The differential equation for wheel speed can be given as:

enter image description here

In here, $T_t$ is driving torque (not braking), $F_x$ is longitudinal force of tyre, $J_w$ is wheel inertia, $\dot{w}$ is angular speed of wheel and $Frolling$ is rolling resistance force.

According the equation, if we increase the resistance force, $\dot{w}$ should increase. Increasing of angular speed may cause higher slip ratios. As long as longitudinal slip ratio increases, it is possible to have higher longitudinal tyre forces according to Magic Formula.

In here i have a confusion: High rolling resistance makes possible to have higher longitudinal force according to this differential equation. How can it be possible? How can $F_x$ increase with higher values of rolling resistance? If $F_x$ becomes high, it is impossible to decrease the vehicle. With the help of the positive and high $F_x$ force, vehicle cannot decelerate.


  • $\begingroup$ Where did you get this equation? At first glance your intuition seems correct, it may be a sign error on rolling resistance. $\endgroup$
    – jko
    Commented May 13, 2021 at 11:34
  • $\begingroup$ I am not entirely clear on what you mean with either "If $T_t$ is constant all of the process, longitudinal slip will be higher. With the effect of high slip ratio, $F_x$ will become higher." OR "How can $F_x$ increase with higher values of rolling resistance? If $F_x$ is high, it is impossible to decrease the vehicle.". Is it possible to rewrite those sentences to clarify your question? $\endgroup$
    – NMech
    Commented May 13, 2021 at 11:56
  • $\begingroup$ Hi jko. Article link: trid.trb.org/view/1375510 $\endgroup$
    – voo
    Commented May 13, 2021 at 12:53
  • $\begingroup$ Hi NMech. According to this differential equation, with the increasing of rolling resistance, angular speed of wheel increases. Because of that, we get higher slip ratios. (en.m.wikipedia.org/wiki/Slip_ratio) and if the slip ratio increases, longitidunal force increases. We use slip ratio and verical tyre force to calculate Fx in MF tyre model as you know. In my opinion, as long as we have high slip ratios, there may be no way to reduce to speed of the vehicle. It is very clear that i cannot comprehend the mechanism of rolling resistance correctly, so i asked the question in here. $\endgroup$
    – voo
    Commented May 13, 2021 at 12:56
  • 1
    $\begingroup$ Additionaly, this link may solve my problem: physics.stackexchange.com/questions/462632/… $\endgroup$
    – voo
    Commented May 13, 2021 at 12:58

2 Answers 2


WIP waiting for clarifications

A way to look at this is the following:

The rotational speed of a wheel will increase if the torque $T_{tij}$ supplied is greater that the total force from the ground.

enter image description here

Figure 1: presents the forces and the deformation of the wheel and the surface source: Physics SE

Which basically means that

if the supplied torque is greater than the total of the resistances then the wheel will accelerate.

In the following images you see the forces in question (wheel is moving towards the right):

enter image description here


  • $T_t$ is the torque on the wheel (that is the way its rotating)
  • $F_x$ is the force on the wheel from the surface. Notice that is in the opposite direction. If you don't consider the rolling resistance this is for summing up all other resistances (mainly aerodynamic)
  • $F_{rr}$ is the rolling resistance. This is a resistance due to the deformation of the wheel and as you can see it is always opposite to the direction of movement. In general it can be considered constant and its calculated as $F_{rr} = C_{rr} N$ (where $C_{rr}$ is the rolling coefficient and $N$ is the vertical reaction).

The key here is that apart from the rolling resistance you have other source of resistance (mainly aerodynamic and maybe hill climb). The main difference, the other resistances and the rolling resistance is that, the rolling resistance is an interaction between the wheel and the road. (While all the other sources of resistance are concerned with forces above above the wheel).

When the vehicle is not accelerating then $F_x$ will be equal to the remaining resistances (e.g. aerodynamic and inclination). Remember that $F_x$ is the force that the surface of the road applies to the wheel (as a reaction).

When the vehicle is accelerating then the total resistances will be less than the torque.

  • $\begingroup$ NMech, your drawing confuses me. You have Frr, a rolling resistance force, acting on the outer tire surface torqueing the wheel in the same direction as the driving torque, which will help the motion forward. Isn't rolling resistance something the driving torque needs to overcome? Wouldn't it be more accurate to represent this resistance as a torque on the wheel opposite to the driving torque? Why should rolling resistance be applied to the outer surface of the tire? $\endgroup$
    – ttonon
    Commented May 13, 2021 at 22:52
  • $\begingroup$ if you think about it, rolling resistance and aerodynamic resistance are both forces that need to be overcome by the torque. Both of them are opposing to the motion (which is to the right in this diagram). So, if you drew the drag it would also be to the left. I think that the confusion arises, from the fact that the torque is pushing to the left in order for its opposite and equal reaction to push to the right. I hope this clarifies my choice of forces' direction convention. Regarding the representation as a torque, any torque can be described as a force at a given distance. $\endgroup$
    – NMech
    Commented May 14, 2021 at 5:37
  • $\begingroup$ NMech, As I understand, your Frr force acts on the bottom outer surface of the tire. This is wrong because it imparts a torque on the wheel in the same direction as the driving torque, and contrary to what you say. $\endgroup$
    – ttonon
    Commented May 15, 2021 at 15:38
  • $\begingroup$ I think the link which i shared may be helpful to understand the effects of rolling resistance. Rolling resistance moment might try to turn wheel in reverse direction. $\endgroup$
    – voo
    Commented May 15, 2021 at 22:13

This is an interesting question. I am not an automotive engineer, so please correct me if I am wrong.

The rolling resistance is a contact force with direction in reverse to the applied force, which in a sense drives up the demand of the driving torque, otherwise, the vehicle will not move forward.

enter image description here

  • $\begingroup$ Hi r13. Yes, rolling resistance is force which occurs in reverse direction of Fx and tends to slow down the car, but the drawing is wrong. Fx and Fr cannot be in same direction. $\endgroup$
    – voo
    Commented May 13, 2021 at 16:58

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