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We’ve been working towards a general understanding of how gear trains work, and today we’ll solve a final piece of the puzzle when we identify how increased gear train torque is gained at the expense of gear train speed. Last time we developed a mathematical relationship between the torque, T, and the rotational speed, n, of the driving and driven gears in a simple gear train. This is represented by equation (8): TDriven ÷ TDriving = nDriving ÷ nDriven (8) For the purpose of our example we’ll assume that the driving gear is mounted to an electric motor shaft spinning at 100 revolutions per minute (RPM) and which produces 50 inch pounds of torque. Previous lab testing has determined that we require a torque of 100 inch pounds to properly run a piece of machinery that’s powered by the motor, and we’ve decided that the best way to get the required torque is not to employ a bigger, more powerful motor, but rather to install a gear train and manipulate its gear sizes until the desired torque is obtained. We know that using this approach will most likely affect the speed of our operation, and we want to determine how much speed will be compromised. So if the torque on the driven gear needs to be 100 inch pounds, then what will be the corresponding speed of the driven gear? To answer this question we’ll insert the numerical information we’ve been provided into equation (8). Doing so we arrive at the following: TDriven ÷ TDriving = nDriving ÷ nDriven (100 inch pounds) ÷ (50 inch pounds) = (100 RPM) ÷ nDriven 2 = (100 RPM) ÷ nDriven nDriven = (100 RPM) ÷ 2 = 50 RPM This tells us that in order to meet our torque requirement of 100 inch pounds, the gear train motor’s speed must be reduced from 100 RPM to 50 RPM, which represents a 50% reduction in speed, hence the tradeoff.
This wraps up our blog series on gears and gear trains. Next time we’ll move on to a new topic: Galileo’s experiments with falling objects.
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Posts Tagged ‘gear teeth’
Determining the Gear Train Tradeoff of Torque vs. Speed, Part Three
Wednesday, August 27th, 2014
The 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 = NDriven ÷ NDriving = nDriving ÷ nDriven (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: TDriving ÷ TDriven = DDriving ÷ DDriven (2) The tradeoff we’ve been alluding to comes in when gear speed, nDriven, represented in equation (1), is decreased, which results in an increase to TDriven 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: TDriving = [HPDriving ÷ nDriving] × 63,025 (3) TDriven = [HPDriven ÷ nDriven] × 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: HPDriving = HPDriven (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. _______________________________________
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Gear Train Torque Equations
Thursday, May 22nd, 2014|
In our last blog we mathematically linked the driving and driven gear Force vectors to arrive at a single common vector F, known as the resultant Force vector. This simplification allows us to achieve common ground between F and the two Distance vectors of our driving and driven gears, represented as D1 and D2. We can then use this commonality to develop individual torque equations for both gears in the train.
In this illustration we clearly see that the Force vector, F, is at a 90º angle to the two Distance vectors, D1 and D2. Let’s see why this angular relationship between them is crucial to the development of torque calculations. First a review of the basic torque formula, presented in a previous blog, Torque = Distance × Force × sin(ϴ) By inserting D1, F, and ϴ = 90º into this formula we arrive at the torque calculation, T1 , for the driving gear in our gear train: T1 = D1 × F × sin(90º) From a previous blog in this series we know that sin(90º) = 1, so it becomes, T1 = D1 × F By inserting D2, F, and ϴ = 90º into the torque formula, we arrive at the torque calculation, T2 , for the driven gear: T2 = D2 × F × sin(90º) T2 = D2 × F × 1 T2 = D2 × F Next week we’ll combine these two equations relative to F, the common link between them, and obtain a single equation equating the torques and pitch circle radii of the driving and driven gears in the gear train. _______________________________________ |
The Mathematical Link Between Gears in a Gear Train
Wednesday, May 14th, 2014|
Last time we analyzed the angular relationship between the Force and Distance vectors in this simple gear train. Today we’ll discover a commonality between the two gears in this train which will later enable us to develop individual torque calculations for them.
From the illustration it’s clear that the driving gear is mechanically linked to the driven gear by their teeth. Because they’re linked, force, and hence torque, is transmitted by way of the driving gear to the driven gear. Knowing this we can develop a mathematical equation to link the driving gear Force vector F1 to the driven gear Force vector F2, then use that linking equation to develop a separate torque formula for each of the gears in the train. We learned in the previous blog in this series that F1 and F2 travel in opposite directions to each other along the same line of action. As such, both of these Force vectors are situated in the same way so that they are each at an angle value ϴ with respect to their Distance vectors D1 and D2. This fact allows us to build an equation with like terms, and that in turn allows us to use trigonometry to link the two force vectors into a single equation: F = [F1 × sin(ϴ)] – [F2 × sin(ϴ)] where F is called a resultant Force vector, so named because it represents the force that results when the dead, or inert, weight that’s present in the resisting force F2 cancels out some of the positive force of F1. Next week we’ll simplify our gear train illustration and delve into more math in order to develop separate torque computations for each gear in the train. _______________________________________ |
Distance and Force Vectors of a Simple Gear Train
Monday, May 5th, 2014|
Last time we examined how torque and force are created upon the driving gear within a simple gear train. Today we’ll see how they affect the driven gear.
Looking at the gear train illustration above, we see that each gear has both distance and force vectors. We’ll call the driving gear Distance vector, D1, and the driven gear Distance vector, D2. Each of these Distance vectors extend from pivot points located at the centers of their respective gear shafts. From there they extend in opposite directions until they meet at the line of action, the imaginary line which represents the geometric path along which Force vectors F1 and F2 are aligned. As we learned last time, the Force vector, F1, results from the torque that’s created at the pivot point located at the center of the driving gear. This driving gear is mounted on a shaft that’s attached to an electric motor, the ultimate powering source behind the torque. F1 follows a path along the line of action until it meets with the driven gear teeth, where it then exerts its pushing force upon them. It’s met by Force vector F2, a resisting force, which extends along the same line of action, but in a direction opposite to that of F1. These two Force vectors butt heads, pushing back against one another. F2 is essentially a negative force manifested by the dead weight of the mechanical load of the machinery components resting upon the shaft of the driving gear. Its unmoving inertia resists being put into motion. In order for the gears in the gear train to turn, F1 must be greater than F2, in other words, it must be great enough to overcome the resistance presented by F2. With the two Force vectors pushing against each other along the line of action, the angle ϴ between vectors F2 and D2, is the same as the angle ϴ between F1 and D1. Next time we’ll use the angular relationship between these four vectors to develop torque calculations for both gears in the gear train. _______________________________________ |
Torque and Force
Tuesday, April 29th, 2014|
We’ve been discussing torque and how it enables more power to be available to applications such as loosening tight nuts with a wrench. Now we’ll see how those same principles apply to another application, a simple gear train. To review, the torque formula is, Torque = Distance × Force × sin(ϴ) where, Distance and Force are vector magnitudes and ϴ is the angle formed between them.
Referring to the gear train illustration above, we see that Force and Distance vectors are present, just as they had been in our previous wrench/nut example. But instead of torque being created by way of force that’s applied to a wrench, things are reversed, and it’s the torque that creates the force. You see, in the wrench/nut example, the force applied to the wrench handle created torque on the nut. In our present gear train example, the torque applied to the motor shaft is created by an electric motor exerting pressure upon the motor shaft, which in turn exerts a force upon the driving gear teeth. The driving gear is also attached to this shaft, so torque causes the driving gear to rotate along with the motor. This rotation results in a force being exerted at the point where the teeth of the driving gear mesh with the teeth of the driven gear. In other words, in the wrench/nut example force created torque, while in the present example torque creates a force. The gear train has a pivot point, as there was in our wrench/nut example, but this time it’s located at the center of the motor shaft rather than at the center of a nut. The pivot point in both examples is where the action takes place. The motor’s shaft and driving gear rotate around it, just as the wrench jaws and handle rotated around the nut’s pivot point. In both examples, the Distance vectors extend out from the pivot points to meet up with the Force vector’s path. In the gear train example, this Force vector path is called a line of action, as introduced earlier in this blog series. This line of action passes through to the point where the driving and driven gear teeth mesh. The force acting upon that point causes the gears in the gear train to rotate, and as they turn mechanical energy is transferred from the motor to whatever machinery component is attached to the shaft of the driven gear. The powered component will then be able to perform useful work such as cutting lumber, mixing frosting for a cake, drilling holes in steel, or propelling vehicles. You will note that there is an angle ϴ which exists between the Distance and Force vectors. Since we have a pivot point, a Force vector, a Distance vector, and an angle ϴ, we are able to apply the torque formula to gear trains exactly as we did in our wrench/nut example. We can then use that formula to calculate how torque is transmitted between gears in the train. Next time we’ll examine the distance and force vectors in a simple gear train. _______________________________________ |
Gear Reduction Worked Backwards
Sunday, March 9th, 2014|
Last time we saw how a gear reduction does as its name implies, reduces the speed of the driven gear with respect to the driving gear within a gear train. Today we’ll see how to work the problem in reverse, so to speak, by determining how many teeth a driven gear must have within a given gear train to operate at a particular desired revolutions per minute (RPM). For our example we’ll use a gear train whose driving gear has 18 teeth. It’s mounted on an alternating current (AC) motor turning at 3600 (RPM). The equipment it’s attached to requires a speed of 1800 RPM to operate correctly. What number of teeth must the driven gear have in order to pull this off? If you’ve identified this to be a word problem, you’re correct.
Let’s first review the gear ratio formulas introduced in my previous two articles: R = nDriving ÷ nDriven (1) R = NDriven ÷ NDriving (2) Our word problem provides us with enough information so that we’re able to use Formula (1) to calculate the gear ratio required: R = nDriving ÷ nDriven = 3600 RPM ÷ 1800 RPM = 2 This equation tells us that to reduce the speed of the 3600 RPM motor to the required 1800 RPM, we need a gear train with a gear ratio of 2:1. Stated another way, for every two revolutions of the driving gear, we must have one revolution of the driven gear. Now that we know the required gear ratio, R, we can use Formula (2) to determine how many teeth the driven gear must have to turn at the required 1800 RPM: R = 2 = NDriven ÷ NDriving 2 = NDriven ÷ 18 Teeth NDriven = 2 × 18 Teeth = 36 Teeth The driven gear requires 36 teeth to allow the gear train to operate equipment properly, that is to say, enable the gear train it’s attached to provide a speed reduction of 1800 RPM, down from the 3600 RPM that is being put out from the driving gear. But gear ratio isn’t just about changing speeds of the driven gear relative to the driving gear. Next time we’ll see how it works together with the concept of torque, thus enabling small motors to do big jobs.
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Gear Reduction
Wednesday, March 5th, 2014|
Last time we learned there are two formulas used to calculate gear ratio, R. Today we’ll see how to use them to calculate a gear reduction between gears in a gear train, a strategy which enables us to reduce the speed of the driven gear in relation to the driving gear. If you’ll recall from last time, our formulas to determine gear ratio are: R = NDriven ÷ NDriving (1) R = nDriving ÷ nDriven (2) Now let’s apply them to this example gear train to see how a gear reduction works.
Here we have a driven gear with 23 teeth, while the driving gear has 18. For our example the electric motor connected to the driving gear causes it to turn at a speed, nDriving, of 3600 revolutions per minute (RPM). Knowing these numerical values we are able to determine the driven gear speed, nDriven. First we’ll use Formula (1) to calculate the gear ratio using the number of teeth each gear has relative to the other: R = NDriven ÷ NDriving R = 23 Teeth ÷ 18 Teeth R = 1.27 In gear design nomenclature, the gear train is said to have a 1.27 to 1 ratio, commonly denoted as 1.27:1. This means that for every tooth on the driving gear, there are 1.27 teeth on the driven gear. Interestingly, the R’s in both equations (1) and (2) are identical, and in our situation is equal to 1.27, although it is arrived at by different means. In Formula (1) R is derived from calculations involving the number of teeth present on each gear, while Formula (2)’s R is derived by knowing the rotational speeds of the gears. Since R is the common link between the two formulas, we can use this commonality to create a link between them and insert the R value determined in one formula into the other. Since we have already determined that the R value is 1.27 using Formula (1), we can replace the R in Formula (2) with this numerical value. As an equation this looks like: R = 1.27 = nDriving ÷ nDriven Now all we need is one more numerical value to solve Formula (2)’s equation. We know that the speed at which the driving gear is rotating, nDriving , is 3600 RPM. We use basic algebra to calculate the driven gear speed, nDriven : 1.27 = 3600 RPM ÷ nDriven nDriven = 3600 RPM ÷ 1.27 = 2834.65 RPM Based on our calculations, the driven gear is turning at a speed that is slower than the driving gear. To determine exactly how much slower we’ll calculate the difference between their speeds: nDriving – nDriven = 3600 RPM – 2834.65 RPM ≈ 765 RPM So in this gear reduction the driven gear turns approximately 765 RPM slower than the driving gear. Next time we’ll apply a gear reduction to a gear train and see how to arrive at a particular desired output speed. _______________________________________ |
Gear Ratio Formulas
Sunday, February 23rd, 2014|
Last time we introduced a way to convert individual gear speeds in relation to one another within a gear train by employing a conversion tool known as the gear ratio. Today we’ll introduce the gear ratio formulas, of which there are two types.
The first formula for determining gear ratio is based on knowing the driving gear revolutions per minute (RPM), notated as nDriving, and the driven gear RPM, nDriven. Given that knowledge we can calculate the gear ratio, R, that exists between them by the formula: R = nDriving ÷ nDriven (1) The other way to determine gear ratio, R, is by knowing the number of teeth on both the driving gear, NDriving, and the driven gear, NDriven. That’s right, it all boils down to simply counting the number of teeth on each gear. In this instance the gear ratio is calculated by the following formula: R = NDriven ÷ NDriving (2) Equations (1) and (2) may look virtually identical, but they’re not. In mechanical engineering calculations, lower case n is typically used to denote the RPM of rotating objects such as shafts, wheels, pulleys, and gears. Upper case N is typically used to denote the number of teeth on a gear. Next time we’ll see how to manipulate these two equations so as to arrive at a particular gear ratio. _______________________________________ |
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, F1 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. F1 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 F1 is generated, which causes gear teeth on the driving gear to push against gear teeth of the driven gear. Force F2 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 at-rest 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, F1, is met by the resisting force of inertia, F2. In order for the train to move, F1 must be greater than F2. If F1 is less than or equal to F2, 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.
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