# Calculating the grip of a tyre for a given surface

Is there a formula to calculate the grip generated by a tyre on a vehicle for a given surface?

I have a 13,000 kg machine pulling nearly 200 N of weight. I need to figure out if we need to put any more weight onto the machine so that it has enough traction to start the initial pull. It's going to use pneumatic tyres and the surface is concrete.

• Are you asking for a formula that combines weather condition, temperature, rubber properties, surface material, tire wear, tire pressure, tire construction, any surface chemicals, surface wear, etc. into an equation that has uniform units and spits out a coefficient of friction? No. I'm guessing that you aren't looking for the effect of all of those combined, but instead are looking for something more specific. Are you just looking for the coefficient of rubber on concrete/asphalt? – hazzey Jun 24 '15 at 2:31
• My guess is they are looking for how the torque on the wheel (or the horsepower on the engine) translates to force on the ground. – Mark Jun 24 '15 at 2:42
• @3kstc It is great to see that you keep returning with your engineering questions. In the future try to include more information about your application in the initial post. That will prevent the need for a long back and forth of clarification comments and get you an answer much more quickly. I've edited this question to include some of the info from the comments for you. – Chris Mueller Jun 24 '15 at 11:16
• You are also missing more detail about the weight that is being pulled. 200N that is on wheels will be easier to pull than 200N of concrete against concrete. – hazzey Jun 24 '15 at 12:37
• What am I missing here? Assuming N stands for "Newtons", 200N is the force required to drag a 20 kg weight across the ground with a coefficient of friction of 1. Why on earth would a 13,000 kg (weight of 127,000 N) machine have a problem with this? – Dave Tweed Jun 24 '15 at 16:28

## 1 Answer

According to the engineering toolbox the coefficient of static friction between a tire and pavement ranges between $\mu_{wet}=0.2$ for a wet surface and crappy tire and $\mu_{dry}=1$ for a dry surface and high quality tire. You can use this to calculate the total amount of force that the tires will be able to apply with $$F_{tire}=\mu F_N=\mu Mg\cos\theta$$ where $\theta$ is the slope of the road, with $\theta=0$ being a flat road. The total amount of force required of the tires to keep your load from rolling backward (with slightly more needed to get it moving forward) is given by $$F_{load}=(W+Mg)\sin\theta$$ where $W$ is the weight of the load. So, the maximum slope that your vehicle will be able to operate on is given by the point at which the backwards force of the load becomes equal to that which the tires can apply; $$\theta=\tan^{-1}\left(\frac{\mu Mg}{W+Mg}\right)=11.3^\circ,$$ assuming that the pavement is wet and your tires aren't particularly great. This doesn't account for any small undulations in the pavement or bumps in the road, but your tractor to load weight ratio is so high that you shouldn't have any problems.

• Does this neglect the friction of the weight? Are you assuming that 200N of pull needs to be developed or are you considering that the pull will need to be 200N X coef. of friction? – hazzey Jun 24 '15 at 12:39
• It does neglect friction in the load. I am assuming that 200 N is the weight of a load which is on a trailer of some sort so that the friction would be small. Upon a re-reading of the question, it isn't clear if that is what was meant though; it could be 200 N sitting on a pallet being dragged across the ground. Perhaps @3kstc will chime in and let us know. – Chris Mueller Jun 24 '15 at 12:44