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Mathematics of Torque and Horsepower |
Written by Craig Fry and Paul Grosse | ||||||||||||||||
Saturday, 10 July 2004 | ||||||||||||||||
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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