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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 thay-tah, is a function found within a field of mathematics known as trigonometry , which concerns itself with the lengths and angles of triangles. ϴ, or thay-tah, 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.
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Archive for March, 2014
Vectors, Sin(ϴ), and the Torque Formula
Wednesday, March 26th, 2014Achieving 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 hand-held 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. _______________________________________ |
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. _______________________________________ |
