Engineering Expert Witness Blog

      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):

T_Driven ÷  T_Driving = n_Driving ÷ n_Driven        (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:

T_Driven ÷ T_Driving =  n_Driving ÷ n_Driven

(100 inch pounds) ÷ (50 inch pounds)  = (100 RPM) ÷ n_Driven

2 =  (100 RPM) ÷ n_Driven

n_Driven = (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.

      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.

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      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.

      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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