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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 ‘force vector’
Determining the Gear Train Tradeoff of Torque vs. Speed, Part Three
Wednesday, August 27th, 2014
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. _______________________________________ |
Manipulating theTorque Formula
Wednesday, April 23rd, 2014|
Last time we introduced the simplified formula for torque: Torque = Distance × Force Today we’ll manipulate it by way of our wrench and nut example to get the torque that we need to loosen a tight nut.
By inserting the numerical values of our illustration into the torque formula, it becomes: Torque = 6 inches × 10 pounds = 60 inch-pounds Inch-pounds may be terminology you’re unfamiliar with, but this notation arises from the fact that torque values are always represented by units of distance and force separated by a hyphen, in our case inch-pounds. This just means that distance and force were multiplied together to calculate torque. In order to manipulate the value for torque all that needs to be done is change either or both numerical values for Distance and Force. Increasing either or both factors produces higher torque, decreasing them less torque. Why manipulate torque? To provide us with a mechanical advantage. Suppose we have a rusted nut that we’re trying to move with a wrench that has a 6 inch handle, and the 10 pounds of force employed by the muscles in our arm just won’t budge it. Put another way, 60 inch-pounds of torque is insufficient to rotate the nut. It’s clear we must increase torque to get things going. Let’s do so by increasing either of the vector magnitudes. First we’ll try increasing the magnitude of the force vector. Instead of simply pushing hard on the wrench handle with our arm, let’s say we push extra hard. The average man can do a bicep curl of between 30 to 40 pounds, but we haven’t been going to the gym lately and we’re really out of shape. So try as we will, we just can’t muster up the bicep strength to apply more than 10 pounds of force to the wrench handle. It’s clear that this approach to increasing torque upon the nut isn’t going to work. The other way to increase torque is to increase the length of the distance vector. We’ll need a wrench with a longer handle, say 9 inches.
By using a wrench with a longer handle we have increased the magnitude of the distance vector from 6 to 9 inches. The torque formula becomes: Torque = 9 inches × 10 pounds = 90 inch-pounds Eureka! The longer handle has provided us with the mechanical advantage needed to increase torque to 90 inch-pounds, thereby overcoming our muscular shortcomings and breaking the nut free. In summary, since torque is the product of the magnitudes of the distance and force vectors, we can increase torque by either increasing the magnitude of the force vector, or as in our example, by increasing the magnitude of the distance vector. Next time we’ll see how to apply the principles of torque to a real world situation involving gear trains in which we need to obtain a mechanical advantage. _______________________________________ |
