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 ‘gears’
Determining the Gear Train Tradeoff of Torque vs. Speed, Part One
Friday, August 15th, 2014The Gear Train Tradeoff
Tuesday, August 5th, 2014
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. _______________________________________

How 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. _______________________________________

The 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. _______________________________________

Gear Trains Are Torque Converters and Why That’s Important
Friday, June 13th, 2014
Today we’ll analyze the relationship between the sizes of the gears within a gear train, as well as their torques, and get an understanding of how gear trains act as torque converters. Last time we developed a single torque equation for a simple gear train: T_{1 } ÷ D_{1} = T_{2 } ÷ D_{2} (1) where T_{1} and T_{2} are the driving and driven gear torques, and D_{1} and D_{2} are the driving and driven gear pitch circle radii. The first thing to be done in order to arrive at a torque conversion analysis is group together like terms in equation (1) so that we end up with terms relating to torque on one side of the equation and terms relating to gear size on the other. We’ll use algebra to divide both sides of the equation by T_{2 }, then multiply both sides by D_{1}. After doing so we get, T_{1} ÷ T_{2} = D_{1} ÷ D_{2} (2) In equation (2), T_{1 } ÷ T_{2} is a torque ratio. Ratio means we’re dividing one of the torques by the other. Likewise, D_{1 ÷} D_{2} is a gear pitch radius ratio, that is to say, it’s the ratio of one gear’s physical size relative to the physical size of the other. What equation (2) tells us is that the individual gears on the gear train will produce torque values which are dependent upon the physical sizes of the two gears with respect to one another. So what’s the practical significance of this? When gear trains are used in industrial applications, they always act as torque converters. One such example would be when a lowtorqueproducing electric motor is used to power a steel cutting lathe. If the motor isn’t tough enough to power the lathe, it itself won’t be modified, but the torque it produces will be. This modification is accomplished by converting the motor’s low torque value, T_{1}, to a higher torque value, T_{2} , and it’s equation (2) that’s used to do it. Next time we’ll delve deeper into the methodology behind gear train torque conversions. _______________________________________

Vectors, Sin(ϴ), and the Torque Formula
Wednesday, March 26th, 2014
Last time we introduced a physics concept known as torque and how it, together with modified gear ratios, can produce a mechanical advantage in devices whose motors utilize gear trains. Now we’ll familiarize ourselves with torque’s mathematical formula, which involves some terminology, symbols, and concepts which you may not be familiar with, among them, vectors, and sin(ϴ). Torque = Distance × Force × sin(ϴ) In this formula, Distance and Force are both vectors. Generally speaking, a vector is a quantity that has both a magnitude — that is, any measured quantity associated with a vector, whether that be measured in pounds or inches or any other unit of measurement — and a direction. Vectors are typically represented graphically in engineering and physics illustrations by pointing arrows. The arrows are indicative of the directionality of the magnitudes involved. Sin(ϴ), pronounced sine thaytah, is a function found within a field of mathematics known as trigonometry , which concerns itself with the lengths and angles of triangles. ϴ, or thaytah, is a Greek symbol used to represent the angle present between the Force and Distance vectors as they interact to create torque. The value of sin(ϴ) depends upon the number of degrees in the angle ϴ. Sin(ϴ) can be found by measuring the angle ϴ, entering its value into a scientific calculator, and pressing the Sin button. We’ll dive into the math behind the vectors next time, when we return to our wrench and nut example and apply vector force quantities.
_______________________________________ 
Achieving Mechanical Advantage Through Torque
Wednesday, March 19th, 2014
Last time we saw how gear train ratios allow us to change the speed of the driven gear relative to the driving gear. Today we’ll extend this concept further and see how gear trains are used to amplify the mechanical power output of small motors and in so doing create a mechanical advantage, an advantage made possible through the physics of torque. Below is an ordinary electric drill. Let’s see what’s inside its shell. There’s a whole lot of mechanical advantage at work here, giving the drill’s small motor the ability to perform big jobs. A motor and gear train are housed within the drill itself. The motor shaft is coupled to the chuck shaft via the gear train, and by extension, the drill bit. A chuck holds the drill bit in place. It’s the drill’s gear train that provides the small motor with the mechanical advantage necessary for this handheld power tool to perform the big job of cutting through a thick steel plate. If the gear train and its properly engineered gear ratio weren’t in place and the chuck’s shaft was connected directly to the motor shaft, the motor would be overwhelmed and would stall or become damaged. Either way, the work won’t get done. To understand how operations like these can be performed, we must first familiarize ourselves with the physics concept of torque. Torque allows us to analyze the rotational forces acting upon rotating objects, such as gears in a gear train and wrenches on nuts and bolts. Manipulating torque allows us to achieve a physical advantage when rotating objects around a pivot point. Let’s illustrate this by using a wrench to turn a nut. The nut is fastened to the bolt with threads, interconnecting spiral grooves formed on both the inside of the nut and the outside of the bolt. A wrench is used to loosen and tighten the nut by rotating it on its mating threads. The nut itself rotates about a pivot point which lies at its center. When you use your arm to manipulate the wrench you apply force, a force which is transmitted at a distance from the pivot point. This in turn creates a torque on the nut. In other words, torque is a function of the force acting upon the handle relative to its distance from the pivot point at the center of the nut. Torque can be increased by changing one or both of its acting factors, force and distance. We’ll see how next time when we examine the formula for torque and manipulate it so that a weak arm can loosen even the tightest nut. _______________________________________ 
Spur Gears In Motion
Wednesday, February 12th, 2014
Last time we learned about forces generated when spur gear teeth mesh and move along a specific line of action. Today we’ll see them in movement. Looking at the illustration below it might appear that there are three teeth in contact, but this isn’t the case. As the gears rotate, only two teeth make contact at any given time, although the third tooth comes very close. The actual point of contact between the teeth is represented by a black dot on the illustration below. This is where two opposing forces, F_{1 }and_{ }F_{2}_{, }meet. Now let’s animate the illustration to see how the line of action remains constant the entire time the gear teeth are in motion. By constant I mean that this imaginary line’s position and angle does not change relative to the gears throughout the course of their movement. In the animation, the point of contact moves along the line of action as the gears turn. Each tooth’s involute profile ensures that smooth contact is maintained along the faces and flanks of the gear teeth. The involute profile’s unique shape facilitates opposing teeth remaining in constant contact along the line of action for the duration of their movement together. If the gear tooth profile wasn’t involute in its shape, say for example it was square or triangular, the forces acting upon the meshed teeth during movement would vary in direction and intensity as a result of uneven contact between the teeth. For example, consider the square shaped tooth profile in the gear train below. As the gears rotate, the pointed tip of one tooth strikes the flat face of another. As they continue to turn, the two flat faces of the teeth slap together, then the pointed tip of one tooth will strike the flat face of the other tooth, and so forth. The result is movement that is jerky and destructive. There would be excessive vibration and wear and tear on the teeth, resulting in rapid gear tooth erosion and decreased efficiency overall. Next time we’ll introduce the gear ratio, a formula which allows us to alter the rotational speed of the driven gear in relation to that of the driving gear, something which comes in handy when designing things that require this differential.
_______________________________________ 
Overcoming Inertia
Monday, February 3rd, 2014
Inertia. It’s the force that keeps us in bed after the alarm has rung. It seems to have a life of its own, and today we’ll see how it comes into play in keeping other stationary objects at rest. Last time we identified a specific point of contact between spur gear teeth in a gear train and introduced the opposing forces, F_{1 } and F _{2}, generated there. Today we’ll see what these forces represent, identifying one of them as inertia. So where do these forces come from? They’re forces generated by different means that converge at the same point of contact, the point at which gear teeth mesh. They follow a very specific geometric path to meet there, an imaginary straight line referred to as the line of action. F_{1} is always generated by a source of mechanical energy. In our locomotive example introduced earlier in this blog series that source is an electric traction motor, upon which a driving gear is mounted. When the motor is energized, a driving force F_{1} is generated, which causes gear teeth on the driving gear to push against gear teeth of the driven gear. Force F_{2} is not as straightforward to understand, because it’s not generated by a motor. Instead, it’s the resisting force that the weight of a stationary object poses against its being moved from an atrest position, known as inertia. The heavier the object, the more inertia it presents with. Trains, of course, are extremely heavy, and to get them to move a great deal of inertia must be overcome. Inertia is also a factor in attempting to stop objects already in motion. To get a stationary locomotive to move, mechanical energy must be transmitted from the driving gear that’s attached to its traction motor, then on to the driven gear attached to its axle. At their point of contact, the driving force of the motor, F_{1,} is met by the resisting force of inertia, F_{2. }In order for the train to move, F_{1} must be greater than F_{2}. If F_{1} is less than or equal to F_{2}, then the train won’t leave the station. Next week we’ll animate our static image and watch the interplay between gear teeth, taking note of the line of action during their movement.
_______________________________________ 
Meshed Gear Teeth and Their Point of Contact
Monday, January 27th, 2014
Last time we learned that the geometric shape specific to spur gear teeth is known as an involute profile. Today we’ll look at the geometry behind this profile and the very specific place at which gear teeth meet, known as the point of contact. The transmission of mechanical energy between meshed gears may seem on its face to be straightforward, after all their gears are interlaced and interact with one another. But their interaction involves some rather complex geometry, because forces are directed in a peculiar fashion between the teeth of the driving and driven gears. Let’s consider the following illustration to get a better understanding.
As we learned previously in this series, the pitch circle of a gear is an imaginary arc passing through each tooth between their top and bottom lands. The pitch circles of the driving and driven gears are represented by heavy red dashed lines in the illustration. To ensure proper alignment and smooth action between gear teeth during rotation, the gears are spaced so that their pitch circles just meet but never intersect. This specific point is known as the point of contact. It is the only point at which gears will come into actual physical contact with one another, and it provides just enough contact so that when the driving gear turns in one direction, say clockwise, its teeth exert pressure upon the driven gear teeth, forcing it to move in the opposite direction, counterclockwise. The forces which come into play at the point of contact are represented in the illustration by a black dot with oppositional blue arrows extending from it. These arrows represent the opposing mechanical forces, F_{1} and F_{2 ,} which act upon the teeth when they make contact. We’ll learn more about the effect of those forces next time when we follow a locomotive from a stationary position into one of movement.
_______________________________________ 