# Starting a vehicle uphill

1. Oct 20, 2015

### rdh288

At work, I am trying to calculate the horsepower required to start a vehicle uphill. We have a hill with a steep grade (14.42%). We are trying to prove/disprove that a delivery truck can stop facing downhill and back into a loading dock. I've seen several calculations of the running speed (page 20 of https://diversity.umn.edu/multicult...edu.multicultural/files/Phys1301CRsolns-2.pdf is the best I've seen), but I can't seem to find the calculations for starting a vehicle. I know the initial friction is much higher then the rolling friction.

2. Oct 20, 2015

### sophiecentaur

Given the load plus truck mass, you can calculate the torque needed to move it uphill - if you know the radius of the wheels etc.. You would then, I think, need to see at what speed the engine would need to be rotating (in reverse gear), to develop that torque. ). Then you would need to look at the spec of the clutch and decide if it could dissipate safely, the energy caused by heating whilst slipping. (Is there, in fact, enough torque? I had a 2CV that would actually not pull itself plus a small trailer up my steep drive to the garage, despite racing the engine and making some very unpleasant burning smells with the clutch.
I don't think that the friction is a significant problem here as there are bearings throughout the transmission and the wheels shouldn't be slipping.

3. Oct 20, 2015

### jack action

First, there is no power if there is no motion. So, you need to find the «starting» torque required first.

Second, before knowing what is required, you have to determine the maximum traction force your truck can support.

Based on this, a RWD 2-axle truck backing uphill would be similar to a FWD going uphill. Therefore, the maximum grade the truck can handle (not even moving, just holding) is:
$$tan\theta = \mu\frac{\frac{l_f}{L}}{1 + \mu\frac{h}{L}}$$
Where $\frac{l_f}{L}$ is the portion of the truck's weight on the rear axle, $\mu$ is the tire-ground friction coefficient (typical: 0.8) and $h$ & $L$ are the height of the center of gravity and wheelbase of the truck, respectively.

If the truck begins to move, then you must add the inertia ($\lambda ma$) and the rolling resistance, which modify the previous equation like this:
$$\frac{sin\theta + \lambda\frac{a}{g} + f_r}{cos\theta} = \mu\frac{\frac{l_f}{L}}{1 + \mu\frac{h}{L}}$$
Where $\frac{a}{g}$ is the acceleration in g's (typical: 0.2-0.3), $\lambda$ is the mass factor to take into account the rotational inertia (not sure what it would be for a truck, but typical for a passenger car in first gear is about 1.2) and $f_r$ is the rolling resistance coefficient (typical: 0.008).

If you find out that you can still go uphill, the traction force ($F_r$) needed at the rear axle would be:
$$F_r = mg\left(sin\theta + \lambda\frac{a}{g} + f_r\right)$$
Where $mg$ is the total weight of the truck.

The engine torque needed is found knowing the rear tire radius, $r_r$, the total gear ratio between the engine and the tires, $GR$, and the transmission efficiency, $\eta$ (typical: 0.85-0.9):
$$T_e = \frac{F_r r_r}{\eta GR}$$
Once the truck starts to move, the power required at the axle is based on its velocity $v$:
$$P = F_r v$$
At the engine, the power will be:
$$P_e = \frac{P}{\eta (1-i)}$$
Where $i$ is the transmission slip (clutch, tire) which is usually around 0.02-0.05.