanical power using:
where is power in watts, is torque in newton metres, and is angular velocity in radians per second. If rotational speed is given in revolutions per minute, convert it using:
| Given information | Power calculation |
|---|---|
| Torque and angular velocity | |
| Force and linear belt speed | |
| Input and output power |
For example, a motor supplies to a gear system. The output shaft produces a torque of at , so:
If output torque and rotational speed are both provided, do not compare either value alone with the input. A transmission can increase torque while reducing angular velocity, or increase angular velocity while reducing torque. In an ideal system, these changes balance so power remains constant; losses make actual output power lower.
For a multi-stage transmission, multiply the efficiencies as decimals. Two gear stages that are each efficient have an overall efficiency of , or .
A common misconception is that the gear ratio directly gives efficiency. Gear ratio determines the theoretical relationship between speed and torque; efficiency measures how much input power reaches the output.
Exam Technique
For B3.3, state the efficiency formula, calculate power consistently, show units, and express the final answer as a percentage. Clearly identify the system boundary so that input and useful output power refer to the same transmission system.
