I have a 43-ton tractor. The tractive effort force [N] (TEF) required for a given rolling resistance (RR) coefficient of 0.06 is determined by the formula:
$${TEF}=(c + Grad\%)mg$$
where Grad% is the gradient percentile of incline, c is the RR coefficient, m is the mass of the tractor and g is the acceleration due to gravity.
At 5% incline the TEF is 46,401.3 N. So the tractor will require 46,401 N of force to move up a 5% incline. As the gradient increases the TEF increases as well - which makes sense.
But say for example I have 275 kW engine delivering max torque at 2300 RPM = 240.8 rad/s (P= τω; τ= P/ω = 275,000/2300 RPM = 275,000/240.8 = 1142.02 Nm <- is this calculation even right?).
How can I find the maximum towing capacity of the tractor at a given gradient? I.e, how much additional mass can I pull?
The weight I can pull will go towards "0" as the angle/gradient of incline increases. This is because the engine will be using more power going up hill then on a flat incline.