“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 (*).

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One thought on ““Seek-Lock-Strike!” Again”