Posts Tagged ‘mechanical power’

Determining Angle of Wrap

Monday, July 17th, 2017

   Sometimes things which appear simple turn out to be rather complex.   Such is the case with the Euler-Eytelwein Formula, a small formula with a big job.   It computes how friction, an omnipresent phenomenon in mechanical assemblies, contributes to the transmission of mechanical power.   Today we’ll determine the value of one of the Euler-Eytelwein Formula’s variables, the angle of wrap.

Determining Angle of Wrap

Determining Angle of Wrap

   Here again is the basis for our calculations, the Euler-Eytelwein Formula.

T1 = T2 × eθ)                                                                 (1)

   To recap what we’ve discussed thus far, T1 is the tight side tension, the maximum the belt can endure before breaking.   T2 is the loose side tension.   It’s just going along for the ride.   The term e is Euler’s Number, a constant equal to 2.718, and the coefficient of friction, μ, for contact points between the belt and pulleys is 0.3 based on their materials.

   The formula introduced last time to calculate the angle of wrap, θ, is,

θ = (1802α) × (π  ÷ 180)                                             (2)

where,

α = sin-1((D1 – D2) ÷ 2x)                                                 (3)

   By direct measurement we’ve determined the pulleys’ diameters, D1 and D2, are equal to 1 foot and 0.25 feet respectively.   The term x is the distance between the two pulley shafts, 3 feet.   The term sin-1 is a trigonometric function known as inverse sine, a button commonly found on scientific calculators.

   Inserting our known values into equation (3) we arrive at,

α = sin-1((1.0 foot – 0.25 feet) ÷ 2 × (3 feet))                  (4)

α = 7.18                                                                        (5)

   We can now incorporate equation (5) into equation (2) to solve for θ,

θ = (180(2 × 7.18)) × (π ÷ 180)                                  (6)

θ = 2.89                                                                         (7)

   Inserting the values for m and θ into equation (1) we arrive at,

T1 = T2 × 2.718(0.3 × 2.89)                                               (8)

T1 = 2.38T2                                                                    (9)

   We have at this point solved for over half of the unknown variables in the Euler-Eytelwein Formula.   We still can’t solve for T1, because we don’t know the value of T2.   But that will change next time when we introduce yet another formula, this one to determine the amount of mechanical power present in our pulley-belt system.

 

 

Copyright 2017 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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The Difference Between Driven and Driving Pulleys

Friday, March 31st, 2017

    Last time we introduced two historical legends in the field of engineering who pioneered the science of mechanical power transmission using belts and pulleysLeonhard Euler and Johann Albert Eytelwein.   Today we’ll build a foundation for understanding their famous Euler-Eytelwein Formula through our example of a simple mechanical power transmission system consisting of two pulleys and a belt, and in so doing demonstrate the difference between driven and driving pulleys.

    Our example of a basic mechanical power transmission system consists of two pulleys connected by a drive belt.   The driving pulley is attached to a source of mechanical power, for example, the shaft of an electric motor.   The driven pulley, which is attached to the shaft of a piece of rotating machinery, receives the mechanical power from the electric motor so the machinery can perform its function.

The Difference Between Driven and Driving Pulleys

The Difference Between Driven and Driving Pulleys 

   

    Next time we’ll see how driven pulleys can be made to spin at different speeds from the driving pulley, enabling different modes of operation in mechanical devices.

Copyright 2017 – Philip J. O’Keefe, PE

Engineering Expert Witness Blog

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The Relationship Between Torque and Horsepower

Tuesday, July 15th, 2014
      We’ve been discussing gear trains for some time now, and last time we posed the question:  Why even bother using a gear train and performing complex computations to arrive at a desired torque for an application?   Why not just use a bigger motor to start with?   Today we’ll see why.

      First, we must acknowledge that sometimes higher torque is achieved by simply using a more powerful motor.   But sometimes this isn’t possible or practical.

      To begin our discussion, we must first understand how torque is related to motor power, the amount of mechanical work a motor can perform.   Torque is in fact a function of how much mechanical power a motor produces.   In the United States motor power is typically measured in units of horsepower.

      The following equation illustrates the relationship between torque, horsepower, and motor speed:

T = [HP ÷  n]  ×  63,025

where T is the motor shaft’s torque in units of inch-pounds, HP is the motor’s horsepower, and n is the speed of the motor shaft in revolutions per minute (RPM).   The number 63,025 in the equation is a constant used to convert the units of horsepower and RPM into units of torque (inch pounds).   This equation applies to all sources of mechanical power.   Its versatility enables design engineers to easily determine if a mechanical power source can deliver the torque required to drive a particular piece of machinery.

      The torque equation above tells us that in order to get a higher torque T for a given speed n, you’ll have to get a motor with a higher HP.   Put another way, if your speed remains constant and you use a motor with higher horsepower, you’ll get more torque for your application simply by increasing the horsepower.

      Next time we’ll plug numbers into our equation and see how it all works.

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