# Tilt-Induced Error Compensation with Vision-Based Method for Polarization Navigation
## 3 Polarization Orientation Error Compensation Method  
Early people found that bees, sand ants, and other animals possess remarkable navigation abilities [33; 34]. Subsequent research revealed that these animals use detected sky polarization information to navigate. The skylight polarization forms a macroscopically stable distribution pattern containing rich directional information, including the degree of polarization (DoP) and the angle of polarization (AoP). The polarization degree and the polarization Angle represent the polarization degree of light and the direction of light vibration (E vector direction), respectively. Polarized light can be described by the Stokes vector. The Stokes vector comprises four components, which is \(S=[I,Q,U,V]\). \(I\) represents the total light intensity. \(Q\) and \(U\) are two mutually perpendicular linearly polarized components, while \(V\) represents circularly polarized light. Given that the sky predominantly exhibits linear polarization, \(V\) is typically a value of zero. The formula for calculating the polarization angle and degree is  
\[AoLP=\frac{1}{2}\arctan\left(\frac{U}{Q}\right) \tag{20}\]  
\[DoLP=\frac{\sqrt{Q^{2}+U^{2}}}{I} \tag{21}\]  
In the polarized light navigation system, the navigation frame and the body frame meet the following conversion relationship:  
\[S_{b}^{p}=C_{n}^{b}S_{n} \tag{22}\]  
where \(S_{b}^{p}\) is the projection of the solar vector within the body frame, \(C_{n}^{b}\) is the transfer matrix, and \(S_{n}\) indicates the sun's position within the navigation frame. When the observed direction aligns with the solar direction, the polarization angle in the body frame can be expressed as  
\[\varphi_{p}^{b}\left(x_{p},y_{p}\right)=\tan^{-1}\left(t\phi_{s}^{b}\right) \tag{23}\]  
where \((x_{p},y_{p})\) is the observation point, and \(\phi_{s}^{b}\) is the solar azimuth in the body frame:  
\[\phi_{s}^{b}=\tan^{-1}(\frac{S_{b}^{p}(2)}{S_{b}^{p}(1)}) \tag{24}\]The transfer matrix from the navigation frame to the body frame can be expressed as  
\[C_{n}^{b}= \begin{bmatrix}\cos\theta\cos\psi&\cos\theta\sin\psi&-\sin\theta\\ \sin\phi\sin\theta\cos\psi-\cos\phi\sin\psi&\sin\phi\sin\theta\sin\psi+\cos\phi \cos\psi&\sin\phi\cos\theta\\ \cos\phi\sin\theta\cos\psi+\sin\phi\sin\psi&\cos\phi\sin\theta\sin\psi-\sin \phi\cos\psi&\cos\phi\cos\theta\end{bmatrix}  \tag{25}\]  
The position of the sun can be solved according to the declination angle of the sun \(\delta_{s}\), the solar hour angle \(T_{S}\), and the latitude of the observation site \(L_{o}\), as per the pertinent astronomical theories. By solving the spherical triangle \(S-O-NP\) in Figure 7, the sun altitude angle \(h_{s}\in[0^{\circ},90^{\circ}]\) and the sun azimuth angle \(\varphi_{gS}\in[0^{\circ},360^{\circ}]\) in the east, north, and up (ENU) geographic coordinate system can be obtained. In Figure 7, the red dot \(S\) represents the sun, \(O\) represents the observation site, \(NP\) represents the North Pole, \(\delta_{s}\) signifies the sun's declination angle, \(T_{S}\) is the sun hour angle, \(L_{o}\) is the latitude of the observation site, \(h_{s}\) is the sun altitude angle, and \(\varphi_{gS}\) is the sun azimuth in the east, north, and up (ENU) geographic coordinate system \(\alpha_{g}y_{g}z_{g}\). The following model can be established in Figure 7:  
\[\left\{ \begin{array}{l}\sin h_{s}=\sin\delta_{s}\sin L+\cos\delta_{s}\cos L \cos T_{s}\\ \cos\varphi_{gS}=\frac{\sin\delta_{s}-\sin h_{s}\sin L}{\cosh h_{s}\cos L} \end{array} \right. \tag{26}\]  
To solve the inverse trigonometric function of Equation (26) and perform quadrant judgment, we can obtain the following:  
\[h_{s}=\arcsin(\sin\delta_{s}\sin L+\cos\delta_{s}\cos L\cos T_{s}) \tag{27}\]  
\[\varphi_{gS}=\arccos\left(\frac{\sin\delta_{s}-\sin h_{s}\sin L}{\cos h_{s} \cos L}\right) \tag{28}\]  
Among them, the calculation formula of the sun's declination angle \(\delta_{s}\) is as follows:  
\[