# Detection and Tracking of Weak Exoatmospheric Target with Elliptical Hough Transform
## 2 Elliptical Characteristic of Exoatmospheric Targets
Furthermore, \(r_{\mathrm{e}}=6378.11\times 10^{3}\,\mathrm{m}\) is the average Earth radius, \(H_{\mathrm{C}}=\mathrm{O}_{\mathrm{e}}\mathrm{A}-r_{\mathrm{e}}\) is the height of turnoff, \(\theta\) is the range angle of the entire midcourse, i.e., \(\theta=\angle\mathrm{A}\mathrm{O}_{\mathrm{e}}\mathrm{B}\). According to classical astrodynamics, these elliptical parameters can be determined by the turnoff parameters (1953)  
\[\left\{ \begin{array}{l}a=-\frac{\mu r_{k}}{r_{k}V_{k}^{2}-2\mu}\\ b=r_{k}\cos\theta_{k}[v_{k}/(2-v_{k})]^{1/2}\end{array} \right. \tag{3}\]  
\[\theta=2\mathrm{arcsin}\left[\frac{v_{k}\cos^{2}\theta_{k}\tan\theta_{k}}{ \sqrt{1+v_{k}(v_{k}-2)\cos^{2}\theta_{k}}}\right] \tag{4}\]  
where \(r_{k}\), \(V_{k}\), and \(\theta_{k}\) are the initial reference (or thrust cutoff) range, velocity, and tilt angle, respectively; \(v_{k}\) is the energy parameter, defined as \(v_{k}\triangleq V_{k}^{2}/(\mu/r_{k})\). Let \(H\) be the radar altitude, note that \(H\ll H_{\mathrm{C}}\), then the radar is located somewhere on Earth's surface, i.e.,  
\[x_{\mathrm{R}}^{2}+y_{\mathrm{R}}^{2}+z_{\mathrm{R}}^{2}=(r_{\mathrm{e}}+H)^{ 2},x_{\mathrm{R}},y_{\mathrm{R}},z_{\mathrm{R}}\in\mathbf{\Omega}_{\mathrm{R}} \tag{5}\]  
where \(\mathbf{\Omega}_{R}\) is a spherical region, related to many limitations, such as geopolitics and the natural environment. From Figure 1, we can draw out some constraint relationships between the elliptical parameters \(a\), \(b\), and \(\theta\). The reentry point \(B\) has the following coordinates  
\[\left\{ \begin{array}{l}x_{\rm B}=(r_{\rm e}+H_{\rm C})\sin(\theta/2)\\ y_{\rm B}=(r_{\rm e}+H_{\rm C})\cos(\theta/2)\end{array} \right. \tag{6}\]  
Substituting Equation (6) into Equation (2), we obtain  
\[\frac{(r_{\rm e}+H_{\rm C})^{2}\sin^{2}(\theta/2)}{b^{2}}+\frac{\left[(r_{\rm e }+H_{\rm C})\cos(\theta/2)-\sqrt{a^{2}-b^{2}}\right]^{2}}{a^{2}}=1 \tag{7}\]  
Expansion of Equation (7) and collection of the terms yield the following constraint  
\[ \begin{array}{ll}\Lambda(a,b,\theta)&\triangleq(a^{2}-b^{2})(r_{\rm e}+H_{ \rm C})^{2}\cos^{2}(\theta/2)\\ &+2b^{2}\sqrt{a^{2}-b^{2}}(r_{\rm e}+H_{\rm C})\cos(\theta/2)\\ &+b^{4}-a^{2}(r_{\rm e}+H_{\rm C})^{2}\\ &=0\end{array}  \tag{8}\]  
Equation (8) is a quadratic equation related to \(\cos(\theta/2)\). Denote the following symbols as  
\[ \begin{array}{ll}A_{1}&\triangleq(a^{2}-b^{2})(r_{\rm e}+H_{\rm C})^{2}\\ B_{1}&\triangleq 2b^{2}\sqrt{a^{2}-b^{2}}(r_{\rm e}+H_{\rm C})\\ C_{1}&\triangleq b^{4}-a^{2}(r_{\rm e}+H_{\rm C})^{2}\end{array}  \tag{9}\]  
Note that we use the subscript to differentiate between symbols in Figure 1. Thus, we have the classic solution as  
\[ \begin{array}{ll}\cos(\theta/2)&=\frac{-B_{1}\pm\sqrt{B_{1}^{2}-4A_{1}C_{1}} }{2A_{1}}\\ &=\frac{-b^{2}\sqrt{a^{2}-b^{2}}(r_{\rm e}+H_{\rm C})\pm\sqrt{a^{2}(a^{2}-b^{2 })(r_{\rm e}+H_{\rm C})^{4}}}{(a^{2}-b^{2})(r_{\rm e}+H_{\rm C})^{2}}\\ &=\frac{-b^{2}}{(r_{\rm e}+H_{\rm C})\sqrt{a^{2}-b^{2}}}\pm\frac{a}{\sqrt{a^{2 }-b^{2}}}\end{array}  \tag{10}\]  
Figure 1: “Target-Radar-Earth” geometry. Next, it can be demonstrated that \(\cos(\theta_{1}/2)=\frac{-b^{2}-a(r_{\rm e}+H_{\rm C})}{(r_{\rm e}+H_{\rm C}) \sqrt{a^{2}-b^{2}}}\) is not a valid solution. In fact, the expansion of \(\cos^{2}(\theta_{1}/2)\) yields  
\[ \begin{array}{ll}\cos^{2}(\theta_{1}/2)&=\frac{\left[-b^{2}-a(r_{\rm e}+H_{ \rm C})\right]^{2}}{\left(r_{\rm e}+H_{\rm C}\right)^{2}\left(a^{2}-b^{2} \right)}\\ &=1+\frac{b^{2}\left[\left(r_{\rm e}+H_{\rm C}\right)^{2}+2a(r_{\rm e}+H_{\rm C })+b^{2}\right]}{\left(r_{\rm e}+H_{\rm C}\right)^{2}\left(a^{2}-b^{2}\right) }>1\end{array}  \tag{11}\]  
Obviously, Equation (11) is a contradiction, since we have a common sense that \(0\leq\cos^{2}(\theta_{1}/2)\leq 1\). It is not difficult to prove that another solution is a feasible one.