We’ve learned several methods to increase the torque of an electric motor through our series of articles on gear trains. One way is to attach a gear train to the motor’s shaft, a relatively simple thing to do. Today we’ll begin our exploration into how this method involves a tradeoff. It comes at the cost of speed. We’ll begin our examination of the tradeoff at play by linking together key elements learned through past blogs on the subject of gear trains. We’ll revisit those lessons through flashbacks. The first flashback we’ll make is to a blog entitled, Gear Ratio Formulas. There we learned that within a simple gear train consisting of two gears, the type most commonly employed to manipulate a motor’s torque, the ratio between the two gears, R, is relative to the ratio of their gear teeth, N. N is determined by the number of teeth each gear has in combination with the speeds, n, that each gear is going: R = N_{Driven} ÷ N_{Driving } = n_{Driving} ÷ n_{Driven} (1) The second flashback we’ll make is to a blog entitled, The Methodology Behind Gear Train Torque Conversions, in which we learned that the ratio of the torque, T, that exists between the gears is relative to the ratio of their respective pitch diameters, D: T_{Driving} ÷ T_{Driven} = D_{Driving} ÷ D_{Driven } (2) The tradeoff we’ve been alluding to comes in when gear speed, n_{Driven}, represented in equation (1), is decreased, which results in an increase to T_{Driven} in equation (2). But in order to see this we’ve got to somehow link the two equations together. In their present form there’s no common link between them. Or is there? There actually is an indirect link between the two equations, which comes by way of the torque equation presented in another past blog. The third flashback we’ll make is to the blog discussing that subject, which is entitled, The Relationship Between Torque and Horsepower. Using facts presented in that blog, the torque equations for our two gears become: T_{Driving} = [HP_{Driving} ÷ n_{Driving}] × 63,025 (3) T_{Driven} = [HP_{Driven} ÷ n_{Driven}] × 63,025 (4) Where’s the link between equations (1) and (2)? To answer that question we must reference a physics law known as The Law of Conservation of Energy . It states that the energy flowing from one gear to another within a gear train remains constant. Energy equates to horsepower, HP, in equations (3) and (4). So if the horsepower flowing through the gears is equal, our working equation becomes: HP_{Driving} = HP_{Driven } (5) Next time we’ll see how equation (5) is key to linking together equations (1) and (2) by way of equations (3) and (4). In so doing we’ll disclose the tradeoff to using gear trains. _______________________________________

Posts Tagged ‘torque equation’
The Gear Train Tradeoff
Tuesday, August 5th, 2014Equations Used to Manipulate Torque Relative to Horsepower
Monday, July 28th, 2014
Last time 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. Today 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 blogs 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] × 63,025 200 = HP × 36.01 HP = 200 ÷ 36.01 = 5.55 horsepower For the purpose of our example today, 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. Using the torque equation and plugging in numbers already provided our equation becomes: 275 inch pounds = [HP ÷1750 RPM] × 63,025 275 = HP × 36.01 HP = 275 ÷ 36.01 = 7.64 horsepower This tells us that we need to replace the 5.55 horsepower motor with a 7.65 horsepower motor. As you might have guessed, the higher the motor’s horsepower, the larger that motor’s size and weight — if you’re not using a gear train, that is. Bigger, bulkier motors cost more to purchase and operate and also take up more space, which often makes them impractical to use. All this translates to the reality that sometimes it just makes more sense to use a gear train to provide more torque. It’s a lot easier and cheaper to attach a gear train to a motor and manipulate its gear sizes to arrive at desired torque than it is to buy a bigger motor. You may have deduced by now that it’s relatively easy to get more torque. Almost too easy. Next time we’ll see how increased torque comes at another type of cost, the cost of speed. _______________________________________
