Archive for August, 2014

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

Wednesday, August 27th, 2014

      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.

Gear expert witness

      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.


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.

Gear engineering expert

      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.

Mechanical Engineering Expert Witness

      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.

Gear engineering expert

      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.