## 3 Results and Discussion
### Effect of the Tyres Deformation on the Vehicle Stability
#### 3.6.1 Effect on the CoG Position of a Load on the Tyres Perpendicular to the Vehicle Support Plane  
The higher the center of gravity is positioned with respect to the tyre sidewall, the more pronounced is the effect of the downstream shift of the CoG as a consequence of the perpendicular displacement of the tyres. To illustrate this effect, it is possible to assume a first simple initial placement of the vehicle in the test rig with the longitudinal axis perpendicular to the direction of the maximum gradient (e.g., in a situation similar to the lateral tilting test). In this case, the equivalent coefficient of elasticity \(k_{\perp}\) in the perpendicular direction for each of the two sides of the tractor is the same, the front and rear wheels being equal in pairs (hence, with the same coefficient of elasticity \(k_{\perp,FL}=k_{\perp,FR}\) and \(k_{\perp,RL}=k_{\perp,RR}\)). In particular, the equivalent coefficient of elasticity \(k_{\perp}\) for each side can be calculated as the sum of the modulus of elasticity of each of the two front wheels \(k_{\perp,FL}=k_{\perp,FR}\) and rear wheels \(k_{\perp,RL}=k_{\perp,RR}\) (the wheels of each side constitute two elastic systems in parallel). With reference to Figure 13 (using the same symbols of Section 3.3), if the distance of H (i.e., the projection of CoG perpendicular to the support plane) from \(L\), the center of left support, is (\(\ell-\Delta\ell\)) (therefore with a possible deviation \(\Delta\ell\) from the centreline of the vehicle, the additional angle \(\Delta\alpha\) of lateral inclination of the vehicle frame and the consequent new distance (\(\ell-\Delta\ell\))\({}_{def}\) of the projection of CoG perpendicular to the support plane from \(L\) can be calculated from the resulting forces on the left side \(R_{\perp,\perp}\) and on the right side \(R_{R,\perp}\) in the absence of deformation through the following equations:  
\[\left\{ \begin{array}{l}k_{\perp}=k_{\perp,FL}+k_{\perp,RL}=k_{\perp,FR}+k_{ \perp,RR}\\ \sin(\Delta\alpha)=\frac{\frac{R_{\perp,L}}{k_{\perp}}-\frac{R_{\perp,L}}{k_{ \perp}}}{2\ell}=\frac{R_{\perp,L}-R_{\perp,L}}{2\ell\lambda_{\perp}}\\ \left(\ell-\Delta\ell\right)_{def}=\left(\ell-\Delta\ell\right)\cdot\cos( \Delta\alpha)-\left(h-h_{t}\right)\cdot\sin(\Delta\alpha)\end{array} \right. \tag{16}\]  
Figure 13: Graphical representation of the effect of a pure compression of the tyres due to the weight component in the direction perpendicular to the support plane, supported by each tyre ((**a**) initial situation–tyres not deformed; (**b**) final situation–tyres deformed). Once the new distance \((\ell-\Delta\ell)_{def}\) has been calculated, it is possible to calculate the new forces applied to the supports (quadrants) of the two sides of the vehicle and then iteratively apply the same equations until numerical convergence. The above-presented scheme may also be applied to a vehicle with a different orientation in space (e.g., during a quasi-static test in which the turntable has brought the longitudinal axis of the vehicle into a generic position; see Figure 14), using the same equations presented above in the direction of the vehicle longitudinal and transverse axis, to calculate the roll and pitch angle increments for the chassis. In this scenario, starting from the initial distribution of the weights on the supports (i.e., corresponding to an infinite perpendicular stiffness for the tyres), it is, therefore, possible to calculate the inclination of the chassis and then, iteratively, the new position of the center of gravity \(CoG_{def}\). In a quasi-static mode test, the effect of the sole vertical compression of the tyres causes a displacement of \(CoG^{\prime}\) towards the outside of the support polygon and, therefore, a widening of the locus described by \(CoG^{\prime}\) during a test.