[su_list icon=”icon: angle-double-right” icon_color=”#3498db”]


If we put together all that we discussed until now, we can recognize how the vertical loads will change very dynamically, above all if we consider a car with significant aero forces. The three main drivers in determining the vertical load acting on each tyre are:

  • Car mass and static weight distribution (or CG position)
  • Aerodynamic forces, which depend on the square of the speed
  • Load transfer, that depends on vehicle acceleration

In our tyre article, we saw how tyres friction circle (or ellipse) change its dimensions and shape depending, among others, on the vertical load. This means that the maximum planar force that each tyre can exchange with the road varies as the vertical load changes and this happens, for a given car (read for a given mass and CG position), depending on vehicle acceleration (load transfer) and speed (downforce).

It follows that even ignoring more complex effects like tyre temperatures and pressure, a car with non-negligible aerodynamic forces will have a grip potential varying constantly during a lap or even during a single manoeuvre.

An example of a similar situation is what happens during a braking phase. The car will approach the braking point still increasing its speed, with the driver keeping the throttle fully open on the straight preceding a corner. At a certain point in time, the driver will push on the brake pedal and start to decelerate. In this instant (we will call it t0), the car will travel at a certain speed V0 and each tyre will experience a vertical load depending on car mass and static weight distribution, on the portion of downforce acting on it and on the load transfer (driven by the braking acceleration that follows brakes action, started by the driver pushing on the brake pedal).

If we focus our attention on a single tyre, we can call Fz0 the vertical load acting on it at t0. This vertical load will drive a maximum planar force that the tyre can exchange with the road, that we will call F0. We say that F0 is a function of the vertical load Fz0, mathematically:

This is typically the moment when car grip potential is at its maximum during the manoeuvre because the downforce acting on the tyres is still at its highest point. Consequently, also the weight transfer normally assumes its maximum magnitude in this instant, loading front tyres even further. This one of the main reasons why a good driver knows this is exactly the moment when he has to apply the highest force on the brake pedal, gradual releasing it as the car slows down (and its grip reduces because its downforce also decreases). As we will see in another article, this is one of the key points in drivers braking technique with high downforce car without braking assistance devices.

In a following instant t1, the car will have a slower speed V1 and our tyre will experience a lower vertical load Fz1. Therefore, the available planar force F1 will also be smaller.

The driver will have to reduce the force he applies on the brake pedal in order to avoid any tyre lock because the overall grip of the car has reduced.

As the reader can realize, the situation would be already complex to handle if each tyre would experience the same grip as all the others because said grip would anyway change during the manoeuvre, depending on travelling speed. To further complicate driver’s life each tyre will experience different vertical loads at any point in time (because of static weight distribution and load transfer, besides downforce) and, hence, different maximum achievable planar forces (moreover very often racecars don’t have same tyre specification at the front and rear axle).

Ideally, we would want our tyres to explore constantly their full potential. Our braking system should be able to vary the braking torque applied to each wheel, depending on the available grip, in order to exchange with the road always the maximum achievable force in each point in time. This, at least in part, what certain anti-lock systems try to achieve.
In many classes, anyway, such systems are forbidden and the driver has always to move between a tyre lock and a non-optimal braking.

In general, for the car to maintain stability it is desirable to reach a lock on the front axle first. The reason for this is easy to explain: if a front tyre is locked, the car loses its ability to change direction by steering and tends to maintain a straight motion or to straighten its trajectory. Inversely, if the rear tyres lock first, the car could incur in a spin or, anyway, in a less controllable situation.

To partially help the drivers and better adapt to a wider range of conditions, racecars often have a device allowing to change how big the braking torque applied to the front wheels is compared to the one applied to the rear wheels. We normally refer to the proportion between the front and total brake torque as “brake balance”. This metric is normally expressed as a percentage and can be calculated using the following equation:

Drivers often change car brake balance either to simply react to tyres locking during the braking phase (at the front or at the rear), for example, because of tyre degradation or changed track conditions, or even to influence car behaviour.

We can now analyze the forces acting on the car during a braking situation. The picture below summarizes all these actions and also indicates with “V” actual car speed.

Fxf and Fxr are front and rear axle overall braking forces (sum of the horizontal forces that the ground applies to the car at the front and rear contact patches).

Fzf and Fzr are the vertical forces acting at the front and rear axle contact patches, reacting car weight (mg), downforce (Fz) and contributing to balance also the moments of horizontal actions like the inertial force -mAx and the aerodynamic drag Fx.

We know that, if the tyres are used at the edge of their performance, Fxf and Fxr depend strictly on Fzf and Fzr through the longitudinal friction coefficient and its load sensitivity.

If we write the equilibrium equations in longitudinal and vertical direction and the rotational balance (all moments about front contact patch), we have:


  • h is the distance of the CG from the ground
  • e is the distance of the CP from the ground
  • a is the horizontal component of the distance of the CG from the front axle
  • b is the horizontal component of the distance of the CG from the rear axle
  • c is the horizontal component of the distance of the CP from the front axle
  • d is the horizontal component of the distance of the CP from the rear axle


The first thing that we can notice is that braking is probably the only situation where the aerodynamic drag helps because it contributes to the braking actions slowing the car down. One thing to keep in mind is that the above equations only consider external forces. If we would break them down to understand which braking torque the brakes should apply to the wheels in order to have an acceleration Ax, we should also consider the engine braking torque, which is the resistance that the engine produces when the throttle is not open. This “negative” torque contributes to the car internal actions that goes into the net braking force.

Also, in this case, the tyre horizontal forces also incorporates front and rear axle tyres rolling resistance, that also contributes to slowing the car down.

Another important point that we can extrapolate from the third equation (rotational equilibrium), is that, while the inertial force “-mAx” produce a load transfer toward the front axle, the aerodynamic drag loads the rear axle.

To have a feeling about the magnitude of the forces acting during braking in race cars, we can consider again the case of an LMP2 car. It is not untypical for these vehicles to produce decelerations exceeding 2.5g (where g is the gravitational constant, equal to about 9.81 m/s2). Since the weight of the car, with the driver on board and some fuel is about 1030kg, we have an inertial force -mAx with a magnitude greater than 25 kN, that need to be balanced by the brakes torques, aerodynamic drag, rolling resistance and engine braking torque.