# A Gift That Keeps On Giving

We see from “Seek-Lock-Strike!” Again that given the missile’s position $(x, y),$ $\frac{dx}{dy}=\frac{a+v_a t-x}{b-y}$

where $x$ and $y$ are themselves functions of time $t.$

It means $\frac{dx}{dy} = \frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{a+v_a t-x}{b-y}\implies \frac{dx}{dt} = \frac{a+v_a t-x}{b-y}\cdot\frac{dy}{dt}.$

That is, let $\kappa(x,y,t) = \frac{a+v_a t-x}{b-y},$ $\frac{dx}{dt}=\kappa\cdot \frac{dy}{dt}.\quad\quad\quad(1)$

We also have (see “Seek-Lock-Strike!”) $v_m = \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\implies v_m = \sqrt{1+\kappa^2}|\frac{dy}{dt}| .$

Since $\frac{dy}{dt} >0,$ $\frac{dy}{dt} = \frac{v_m}{\sqrt{1+\kappa^2}}.\quad\quad\quad(2)$

Substitute (2) into (1) yields $\frac{dx}{dt} = \frac{v_m\cdot \kappa}{\sqrt{1+\kappa^2}}.\quad\quad\quad(3)$

It follows that $(x(t), y(t))$, the position of the missile satisfies the initial-value problem $\begin{cases} \frac{dx}{dt} = \frac{\kappa\cdot v_m}{\sqrt{1+\kappa^2}} \\ \frac{dy}{dt} = \frac{v_m}{\sqrt{1+\kappa^2}} \\x(0)=0, y(0)=0\end{cases}\quad\quad\quad(4)$

To obtain the missile’s trajectory, we solve (4) numerically using the Runge-Kutta algorithm. It integrates (4) from $t=0$ to $t= t_*$ (see “Seek-Lock-Strike!”).

Fig. 1 $a=100\;m, b=3000\;m, v_a=1500\;ms^{-1},v_m=2000\;ms^{-1}$

The missile strike is illustrated in Fig. 1 and 2.

Fig. 2 $a=100\;m, b=3000\;m, v_a=1500\;ms^{-1},v_m=2000\;ms^{-1}$

Fig. 3 $a=-1200\;m, b=3000\;m, v_a=1500\;ms^{-1},v_m=2000\;ms^{-1}$

The trajectories shown are much smoother than those in “Seek-Lock-Strike!” Animated.

# “Seek-Lock-Strike!” Animated

In “Seek-Lock-Strike!” Again, we obtained the missile’s trajectory. Namely, $x = \frac{1}{2} \left(\frac{(b-y)^{r+1}}{b^r \cdot f \cdot (r+1)}-\frac{b^r \cdot f \cdot (b-y)^{1-r}}{1-r}\right) -k\quad\quad\quad(*)$

where $f = \frac{a}{b}+\sqrt{1+(\frac{a}{b})^2},$ $k = \frac{b}{2}\left(\frac{1}{f \cdot (r+1)}-\frac{f}{1-r}\right).$

Since the fighter jet maintains its altitude ( $y= b$), the missile must strike it at $(x_*, b)$. Setting $y=b$ in $(*)$ gives $x_* = -k.$

Hence, we can plot $(*):$

Fig. 1 $a = 100\;m, b=3000\;m, v_a=1500\;ms^{-1}, v_m=2000\;ms^{-1}$

We can also illustrate “Seek-Lock-Strike” in animations:

Fig. 2 $a = 100\;m, b=3000\;m, v_a=1500\;ms^{-1}, v_m=2000\;ms^{-1}$

Fig. 3 $a = -2500\;m, b=3000\;m, v_a=1500\;ms^{-1}, v_m=2000\;ms^{-1}$

# “Seek-Lock-Strike!” Again

We can derive a different governing equation for the missile in “Seek-Lock-Strike!“.

Fig. 1

Looking from a different viewpoint (Fig. 1), we see $\frac{dx}{dy} = \frac{a+v_at-x}{b-y}.\quad\quad\quad(1)$

Solving (1) for $t$, $t = -\frac{\frac{dx}{dy}y-b\frac{dx}{dy}-x+a}{v_a}.\quad\quad\quad(2)$

We also have $v_m t = \int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2} \implies t = \frac{\int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2}\;dy}{v_m}.\quad\quad\quad(3)$

Equate (1) and (2) gives $-\frac{\frac{dx}{dy}y-b\frac{dx}{dy}-x+a}{v_a}-\frac{\int\limits_{0}^{y}\sqrt{1+(\frac{dx}{dy})^2}\;dy}{v_m} = 0.\quad\quad\quad(4)$

The governing eqaution emerges after differentiate (4) with respect to $x:$ $-\frac{d^2x}{dy^2}y+b\frac{d^2x}{dy^2}-\frac{v_a\sqrt{1+(\frac{dx}{dy})^2}}{v_m} = 0.\quad\quad\quad(5)$

We let $p = \frac{dx}{dy}$ so $\frac{d^2x}{dy^2} = \frac{d}{dy}\left(\frac{dx}{dy}\right)=\frac{dp}{dy}$ and express (5) as $\frac{dp}{dy}(b-y) -r\sqrt{1+(\frac{dx}{dy})^2} = 0\quad\quad\quad(*)$

where $r = \frac{v_a}{v_m}.$

Fig. 2

Using Omega CAS Explorer, we compute the missile’s striking time $t_*$ (see Fig. 3). It agrees with the result obtained previously.

Fig. 3

Exercise-1 Obtain the missile’s trajectory from (*).

# “Seek-Lock-Strike!” Simplified

There is an easier way to derive the governing equation ((5), “Seek-Lock-Strike!“) for the missile.

Solving $\frac{dy}{dx}(a+v_at-x) = b-y$

for $t,$ we have $t = \frac{(x-a)\frac{dy}{dx}-y+b}{v_a\frac{dy}{dx}}.\quad\quad\quad(1)$

From $v_m t = \int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx,$

we also have $t = \frac{\int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx}{v_m}.\quad\quad\quad(2)$

Equate (1) and (2) gives $\frac{\int\limits_{0}^{x}\sqrt{1+(\frac{dy}{dx})^2}\;dx}{v_m}-\frac{(x-a)\frac{dy}{dx}-y+b}{v_a\frac{dy}{dx}}=0.\quad\quad\quad(3)$

Differentiate (3) with repect to $x,$ we obtain $(bv_m-v_my)\frac{d^2y}{dx^2} + v_a(\frac{dy}{dx})^2\sqrt{1+(\frac{dy}{dx})^2} = 0.\quad\quad\quad(4)$

(see Fig. 1)

Fig. 1

Let $r=\frac{v_a}{v_m}, p=\frac{dy}{dx} \implies \frac{d^2y}{dx^2} = p\frac{dp}{dy},$ (4) bcomes $(b-y)p\frac{dp}{dy} + r p^2 \sqrt{1+p^2} = 0.$

Since $p = \frac{dy}{dx} \ne 0$, dividing $p$ through yields $(b-y)\frac{dp}{dy} + rp\sqrt{1+p^2}=0,$

the governing equation for the missile.