Hold onto your hats, we’re going to deal with a lot of equations today! Last time we used flashbacks to previous blogs in this series to revisit key equations in our ongoing discussion of gear trains and torque. We also introduced The Law of Conservation of Energy in conjunction with five equations that together demonstrate how when increasing torque by use of a simple gear train, we do so at the cost of speed. Those five equations are:
where, R is the gear ratio, N the number of gear teeth, n the gear’s rotational speed, T the torque, D the gear pitch radius, and HP is the horsepower transmitted by the gears. As we work the equations, keep in mind that our ultimate objective is to find a way to link together (1) and (2), the equations dealing with gear torque and speed. Once we accomplish this we’ll see how increased torque is obtained at the cost of speed. But because there are no common terms between equations (1) and (2), our first step is to develop one. Developing a link between equations (1) and (2) is a process that begins with combining equations (2), (3), and (4).
The common terms in these three equations are T_{Driving} and T_{Driven, }so we’ll manipulate things in order to group them together. We’ll substitute equation (3) for the T_{Driving} term in equation (2), and substitute equation (4) for the T_{Driven} term in equation (2). We are now able to link all three equations to get: {[HP_{Driving} ÷ n_{Driving}] × 63,025} ÷ {[HP_{Driven} ÷ n_{Driven}] × 63,025} = D_{Driving} ÷ D_{Driven} (6) Now let’s go a step further to simplify equation (6). From equation (5) we know that the driving and driven gear horsepowers are equal. So, in equation (6), the HP_{Driving} and HP_{Driven} cancel out, along with the two 63,025 terms, allowing us to arrive at equation (7): {[HP_{Driving} ÷ n_{Driving}] × 63,025} ÷ {[HP_{Driven} ÷ n_{Driven}] × 63,025} = D_{Driving} ÷ D_{Driven}
Next week we’ll use equation (7) to link together R, N, n, of equation (1) with D and T of equation (2) and in so doing disclose mathematically the tradeoff between torque and speed, then apply our findings to an example. _______________________________________

Posts Tagged ‘pitch diameter’
Determining the Gear Train Tradeoff of Torque vs. Speed, Part One
Friday, August 15th, 2014How to Increase Gear Train Torque
Thursday, July 10th, 2014
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. _______________________________________

Determining Torque Within a Gear Train
Monday, June 30th, 2014
Last time we set up an example where an electric motor is connected to a lathe via a gear train. Today we’ll take the numerical values present on that gear train and plug them into the torque ratio equation we’ve been working with for the past few blogs. In the illustration below the electric motor exerts 200 inch pounds of torque upon the driving gear. The driving gear pitch circle diameter is 6 inches, while the driven gear pitch circle diameter is 8 inches. It’s been determined through previous lab testing that the lathe we’ll be using requires at least 275 inch pounds of torque to be exerted upon the driven gear shaft in order to operate properly. Will the gear train shown below meet this requirement? First, a review of the torque ratio equation: T_{1 }÷ T_{2} = D_{1 }÷ D_{2} Now we’ll crunch numbers. T_{1} is equal to 200 inch pounds, D_{1} is equal to 3 inches (pitch radius equals pitch diameter divided by two), and D_{2} is equal to 4 inches. This gives us: (200 inch pounds) ÷ T_{2} = (3 inches) ÷ (4 inches) T_{2} = (200 inch pounds) ÷ (0.75) = 266.67 inch pounds So, does the gear train as presented here supply enough torque to power the lathe properly? No, it does not. It provides only 266.67 inch pounds, not the 275 inch pounds of torque required. Next time we’ll see how to manipulate gear sizes within a gear train in order to meet a given torque requirement. _______________________________________
