Posts Tagged ‘gear expert witness’

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.


Determining Torque Within a Gear Train

Monday, June 30th, 2014

      Last time we set up an example where an electric motor is connected to a lathe via a gear train.   Today we’ll take the numerical values present on that gear train and plug them into the torque ratio equation we’ve been working with for the past few blogs.

      In the illustration below the electric motor exerts 200 inch pounds of torque upon the driving gear.   The driving gear pitch circle diameter is 6 inches, while the driven gear pitch circle diameter is 8 inches.   It’s been determined through previous lab testing that the lathe we’ll be using requires at least 275 inch pounds of torque to be exerted upon the driven gear shaft in order to operate properly.   Will the gear train shown below meet this requirement?

gear train torque calculations

      First, a review of the torque ratio equation:

T1 ÷  T2 = D1 ÷ D2

      Now we’ll crunch numbers.  T1 is equal to 200 inch pounds, D1 is equal to 3 inches (pitch radius equals pitch diameter divided by two), and D2 is equal to 4 inches.   This gives us:

(200 inch pounds) ÷  T2 = (3 inches) ÷ (4 inches)

T2 = (200 inch pounds) ÷ (0.75) = 266.67 inch pounds

      So, does the gear train as presented here supply enough torque to power the lathe properly?   No, it does not.   It provides only 266.67 inch pounds, not the 275 inch pounds of torque required.

      Next time we’ll see how to manipulate gear sizes within a gear train in order to meet a given torque requirement.


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.

Gear expert witness

      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.