Gravity-Turn-Assisted Optimal Guidance Law

Abstract

This chapter proposes a new optimal guidance law that directly utilizes, instead of compensating, the gravity for accelerating missiles. The desired collision triangle that considers both gravity and vehicle’s axial acceleration is analytically derived based on geometric conditions. The concept of instantaneous zero-effort-miss is introduced to allow for analytical guidance command derivation. The proposed optimal guidance law is derived by using the optimal error dynamics proposed in Chap. 2. The relationships of the proposed formulation with conventional PNG and guidance-to-collision (G2C) are analyzed and the results show that the proposed guidance law encompasses previously suggested approaches. The significant contribution of the proposed guidance law lies in that it ensures zero final guidance command and enables energy saving with the aid of utilizing gravity turn. Nonlinear numerical simulations clearly demonstrate the effectiveness of the proposed approach.

Appendix. Closed-Form Solution of Eqs. (6.16)–(6.18)

Noting from Eqs. (6.16)–(6.18) that the solution requires the integration of

$$\begin{aligned} h\left( \gamma _{M} \right) \left| \sec \gamma _{M} +\tan \gamma _{M} \right| ^{n} \end{aligned}$$

(6.60)

where $h\left( \gamma _{M} \right)$ is a function of secant and tangent functions. For an arbitrary function $\beta \left( \gamma _{M} \right)$, we have

$$\begin{aligned} \begin{aligned}&\frac{d}{d\gamma _{M} } \left[ \beta \left( \gamma _{M} \right) \left| \sec \gamma _{M} +\tan \gamma _{M} \right| ^{n} \right] \\&\quad \quad \quad =\frac{d\beta \left( \gamma _{M} \right) }{d\gamma _{M} } \left| \sec \gamma _{M} +\tan \gamma _{M} \right| ^{n} +n \beta \left( \gamma _{M} \right) \sec \gamma _{M} \left| \sec \gamma _{M} +\tan \gamma _{M} \right| ^{n}\\&\quad \quad \quad =\left( \frac{d\beta \left( \gamma _{M} \right) }{d\gamma _{M} } +n\beta \left( \gamma _{M} \right) \sec \gamma _{M} \right) \left| \sec \gamma _{M} +\tan \gamma _{M} \right| ^{n} \end{aligned} \end{aligned}$$

(6.61)

which reveals that the general solution of $\int h\left( \gamma _{M} \right) \left| \sec \gamma _{M} +\tan \gamma _{M} \right| ^{n} d\gamma _{M}$ can be obtained by equalizing $h\left( \gamma _{M} \right)$ with $\frac{d\beta \left( \gamma _{M} \right) }{d\gamma _{M} } +n\beta \left( \gamma _{M} \right) \sec \gamma _{M}$.

Based on the properties of secant and tangent functions, the function $\beta \left( \gamma _{M} \right)$ in solving Eq. (6.16)–(6.18) can be formulated as a general form as

$$\begin{aligned} \beta \left( \gamma _{M} \right) =a_{1} \sec \gamma _{M} +a_{2} \tan \gamma _{M} +a_{3} \sec ^{2} \gamma _{M} +a_{4} \tan ^{2} \gamma _{M} +a_{5} \sec \gamma _{M} \tan \gamma _{M} \end{aligned}$$

(6.62)

where $a_{i}$, $i=1,2,3,4,5$, are unknown constant coefficients to be determined.

Differentiating Eq. (6.62) with respect to $\gamma _{M}$ gives

$$\begin{aligned} \begin{aligned} \frac{d\beta \left( \gamma _{M} \right) }{d\gamma _{M} } = \,\,&a_{1} \sec \gamma _{M} \tan \gamma _{M} +a_{2} \sec ^{2} \gamma _{M} +2a_{3} \sec ^{2} \gamma _{M} \tan \gamma _{M} \\&+2a_{4} \sec ^{2} \gamma _{M} \tan \gamma _{M} +a_{5} \left( \sec \gamma _{M} \tan ^{2} \gamma _{M} +\sec ^{3} \gamma _{M} \right) \end{aligned} \end{aligned}$$

(6.63)

Substituting Eqs. (6.62) and (6.63) into Eq. (6.61) yields

$$\begin{aligned} \begin{aligned}&\frac{d}{d\gamma _{M} } \left[ \beta \left( \gamma _{M} \right) \left| \sec \gamma _{M} +\tan \gamma _{M} \right| ^{n} \right] \\&\quad \quad \quad =\left[ \left( na_{1} +a_{2} \right) \sec ^{2} \gamma _{M} +\left( na_{2} +a_{1} \right) \sec \gamma _{M} \tan \gamma _{M}\right. \\&\quad \quad \quad \quad +\left( na_{3} +a_{5} \right) \sec ^{3} \gamma _{M} +\left( na_{4} +a_{5} \right) \sec \gamma _{M} \tan ^{2} \gamma _{M} \\&\quad \quad \quad \quad \left. +\left( na_{5} +2a_{3} +2a_{4} \right) \sec ^{2} \gamma _{M} \tan \gamma _{M}\right] \left| \sec \gamma _{M} +\tan \gamma _{M} \right| ^{n} \end{aligned} \end{aligned}$$

(6.64)

For Eq. (6.16), we have $h\left( \gamma _{M} \right) =\sec ^{2} \gamma _{M}$, $n=\kappa$. Comparing the coefficients results in

$$\begin{aligned} \left\{ \begin{array}{l} {\kappa a_{1} +a_{2} =1} \\ {\kappa a_{2} +a_{1} =0} \\ {\kappa a_{3} +a_{5} =0} \\ {\kappa a_{4} +a_{5} =0} \\ {\kappa a_{5} +2a_{3} +2a_{4} =0} \end{array}\right. \end{aligned}$$

(6.65)

Solving Eq. (6.65) gives the unknown coefficients as

$$\begin{aligned} a_{1} =\frac{\kappa }{\kappa ^{2} -1} ,\quad a_{2} =-\frac{1}{\kappa ^{2} -1} ,\quad a_{3} =a_{4} =a_{5} =0 \end{aligned}$$

(6.66)

Then, the closed-form solution of Eq. (6.16) is given by

$$\begin{aligned} t\left( \gamma _{M} \right) =t\left( \gamma _{M0} \right) -\frac{C}{g} \left[ f_{t} \left( \gamma _{M} \right) -f_{t} \left( \gamma _{M0} \right) \right] \end{aligned}$$

(6.67)

For Eq. (6.17), we have $h\left( \gamma _{M} \right) =\sec ^{2} \gamma _{M}$, $n=2\kappa$. Replacing $\kappa$ with $2\kappa$ in Eq. (6.65) leads to the closed-form solution of Eq. (6.17) as

$$\begin{aligned} x_{M} \left( \gamma _{M} \right) =x_{M} \left( \gamma _{M0} \right) -\frac{C^{2} }{g} \left[ f_{x} \left( \gamma _{M} \right) -f_{x} \left( \gamma _{M0} \right) \right] \end{aligned}$$

(6.68)

For Eq. (6.18), we have $h\left( \gamma _{M} \right) =\sec ^{2} \gamma _{M} \tan \gamma _{M}$, $n=2\kappa$. Comparing the coefficients gives the following coupled equations

$$\begin{aligned} \left\{ \begin{array}{l} {\kappa a_{1} +a_{2} =0} \\ {\kappa a_{2} +a_{1} =0} \\ {\kappa a_{3} +a_{5} =0} \\ {\kappa a_{4} +a_{5} =0} \\ {\kappa a_{5} +2a_{3} +2a_{4} =1} \end{array}\right. \end{aligned}$$

(6.69)

Solving Eq. (6.69) yields

$$\begin{aligned} a_{1} =a_{2} =0,\quad a_{3} =-\frac{1}{\kappa ^{2} -4} ,\quad a_{4} =-\frac{1}{\kappa ^{2} -4} ,\quad a_{5} =\frac{\kappa }{\kappa ^{2} -4} \end{aligned}$$

(6.70)

which gives the closed-form solution of Eq. (6.18) as

$$\begin{aligned} y_{M} \left( \gamma _{M} \right) =y_{M} \left( \gamma _{M0} \right) -\frac{C^{2} }{g} \left[ f_{y} \left( \gamma _{M} \right) -f_{y} \left( \gamma _{M0} \right) \right] \end{aligned}$$

(6.71)