## 2 Mathematical Formulation
\begin{array}{c}\frac{\mathrm{p}(\mathrm{b}_{1})}{\mathrm{p}}\ =\ \mathrm{A}_{\mathrm{B1}}v_{ \mathrm{r}}\omega\sin(\alpha)\sin(\beta_{1})\sqrt{\mathrm{c}^{2}-\left(-\mu+ \mathrm{b}_{1}\right)^{2}}+\mathrm{A}_{\mathrm{B1}}w_{1}\frac{\mathrm{c}\dot{ c}+\dot{\mu}\left(-\mu+\mathrm{b}_{1}\right)}{\sqrt{c^{2}-\left(-\mu+\mathrm{b}_{1} \right)^{2}}}\\ -\frac{1}{2}\left(\frac{\mathrm{A}_{\mathrm{B1}}w_{1}\left(-\mu+\mathrm{b}_{1} \right)}{\sqrt{c^{2}-\left(-\mu+\mathrm{b}_{1}\right)^{2}}}+v_{\mathrm{t1}} \right)^{2}\ =\ 0\end{array}  \tag{33}\]  
where \(\mathrm{b}_{1}\) is the horizontal distance between the keel and knuckle in the right side, the one wet chine-phase starts when \(\mathrm{c}_{1}>\mathrm{b}_{1}\). In Equation (33), the effect of the acceleration is not considered. The half beam in the left side is evaluated according with Toyama [3]:  
\[\mathrm{c}_{2}\ =\ \frac{\pi\mathrm{R}\mathrm{z}_{0}}{\left(\mathrm{R}+1\right) \left(1+\mu_{0}\right)\sqrt{1-\mu_{0}^{2}}}\left(\frac{1}{\tan(\beta_{2})}\right) \tag{34}\]  
\[\dot{\mathrm{c}}_{2}\ =\ \frac{\pi\mathrm{R}\mathrm{w}}{\left(\mathrm{R}+1\right) \left(1+\mu_{0}\right)\sqrt{1-\mu_{0}^{2}}}\left(\frac{1}{\tan(\beta_{2})}\right) \tag{35}\]  
\[\dot{\mathrm{c}}_{2}\ =\ -\dot{\mu}+\dot{\mathrm{c}}. \tag{36}\]  
Solving Equations (33)-(36) the values for \(\mathrm{c}\), \(\dot{\mathrm{c}}\), \(\mu\), and \(\dot{\mu}\) are calculated for the flow separation in the right side. When the two dry chines-phase occurs the boundary conditions are the set presented in Equation (37). The equations are solved in a coupled form to find the values of \(\mathrm{c}\), \(\dot{\mathrm{c}}\), \(\mu\), and \(\dot{\mu}\). \[\left\{\begin{array}{c}\frac{\mathrm{p}(\mathrm{b}_{1})}{\mathrm{p}}\ =\ \mathrm{A}_{\mathrm{B1}}v_{ \mathrm{r}}\omega\sin(\alpha)\sin(\beta_{1})\sqrt{\mathrm{c}^{2}-\left(-\mu+ \mathrm{b}_{1}\right)^{2}}+\frac{\mathrm{A}_{\mathrm{B1}}w_{1}\left(\mathrm{c} \dot{c}+\dot{\mu}\left(-\mu+\mathrm{b}_{1}\right)\right)}{\sqrt{c^{2}-\left(- \mu+\mathrm{b}_{1}\right)^{2}}}\\ -\frac{1}{2}\left[\frac{\mathrm{A}_{\mathrm{B1}}w_{1}\left(-\mu+\mathrm{b}_{1} \right)}{\sqrt{c^{2}-\left(-\mu+\mathrm{b}_{1}\right)^{2}}}+v_{\mathrm{t1}} \right]^{2}\where \(\rm{b_{2}}\) is the horizontal distance between the keel and knuckle in the left side, the two wet chine-phase starts when \(\rm{c_{2}}>\rm{b_{2}}\). The sectional forces can be computed as:  
\[\rm{f_{z}}\ =\ \int_{-y_{2}}^{y_{1}}\rm{pdy} \tag{38}\]  
\[\rm{f_{y}}\ =\ \tan\beta_{2}\int_{-y_{2}}^{0}\rm{pdy}-\tan\beta_{1}\int_{0}^{y_{ 1}}\rm{pdy} \tag{39}\]  
\[\rm{m_{x}}\ =\ \frac{1}{\cos^{2}\beta_{2}}\int_{-y_{2}}^{0}\rm{ypdy}+\frac{1}{ \cos^{2}\beta_{1}}\int_{0}^{y_{1}}\rm{ypdy} \tag{40}\]  
\[\rm{m_{xcg}}\ =\ \rm{m_{x}}-\rm{V_{cg}}\sin\phi\rm{f_{z}}\ +\rm{V_{cg}}\cos \phi\rm{f_{y}} \tag{41}\]  
where \(\rm{f_{z}}\) is the vertical force, \(\rm{f_{y}}\) is the horizontal force, \(\rm{m_{x}}\) is the roll moment in the keel, \(\rm{m_{xcg}}\) is the roll moment in the CG of the section, and \(\rm{y_{1}}\) and \(\rm{y_{2}}\) are the horizontal wetted coordinate in the right side and left side respectively. In case of prescribed motions, these equations can be used to evaluate the pressure and forces in the section.