## 2 Algorithm Principle
### The Principle of Barometric Height Measurement and Its Auxiliary Positioning Algorithm
#### 2.1.1 The Principle of Barometric Pressure Measurement
\(\square\)  
From the definition:  
\[e^{2}=\frac{R_{E}^{2}-R_{P}^{2}}{R_{E}^{2}}=\frac{R_{e}^{2}-R_{P}^{2}}{R_{e}^{2}}. \tag{3}\]  
So we can get:  
\[\frac{R_{E}}{R_{e}}=\frac{R_{P}}{R_{P}}. \tag{4}\]  
Since the moving carrier is located at point A on the large ellipsoid and the plane coordinate is \((x_{c},z_{c})\), the slope of the straight line AO\({}_{1}\) is [16]:  
\[\tan\theta_{Lat}=\frac{R_{e}^{2}}{R_{P}^{2}}\cdot\frac{z}{x}. \tag{5}\]  
We cross point A\({}_{1}\) to make the normal line of the small ellipse and the polar axis intersects at point O\({}_{2}\), the angle between the straight line A\({}_{1}\)O\({}_{2}\) and the \(x\)-axis is \(\theta^{\prime}_{Lat}\), and now cross point A\({}_{1}\) to make the tangent of the small ellipse and the \(x\)-axis intersect at point \(Q_{1}\), and the angle with the \(x\)-axis is \(\phi\). The slope of the tangent A\({}_{1}\)Q\({}_{1}\) is [16]:  
\[-\tan\phi=-\tan(\frac{\pi}{2}-\theta^{\prime}_{Lat})=-\cot\theta^{\prime}_{Lat }=-\frac{R_{p}^{2}}{R_{e}^{2}}\cdot\frac{\chi^{\prime}c}{z^{\prime}c}=-\frac{1 }{\frac{R_{e}^{2}}{R_{p}^{2}}\cdot\frac{\cdot\epsilon}{x^{\prime}}}. \tag{6}\]  
We use A and A\({}_{1}\), respectively, to make the vertical line of the \(x\)-axis and cross the \(x\)-axis at the two points of \(M\) and \(N\). According to the principle of triangle similarity, we can obtain \(\Delta\)APM \(\sim\)\(\Delta\)A\({}_{1}\)PN, then:  
\[\frac{x_{c}}{z_{c}}=\frac{x^{\prime}c}{z^{\prime}c}. \tag{7}\]  
Then from Equations (4)-(7), we can obtain:  
\[\tan\theta_{Lat}\cdot\tan\phi=-1. \tag{8}\]  
From the Formula (8), it can be seen that A\({}_{1}\)Q\({}_{1}\) and the straight line AO\({}_{1}\) are perpendicular to each other. Furthermore, we cross point A to make the tangent of the large ellipse and the \(x\)-axis intersect at point \(Q\); then from Equation (8) and its conclusion, we can see  
Figure 1: 2-dimensional schematic diagram of similar ellipsoid. that the straight line A\({}_{1}\)Q\({}_{1}\) is parallel to the straight line AQ, that is to say, the straight lines of the primary unitary circle radii passing through points A\({}_{1}\) and A are coincident. According to the formula of the primary unitary circle radius of the small ellipse and the large ellipse at latitude \(\theta_{Lat}\), we can obtain [24]:  
\[\left\{ \begin{array}{c}R_{S}=\frac{R_{e}}{\sqrt{1-e^{2}\sin^{2}\theta_{Lat}}} \\ R_{L}=\frac{R_{E}^{2}}{\sqrt{1-e^{2}\sin^{2}\theta_{Lat}}}=R_{e}+H_{k}\\ \end{array} \right.. \tag{9}\]  
Then there are:  
\[\frac{R_{L}}{R_{S}}=\frac{R_{E}}{R_{e}}=\frac{R_{S}+H_{k}}{R_{S}}. \tag{10}\]  
From Equations (4) and (10), we can obtain: \(\frac{R_{F}}{R_{e}}=\frac{R_{P}}{R_{P}}=\frac{R_{L}}{R_{S}}\); at this point, the proof process is complete. Since the height of the moving carrier is changing, the large ellipsoid also changes momentarily, but this does not affect the relative properties of similar ellipsoids. According to the point A\((x_{c},y_{c},z_{c})\) on the large ellipsoid where the moving carrier is located and the measured height information is \(H_{k}\), an ellipsoid equation can be approximately established [25]:  
\[\frac{x_{c}^{2}}{\rho_{0k}^{2}}+\frac{y_{c}^{2}}{\rho_{0k}^{2}}+\frac{z_{c}^{ 2}}{\rho_{Pk}^{2}}=1, \tag{11}\]  
where \(\rho_{0k}=R_{e}+H_{k}\), \(\rho_{Pk}=R_{p}+H_{k}\), \(R_{e}=6378.137\) km, and \(R_{p}=6356.752\) km. In addition, since the clock bias between the receiver carried by the moving carrier and the LEO satellite was considered to be completely eliminated, that is, we do not need to consider the solution of the clock bias at this time, we only need the pseudorange measurement information of the two satellites in order to complete the positioning solution.