Already we learned that we can get more torque out of a motor by using one of two methods. In the first method we attach a gear train to the motor, then try different gear sizes until we arrive at the desired torque for the application. In the second method we eliminate the gear train altogether and simply use a higher horsepower motor to give us the torque we need. Here we’ll explore the second method. We’ll use the equation presented in our last blog to determine torque, T, relative to a motor’s horsepower, HP, when the motor operates at a speed, n:T = [HP ÷ n] × 63,025 In earlier topic in this series we employed a gear train attached to an electric motor to power a lathe. It provided an insufficient 200 inch pounds of torque when 275 is required. Let’s use the equation and a little algebra to see how much horsepower this motor develops if it turns at a speed of 1750 RPM, a common speed for alternating current (AC) motors:200 inch pounds = [HP ÷ 1750 RPM] × 63,025200 = HP × 36.01HP = 200 ÷ 36.01 = 5.55 horsepower For the purpose of our example here, let’s say we’ve nonsensically decided not to use a gear train, leaving us with no choice but to replace the underpowered motor with a more powerful one. So let’s see what size motor we’ll need to provide us with the required horsepower of 275 inch pounds. |
Posted inStrength of Materials