Fast Car Physics

Chapter 4. Basic Vehicle Dynamics: Load Transfer and Tires

Chapter 4 Basic Vehicle Dynamics: Load Transfer and Tires 4.1 CENTER OF GRAVITY The most fundamental physics quantity involved in vehicle dynamics is the center of gravity. In physics, we tend to talk about the center of mass, while engineers talk about the center of gravity (CG). The good news is that as long as the local gravitational field is uniform, the two are identical. Like everyone else, we’ll use the terms interchangeably. The center of mass is the mass-weighted average position of the mass of a system. As far as gravity is concerned, we can treat the system as if all of the mass were located at this single point. For a string of small masses strung out in a line, we add up the product of the particle mass, mi , times the x-coordinate, xi , for its position and divide by the total mass, MT : xCM  mixi MT . Of course, a car is a three-dimensional object, so we will need to add to this a calculation to the y and z centers of mass: basic vehicle dynamics 83 yCM  miyi MT zCM  mizi MT . For the parts that aren’t small, we will need to do an integral. Since most of us are going to use production cars and not design them, both the sum and the integration seem a bit challenging. Instead, we will measure a few parameters for the car and do an equilibrium calculation to find the center of gravity. We will do three separate calculations to find the longitudinal CG, the lateral CG, and the height of CG. Let’s assume that all the fluids are full and the tires are at operating pressure. 4.2 LONGITUDINAL AND LATERAL CENTER OF GRAVITY Figure 4.1 is a side-view free-body diagram for our car at rest. The dimension labeled L is the wheelbase of the car and our unknown quantity is d, the horizontal distance from the front wheel hub center to the center of gravity. Five Figure 4.1 Side-view free-body diagram for a car at rest. NF and NR are the front and rear normal forces at the tires, respectively. W is the weight. L is the wheelbase of the car, and d is the unknown location of the CG behind the midpoint of the front wheel. x CG NF NR L d W z y 84 fast car physics external forces act on the car, gravity and a normal force at each tire. For the purpose of finding the distance d, we will combine the two front normal forces as well as the two rear normal forces. For the equilibrium calculation, we will sum the forces by component and set them equal to zero, as well as summing the torques and setting those equal to zero. Since the forces act only in the z-direction, two of the sums are eliminated: (4.1) Fz  0 NF NR  W  0   0 NF d  NR (Ld)  0 d  L NR . W How do we determine values for the forces to solve this equation? We measure the four normal forces by placing a scale under each wheel on a level floor. Figure 4.2 Rear view of the static forces acting on the car. T is the wheel track and t is the distance from the left wheel center to the center of gravity. x NRight NLeft W CG z x T t basic vehicle dynamics 85 The sum of the normal forces is equal to the weight. The distance d will then help us to evaluate the load transfer under braking. Figure 4.2 is the back view of the car. The quantity T is called the wheel track or simply the track. NLeft is the sum of the normal forces on the two lefthand tires, and NRight is for the right side. We can apply an identical procedure to the longitudinal case for this problem and solve for t, the distance to the centerline of the left-hand wheels. For a symmetric car t = T/2. A driver is probably the largest asymmetry in most cars. The battery is next in mass. It still should be nearly T/2. The resulting expression for t is t  T NLeft . W  (4.2) 4.3 HEIGHT OF THE CENTER OF GRAVITY To find the height of the center of gravity, we will again perform...