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If Imperial units are used, and if torque is in pounds-force feet and rotational speed in revolutions per minute, the above equation gives power in foot pounds-force per minute. The horsepower form of the equation is then derived by applying the conversion factor 33,000 ft⋅lbf/min per horsepower:
Engine power is the power that an engine can put out. It can be expressed in power units, most commonly kilowatt, pferdestärke (metric horsepower), or horsepower.In terms of internal combustion engines, the engine power usually describes the rated power, which is a power output that the engine can maintain over a long period of time according to a certain testing method, for example ISO 1585.
Two common definitions used today are the imperial horsepower as in "hp" or "bhp" which is about 745.7 watts, and the metric horsepower as in "cv" or "PS" which is approximately 735.5 watts. The electric horsepower "hpE" is exactly 746 watts, while the boiler horsepower is 9809.5 or 9811 watts, depending on the exact year.
One metric horsepower is needed to lift 75 kilograms by 1 metre in 1 second. Power in mechanical systems is the combination of forces and movement. In particular, power is the product of a force on an object and the object's velocity, or the product of a torque on a shaft and the shaft's angular velocity.
The torque is then related to the lever length, shaft diameter and measured force. The device is generally used over a range of engine speeds to obtain power and torque curves for the engine, since there is a non-linear relationship between torque and engine speed for most engine types. Power output in SI units may be calculated as follows:
Speed has dropped out of the equation, and the only variables are the torque and displacement volume. Since the range of maximum brake mean effective pressures for good engine designs is well established, we now have a displacement-independent measure of the torque-producing capacity of an engine design – a specific torque of sorts.
The torque on shaft is 0.0053 N⋅m at 2 A because of the assumed radius of the rotor (exactly 1 m). Assuming a different radius would change the linear K v {\displaystyle K_{\text{v}}} but would not change the final torque result.
The equation for torque is very important in angular mechanics. Torque is rotational force and is determined by a cross product. This makes it a pseudovector. = where is torque, r is radius, and is a cross product. Another variation of this equation is: