Posts Tagged ‘gear train’

Equating Torques and Pitch Circle Radii Within a Gear Train

Thursday, May 29th, 2014

      Last time we developed torque equations for the driving and driven gears within a simple gear train.  They are,

T1 = D1 × F

T2 = D2 × F

where, T1 and T2 are the driving and driven gear torques, D1 and D2 are the driving and driven gear pitch radii, and F is the resultant Force vector, the common factor between the two equations.

gear train exoert

      Now we’ll combine these two equations relative to F to arrive at a single equation which equates the torques and pitch circle radii of the driving and driven gears in the gear train.   This type of computation is commonly used to design gear trains to ensure they perform at a given level.

      As a first step we’ll use algebra to rearrange terms and place the two equations equal to F.    First we’ll do it for the driving gear, dividing both sides of the equation by the pitch circle radius, D1.

T1 ÷  D1 = D1 ÷ D1 × F

T1 ÷ D1= 1 × F

F = T1 ÷ D1

      In a similar fashion, we’ll do it for the driven gear by dividing both sides of the equation by the pitch circle radius, D2.

T2 = D2 × F   →   F = T2 ÷ D2

      Since F is the common term between the two equations, we can set them up as equal to each other,

F = T1 ÷ D1 = T2 ÷ D2

which means that,

T1 ÷ D1 = T2 ÷ D2

      Next time we’ll see how to use this equation to manipulate our gear train so that it acts as a torque converter by increasing T2 with respect to T1 and the ratio of D1 to D2, thus providing a mechanical advantage to the electric motor the gear train is attached to.

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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.

gear train exoert

      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.

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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.

gears and torque

      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.

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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.

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      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.

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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.

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      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.

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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 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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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.

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      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.

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      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.

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      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.

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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.

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      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.

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      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.

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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.

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      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.

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