## 3 Results and Discussion
### Possible Operative Modes of the Test Rig
This would allow exploring the stability of vehicles at a limited number of angles, equally-spaced by a fixed angle (e.g., \(10^{\circ}\) or \(20^{\circ}\)) until the turntable performs a complete rotation with a low number of stop-and-go not to lengthen a test too much. Then, it will be possible to thicken the measurements close to the angle of rotation where the stability conditions will be more critical. In each of the two described modes (completely static, quasi-static), a condition of incipient overturning for the vehicle corresponds to a condition of null sustained force on at least one support of the vehicle i.e., on one wheel [4] or, in this case, on a quadrant. If the chassis and each element involved in supporting the weight of the vehicle (e.g., tires, suspensions) are considered to have an infinite stiffness, the above-enunciated condition regards two supports (or quadrants). In a real situation, considering the elasticity of all the elements of the vehicle involved in supporting its weight, at least only the tires, a critical condition for the stability is considered to be already reached when one of the quadrants registers a null sustained force. Possibility of Using the Position of the CoG Projection to Express the Incipient Roll-over Condition  
In the situation described in Figure 4, a tractor is placed motionless on an inclined slope with the longitudinal axis perpendicular to the maximum slope direction (i.e., \(\beta^{\prime}=90^{\circ}\)); the following hypotheses are valid for this explanation:  
* the front and rear tracks have the same width, hence L and R (i.e., the centers of the left/right tires) are limit points for the rollover also in the transversal plane containing the \(CoG\);
* the tractor \(CoG\) is exactly in the median longitudinal plane of the vehicle (hence: \(\overline{LH}=\overline{HR}\)). The same figure shows also the weight force and the boundary reactions that occur in correspondence to the tractor points of supports on the ground (i.e., the centers of the tires). By imposing the equilibrium of the system (the vehicle) on the vertical translation and on the rotation around a point (in this case, the point \(L\)), i.e., the first and second condition for the static equilibrium of rigid bodies, the following expressions are obtained for the modules of the binding reactions (and of their components perpendicular to the support plane) in the points \(L\) and \(R\):  
\[\left\{ \begin{array}{l}R_{R}=\frac{p}{2}\cdot\left(1-\frac{h}{\ell}\tan \alpha\right)\\ R_{L}=\frac{p}{2}\cdot\left(1+\frac{h}{\ell}\tan\alpha\right)\end{array} \right. \Rightarrow\left\{ \begin{array}{l}R_{R\perp}=R_{R}\cos\alpha=\frac{p}{2} \cdot\left(\cos\alpha-\frac{h}{\ell}\sin\alpha\right)\\ R_{L\perp}=R_{L}\cos\alpha=\frac{p}{2}\cdot\left(\cos\alpha+\frac{h}{\ell}\sin \alpha\right)\end{array} \right. \tag{2}\]  
The ratio between these two components of the binding reactions perpendicular to the support plane at the points \(L\) and \(R\) is:  
\[r_{BR}=\frac{R_{R\perp}}{R_{L\perp}}=\frac{R_{R}\cos\alpha}{R_{L}\cos\alpha}= \frac{R_{R}}{R_{L}}=\frac{\ell-h\tan\alpha}{\ell+h\tan\alpha} \tag{3}\]  
In the same situation it is possible to calculate the distance of \(CoG^{\prime}\) from \(L\) and \(R\), and their ratio:  
\[\left\{ \begin{array}{l}\frac{\overline{CoG^{\prime}L}}{\overline{CoG^{\prime }R}}=\ell-h\tan\alpha\\ \overline{CoG^{\prime}L}=2\ell-\overline{CoG^{\prime}L}=\ell+h\tan\alpha \end{array} \right.\Rightarrow r_{dist}=\frac{\overline{CoG^{\prime}L}}{ \overline{CoG^{\prime}R}}=\frac{\ell-h\sin\alpha}{\ell+h\sin\alpha} \tag{4}\]  
As it is possible to observe from the previous equations, we have that \(r_{BR}=r_{dist}\) and therefore, starting from the position of the \(CoG\) projection on the support plane, it is possible to calculate both the modules of the binding reactions \(R_{R}\) and \(R_{L}\), and the modules of their components perpendicular to the support plane \(R_{R\perp}\) and \(R_{L\perp}\).