[quote="flyingace"](quoted from post at 16:24:54 03/03/12)
....."See, I don't agree here. In theory, a little electric motor that spins at ridiculously high RPM could make as much HP as a 40 HP JD. Say it has only 1 ft/lb of torque, but spins a 210,000 RPM. According to the formula I used 40 HP = (1 * X) / 5252 where X is RPM. 1 ft/lb of torque will never be enough to pull a plow now matter how fast it spins or how low it's geared, right? I'm guessing this would also be true for any motorcycle engine making 40 Hp".....
The formula is correct and works for low torque and very high RPM. A turboshaft (gas turbine) would be a good example, the turbine spins at very high speed but has very little torque. The Honeywell T55 gas turboshaft puts out 4115 maximum continous HP. Two of these engines are used to power the Chinook helicoptor. The M1 tank is powered by a turboshaft (gas turbine) engine. During developement both diesel and turboshaft versions were built - the army selected the turbine. Large ships can be powered by large slow turning diesels or a gas turbine turboshaft - it takes big time torque numbers to spin those props.
At the farm progress show this summer New Holland had a cut away of their tubo-compounded engine. The engine had an addtional turbo charger down stream of the main turbo charger. Instead of driving a compressor wheel the down stream tubine was geared to the engine. New Holland claimed it returned 40 HP to the engine. Again the turbine spun at very high RPM but had very low torque.
The formula is correct and 1 ft-lb can certainly pull a plow.
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