You will also recall that a tyre has a limited ability to stick to the ground. Apply a force that is too large, and the tyre slides. The maximum force that a tyre can take depends on the weight applied to the tyre: F   µW where F is the force on the tyre, µ is the coefficient of adhesion (and depends on tyre compound, ground characteristics, temperature, humidity, phase of the moon, etc.), and W is the weight or load on the tyre.

By Newton's second law, the weight on the tyre depends on the fraction of the car's mass that the tyre must support and the acceleration of gravity, g = 32.1 ft / sec2. The fraction of the car's mass that the tyre must support depends on geometrical factors such as the wheelbase and the height of the centre of gravity. It also depends on the acceleration of the car, which completely accounts for weight transfer.

It is critical to separate the geometrical, or kinematic, aspects of weight transfer from the mass of the car. Imagine two cars with the same geometry but different masses (weights). In a one g braking manoeuvre, the same fraction of each car's total weight will be transferred to the front. In the example of Part 1 of this series, we calculated a 20% weight transfer during one g braking because the height of the CG was 20% of the wheelbase. This weight transfer will be the same 20% in a 3500 pound, stock Corvette as in a 2200 pound, tube-frame, Trans-Am Corvette so long as the geometry (wheelbase, CG height, etc.) of the two cars is the same. Although the actual weight, in pounds, will be different in the two cases, the fractions of the cars' total weight will be equal.

Separating kinematics from mass, then, we have for the weight W = f(a)mg where f(a) is the fraction of the car's mass the tyre must support and also accounts for weight transfer, m is the car's mass, and g is the acceleration of gravity.

Finally, by Newton's second law again, the acceleration of the tyre due to the force F applied to it is a = F / f(a)m We can now combine the expressions above to discover a fascinating fact:

a = F / f(a)mamax

The maximum acceleration a tyre can take is µg, a constant, independent of the mass of the car! While the maximum force a tyre can take depends very much on the current vertical load or weight on the tyre, the acceleration of that tyre does not depend on the current weight. If a tyre can take one g before sliding, it can take it on a lightweight car as well as on a heavy car, and it can take it under load as well as when lightly loaded. We hinted at this fact in Part 2, but the analysis above hopefully gives some deeper insight into it. We note that amax being constant is only approximately true, because µ changes slightly as tyre load varies, but this is a second-order effect (covered in a later article).

So, in an approximate way, we can consider the available acceleration from a tyre independently of details of weight transfer. The tyre will give you so many gees and that's that. This is the essential idea of the traction budget. What you do with your budget is your affair. If you have a tyre that will give you one g, you can use it for accelerating, braking, cornering, or some combination, but you cannot use more than your budget or you will slide. The front-back component of the budget measures accelerating and braking, and the right-left component measures cornering acceleration. The front-back component, call it ay, combines with the left-right component, ax, not by adding, but by the Pythagorean formula:

Rather than trying to deal with this formula, there is a convenient, visual representation of the traction budget in the circle of traction. Figure 2  shows the circle. It is oriented in the same way as the X-ray view of the contact patch, Figure 1, so that up is forward and right is rightward. The circular boundary represents the limits of the traction budget, and every point inside the circle represents a particular choice of how you spend your budget. A point near the top of the circle represents pure, forward acceleration, a point near the bottom represents pure braking. A point near the right boundary, with no up or down component, represents pure rightward cornering acceleration. Other points represent Pythagorean combinations of cornering and forward or backward acceleration.