## 2 Mathematical Model  
### Dynamic Model of Water Surface Take-Off in Body Coordinate System  
The take-off of an HAUQ on the water surface mainly refers to the process that as the HAUQ body is totally or partially submerged in the water medium, the pulling force generated by the rotation of the quadrotor pulls it out of the water and drains the residual water in the body. In order to establish the dynamic equation of an HAUQ, the inertial coordinates \(\{\mathcal{F}^{I}\}=\{O,X,Y,Z\}\) of water-air integration are established. In this coordinate system, a point on the water surface where the flight path is located is the reference system origin. Any direction of the water surface is taken as the positive \(OX\) direction. The \(OZ\)-axis is perpendicular to the water surface and upward. The water depth is negative, and the height is positive. The \(OY\) is perpendicular to the \(OXZ\) plane and determined by the right-hand rule. Define the body coordinate system as \(\{\mathcal{F}^{b}\}=\{O^{b},X^{b},Y^{b},Z^{b}\}\). \(R_{v}\in SO(3)\) is the velocity transform matrix from coordinate system \(\mathcal{F}^{b}\) to coordinate system \(\mathcal{F}^{I}\). Define the HAUQ position as \(P^{I}=[x,y,z]^{T}\) which represents the position of the trans-media UAV in the inertial coordinates \(\{\mathcal{F}^{I}\}=\{O,X,Y,Z\}\). The flight attitude is \(\Theta=[\phi,\theta,\psi]^{T}\) and represents the roll angle, the pitch angle, and the yaw angle, respectively. \(V^{b}\in\mathbb{R}^{3}\) is the velocity component in the body coordinate system. \(\omega^{b}\) is the angular velocity in the body coordinate system. Let the mass of the HAUQ without water be \(M\), and the mass of the HAUQ is \(M+M_{\Delta}(t)\) during the take-off process on the water surface. \(M_{\Delta}(t)\to 0\) caused by the fast drainage holds and changes rapidly. It is worth noting that the \(M_{\Delta}(t)\) does not include the added mass caused by underwater navigation, which is considered as the uncertainty. The rapid mass change also brings about the change in the center of mass position and the moment of inertia. The moment of inertia matrix is defined as \(J+J_{\Delta}(t)\). We have  
\[\dot{P}^{I}=R_{v}V^{b},\;\dot{R}_{v}=R_{v}S(\omega^{b}) \tag{1}\]  
where  
\[S(x)= \begin{bmatrix}0&-x_{3}&x_{2}\\ x_{3}&0&-x_{1}\\ -x_{2}&x_{1}&0\end{bmatrix} ,\;\forall x= \begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix} \]  
From [PERSON]'s laws of motion, we have Equation (3),  
\[(M+M_{\Delta}(t))\dot{V}^{b}=-(M+M_{\Delta}(t))S(\omega^{b})V^{b} +A_{v}(V^{b})+ \begin{bmatrix}0\\ 0\\ -(F_{1}+F_{2}+F_{3}+F_{4})\end{bmatrix} +R_{v}^{T} \begin{bmatrix}0\\ 0\\ (M+M_{\Delta}(t))g\end{bmatrix} +\Delta_{1}\] \[(J+J_{\Delta}(t))\dot{\omega}^{b}= \begin{bmatrix}-(l+\Delta I_{1 }(t))F_{1}+(l+\Delta I_{2}(t))F_{2}-(l+\Delta I_{3}(t))F_{3}+(l+\Delta I_{4}(t ))F_{4}\\ (l+\Delta I_{1}(t))F_{1}+(l+\Delta I_{2}(t))F_{2}-(l+\Delta I_{3}(t))F_{3}-(l+ \Delta I_{4}(t))F_{4}\\ -C(F_{1}-F_{2}+F_{3}-F_{4})\end{bmatrix} \] \[-S(\omega^{b})(J+J_{\Delta}(t))\omega^{b}+B_{\omega}(\omega^{b})+ \Delta_{2} \tag{2}\]  
where \(\Delta M(t)\) represents the change in the extra mass during the drainage process of cross-medium flight, \(\Delta I_{i}(t),i=1,...,4\) is the change in the arm of force caused by the change in the mass center, \(\Delta J(t)\) is the change moment of inertia with time \(t\), and \(A_{v}(V^{b})\) and \(B_{\omega}(\omega^{b})\) are the unknown terms including forces and moments caused by the complex multi-media dynamics and the fast-changing buoyancy. \(\Delta M(t)\to 0\), \(\Delta I_{i}(t)\to 0,i=1,...,4\)and \(\Delta J(t)\to 0\) during the fast drainage. \(\Delta_{i},\ i=1,2\) is the external disturbance. Define \(F^{b}=[F_{1},F_{2},F_{3},F_{4}]^{T}\) and  
\[N= \begin{bmatrix}0&0&0&0\\ 0&0&0&0\\ -1&-1&-1&-1\end{bmatrix} ,\ G= \begin{bmatrix}0\\ 0\\ M\end{bmatrix} ,\ G_{\Delta}(t)= \begin{bmatrix}0\\ 0\\ M_{\Delta}(t)\end{bmatrix} ,\ L= \begin{bmatrix}-l&l&-l&l\\ l&l&-l&-l\\ -C&C&-C&C\end{bmatrix} ,\] \[L_{\Delta}(t)=