## 3 Mathematical Solution of the Problem  
### Transformation and Solving the Established Equation System  
The angle \(\beta_{d}\) could be calculated by taking the right sides from Equation (13) as equal:  
\[\frac{d-R\cdot sin\beta_{d}}{cos\beta_{d}}=d\cdot ln\left[\frac{tan\left(\frac{ \beta_{max}}{2}\right)}{tan\left(\frac{\beta_{d}}{2}\right)}\right] \tag{14}\]  
Both sides of the newly formed Equation (14) can be divided by the datum length \(d\), and then transformed:  
\[d\cdot\frac{1-\frac{R}{d}\cdot sin\beta_{d}}{cos\beta_{d}}=d\cdot ln\left[ \frac{tan\left(\frac{\beta_{max}}{2}\right)}{tan\left(\frac{\beta_{d}}{2} \right)}\right] \tag{15}\]  
\[\frac{1}{cos\beta_{d}}-\frac{R}{d}\frac{sin\beta_{d}}{cos\beta_{d}}=ln\left[ \frac{tan\left(\frac{\beta_{max}}{2}\right)}{tan\left(\frac{\beta_{d}}{2} \right)}\right] \tag{16}\]  
\[\frac{1}{cos\beta_{d}}-\frac{R}{d}\cdot tg\beta_{d}=ln\left[\frac{tan\left( \frac{\beta_{max}}{2}\right)}{tan\left(\frac{\beta_{d}}{2}\right)}\right] \tag{17}\]  
The maximum steering angle \(\beta_{max}\) can be calculated applying Equation (1) from the ICAO mathematical model, described in the previous section:  
\[tan\left(\frac{\beta_{max}}{2}\right)=\frac{1-e^{\text{K}\Theta_{\text{S}}}}{ X-K-X\cdot e^{\text{K}\Theta_{\text{S}}}-K\cdot e^{\text{K}\Theta_{\text{S}}}} \tag{18}\]  
Considering Equations (2) and (3), the angle \(\beta_{max}\) directly depends on the steering path circular radii \(R\), the datum length \(d\), and the turning angle \(\theta_{s}\) which are constant when performing the turning maneuver. In accordance with that, the angle \(\beta_{max}\) also represents a constant value. Therefore, in order to facilitate the solving of Equation (17), the constant \(c\) is introduced:  
\[c=tan\left(\frac{\beta_{max}}{2}\right) \tag{19}\]  
To further simplify the Equation (17), the following trigonometry half-angle identities are used:  
\[cos\beta_{d}=\frac{1-tg^{2}\left(\frac{\beta_{d}}{2}\right)}{1+tg^{2}\left( \frac{\beta_{d}}{2}\right)} \tag{20}\]  
\[tg\beta_{d}=\frac{2\cdot tg\left(\frac{\beta_{d}}{2}\right)}{1-\text{tg}^{2} \left(\frac{\beta_{d}}{2}\right)} \tag{21}\]  
and Equation (17) takes the following form:  
\[\frac{1}{\frac{1-tg^{2}\left(\frac{\beta_{d}}{2}\right)}{1+tg^{2}\left(\frac {\beta_{d}}{2}\right)}}-\frac{R}{d}\cdot tg\beta_{d}=ln\left[\frac{c}{tan\left( \frac{\beta_{d}}{2}\right)}\right] \tag{22}\]  
\[\frac{1+tg^{2}\left(\frac{\beta_{d}}{2}\right)}{1-tg^{2}\left(\frac{\beta_{d} }{2}\right)}-\frac{R}{d}\cdot\frac{2\cdot tg\left(\frac{\beta_{d}}{2}\right)} {1-tg^{2}\left(\frac{\beta_{d}}{2}\right)}=ln\left[\frac{c}{tan\left(\frac{ \beta_{d}}{2}\right)}\right] \tag{23}\]Finally, by introducing the substitution \(t=tg\Big{(}\frac{\beta_{d}}{2}\Big{)}\), Equation (23) is transformed into the following:  
\[d\cdot\frac{1-\frac{R}{d}\cdot sin\beta_{d}}{cos\beta_{d}}=d\cdot ln\left[ \frac{tan\left(\frac{\beta_{max}}{2}\right)}{tan\left(\frac{\beta_{d}}{2} \right)}\right] \tag{15}\]0 \[d\cdot\frac{1-\frac{R}{d}\cdot sin\beta_{d}}{cos\beta_{d}}=d\cdot ln\left[ \frac{tan\left(\frac{\beta_{max}}{2}\right)}{tan\left(\frac{\beta_{d}}{2} \right)}\right] \tag{15}\]1  
The last equation represents the most important contribution to the exact mathematical solution of the addressed problem. Its solution first returns the value of the substitution \(t\), and then consequently the required parameters \(F_{d}\) and \(\beta_{d}\) which precisely determine the maximum transient offtracking vehicle position. Nevertheless, Equation (25) belongs to the group of transcendental equations that cannot be solved by classic algebraic methods alone and often require the application of various numerical techniques, i.e., the Newton-Raphson method, bisection method, secant method, regula falsi methods, etc. Additionally, these numerical methods are useful for approximating the equation solution to any desired degree of accuracy [32].