Coby, curious how you are calculating this. Simple ratio of cutting width to traveled speed, with all other variables remaining identical? If so, I hadn't thought of it that way. Not sure about that approach. Won't other factors change with the increase in cutting width, e.g. implement weight, maybe gearbox size, friction coefficients... :scratchchin:

Within limits, (keeping it simple)

**It takes a certain amount of total energy to cut a given area.** No matter if you do it quickly or slowly.

Power is a measure of how fast you can apply that energy (to reach that total).

So at a constant (horse) power output over a set amount of time the energy you can apply (and area you can cut) is fixed/limited (it is what it is). (That is, it doesn't matter if your going faster with a narrow cutter, or slower with a wide cutter, at the same constant horsepower output, the area cut (i.e. the total energy output) will be the same).

You can think of it in a lot of different ways, but horsepower required is a function of how quickly do you want to cut a given area.

Just like riding a 10 speed bike up a hill. It will take double the (horse) power to do it in 2 minutes versus 4 minutes, but both riders spent the same amount of total energy to get to the top of the hill.

(As an aside: If you do it in 2 minutes in low gear you need a lot of rpms and a little torque, if you also do it in 2 minutes, but in high gear you need a lot of torque, little rpms. Here, there's the same amount of energy

**and power output** in both cases. (but now we're into torque curves, optimum rpm's, and engine's limits on rpm and torque, etc...))

Yes, in reality, implement weight, and gearbox differences, gearing of the tractor, even the torque of a 5' blade versus a 6' blade comes into play, but these variables are probably insignificant compared to the variables of the grass/brush your cutting and the speed you're traveling.