The original inquirer was actually speaking about turbojet engines, which often are rated in pounds of thrust. Most are actually turbofan engines, with a large ducted fan in front that blows only air. This increases efficiency and reduces noise.
Carleton -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of Bill Hooper Sent: Tuesday, May 11, 2004 11:47 To: U.S. Metric Association Subject: [USMA:29746] Re: power, thrust and force On 2004 May 11 , at 12:18 AM, J. Ward wrote: > If the force of thrust is in newtons, and the speed is in m/s, then > the power > in watts is simply > Power = Force * Speed. > While you equation is undoubtably correct, you have overlooked an important point. The force must be acting on the object which is moving at the speed. That will give the power of produced (or expended) by that object. Since the original question, I believe, asked about the thrust of the propellor on the air (equivalent to the thrust of the air that pushes the plane) and the power output of the engine, the above equation could not be used (not directly anyway). In a plane, one has to be concerned with the power generated by the engine and the power expended by the propellor against the air. It is the power of the propellor against the air which you get if you multiply the thrust (force) of the propellor against the air times the speed of the propellor through the air*. That power may not be the same as the power generated by the engine. The power output of the engine would be the force of engine against the shaft of the propellor multiplied times the speed of the propellor shaft. (Since this is a rotational situation, the appropriate variation of the above equation would be: power equals the "rotational force" (called the torque, in newton-metres in SI) multiplied by the rotational speed of the shaft (the angular speed, or angle turned per unit time, measured in radians per second in SI). Regards, Bill Hooper Fernandina Beach, Florida, USA *PS I may not have been completely correct above when I referred to the speed of the propellor through the air. The air arrives at the propellor with one speed (the speed of the plane) and leaves with a different speed, because it has been pushed toward the rear by the propellor. This change in speed of the air as it goes past the propellor would have to be taken into account for my description above to be accurate. The science is easy; the engineering is where it gets difficult ... as always. :-) ======================== SIMPLIFICATION begins with SI ========================
