The Mathematics to Understanding the Relationship of an Engines Torque and Horsepower
Reprinted by permission of Craig Fry, from http://home.fuse.net/pagrosse/
When reading any sort of article or information concerning an engine, a reader is confronted with various terms such as horsepower, torque, and rpm. The terms horsepower torque, and rpm are used in engineering to design transmission shafts as well as . When designing a transmission shaft one must consider the power (watts or horsepower), and speed of rotation (rpm or frequency) to choose the proper material so that the maximum shearing stress allowable will not be exceeded. When speaking of automobiles torque is generally gauged by a dynamometer placed on the wheel axle or engine flywheel, depending on whether one wants to know the torque of the engine on the drive shaft, or the torque of the axle on the wheel. This is the simplest set of equations anyone can use in order to find the forces and power produced by an engine.
Simple equations to understand engine output:
|T = F * X
Torque is a force times a distance. T is the Greek letter tau, representing torque in (Newton meters also called joules) and is equal to force in (Newton's) multiplied by the distance x in (meters).|
|w = 2 * p * f
||Where w is the Greek letter omega representing angular velocity in (radians per second) and is equal to two times pi multiplied by the frequency.|
|P =T * w
||Where P is power rated in (watts) and is equal to the torque in (Newton meters) multiplied by w which is angular velocity (radians per second)|
|Watts = Newton meters /sec
||The power (watts) is simplified into work per unit time rated in Newton meters per second. |
|1 hp = 745.7 watts
|| The power of an engine can be rated in watts or horsepower.|
|f = rev./ sec
||The frequency of an engine is the rpm quantity which in this application will be 2,500 rpm (middle of the power band for an ordinary engine |
The hypothetical engine to be considered below produces 300 foot pounds of torque at 5,000 rpm. In order to find this engine's horsepower one would use the above equations as such:
There is 3.28 feet in a meter, and 4.448 Newtons in a pound, which will be needed since the experimental engine's torque is rated in foot pounds force. The frequency of the engine is calculated by:
f = (5000 revolutions per minute) / (60 seconds)
f = 83.33 revolutions per sec.
Implementing the engine's frequency into the equation of angular velocity w = 2 * p * f
w = 2 * p * (83.33 revolutions per sec.)
w = 523.6 radians per second
In order to convert foot pounds of torque to Newton meters of force one uses this equation.
T = 300 foot pounds of torque
T = (300 foot * pounds) * (4.448 Newton/pound) * (1 meter/ 3.28 feet)
T = 406.83 Newton meters
Then the power is found by multiplying the angular velocity times torque
P = T * w
P = T * (523.6 radians per sec.)
P = (406.83 Newton meters) x (523.6 radians per sec.)
P = 213,015.69 watts
Watts is then converted into the American standard for power called horsepower by dividing watts by 745.7.
213,015.69 watts/ 745.7 watts = 285.66 horsepower
Beer, Ferdinand P., Johnston, E. Russell Jr. Mechanics of Materials. New York: McGraw-Hill, Inc., 1981.
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