鲁 鹏，北京理工大学宇航学院
本文主要参考文献[1-2]学习迭代制导技术。文献[3]是迭代制导技术的原始文献，写得很乱，不适入门者学习。
一般情况下，对火箭在大气层内（低于90 k m 90km 9 0 k m O O O O x y z Oxyz O x y z O x o c f y o c f z o c f Ox_{ocf}y_{ocf}z_{ocf} O x o c f y o c f z o c f O x r c f y r c f z r c f Ox_{rcf}y_{rcf}z_{rcf} O x r c f y r c f z r c f
使用迭代制导时，一般会把火箭运动方程表示在制导坐标系下，考虑到运载火箭在实际飞行过程中，其控制系统是可以保证滚转角γ o c f ≈ 0 \gamma_{ocf} \approx 0 γ o c f ≈ 0 (1) [ x ¨ o c f y ¨ o c f z ¨ o c f ] = F m [ cos φ o c f cos ψ o c f sin φ o c f cos ψ o c f − sin ψ o c f ] + [ g o c f x g o c f y g o c f z ]
\left[ \begin{array}{c}
{\ddot{x}_{ocf}} \\ {\ddot{y}_{ocf}} \\ {\ddot{z}_{ocf}}
\end{array}\right] =
\frac{F}{m} \left[ \begin{array}{c}
{\cos \varphi_{ocf} \cos \psi_{ocf}} \\
{\sin \varphi_{o c f} \cos \psi_{ocf}} \\
{-\sin \psi_{ocf}}\end{array}\right] +
\left[ \begin{array}{l}
{g_{ocfx}} \\
{g_{ocfy}} \\
{g_{ocfz}}
\end{array}\right] \tag{1}
⎣ ⎡ x ¨ o c f y ¨ o c f z ¨ o c f ⎦ ⎤ = m F ⎣ ⎡ cos φ o c f cos ψ o c f sin φ o c f cos ψ o c f − sin ψ o c f ⎦ ⎤ + ⎣ ⎡ g o c f x g o c f y g o c f z ⎦ ⎤ ( 1 ) φ o c f \varphi_{ocf} φ o c f ψ o c f \psi_{ocf} ψ o c f F F F m m m O x o c f y o c f z o c f Ox_{ocf}y_{ocf}z_{ocf} O x o c f y o c f z o c f O x o c f y o c f z o c f Ox_{ocf}y_{ocf}z_{ocf} O x o c f y o c f z o c f
取状态变量x = [ x ˙ o c f x o c f y ˙ o c f y o c f z ˙ o c f z o c f ] T \boldsymbol{x}=\left[ \begin{array}{lllll}{\dot{x}_{o c f}} & {x_{o c f}} & {\dot{y}_{ocf}} & {y_{ocf}} & {\dot{z}_{ocf}} & {z_{ocf}}\end{array}\right]^{T} x = [ x ˙ o c f x o c f y ˙ o c f y o c f z ˙ o c f z o c f ] T (2) J = ∫ 0 t g d t x ˙ = A x + B u + C x 0 = [ x ˙ o c f i x o c f i y ˙ o c f i y o c f i z ˙ o c f i z o c f i ] T x f = [ x ˙ o c f f x o c f f y ˙ o c f f y o c f f z ˙ o c f f z o c f f ] T
\begin{gathered}
J = \int_{0}^{t_g} dt \\
\dot{\boldsymbol{x}}= A\boldsymbol{x} + B \boldsymbol{u}+C \\
\boldsymbol{x}_{0}=\left[\dot{x}_{ocfi} \quad x_{ocfi} \quad \dot{y}_{ocfi} \quad y_{ocfi} \quad \dot{z}_{ocfi} \quad z_{ocfi}\right]^{T} \\
\boldsymbol{x}_{f}=\left[{}{\dot{x}_{ocff}} \quad {x_{ocff}} \quad {\dot{y}_{ocff}} \quad {y_{ocff}} \quad {\dot{z}_{ocff}} \quad {z_{ocff}}\right]^{T} \\
\end{gathered}\tag{2}
J = ∫ 0 t g d t x ˙ = A x + B u + C x 0 = [ x ˙ o c f i x o c f i y ˙ o c f i y o c f i z ˙ o c f i z o c f i ] T x f = [ x ˙ o c f f x o c f f y ˙ o c f f y o c f f z ˙ o c f f z o c f f ] T ( 2 ) A = [ 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 ] ， B = F / m ， u = [ cos φ o c f cos ψ o c f 0 sin φ o c f cos ψ o c f 0 − sin ψ o c f 0 ] ， C = [ g o c f x 0 g o c f y 0 g o c f z 0 ] A=\left[ \begin{array}{cccccc}{0} & {0} & {0} & {0} & {0} & {0} \\ {1} & {0} & {0} & {0} & {0} & {0} \\ {0} & {0} & {0} & {0} & {0} & {0} \\ {0} & {0} & {1} & {0} & {0} & {0} \\ {0} & {0} & {0} & {0} & {0} & {0} \\ {0} & {0} & {0} & {0} & {1} & {0}\end{array}\right]，
B={F}/{m}，
\boldsymbol{u}=\left[ \begin{array}{c}{\cos \varphi_{o c f} \cos \psi_{ocf}} \\ {0} \\ {\sin \varphi_{ocf} \cos \psi_{ocf}} \\ {0} \\ {-\sin \psi_{ocf}} \\ {0}\end{array}\right]，
C=\left[ \begin{array}{c}{g_{o c f x}} \\ {0} \\ {g_{o c f y}} \\ {0} \\ {g_{o c f z}} \\ {0}\end{array} \right] A = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ ， B = F / m ， u = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ cos φ o c f cos ψ o c f 0 sin φ o c f cos ψ o c f 0 − sin ψ o c f 0 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ ， C = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ g o c f x 0 g o c f y 0 g o c f z 0 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤
根据连续系统最优控制泛函求极值的必要条件可知[4]
极值条件：(3) [ ∂ H ∂ φ o c f ∂ H ∂ ψ o c f ] = 0   ⟹   [ λ 1 sin φ o c f cos ψ o c f − λ 3 cos φ o c f cos ψ o c f λ 1 cos φ o c f sin ψ o c f + λ 3 sin φ o c f sin ψ o c f + λ 5 cos ψ o c f ] = 0
\begin{bmatrix}{\frac{\partial H}{\partial \varphi_{ocf}}} \\ {\frac{\partial H}{\partial \psi_{ocf}}}\end{bmatrix}=\boldsymbol{0} \implies
\begin{bmatrix}{\lambda_{1} \sin \varphi_{ocf} \cos \psi_{ocf}-\lambda_{3} \cos \varphi_{ocf} \cos \psi_{ocf}} \\ {\lambda_{1} \cos \varphi_{ocf} \sin \psi_{ocf}+\lambda_{3} \sin \varphi_{ocf} \sin \psi_{ocf}+\lambda_{5} \cos \psi_{ocf}}\end{bmatrix} = \boldsymbol{0} \tag{3}
[ ∂ φ o c f ∂ H ∂ ψ o c f ∂ H ] = 0 ⟹ [ λ 1 sin φ o c f cos ψ o c f − λ 3 cos φ o c f cos ψ o c f λ 1 cos φ o c f sin ψ o c f + λ 3 sin φ o c f sin ψ o c f + λ 5 cos ψ o c f ] = 0 ( 3 ) H H H
伴随条件：(4) λ ˙ = − ∂ H ∂ x ⟹   [ λ ˙ 1 λ ˙ 2 λ ˙ 3 λ ˙ 4 λ ˙ 5 λ ˙ 6 ] = [ − λ 2 0 − λ 4 0 − λ 6 0 ]
\dot{\boldsymbol{\lambda}}=-\frac{\partial H}{\partial \boldsymbol{x}} \implies
\left[ \begin{array}{c}{\dot{\lambda}_{1}} \\ {\dot{\lambda}_{2}} \\ {\dot{\lambda}_{3}} \\ {\dot{\lambda}_{4}} \\ {\dot{\lambda}_{5}} \\ {\dot{\lambda}_{6}}\end{array}\right]=\left[ \begin{array}{c}{-\lambda_{2}} \\ {0} \\ {-\lambda_{4}} \\ {0} \\ {-\lambda_{6}} \\ {0}\end{array}\right] \tag{4}
λ ˙ = − ∂ x ∂ H ⟹ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ λ ˙ 1 λ ˙ 2 λ ˙ 3 λ ˙ 4 λ ˙ 5 λ ˙ 6 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ − λ 2 0 − λ 4 0 − λ 6 0 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ ( 4 ) (5) λ T δ x ∣ t f = 0   ⟹   [ λ 1 δ x ˙ o c f λ 2 δ x o c f λ 3 δ x ˙ o c f λ 4 δ y o c f λ 5 δ z ˙ o c f λ 6 δ z o c f ] ∣ t f = 0
\boldsymbol{\lambda}^{T} \delta \boldsymbol{x} \vert_{t_{f}}=\boldsymbol{0} \implies
\left[ \begin{array}{c}{\lambda_{1} \delta \dot{x}_{ocf}} \\ {\lambda_{2} \delta x_{ocf}} \\ {\lambda_{3} \delta \dot{x}_{ocf}} \\ {\lambda_{4} \delta y_{ocf}} \\ {\lambda_{5} \delta \dot{z}_{ocf}} \\ {\lambda_{6} \delta z_{ocf}}\end{array}\right]\vert_{t_{f}}=\boldsymbol{0} \tag{5}
λ T δ x ∣ t f = 0 ⟹ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ λ 1 δ x ˙ o c f λ 2 δ x o c f λ 3 δ x ˙ o c f λ 4 δ y o c f λ 5 δ z ˙ o c f λ 6 δ z o c f ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ ∣ t f = 0 ( 5 ) (6) { tan φ o c f = λ 30 − λ 40 t λ 10 tan ψ o c f = − ( λ 50 − λ 60 t ) λ 10 cos φ o c f + ( λ 30 − λ 40 t ) sin φ o c f
\left\{\begin{aligned}
\tan \varphi_{ocf} =& \frac{\lambda 30-\lambda 40 t}{\lambda 10} \\
\tan \psi_{ocf} =& \frac{-(\lambda 50-\lambda 60 t)}{\lambda 10 \cos \varphi_{ocf}+(\lambda 30-\lambda 40 t) \sin \varphi_{ocf}}
\end{aligned}\right. \tag{6}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ tan φ o c f = tan ψ o c f = λ 1 0 λ 3 0 − λ 4 0 t λ 1 0 cos φ o c f + ( λ 3 0 − λ 4 0 t ) sin φ o c f − ( λ 5 0 − λ 6 0 t ) ( 6 ) φ o c f \varphi_{ocf} φ o c f ψ o c f \psi_{ocf} ψ o c f (7) { φ o c f ≈ φ ‾ o c f + K φ 2 t − K φ 1 ψ o c f ≈ ψ ‾ o c f + K ψ 2 t − K ψ 1
\left\{\begin{array}{c}{\varphi_{ocf} \approx \overline{\varphi}_{ocf}+K_{\varphi 2} t-K_{\varphi 1}} \\ {\psi_{ocf} \approx \overline{\psi}_{ocf}+K_{\psi 2} t-K_{\psi 1}}\end{array}\right. \tag{7}
{ φ o c f ≈ φ o c f + K φ 2 t − K φ 1 ψ o c f ≈ ψ o c f + K ψ 2 t − K ψ 1 ( 7 ) φ ‾ o c f \overline{\varphi}_{ocf} φ o c f ψ ‾ o c f \overline{\psi}_{ocf} ψ o c f K φ 2 K_{\varphi 2} K φ 2 K φ 1 K_{\varphi 1} K φ 1 K ψ 2 K_{\psi 2} K ψ 2 K ψ 1 K_{\psi 1} K ψ 1 φ ‾ o c f \overline{\varphi}_{ocf} φ o c f K φ 2 K_{\varphi 2} K φ 2 K φ 1 K_{\varphi 1} K φ 1
本节将计算剩余飞行时间以及俯仰角φ o c f \varphi_{ocf} φ o c f ψ o c f \psi_{ocf} ψ o c f
利用雅克比迭代法求解下面这个非线性方程，可得剩余飞行时间[2](8) t g = t h ( 1 − e − Δ V V e x )
t_{g}=t_{h}\left(1-e^{-\frac{\Delta V}{V e x}}\right) \tag{8}
t g = t h ( 1 − e − V e x Δ V ) ( 8 ) Δ V = ( x ˙ o c f f − x ˙ o c f i − g o c f x t g ) 2 + ( y ˙ o c f f − y ˙ o c f i − g o c f y t g ) 2 + ( z ˙ o c f f − z ˙ o c f i − g o c f z t g ) 2 \Delta V= \sqrt{\left(\dot{x}_{ocff}-\dot{x}_{ocfi}-g_{ocfx} t_{g}\right)^{2} + \left(\dot{y}_{ocff}-\dot{y}_{ocfi}-g_{ocfy} t_{g}\right)^{2} + \left(\dot{z}_{ocff}-\dot{z}_{ocfi}-g_{ocfz} t_{g}\right)^{2}} Δ V = ( x ˙ o c f f − x ˙ o c f i − g o c f x t g ) 2 + ( y ˙ o c f f − y ˙ o c f i − g o c f y t g ) 2 + ( z ˙ o c f f − z ˙ o c f i − g o c f z t g ) 2
传统的迭代制导方法使用下式计算引力[1-3](9) g o c f = [ g o c f x g o c f y g o c f z ] = 1 2 ( [ g o c f i x g o c f i y g o c f i z ] + [ g o c f f x g o c f f y g o c f f z ] )
\boldsymbol{g}_{ocf}=\left[\begin{array}{l}{g_{ocfx}} \\ {g_{ocfy}} \\ {g_{ocfz}}\end{array}\right]=
\frac{1}{2}(\left[ \begin{array}{l}{g_{ocfix}} \\ {g_{ocfiy}} \\ {g_{ocfiz}}\end{array}\right]+
\left[\begin{array}{l}{g_{ocffx}} \\ {g_{ocffy}} \\ {g_{ocffz}}\end{array}\right]) \tag{9}
g o c f = ⎣ ⎡ g o c f x g o c f y g o c f z ⎦ ⎤ = 2 1 ( ⎣ ⎡ g o c f i x g o c f i y g o c f i z ⎦ ⎤ + ⎣ ⎡ g o c f f x g o c f f y g o c f f z ⎦ ⎤ ) ( 9 ) g o c f i = [ g o c f i x g o c f i y g o c f i z ] T \boldsymbol{g}_{ocfi} = [g_{ocfix}\quad g_{ocfiy}\quad g_{ocfiz}]^{T} g o c f i = [ g o c f i x g o c f i y g o c f i z ] T g o c f f = [ g o c f f x g o c f f y g o c f f z ] T \boldsymbol{g}_{ocff} = [g_{ocffx}\quad g_{ocffy}\quad g_{ocffz}]^{T} g o c f f = [ g o c f f x g o c f f y g o c f f z ] T
地心角ϕ \phi ϕ
为了满足轨道倾角i i i Ω \Omega Ω z o c f f = 0 z_{ocff}=0 z o c f f = 0 z ˙ o c f f = 0 \dot{z}_{ocff}=0 z ˙ o c f f = 0 x o c f f = 0 x_{ocff} = 0 x o c f f = 0 a a a e e e x ˙ o c f f \dot{x}_{ocff} x ˙ o c f f y ˙ o c f f \dot{y}_{ocff} y ˙ o c f f y o c f f y_{ocff} y o c f f
(10) x ˙ o c f f = ( 1 + e cos f ) μ a ( 1 − e 2 )
\dot{x}_{ocff} = (1 + e\cos f) \sqrt{\frac{\mu}{a\left(1-e^{2}\right)}} \tag{10}
x ˙ o c f f = ( 1 + e cos f ) a ( 1 − e 2 ) μ ( 1 0 ) (11) y ˙ o c f f = e sin f μ a ( 1 − e 2 )
\dot{y}_{ocff} = e \sin f \sqrt{\frac{\mu}{a\left(1-e^{2}\right)}} \tag{11}
y ˙ o c f f = e sin f a ( 1 − e 2 ) μ ( 1 1 )
(12) y o c f f = a ( 1 − e 2 ) 1 + e cos f
y_{ocff} = \frac{a\left(1-e^{2}\right)}{1+e \cos f} \tag{12}
y o c f f = 1 + e cos f a ( 1 − e 2 ) ( 1 2 )
火箭在入轨点的速度V o c f f = [ x ˙ o c f f y ˙ o c f f z ˙ o c f f ] T V_{ocff} = \left[ \dot{x}_{ocff} \quad\dot{y}_{ocff} \quad\dot{z}_{ocff} \right]^{T} V o c f f = [ x ˙ o c f f y ˙ o c f f z ˙ o c f f ] T (13) V o c f f = μ ( 2 y o c f f − 1 a )
V_{ocff}=\sqrt{\mu\left( \frac{2}{y_{ocff}} - \frac{1}{a} \right)} \tag{13}
V o c f f = μ ( y o c f f 2 − a 1 ) ( 1 3 ) β \beta β β \beta β f f f cos β = r v f ˙ = 1 + e cos f 1 + e 2 + 2 e cos f sin β = 1 v r ˙ = e sin f 1 + e 2 + 2 e cos f }
\left.
\begin{aligned}{\cos \beta=\frac{r}{v} \dot{f}=\frac{1+e \cos f}{\sqrt{1+e^{2}+2 e \cos f}}} \\
{\sin \beta=\frac{1}{v} \dot{r}=\frac{e \sin f}{\sqrt{1+e^{2}+2 e \cos f}}}\end{aligned}
\right\}
cos β = v r f ˙ = 1 + e 2 + 2 e cos f 1 + e cos f sin β = v 1 r ˙ = 1 + e 2 + 2 e cos f e sin f ⎭ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎫
(14) x ˙ o c f f = V o c f f cos β = ( 1 + e cos f ) μ a ( 1 − e 2 )
\dot{x}_{ocff} = V_{ocff}\cos{\beta} = (1 + e\cos f) \sqrt{\frac{\mu}{a\left(1-e^{2}\right)}} \tag{14}
x ˙ o c f f = V o c f f cos β = ( 1 + e cos f ) a ( 1 − e 2 ) μ ( 1 4 )
(15) y ˙ o c f f = V o c f f sin β = e sin f μ a ( 1 − e 2 )
\dot{y}_{ocff} = V_{ocff}\sin{\beta} = e \sin f \sqrt{\frac{\mu}{a\left(1-e^{2}\right)}} \tag{15}
y ˙ o c f f = V o c f f sin β = e sin f a ( 1 − e 2 ) μ ( 1 5 )
如图 2 所示，真近点角f f f ϕ \phi ϕ ω \omega ω (16) f = ϕ − ω
f=\phi-\omega \tag{16}
f = ϕ − ω ( 1 6 ) 计算地心角 ϕ \phi ϕ
地心角ϕ \phi ϕ ϕ 1 \phi_{1} ϕ 1 ϕ 2 \phi_{2} ϕ 2 ϕ 1 \phi_{1} ϕ 1 (17) ϕ 1 = arctan x r c f y r c f
\phi_{1} = \arctan{\frac{x_{rcf}}{y_{rcf}}} \tag{17}
ϕ 1 = arctan y r c f x r c f ( 1 7 ) x r c f x_{rcf} x r c f y r c f y_{rcf} y r c f x x x y y y
ϕ 2 \phi_{2} ϕ 2 ϕ 2 \phi_{2} ϕ 2 (18) ϕ 2 ≈ [ V t g + S t h r u s t − k t g ( V − Δ V − V o c f f ) ] cos θ y o c f f
\phi_{2} \approx \frac{[Vt_{g} + S_{thrust} - kt_g(V - \Delta V - V_{ocff})]\cos{\theta}}{y_{ocff}} \tag{18}
ϕ 2 ≈ y o c f f [ V t g + S t h r u s t − k t g ( V − Δ V − V o c f f ) ] cos θ ( 1 8 ) V V V Δ V \Delta V Δ V V o c f f V_{ocff} V o c f f S t h r u s t S_{thrust} S t h r u s t [ 0 t g ] [0 \quad t_{g}] [ 0 t g ] k k k k k k θ \theta θ θ = 0 \theta = 0 θ = 0 ϕ 2 \phi_{2} ϕ 2 y o c f f y_{ocff} y o c f f y o c f f y_{ocff} y o c f f ϕ \phi ϕ ϕ 2 \phi_{2} ϕ 2 y o c f f y_{ocff} y o c f f
对推力的一次积分[2](19) V t h r u s t = [ F 0 ( t g ) cos ( φ ‾ o c f ) + F 0 ( t g ) K φ 1 sin ( φ ‾ o c f ) − F 1 ( t g ) K φ 2 sin ( φ ‾ o c f ) F 0 ( t g ) sin ( φ ‾ o c f ) − F 0 ( t g ) K φ 1 cos ( φ ‾ o c f ) + F 1 ( t g ) K φ 2 cos ( φ ‾ o c f ) F 0 ( t g ) K ψ 1 cos ( ψ ‾ o c f ) − F 1 ( t g ) K ψ 2 cos ( ψ ‾ o c f ) + F 0 ( t g ) sin ( ψ ‾ o c f ) ]
\boldsymbol{V}_{thrust} = \left[\begin{array}{l}
{F0\left(t_{g}\right) \cos \left(\overline{\varphi}_{ocf}\right)+F0\left(t_{g}\right) K_{\varphi 1} \sin \left(\overline{\varphi}_{ocf}\right)} {-F1\left(t_{g}\right) K_{\varphi 2} \sin \left(\overline{\varphi}_{ocf}\right)} \\
{F0\left(t_{g}\right) \sin \left(\overline{\varphi}_{ocf}\right)-F 0\left(t_{g}\right) K_{\varphi 1} \cos \left(\overline{\varphi}_{ocf}\right)} {+F1\left(t_{g}\right) K_{\varphi 2} \cos \left(\overline{\varphi}_{ocf}\right)} \\
{F0\left(t_{g}\right) K_{\psi 1} \cos \left(\overline{\psi}_{ocf}\right)-F 1\left(t_{g}\right) K_{\psi 2} \cos \left(\overline{\psi}_{ocf}\right)+F0\left(t_{g}\right) \sin \left(\overline{\psi}_{ocf}\right)}
\end{array}\right] \tag{19}
V t h r u s t = ⎣ ⎡ F 0 ( t g ) cos ( φ o c f ) + F 0 ( t g ) K φ 1 sin ( φ o c f ) − F 1 ( t g ) K φ 2 sin ( φ o c f ) F 0 ( t g ) sin ( φ o c f ) − F 0 ( t g ) K φ 1 cos ( φ o c f ) + F 1 ( t g ) K φ 2 cos ( φ o c f ) F 0 ( t g ) K ψ 1 cos ( ψ o c f ) − F 1 ( t g ) K ψ 2 cos ( ψ o c f ) + F 0 ( t g ) sin ( ψ o c f ) ⎦ ⎤ ( 1 9 ) F 0 ( t ) F0(t) F 0 ( t ) F 1 ( t ) F1(t) F 1 ( t ) (20) F 0 ( t ) ≜ ∫ 0 t V e x t h − t d t = V e x ln t h t h − t
F0(t) \triangleq \int_{0}^{t} \frac{V_{ex}}{t_{h}-t} d t=V_{ex} \ln \frac{t_{h}}{t_{h}-t} \tag{20}
F 0 ( t ) ≜ ∫ 0 t t h − t V e x d t = V e x ln t h − t t h ( 2 0 ) (21) F 1 ( t ) ≜ ∫ 0 t V e x t h − t t d t = t h F 0 ( t ) − V e x t
F1(t) \triangleq \int_{0}^{t} \frac{V_{ex}}{t_{h}-t} t d t=t_{h} F 0(t)-V_{ex} t \tag{21}
F 1 ( t ) ≜ ∫ 0 t t h − t V e x t d t = t h F 0 ( t ) − V e x t ( 2 1 )
因为φ ‾ o c f \overline{\varphi}_{ocf} φ o c f ψ ‾ o c f \overline{\psi}_{ocf} ψ o c f K φ 2 t − K φ 1 K_{\varphi 2} t-K_{\varphi 1} K φ 2 t − K φ 1 K ψ 2 t − K ψ 1 K_{\psi 2} t-K_{\psi 1} K ψ 2 t − K ψ 1 (22) V t h r u s t = [ F 0 ( t g ) cos φ ‾ o c f cos ψ ‾ o c f F 0 ( t g ) sin φ ‾ o c f cos ψ ‾ o c f 0 ]
\boldsymbol{V}_{thrust} =
\left[ \begin{array}{c}
F0\left(t_{g}\right) \cos \overline{\varphi}_{ocf} \cos \overline{\psi}_{ocf} \\
F0\left(t_{g}\right) \sin \overline{\varphi}_{ocf}\cos \overline{\psi}_{ocf} \\
0
\end{array}\right] \tag{22}
V t h r u s t = ⎣ ⎡ F 0 ( t g ) cos φ o c f cos ψ o c f F 0 ( t g ) sin φ o c f cos ψ o c f 0 ⎦ ⎤ ( 2 2 ) 与 与 与 以 及 以及 以 及 与 与 与 (23) F 0 ( t g ) K ρ 1 = F 1 ( t g ) K φ 2 F 0 ( t g ) K ψ 1 = F 1 ( t g ) K ψ 2
\begin{array}{l}
{F 0\left(t_{g}\right) K_{\rho 1}=F 1\left(t_{g}\right) K_{\varphi 2}} \\
{F 0\left(t_{g}\right) K_{\psi 1}=F 1\left(t_{g}\right) K_{\psi 2}}
\end{array} \tag{23}
F 0 ( t g ) K ρ 1 = F 1 ( t g ) K φ 2 F 0 ( t g ) K ψ 1 = F 1 ( t g ) K ψ 2 ( 2 3 ) (24) R o c f f = R t h r u s t + R g r a v i t y + V o c f i t g + R o c f i
\boldsymbol{R}_{ocff}=\boldsymbol{R}_{thrust}+\boldsymbol{R}_{gravity}+\boldsymbol{V}_{ocfi} t_{g}+\boldsymbol{R}_{ocfi} \tag{24}
R o c f f = R t h r u s t + R g r a v i t y + V o c f i t g + R o c f i ( 2 4 ) (25) [ x o c f f y o c f f z o c f f ] = [ F 2 ( t g ) cos ( φ ‾ o c f ) + F 2 ( t g ) K φ 1 sin ( φ ‾ o c f ) − F 3 ( t g ) K φ 2 sin ( φ ‾ o c f ) F 2 ( t g ) sin ( φ ‾ o c f ) − F 2 ( t g ) K φ 1 cos ( φ ‾ o c f ) + F 3 ( t g ) K φ 2 cos ( φ ‾ o c f ) F 2 ( t g ) K ψ 1 cos ( ψ ‾ o c f ) − F 3 ( t g ) K ψ 2 cos ( ψ ‾ o c f ) + F 2 ( t g ) sin ( ψ ‾ o c f ) ] + [ 1 2 g o c f x t g 2 1 2 g o c f y t g 2 1 2 g o c f z t g 2 ] + [ x ˙ o c f i t g y ˙ o c f i t g z ˙ o c f i t g ] + [ x o c f i y o c f i z o c f i ]
\left[ \begin{gathered}{x_{ocff}} \\ {y_{ocff}} \\ {z_{ocff}}\end{gathered}\right] =
\left[ \begin{gathered}
{F 2\left(t_{g}\right) \cos \left(\overline{\varphi}_{ocf}\right)+F 2\left(t_{g}\right) K_{\varphi 1} \sin \left(\overline{\varphi}_{ocf}\right) -F3 \left(t_{g}\right) K_{\varphi 2} \sin \left(\overline{\varphi}_{ocf}\right)} \\
{F 2\left(t_{g}\right) \sin \left(\overline{\varphi}_{ocf}\right)-F 2\left(t_{g}\right) K_{\varphi 1} \cos \left(\overline{\varphi}_{ocf}\right)+F 3\left(t_{g}\right) K_{\varphi 2} \cos \left(\overline{\varphi}_{ocf}\right)} \\
{F 2\left(t_{g}\right) K_{\psi 1} \cos \left(\overline{\psi}_{ocf}\right)-F 3\left(t_{g}\right) K_{\psi 2} \cos \left(\overline{\psi}_{ocf}\right)+F 2\left(t_{g}\right) \sin \left(\overline{\psi}_{ocf}\right)}
\end{gathered}\right] \\
+\left[ \begin{gathered}
{\frac{1}{2} g_{ocfx} t_{g}^{2}} \\ {\frac{1}{2} g_{ocfy} t_{g}^{2}} \\ {\frac{1}{2} g_{ocfz} t_{g}^{2}}
\end{gathered}\right]
+\left[ \begin{gathered}
{\dot{x}_{ocfi} t_{g}} \\ {\dot{y}_{ocfi} t_{g}} \\ {\dot{z}_{ocfi} t_{g}}
\end{gathered}\right]
+\left[ \begin{gathered}
{x_{ocfi}} \\ {y_{ocfi}} \\ {z_{ocfi}}
\end{gathered}\right] \tag{25}
⎣ ⎢ ⎡ x o c f f y o c f f z o c f f ⎦ ⎥ ⎤ = ⎣ ⎢ ⎡ F 2 ( t g ) cos ( φ o c f ) + F 2 ( t g ) K φ 1 sin ( φ o c f ) − F 3 ( t g ) K φ 2 sin ( φ o c f ) F 2 ( t g ) sin ( φ o c f ) − F 2 ( t g ) K φ 1 cos ( φ o c f ) + F 3 ( t g ) K φ 2 cos ( φ o c f ) F 2 ( t g ) K ψ 1 cos ( ψ o c f ) − F 3 ( t g ) K ψ 2 cos ( ψ o c f ) + F 2 ( t g ) sin ( ψ o c f ) ⎦ ⎥ ⎤ + ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 2 1 g o c f x t g 2 2 1 g o c f y t g 2 2 1 g o c f z t g 2 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ + ⎣ ⎢ ⎡ x ˙ o c f i t g y ˙ o c f i t g z ˙ o c f i t g ⎦ ⎥ ⎤ + ⎣ ⎢ ⎡ x o c f i y o c f i z o c f i ⎦ ⎥ ⎤ ( 2 5 )
其中F 2 ( t ) F2(t) F 2 ( t ) F 3 ( t ) F3(t) F 3 ( t ) (26) F 2 ( t ) ≜ ∫ 0 t ∫ 0 s V e x t h − t d t d s = F 0 ( t ) ⋅ t − F 1 ( t )
F 2(t) \triangleq \int_{0}^{t} \int_{0}^{s} \frac{V_{ex}}{t_{h}-t} d t d s=F 0(t) \cdot t-F 1(t) \tag{26}
F 2 ( t ) ≜ ∫ 0 t ∫ 0 s t h − t V e x d t d s = F 0 ( t ) ⋅ t − F 1 ( t ) ( 2 6 )
(27) F 3 ( t ) ≜ ∫ 0 t ∫ 0 s V e x t h − t t d t d s = F 2 ( t ) ⋅ t h − V e x 2 t 2
F 3(t) \triangleq \int_{0}^{t} \int_{0}^{s} \frac{V_{ex}}{t_{h}-t} t d t d s=F 2(t) \cdot t_{h}-\frac{V_{ex}}{2}t^{2} \tag{27}
F 3 ( t ) ≜ ∫ 0 t ∫ 0 s t h − t V e x t d t d s = F 2 ( t ) ⋅ t h − 2 V e x t 2 ( 2 7 )
根据等式（23）和（25）可得[2][5](28) K φ 1 = y o c f f − F 2 ( t g ) sin ( φ ‾ o c f ) − 1 2 g o c f y t g 2 − y ˙ o c f i t g − y o c f i [ − F 2 ( t g ) + F 3 ( t g ) F 0 ( t g ) F 1 ( t g ) ] cos ( φ ‾ o c f )
K_{\varphi 1}=\frac{y_{ocff}-F2\left(t_{g}\right) \sin \left(\overline{\varphi}_{ocf} \right)- \frac{1}{2} g_{ocfy} t_{g}^{2}-\dot{y}_{ocfi} t_{g}-y_{ocfi}}{\left[ -F2 \left(t_{g}\right)+\frac{F 3\left(t_{g}\right) F 0\left(t_{g}\right)}{F 1\left(t_{g}\right)}\right] \cos \left(\overline{\varphi}_{o c f}\right)} \tag{28}
K φ 1 = [ − F 2 ( t g ) + F 1 ( t g ) F 3 ( t g ) F 0 ( t g ) ] cos ( φ o c f ) y o c f f − F 2 ( t g ) sin ( φ o c f ) − 2 1 g o c f y t g 2 − y ˙ o c f i t g − y o c f i ( 2 8 )
(29) K φ 2 = [ y o c f f − F 2 ( t g ) sin ( φ ‾ o c f ) − 1 2 g o c f y t g 2 − y ˙ o c f i t g − y o c f i ] F 0 ( t g ) [ − F 2 ( t g ) F 1 ( t g ) + F 3 ( t g ) F 0 ( t g ) ] cos ( φ ‾ o c f )
K_{\varphi 2}=\frac{\left[y_{ocff}-F2\left(t_{g}\right) \sin \left(\overline{\varphi}_{ocf}\right)-\frac{1}{2} g_{ocfy} t_{g}^{2}-\dot{y}_{ocfi} t_{g}-y_{ocfi}\right] F0\left(t_{g}\right)}{\left[-F2\left(t_{g}\right) F1\left(t_{g}\right)+F3\left(t_{g}\right) F0\left(t_{g}\right)\right] \cos \left(\overline{\varphi}_{ocf}\right)} \tag{29}
K φ 2 = [ − F 2 ( t g ) F 1 ( t g ) + F 3 ( t g ) F 0 ( t g ) ] cos ( φ o c f ) [ y o c f f − F 2 ( t g ) sin ( φ o c f ) − 2 1 g o c f y t g 2 − y ˙ o c f i t g − y o c f i ] F 0 ( t g ) ( 2 9 )
(30) K ψ 1 = z o c f f − 1 2 g o c f z t g 2 − z ˙ o c f i t g − z o c f i + F 2 ( t g ) sin ( ψ ‾ o c f ) [ F 2 ( t g ) − F 3 ( t g ) F 0 ( t g ) F 1 ( t g ) ] cos ( ψ ‾ o c f )
K_{\psi 1}=\frac{z_{ocff}-\frac{1}{2} g_{ocfz} t_{g}^{2}-\dot{z}_{ocfi} t_{g}-z_{ocfi}+F 2\left(t_{g}\right) \sin \left(\overline{\psi}_{ocf}\right)}{\left[ F2\left(t_{g}\right)-\frac{F 3\left(t_{g}\right) F 0\left(t_{g}\right)}{F 1\left(t_{g}\right)}\right] \cos \left(\overline{\psi}_{o c f}\right)} \tag{30}
K ψ 1 = [ F 2 ( t g ) − F 1 ( t g ) F 3 ( t g ) F 0 ( t g ) ] cos ( ψ o c f ) z o c f f − 2 1 g o c f z t g 2 − z ˙ o c f i t g − z o c f i + F 2 ( t g ) sin ( ψ o c f ) ( 3 0 )
(31) K ψ 2 = [ z o c f f − 1 2 g o c f z t g 2 − z ˙ o c f i t g − z o c f i + F 2 ( t g ) sin ( ψ ‾ o c f ) ] F 0 ( t g ) [ F 2 ( t g ) F 1 ( t g ) − F 3 ( t g ) F 0 ( t g ) ] cos ( ψ ‾ o c f )
K_{\psi 2}=\frac{\left[z_{ocff}-\frac{1}{2} g_{ocfz} t_{g}^{2}-\dot{z}_{ocfi} t_{g}-z_{ocfi} + F2\left(t_{g}\right) \sin \left(\overline{\psi}_{ocf}\right)\right] F0\left(t_{g}\right)}{\left[ F 2\left(t_{g}\right) F 1\left(t_{g}\right)-F 3\left(t_{g}\right) F 0\left(t_{g}\right)\right] \cos \left(\overline{\psi}_{ocf}\right)} \tag{31}
K ψ 2 = [ F 2 ( t g ) F 1 ( t g ) − F 3 ( t g ) F 0 ( t g ) ] cos ( ψ o c f ) [ z o c f f − 2 1 g o c f z t g 2 − z ˙ o c f i t g − z o c f i + F 2 ( t g ) sin ( ψ o c f ) ] F 0 ( t g ) ( 3 1 )
计算重力损失修正常数 k k k
火箭瞬时速度矢量在目标轨道平面的投影记为V x y \boldsymbol{V}_{xy} V x y V x y \boldsymbol{V}_{xy} V x y ∥ V x y ∥ t \Vert \boldsymbol{V}_{xy}\Vert t ∥ V x y ∥ t F 2 ( t g ) F2(t_{g}) F 2 ( t g ) k t g 2 kt^{2}_{g} k t g 2 k k k S 1 S_{1} S 1 S 1 = ∥ V x y ∥ t g + F 2 ( t g ) + k t g 2 S_{1}=\Vert \boldsymbol{V}_{xy}\Vert t _{g} + F2(t_{g}) + kt^{2}_{g} S 1 = ∥ V x y ∥ t g + F 2 ( t g ) + k t g 2
文献[1]使用了另一种方法估计火箭轨迹在轨道平面内投影的长度。求视加速度 F / m F/m F / m W r c f x 、 W r c f y 、 W r c f z W_{rcfx}、W_{rcfy}、W_{rcfz} W r c f x 、 W r c f y 、 W r c f z t h o r b i t = V e x / W r c f x 2 + W r c f y 2 t_{horbit}=V_{ex}/\sqrt{W^2_{rcfx}+W^2_{rcfy}} t h o r b i t = V e x / W r c f x 2 + W r c f y 2 t h o r b i t t_{horbit} t h o r b i t F 0 ( t ) F0(t) F 0 ( t ) F 1 ( t ) F1(t) F 1 ( t ) F 2 ( t ) F2(t) F 2 ( t ) t h t_{h} t h (32) F 0 ( t ) o r b i t ≜ ∫ 0 t V e x t h o r b i t − t d t = V e x ln t h o r b i t t h o r b i t − t
F0(t)_{orbit} \triangleq \int_{0}^{t} \frac{V_{ex}}{t_{horbit}-t} d t=V_{ex} \ln \frac{t_{horbit}}{t_{horbit}-t} \tag{32}
F 0 ( t ) o r b i t ≜ ∫ 0 t t h o r b i t − t V e x d t = V e x ln t h o r b i t − t t h o r b i t ( 3 2 )
(33) F 1 ( t ) o r b i t ≜ ∫ 0 t V e x t h o r b i t − t t d t = t h o r b i t F 0 ( t ) o r b i t − V e x t
F1(t)_{orbit} \triangleq \int_{0}^{t} \frac{V_{ex}}{t_{horbit}-t} t d t=t_{horbit} F0(t)_{orbit}-V_{ex} t \tag{33}
F 1 ( t ) o r b i t ≜ ∫ 0 t t h o r b i t − t V e x t d t = t h o r b i t F 0 ( t ) o r b i t − V e x t ( 3 3 )
(34) F 2 ( t ) o r b i t ≜ ∫ 0 t ∫ 0 s V e x t h o r b i t − t d t d s = F 0 ( t ) o r b i t ⋅ t − F 1 ( t ) o r b i t
F 2(t)_{orbit} \triangleq \int_{0}^{t} \int_{0}^{s} \frac{V_{ex}}{t_{horbit}-t} dtds=F0(t)_{orbit} \cdot t-F1(t)_{orbit} \tag{34}
F 2 ( t ) o r b i t ≜ ∫ 0 t ∫ 0 s t h o r b i t − t V e x d t d s = F 0 ( t ) o r b i t ⋅ t − F 1 ( t ) o r b i t ( 3 4 )
下图中，$\boldsymbol{r}{xy} 表 示 火 箭 瞬 时 位 置 矢 量 在 目 标 轨 道 平 面 的 投 影 ， 表示火箭瞬时位置矢量在目标轨道平面的投影， 表 示 火 箭 瞬 时 位 置 矢 量 在 目 标 轨 道 平 面 的 投 影 ， {xy}表 示 火 箭 瞬 时 速 度 矢 量 在 目 标 轨 道 平 面 的 投 影 ， 火 箭 瞬 时 的 当 地 轨 道 倾 角 记 为 表示火箭瞬时速度矢量在目标轨道平面的投影，火箭瞬时的当地轨道倾角记为 表 示 火 箭 瞬 时 速 度 矢 量 在 目 标 轨 道 平 面 的 投 影 ， 火 箭 瞬 时 的 当 地 轨 道 倾 角 记 为 cos ( 9 0 ∘ − θ H ) = r x y T V x y ∥ r x y ∥ ∥ V x y ∥
\cos(90^{\circ}-\theta_{H}) = \frac{\boldsymbol{r}_{xy}^{T}\boldsymbol{V}_{xy}}{\Vert\boldsymbol{r}_{xy}\Vert \Vert\boldsymbol{V}_{xy}\Vert}
cos ( 9 0 ∘ − θ H ) = ∥ r x y ∥ ∥ V x y ∥ r x y T V x y
(35)   ⟹   cos ( θ H ) = ∣ x y ˙ − y x ˙ ∣ ∥ r x y ∥ ∥ V x y ∥
\implies \cos(\theta_{H}) = \frac{\vert x\dot{y}-y\dot{x}\vert}{\Vert\boldsymbol{r}_{xy}\Vert \Vert\boldsymbol{V}_{xy}\Vert} \tag{35}
⟹ cos ( θ H ) = ∥ r x y ∥ ∥ V x y ∥ ∣ x y ˙ − y x ˙ ∣ ( 3 5 )
其中r x y = [ x y ] T \boldsymbol{r}_{xy} = \lbrack x \quad y\rbrack^{T} r x y = [ x y ] T V x y = [ x ˙ y ˙ ] T \boldsymbol{V}_{xy}=[\dot{x}\quad \dot{y}]^{T} V x y = [ x ˙ y ˙ ] T
已知入轨点当地的轨道倾角θ \theta θ S 2 S_{2} S 2 (36) S 2 ≈ ∥ V x y ∥ t g cos θ H + F 2 ( t g ) o r b i t cos θ
S_{2} \approx \Vert \boldsymbol{V}_{xy}\Vert t_{g}\cos{\theta_{H}} + F2(t_{g})_{orbit}\cos{\theta} \tag{36}
S 2 ≈ ∥ V x y ∥ t g cos θ H + F 2 ( t g ) o r b i t cos θ ( 3 6 ) S 2 S_{2} S 2 S 1 S_{1} S 1 S 1 ≈ S 2 S_{1} \approx S_{2} S 1 ≈ S 2 ∥ V x y ∥ t g + F 2 ( t g ) + k t g 2 ≈ ∥ V x y ∥ t g cos θ H + F 2 ( t g ) o r b i t cos θ
\Vert \boldsymbol{V}_{xy}\Vert t_{g} + F2(t_{g}) + kt^{2}_{g}
\approx \Vert \boldsymbol{V}_{xy}\Vert t_{g}\cos{\theta_{H}} + F2(t_{g})_{orbit}\cos{\theta}
∥ V x y ∥ t g + F 2 ( t g ) + k t g 2 ≈ ∥ V x y ∥ t g cos θ H + F 2 ( t g ) o r b i t cos θ
(37)   ⟹   k ≈ ∥ V x y ∥ t g ( cos θ H − 1 ) + F 2 ( t g ) o r b i t cos θ − F 2 ( t g ) t g 2
\implies k \approx \frac{\Vert \boldsymbol{V}_{xy}\Vert t_{g}(\cos{\theta_{H}-1}) + F2(t_{g})_{orbit}\cos{\theta} - F2(t_{g})} {t^2_{g}} \tag{37}
⟹ k ≈ t g 2 ∥ V x y ∥ t g ( cos θ H − 1 ) + F 2 ( t g ) o r b i t cos θ − F 2 ( t g ) ( 3 7 )
用低地球轨道（LEO）任务进行仿真验证，任务简图如图 6，仿真参数如下表[8]
装订项
值
制导开始时刻的质量m 0 m_{0} m 0
86500 k g 86500kg 8 6 5 0 0 k g
发动机真空推力T v a c T_{vac} T v a c
8 × 1 0 5 N 8\times 10^5 N 8 × 1 0 5 N
秒耗量m ˙ \dot{m} m ˙
272 k g / s 272kg/s 2 7 2 k g / s
发射点赤经
1 0 ∘ 10^{\circ} 1 0 ∘
发射点经度
10 0 ∘ E 100^{\circ}\text{E} 1 0 0 ∘ E
发射点地理维度B 0 B_{0} B 0
41.190 6 ∘ N 41.1906^{\circ}\text{N} 4 1 . 1 9 0 6 ∘ N
目标轨道半长轴a a a
6622785.34 m 6622785.34m 6 6 2 2 7 8 5 . 3 4 m
目标轨道偏心率e e e
0 0 0
目标轨道倾角i i i
68.84 6 ∘ 68.846^{\circ} 6 8 . 8 4 6 ∘
目标轨道升交点赤经Ω \Omega Ω
− 9.96 4 ∘ -9.964^{\circ} − 9 . 9 6 4 ∘
目标轨道近地点幅角ω \omega ω
0 ∘ 0^{\circ} 0 ∘
目标入轨点对应的真近点角f f f
49.682 3 ∘ 49.6823^{\circ} 4 9 . 6 8 2 3 ∘
制导开始时刻火箭相对于发射惯性坐标系的位置r 0 \boldsymbol{r}_{0} r 0
[ 2.9058 × 1 0 5 92126 − 33401 ] m \left[2.9058\times 10^{5} \enspace 92126\enspace -33401\right]m [ 2 . 9 0 5 8 × 1 0 5 9 2 1 2 6 − 3 3 4 0 1 ] m
制导开始时刻火箭相对于发射惯性坐标系的速度v 0 \boldsymbol{v}_{0} v 0
[ 2815.75 494.703 70.953 ] m / s \left[2815.75 \enspace 494.703 \enspace 70.953 \right]m/s [ 2 8 1 5 . 7 5 4 9 4 . 7 0 3 7 0 . 9 5 3 ] m / s
制导周期Δ t \Delta t Δ t
2 s 2s 2 s
制导开始时刻火箭俯仰角
61.338 3 ∘ 61.3383^{\circ} 6 1 . 3 3 8 3 ∘
制导开始时刻火箭偏航角
0.310 5 ∘ 0.3105^{\circ} 0 . 3 1 0 5 ∘
仿真结果
入轨点轨道根数如下表
入轨点轨道六根数
仿真得到的数值
轨道半长轴a a a
6.5608 × 1 0 6 6.5608\times 10^{6} 6 . 5 6 0 8 × 1 0 6
偏心率e e e
0.0162
轨道倾角i i i
68.834 0 ∘ 68.8340^{\circ} 6 8 . 8 3 4 0 ∘
升交点赤经Ω \Omega Ω
− 9.984 0 ∘ -9.9840^{\circ} − 9 . 9 8 4 0 ∘
近地点幅角ω \omega ω
− 68.361 0 ∘ -68.3610^{\circ} − 6 8 . 3 6 1 0 ∘
入轨点真近点角f f f
57.589 9 ∘ 57.5899^{\circ} 5 7 . 5 8 9 9 ∘
从图7可看出，火箭俯仰角φ \varphi φ
count
f / d e g f/deg f / d e g φ ˉ o c f / d e g \bar{\varphi}_{ocf}/deg φ ˉ o c f / d e g t g 0 / s t_{g0}/s t g 0 / s t g / s t_g/s t g / s ϕ / d e g \phi/deg ϕ / d e g
1
57.6823 14.9774
260
261.2839
57.6823
2
58.0049 14.8186 258
259.4437
58.0049
3
57.9859
14.4737 256
257.5097
57.9859
4
579736
14.3217
254
255.6799
57.9736
5
57.9615
14.1649
252
253.8467
57.9610
…
…
…
…
…
…
从表中数据可以看出入轨点真近点角f f f f f f f = 58.014 9 ∘ f = 58.0149^{\circ} f = 5 8 . 0 1 4 9 ∘
[1] 韩祝斋. 用于大型运载火箭的迭代制导方法[J]. 宇航学报, 1983, 4(1):12-24.
