Work does not have a unit of time associated with it. A 120 HP tractor or a 50 HP tractor can both plow a 100 acre field. The amount work that each tractor does is the same and is equal to the force required to pull the plow times the distance required to cover the 100 acres. The answer to your question is yes - actually most anything could do the work of a 120 HP tractor if given enough time.
Power as in HP is a much better measure of tractor performance since power has a unit of time associated with it. Power is defined as work performed per unit time and this is why a 120 HP tractor can plow the 100 acres in less time than the 50 HP tractor.
Allow me to reword your question: Can a small low torque high revving engine pull a plow as fast as a high torque low revving tractor engine if they have similar HP ratings. The answer to this question is also yes.
Ok.... now the hard part, to explain the answer: We want to know if we swapped engines in a tractor could we still plow the same amount of acres in the same amount of time. To do this lets assume we want to plow at 4 MPH and that the rear tires do not slip. Therefore, the tractor engine combinations have to develop the same drawbar pull at 4 MPH to plow the 100 acres in the same amount of time. Or, the tractor that develops a higher drawbar pull could pull a larger plow and therefore would take less time to complete the plowing. If we assume our tractor has rear tires with a overall diameter of 58", to plow at 4 MPH we will need to select a gear ratio such that the tires rotate at 23.2 RPM. We will also assume that the tractor transmission has the exact ratio we need. Since we assumed the tires do not slip, drawbar pull is proportional to the torque at the rear axle. Therefore, the engine that can send the greatest amount of torque to the rear axle will develop greatest drawbar pull.
I have in front of me the latest issue of Cycle World. The Kawasaki XZ-14 puts out 191.7 HP @ 10,000 RPM as measured at the rear wheel. Maxmum torque is 113.2 ft-lbs @ 7600 RPM however we want the torque at the HP peak which is 100.7 ft-lbs @ 10,000 RPM. To turn the rear tractor tires at 23.2 RPM with the engine at it's HP peak we need a 431:1 (10,000/23.2) gear ratio. The engine torque (100.7 ft-lbs) is multiplied by the gear ratio (431) resulting in 43,402 ft-lbs of torque available at the rear alxe to pull a plow at 4 MPH.
I will leave the rest to you - to compare tractors/engines simply take the engine torque at the HP peak, determine the gear ratio for 4 MPH (or any desired MPH) and multiply the two numbers together. The resulting torque at the rear alxe for a given speed is the bottom line - not engine torque. You will find the engine with the highest HP always wins. Yes, there is some loss in the gear trains but it is not significant to what we are doing here.
What we are doing is comparing the tractors drawbar power. Power is work per unit time - the tractor with the highest drawbar power can do the most work in the shortest amount of time.
Sorry this got so long - hope it helps. Got to go the myple syrup is ready for the bottles.
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