created at September 5, 2021

When the car accelerates, its rear axle is being loaded more than it does in the resting state, whereas the front axle becomes underloaded. But why does it happen?

It's enough to analyze the car in the plane (in-plane problem), because the forces will be symmetric to the car's symmetry plane when moving straight forward. Let's display the main forces acting on the car at uniformly accelerated forward movement:

These are: the gravity force mg (the center of gravity is marked with black dot C), axle reaction forces N1 and N2, the thrust force Ft, which is acting in the road's surface plane. It's not so important in this case what's the type of transmission, the thrust for will act in the road's surface plane anyway, because in essence it is the friction force between the tire and road surface.

Let's also introduce the necessary geometrical parameters. These are: the wheelbase L, the horizontal distances between the axles and the center of gravity L1 and L2 (so that L = L1 + L2), and also the height of the center of gravity point over the road surface. The last one is quite tricky to figure out, but we need to consider some value for calculation.

The vehicle being accelerated is quite a dynamic process, although if we consider the acceleration to be nearly constant, we can investigate the vehicle's static equilibrium during acceleration. The thrust force Ft doesn't go through the center of gravity, thus it creates a torque, which tries to twist a car counter-clockwise (when looking on the image above). However, in fact we know this doesn't happen, because the torque is being compensated by another torque, produced by the axle reactions N1 and N2. In other words, the reactions N1 and N2 are readjusting during acceleration to prevent vehicle from twisting in its symmetry plane.

Let's write down two equations of statics: displacement equation for y axis, and torque equations for center of gravity point:

$\begin{cases} Y:\quad N1 + N2 - m\cdot g = 0 \\ M_{C}:\quad N1\cdot L1 + F_t\cdot h - N2\cdot L2 = 0 \end{cases}$

Expressing the N1 from the first equation and putting to the second equation:

$N1 = m\cdot g - N2;$
$(m\cdot g - N2)\cdot L1 + m\cdot a\cdot h - N2\cdot L2 = 0$
$N2 = \frac{m}{L}(g\cdot L1 + a\cdot h)$

We got N2, putting it back to the first equation:

$N1 = m\cdot \left (g - \frac{g\cdot L1 + a\cdot h}{L} \right ) =$
$m\cdot \left( \frac{g\cdot L}{L} - \frac{g\cdot L1}{L} - \frac{a\cdot h}{L} \right) =$
$\frac{m}{L} \left( g\cdot L2 - a\cdot h\right)$

## What conclusions can be drawn?

When there is no acceleration (a = 0), but the uniform movement only, the N1 and N2 reactions can be found from the simple proportion - the closer the center of gravity is to the axle, the more load it takes. When acceleration is positive (car accelerates), N1 decreases (the forward axle is underloaded) and N2 increases (the rear axle is overloaded). When the car stops (negative acceleration), it's all happens vice-versa - to front axle is overloaded and the rear axle is underloaded. This is the reason why the front brakes are usually more massive to be able to take higher load.

Also we see, that when a·h reaches G·L2, the load is fully drawn away from the front axle and the rear axle takes the full vehicle's weight. When increasing the a·h further the car will topple back. In the real life it's hard to imagine, because the car likely won't reach so high acceleration a, and also the height h is relatively small comparing to the L2 distance. If summarize, to make it easier to topple the car back we can: increase acceleration a, increase the height h, decrease the distance L2. For example, the sports motorcycle has a higher h value comparing to the L2, so it's much easier to rear the motorcycle.

Another conclusion is about the advantage of the rear (full) wheel drive vehicles comparing to the front wheel drive vehicles in accelerating state. RWD vehicles are transferring the torque to the rear axle, which is overloaded, thus the tire-asphalt friction force also increases, allowing us to transfer more power. While designing the RWD cars, it's usually the goal to move the center of gravity back to the rear axle as much as possible. The weight distribution of BMW M3 is close to 50/50, whereas the front wheel drive vehicles could typically have the weight distribution around 60/40 (the center of gravity is closer to the front axle).

In the real-life scenario the acceleration won't be constant. It's usually much higher at the very start and decreases while the speed becomes higher. From the second Newton's law we can estimate the starting acceleration as:

$a = \frac{T\cdot r1\cdot r2}{R\cdot m}$

where T - engine torque, r1 - gearbox ratio, r2 - the final gear ratio, R - wheel radius, m - vehicle mass.