Three gears with a molecule m 10 mm and a pressure angle of
Solution
Speed ratio of a gear system = No. of teeth on the output gear/No. of teeth on the input gear.
Since we have a train of gears, we have the input-idler system and the idler output system and the final ratio being the multiplication of them both.
Hence, speed ratio of the system = 48/12 *12/24 = 2
Hence for every two rotations of the input shaft, the output shaft rotates once.
Given input speed = 1800 rpm
Therefore, the output speed is 900 rpm = 2pi x 900/60 =30pi
But the power transmitted = 7460 watt = torque x angular velocity
Hence, torque = 7460/30pi = 79.12 N-m
Given module of the gear = 10 mm = D/T
where D = diameter and T = No. of teeth
Hence, diameter of pinion = 10x24 = 240 mm
Diameter of idler gear = 10x12 = 120 mm
Diameter of final gear = 10x48 = 480 mm
Hence, the center distance = 240/2 + 120 + 480/2 = 480 mm
