Last week we worked with a gear train equation and found that the gears under consideration were not sized properly to run a lathe. Today we’ll increase the gear train torque and solve that problem. How do we manipulate things to obtain the 275 inch pounds of torque required to drive the lathe? Last week we tried using a driven gear with a diameter of 8 inches and found that to be insufficient in size. So today the first thing we’ll try is a bigger driven gear, one with a pitch diameter of 8.5 inches. That’s 0.5 inches larger in diameter than the gear used in last week’s equation, and this just so happens to be the next size up in the gear manufacturer’s catalog. As we did last week, we’ll begin our calculations with the torque ratio equation: T_{1 } ÷ T_{2} = D_{1 } ÷ D_{2} We’ll use the same values as last week for T_{1}, and D_{1}, 200 inch pounds and 3 inches respectively, but we’ll increase the new value for D_{2}, the driven gear pitch radius, to 4.25 inches (the new pitch diameter divided by two). Inserting these values into the torque equation, the only variable remaining without a value is torque T_{2}. Let’s determine that value now by using algebra to rearrange terms. (200 inch pounds) ÷ T_{2} = (3 inches) ÷ (4.25 inches) (200 inch pounds) ÷ T_{2} = 0.70 T_{2} = (200 inch pounds) ÷ (0.70) = 283.33 inch pounds The value of T_{2 }is found to be 283.33 inch pounds, which meets the torque requirement required to run the lathe. We were able to arrive at this torque by simply increasing the size of the driven gear relative to the size of the driving gear. In the world of Newtonian physics, this is a rather straightforward arrangement. It all boils down to this simple dynamic: When the motor’s force is acting upon a wider gear, the force is located a longer distance from the center of the driving gear shaft, which results in more torque on the shaft. As borne out by the example provided today, the larger the driven gear is in comparison to the driving gear, the more the gear train amplifies the torque that’s delivered by the motor. The principle at play here is exactly the same as that presented in a previous blog article where, for a given force exerted upon a wrench, torque was increased by simply increasing the length of the wrench handle. Some of you may be wondering why we didn’t just use a bigger, more powerful motor to begin with, thereby eliminating the need for a gear train and all the calculations we’ve been running? We’ll see why that’s not always possible or practical next time. _______________________________________

Posts Tagged ‘engine lathe’
How to Increase Gear Train Torque
Thursday, July 10th, 2014The Methodology Behind Gear Train Torque Conversions
Sunday, June 22nd, 2014
Last time we learned that gear trains are torque converters, and we developed a torque ratio equation which mathematically ties the two gears in a gear train together. That equation is: T_{1 } ÷ T_{2} = D_{1} ÷ D_{2} Engineers typically use this equation knowing only the value for T_{2}, the torque required to properly drive a piece of machinery. That knowledge is acquired through trial testing during the developmental phase of manufacturing. Once T_{2} is known, a stock motor is selected from a catalog with a torque value T_{1} which closely approximates that of the required torque, T_{2}. Then calculations are performed and lab tests are run to determine the driving and driven gear sizes, D_{1} and D_{2 } which will enable the gear train to convert T_{1} into the required value of T_{2}. This series of operations are often a time consuming and complex process. To simplify things for the purpose of our example, we’ll say we’ve been provided with all values required for our equation, except one, the value of T_{2}. In other words, we’ll be doing things in a somewhat reverse order, because our objective is simply to see how a gear train converts a known torque T_{1} into a higher torque T_{2}. We’ll begin by considering the gear train illustration above. For our purposes it’s situated between an electric motor and the lathe it’s powering. The motor exerts a torque of 200 inch pounds upon the driving gear shaft of the lathe, a torque value that’s typical for a mid sized motor of about 5 horsepower. Asis, this motor is unable to properly drive the lathe, which is being used to cut steel bars. We know this because lab testing has shown that the lathe requires at least 275 inch pounds of torque in order to operate properly. Will the gears on our gear train be able to provide the required torque? We’ll find out next time when we insert values into our equation and run calculations. _______________________________________
