Force = kg.m/s^2 = N (newton) Energy, work, quantity of heat = N.m = J (Joule) Power, heat flow rate = J/s = W (watt)
Just thought you'd like to know the SI. Stan Doore ----- Original Message ----- From: "Bill Hooper" <[EMAIL PROTECTED]> To: "U.S. Metric Association" <[EMAIL PROTECTED]> Sent: Tuesday, May 11, 2004 11:46 AM 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 > ======================== >
