# How much load can a tractor tow uphill?

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.

• Torque divided by wheel radius is your max force you can transmit. Subtract your TEF, and make it 0. Solve for M. That's the most mass you can move at that incline. If M equals your base mass (before adding the trailer), then that's the maximum incline you can go.
– Mark
Jun 30 '15 at 3:20
• @Mark so my wheel radius is 0.66m; 1142.02Nm/0.66m=1730N; this 1730N is roughly 176kg.... can you show me a sample calculation, or a reference? Jun 30 '15 at 4:17
• Note that you needed 46,401 N of force to move, and only have 1730N at the wheel - so your tires would just spin out at that incline.
– Mark
Jun 30 '15 at 13:31
• To get torque at the wheels you need to divide by the wheel RPM not the engine RPM because your drive train increases torque by decreasing speed.
– Rick
Aug 4 '15 at 16:26
• @Mark too low of a force available to drive the wheels would mean the engine would stall (or the clutch would disengage), it certainly wouldn't cause the wheels to loose traction and spin out. But that force would be the force without a gear box, the engine directly connected to the wheel.
– Rick
Aug 4 '15 at 16:28

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.