How much load can a tractor tow uphill?

Question

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.

Answer 1

The power that the engine can deliver goes through a transmission before reaching the wheels. I believe most tractors are geared such that the maximum torque that can be applied to the wheels exceeds the maximum torque a road surface can apply to the wheel. This means it is likely that your tractor is traction limited rather than power limited.

The maximum tractive force can be approximated by:

$$F_t=\mu W \sqrt{1-\frac1{Grad\%}}$$

Where $W$ is the weight supported by the driving wheels and $\mu$ is the friction coefficient.

For two wheel drive tractors typically $55\%$ to $70\%$ of the weight will be on the drive tires when on level ground without pulling anything. However, the weight will shift when pulling and with changes in incline. The pulling alone can sometimes shift up to $90\%$ of the weight to the rear tires. With a steep incline this may actually cause the tractor to nose up. In that case, the limiting factor is nosing up, rather than traction.

The friction coefficient for tires on asphalt is around $1$. For dirt, especially loose dirt, the dirt actually shears internally before the tire/dirt interface shears. This results in a "friction coefficient" that is dependent on contact area, and is why tractors typically have large tires.

The force required to pull something can be calculated from the formula you gave in your question, where $m$ is the mass of both the tractor and the thing you'd like to pull. (this requires the thing your pulling to have a similar rolling resistance)

Note however, that that formula is valid only for shallow hills, and become less and less accurate as the hills get steeper. A $5\%$ grade is still well within that accurate range though.