# Correction Strategy of Online Midcourse Guidance for High-Speed Gliding Target Interceptor
## 2 Model Description
### Calculation of Required Velocity Based on Lambert Guidance  
With the required velocity, the interceptor can reach the target in a certain terminal constraint according to the current state of inertial motion without any additional orbital maneuver. Therefore, the first step is to solve the required velocity of the interceptor. Most interception tasks require the interceptor to reach the target PIP at a fixed time, as arriving too early or too late will result in a failed interception. Solving the required velocity in the interceptor's trajectory transformation is the classical Lambert problem, which belongs to solving the two-point boundary value problem under the two-body theory.  
The essence is to solve the speed problem of fixed time flight. As shown in Figure 3, the spacecraft is required to fly from point \(A\) to point \(P\), where point \(O\) represents the Earth's center and \(\varphi\) is the transfer angle. At the initial time, the spacecraft is at point \(A\) and the velocity is \(V_{A}\); if not adjusted, the interceptor will fly to the point \(P^{\prime}\) after time \(t_{f}\). When the velocity is \(V_{r}\) at point \(A\), the spacecraft will fly to the target point \(P\) after time \(t_{f}\). The solution of \(\mathbf{V}_{r}\) is the problem solved by [PERSON].  
Referring to the required velocity solution for a ballistic missile in-flight, when given the initial track angle and required distance, the missile's initial speed \(v_{0}\) must be sufficient to hit the target on the ground.  
\[v_{0}=\sqrt{\frac{gm(1-\cos\varphi)}{r_{0}\cos\gamma\left[\frac{r_{0}\cos \varphi}{a}-\cos(\varphi+\gamma)\right]}} \tag{8}\]  
where  
\[r_{0}=a+alt \tag{9}\]\(alt\) is the initial height of the missile relative to the surface. Extend the speed equation  
\[v_{0}=\sqrt{\frac{gm(1-\text{cos}\varphi)}{r_{0}\text{cos}\gamma\Big{[}\frac{r_{0 }\text{cos}\gamma}{r_{f}}-\text{cos}(\varphi+\gamma)\Big{]}}} \tag{10}\]  
where \(r_{f}\) is  
\[r_{f}=a+alt_{f} \tag{11}\]  
where \(alt_{f}\) refers to the height of the target.  
Under elliptical trajectory, the time required for missile and target encounter is  
\[t_{f}=\frac{r_{0}}{\text{cos}\gamma}\left\{\frac{\tan\gamma(1-\text{cos}\varphi )+(1-\lambda)\text{sin}\varphi}{(2-\lambda)\Big{[}\frac{1-\text{cos}\varphi}{ \text{cos}\gamma}+\frac{\text{cos}(\gamma+\varphi)}{\text{cos}\gamma}\Big{]} }+\frac{2\text{cos}\gamma}{\lambda[(2/\lambda)-1]^{1.5}}\tan^{-1}\left(\frac {\sqrt{\frac{2}{\lambda}-1}}{\text{cos}\gamma\text{cot}(\varphi/2)-\text{sin} \gamma}\right)\right\} \tag{12}\]  
\[\lambda=\frac{r_{0}v^{2}}{gm} \tag{13}\]  
where \(v\) is the speed required to intercept the target and \(\lambda\) represents the energy parameter. The transfer angle can be obtained when the initial position vector \(\mathbf{r}_{0}\) and the end position vector \(\mathbf{r}_{f}\) are known in the geocentric coordinate system. \(\varphi\) is the flight transfer angle, which can be expressed as  
\[\varphi=\text{cos}^{-1}\frac{\mathbf{r}_{0}\cdot\mathbf{r}_{f}}{|\mathbf{r}_{0}|\Big{|} \mathbf{r}_{f}\Big{|}} \tag{14}\]  
According to Equations (10), Equation (12), and Equation (14), we can obtain  
\[\varphi=f\Big{(}\mathbf{r}_{0},\mathbf{r}_{f}\Big{)} \tag{15}\]  
\[\mathbf{V}=f\Big{(}\mathbf{r}_{0},\mathbf{r}_{f},\varphi,\gamma\Big{)} \tag{16}\]  
\[t_{f}=f(v_{r},\varphi,\gamma) \tag{17}\]  
where \(v_{r}\) is the required speed of trajectory transformation. Since the required speed cannot be directly obtained, the trajectory angle iteration method can be used to quickly solve the required speed in the Lambert problem, and the solution process is detailed in reference [33].  
Figure 3: Required velocity diagram.