Posts Tagged ‘gear speed’

Determining the Gear Train Tradeoff of Torque vs. Speed, Part Two

Wednesday, August 20th, 2014

      In this blog series we’ve been examining gear train usefulness, specifically in terms of increasing torque.   Equations presented last week began us on the final leg of our journey, and we’ve arrived at the point where the closing combination of equations will demonstrate the loss of speed that takes place when torque is increased within a gear train.

     To that end, the two main equations under consideration as presented last week, are:

R = NDriven ÷  NDriving =  nDriving ÷ nDriven (1)
TDriving ÷ TDriven =  DDriving ÷ DDriven (2)

where R is the gear ratio of the gear train, N is the number of gear teeth, n is the gear rotational speed in revolutions per minute (RPM), T is the torque, and D is the gear pitch radius.

      We were able to link these two equations by working through five key design equations applicable to simplified gear trains.   For the full step-by-step progression see last week’s blog.

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      After working through the equations presented last time we were able to arrive at an equation which links equations (1) and (2).   Here it is:

nDriven ÷ nDriving = DDriving ÷ DDriven (7)

      If you follow the color coding, you’ll see the elements of equations (1) and (2) which come together in equation (7).   Because equation (7) links the gear speed ratios (red) with the gear pitch radii ratios (green), we can set the ratios in equation (1) equal to those in equation (2).   Doing so, we get:

R = NDriven ÷  NDriving = nDriving ÷ nDriven = DDriven ÷ DDrivng = TDriven ÷ TDriving

      In order to see the tradeoff between speed and torque, we need only consider the parts of the equation which concern themselves with factors relating to speed and torque.   Removing the other unnecessary factors, we arrive at:

TDriven ÷  TDriving = nDriving ÷  nDriven (8)

      Next week we’ll plug numbers into equation (8) and disclose the tradeoff of speed for torque.


Determining the Gear Train Tradeoff of Torque vs. Speed, Part One

Friday, August 15th, 2014

      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:

R = NDriven ÷  NDriving =  nDriving ÷ nDriven (1)
TDriving ÷ TDriven =  DDriving ÷ DDriven (2)
TDriving = [HPDriving ÷ nDriving] × 63,025 (3)
TDriven = [HPDriven ÷ nDriven] × 63,025 (4)
HPDriving =  HPDriven (5)

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.

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      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).

TDriving ÷ TDriven =  DDriving ÷ DDriven (2)
TDriving = [HPDriving ÷ nDriving] × 63,025 (3)
TDriven = [HPDriven ÷ nDriven] × 63,025 (4)

      The common terms in these three equations are TDriving and TDriven, so we’ll manipulate things in order to group them together.   We’ll substitute equation (3) for the TDriving term in equation (2), and substitute equation (4) for the TDriven term in equation (2).   We are now able to link all three equations to get:

{[HPDriving ÷ nDriving] × 63,025} ÷  {[HPDriven ÷ nDriven] × 63,025}

                                                                                        = DDriving ÷ DDriven         (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 HPDriving and HPDriven cancel out, along with the two 63,025 terms, allowing us to arrive at equation (7):

{[HPDriving ÷ nDriving] × 63,025} ÷  {[HPDriven ÷ nDriven] × 63,025}

                                                                             = DDriving ÷ DDriven

nDriven ÷ nDriving = DDriving ÷ DDriven (7)

      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.


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.

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      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.


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.


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.


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.


When Do You Need To Modify Gear Ratio?

Wednesday, February 19th, 2014

      Last time we saw how the involute profile of spur gear teeth ensures smooth contact between gears when they rotate.   Today we’ll see why it’s important to be able to change the rotational speed of the driven gear in relation to that of the driving gear by modifying their gear ratio, the speeds at which gears move relative to one another.

      Why would we want to modify the rotational speeds of gears relative to one another?   One reason is to compensate for the fact that alternating electric current (AC) motors drive most modern machinery, and these motors operate at a fixed speed determined by the 60 cycles per second frequency of electricity provided by the utility power grids of North America.   By fixed speed I mean that the motor’s shaft revolves at a single, fixed rate.  It can’t run any faster or slower.   This is fine for some motorized applications, but not others.

      Basic machinery such as wood cutting saws, grinders, and blowers function well within the parameters of the AC motor’s fixed speed, because their working parts are intended to rotate at the same rate as the motor’s shaft.   As a matter of fact, in this instance there’s often no need for a gear train, because the working parts can be connected directly to the motor’s shaft, and the machinery will be powered and function correctly.   There are many instances however in which a fixed speed does not match the speed required for more complex machinery to correctly perform precise, specialized tasks.

      Take a machine tool meant to cut steel bars, for example.   It has a rotating part meant to cut through the steel during machining, and to properly do so its cutting tool bit must turn at 400 revolutions per minute (RPM).   If it turns any faster, the cut won’t be smooth and the tool bit will overheat and break due to increased friction.   If the AC motor driving the machine tool turns at 1750 RPM, a common speed for such motors, then the tool bit will be turning at a much faster rate than the desired 400 RPM, and this presents a problem.

      To solve the problem we need only add a gear train between the motor and the part containing the tool bit, meaning, we must connect the gear train’s driving gear to the motor’s shaft and a driven gear to the part’s shaft.   But in order for this arrangement to work a conversion must take place, that is, we must design the gear train to operate at a specific gear ratio.   By gear ratio, I mean the speeds at which the two gears will rotate relative to one another.

      Next time we’ll introduce the gear ratio formulas that make it all work.


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