# What is the relationship between rolling resistance and velocity?

I'm a games programmer, trying to write a simple car physics simulation. I'm aware that a car travelling in a straight line will exert a traction force that drives it forwards (by turning the wheels and pushing back against the ground), and that the main forces that act against this acceleration are aerodynamic drag and rolling resistance. But I can't seem to find a sensible way of calculating the rolling resistance for a given vehicle.

$F_{rr} = C_{rr} * v$

That is, the force is proportional to a rolling resistance coefficient multiplied by the velocity. In my tests, this seems wrong, and the page itself (and the sources it cites) admit that they're on shaky ground in terms of explaining or proving that formula.

This page can't make up its mind. For most of the page it says that

$F_{rr} = C_{rr} * W$

That is, the force is equal to a coefficient multiplied by the weight ($mg$) of the vehicle - i.e. the force is the same regardless of velocity. It even provides a table of coefficients for different circumstances. But if the force is constant, won't a car in neutral with the engine off be accelerated backwards by this force? What is rolling resistance at velocity 0?

Then, for a bit of that page it claims that velocity is a factor in calculating the coefficient:

The rolling coefficients for pneumatic tyres on dry roads can be calculated as

$c = 0.005 + 1/p (0.01 + 0.0095(v/100)^2)$

where $c$ = rolling coefficient

$p$ = tyre pressure (bar)

$v$ = velocity (km/h)

This makes no attempt to explain what all those "magic numbers" mean, and still produces a coefficient of ~0.0088, which in a 1500 kg car would yield a force of 129 N whilst the car was standing still. That can't be right...

So, which is right? Given basic information about a vehicle (mass, velocity, information about the wheels and the surface they're rolling over), and ignoring aerodynamic drag, what's a sensible way to come up with a broadly plausible rolling resistance?

• Mathjax your question properly please. Commented Nov 26, 2015 at 14:14
• Like this? Sorry, I'm new here... Commented Nov 26, 2015 at 14:31

If you are still interested in this question (and hopefully still have uses for the answer), I will try to answer it in an efficient manner. First of all, The rolling resistance force is an interaction between the ground and the wheel, which is independent of speed ONLY when the ground surface is completely flat and rigid. If the terrain is bumpy/hilly, the rolling resistance does depend on the speed.

On bumpy ground: It would usually be best to classify the resistance into a few different stages... At low speeds the wheels stay in steady contact with the ground and do not suffer from impact resistances with the tiny hills (like stones) on the ground. At medium speeds the wheels bounce off the top of each bump and land in the valley between the bumps and impact at the base of the next bump. At high speeds the wheels ski over the bumps, only hitting the very tops of them. As a general rule of thumb, on rough ground, the rolling resistance is least at a low speed, is most at a medium speed, and becomes less once the speed is high enough to reach the last stage (which I call rockoplaning... like hydroplaning on rocks).

On smooth/flat and rigid ground: the rolling resistance force is greatest when the wheel is not rolling... but this does not cause the vehicle to roll backward because the rolling resistance is acting in two directions as once... i.e. it is pressing forward and backward with equal force. The rolling resistance force decreases once the wheel begins rolling, but it changes so that it is only acting against the direction of motion. Once the wheel is in steady motion, there is a rolling resistance force acting against the wheel which is not dependent on speed whatsoever. The aerodynamics are greatly dependent on speed, but the rolling resistance is not... which is why the rolling resistance matters a lot at low speeds, but barely at all while driving at high speeds (the aerodynamics are so much more powerful than the rolling resistance when at high speeds). I can vouch for the fact that the rolling resistance is best expressed as Crr*Fn, where Fn is the force pressing the wheel against the ground. The force Fn may not be equal to the weight, if something makes the wheels press on the ground more or less force (like aerodynamic lift/downforce). I have researched, experimented and studied rolling resistance sufficiently to verify that as a general rule, all you need to know is the load force and the coefficient of rolling resistance. The coefficient should be easy to find online, and for a typical car on asphalt, it should be between 0.015 and 0.02.

Some online sources will tell you that rolling resistance is dependent upon speed, but in those cases, they are usually using rolling resistance as a broad term that applies to more than just the wheels rolling... i.e. they mean either the total resistance or the rolling resistance + the resistance of the axles. Because the axles on a train cause more resistance than the wheels, engineers consider the axle resistance to be part of the rolling resistance. Because axles tend to use fluid lubrication, axle resistance does depend upon speed. While there are exceptions, generally fluids cause friction that depends on speed, while solids do not.

It is incorrect to relate rolling resistance to aerodynamic resistance by some arbitrary factor (like thinking 50% of the resistance is aero, 50% is rolling), because the fact that aerodynamics get stronger and rolling resistance doesn't... So at a very low speed rolling resistance may be >95% of the total resistance, but at a very high speed rolling resistance may be <5% of the total resistance.

Suggestions for your program: The key is that the rolling resistance should be a force trying to make the vehicle stop... not trying to make it go backwards. And the force should still exist when it is not driving, but the force should just try to keep it from moving, like it is held still by a magnet. Crr should be around 0.02?

Rolling resistance is indeed a function of vehicle speed (tire rotational velocity, to be precise). The SAE J2452 tire test standard for measuring tire rolling resistance even provides a standard formula for this: $$F_{rr}=P^αF_z^β(a+bv+cv^2)$$ You'll note the dependence on $$v$$ and $$v^2$$ in this formula. The dependence on velocity is NOT due to road roughness, NOR on aerodynamic drag, but is due to the hysteretic loss of rubber being a function of deformation frequency: higher frequency deformation increases the hysteretic loss of rubber. A tire rotating faster deforms the tread rubber at higher frequency, therefore increasing the hysteretic loss.

Tire manufacturers (and tire standards, for the most part) are careful to subtract any possible aerodynamic loss of the tire when measuring rolling resistance, because the aerodynamic loss of the tire is highly dependent on the vehicle airflow and aerodynamic design, which cannot be predicted by the tire manufacturers. The $$v^2$$ term in the above equation is therefore NOT due to aerodynamic drag (even though aerodynamic drag, in general, is indeed proportional to $$v^2$$).

I have also come across the equation $$C_{rr}=0.005+\frac{1}{P}\left(0.01+0.0095\left(\frac{v}{100}\right)^2\right)$$ in the Engineering Toolbox website (generally a really useful website for engineering nerds like me) Engineering Toolbox Rolling Resistance. I don't think you will find a derivation or explanation of the $$0.0095$$ coefficient. for the same reason that SAE J2452 doesn't offer one: the coefficient is experimentally determined by regression fit to actual measurement of a given set of tires, and will vary depending on which tires are measured.

The formulas are both somewhat unsatisfactory, as you point out, because at v=0 and $$F_z >0$$, $$F_{rr} = C_{rr}F_z$$ would be non-zero, which is counterintuitive.

Rolling resistance is not, strictly speaking, an actual force. This is why a stationary tire (or one rotating at an angular velocity near zero) has a non-zero rolling resistance, but never experiences a force that causes a tire to spontaneously rotate backwards. Rolling resistance is merely an attempt to ascribe the energy loss of a loaded tire travelling across some surface to the distance, loading, (and velocity) travelled. Because rolling resistance describes an energy loss, it can never result in a net backward rolling force on a tire. This is why pushing a heavily loaded wheelbarrow requires more forward force than pushing a lightly loaded one, but no wheelbarrow has ever spontaneously rolled backward (except in "Fail" videos on YouTube). But it also requires some minimum forward force before any forward motion results, and I think this is the $$F_{rr}$$ at v=0. I know this sounds hairsplitting, but learned papers describe it so.

This problem requires a free-body diagram, where there is a friction force pushing the car forward, and air resistance and rolling resistance resisting the forward motion of the car. If there is a net force on the car (a non-zero force remaining after adding up all of the propulsive and retarding forces) the car will accelerate, either in the forward direction or the reverse direction. If the forces cancel each other, the car will travel at constant velocity.

I collected actual coast-down data for a Datsun 280 ZX many years ago. When I went through a fairly detailed analysis of the data, I eventually determined that the retarding forces on this automobile were approximately 50% wind resistance and 50% rolling resistance at highway speeds.

If you are intending to model these forces for a range of speeds, you may want to see this link for modeling air resistance: https://en.wikipedia.org/wiki/Drag_(physics)

For rolling resistance, you may want to see this link: https://en.wikipedia.org/wiki/Rolling_resistance

When you use the referenced sources, use the real-world data given above for a boundary condition to weed out the stuff that is "on shaky ground". For an exact value of "highway" speed, use 60 mph.

Good luck - your simulations will probably take some trial and error to arrive at reasonable results.

Friction is awfully complicated because you're trying to model a system with infinite microscopic interaction with few macroscopic variables, It ought go haywire theoretically and hence all those "empiric" and "adhoc" stuff you have already found out. It's not possible to find a "unified theory of friction"(Well formally we have one, that is QED). That's why it's always better to start with a simple model(let's say second order in velocity) and then use some real data(I think there ought to be tons of open data available ) and iterate for a realistic model.

If you have an access to a "MATLAB" copy then you could use their models for vehicle do these simulations ...(I've never used it though, but a quick look states they have a lot of parameters for fine tuning)

http://www.mathworks.com/help/physmod/sdl/ug/complete-car-model-and-simulation.html