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A stealth missile $M$ is launched from space station. You, at another space station far away, are trusted with the mission of intercepting $M$ using a single cruise missile $C$ at your disposal .

You know the target missile is traveling in straight line at constant speed $v_m$. You also know the precise location and time at which it was launched. $M$, built by state-of-the-art stealth technology however, is invisible (to you or your $C$). So you have no idea in which direction it is going. Your $C$ has a maximum speed $v_c>v_m$.

Can you control trajectory of $C$ so that it is guaranteed to intercept $M$ in finite time? Is this possible?


I can think of 3 apparent possibilities:

  1. Precise interception is possible. (It is possible in two dimensions, by calibrating your missile's trajectory to the parameters of certain logarithm spiral).
  2. Precise interception is impossible, but for any $\epsilon\gt 0$, paths can be designed so that $C$ can get as close to $M$ as $\epsilon$ in finite time.
  3. There's no hope, and your chance of intercepting or getting close to $M$ diminishes as time goes by.

This question is inspired by a similar problem in two dimensions by Louis A. Graham in his book Ingenious Mathematical Problems and Methods.

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    $\begingroup$ for the two-dimensional problem, see Tanvolutes: Generalized involutes $\endgroup$ – Carlo Beenakker Apr 18 '20 at 9:38
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    $\begingroup$ What do you mean by a "cruise missile"? Cruise missiles have wings and travel in the atmosphere. $\endgroup$ – Alexandre Eremenko Apr 18 '20 at 12:10
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    $\begingroup$ I'm probably missing something here, but I have to ask. It is stated that the stealth missile $M$ is traveling in a straight line, and we know its location and time at launch. If, for instance, both space stations are maintaining same orbit (i.e. their distance is constant); shouldn't that mean that we know the (shortest or longest) path the stealth missile will take assuming the sphere is symmetrical around its x-axis? $\endgroup$ – litmus Apr 19 '20 at 19:07
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There is no hope. Timothy Budd already explained why staying on the sphere $S_t$ of possible locations of $M$ will not work; so this is to explain why leaving it will not help. What is enough to show is the following claim: at large times, if $t_1<t_2$ and $C_{t_{1,2}}$ are at $\epsilon$-neighborhoods of $S_{t_{1,2}}$ respectively, then the radial projections of $C_{t_{1,2}}$ onto the unit sphere are at distance at most $c\cdot \log {t_2/t_1}$ from each other. With the claim, essentially the same argument will work.

(Details: think about the task in terms of gradually ruling out possible directions in which $M$ can go. One can only exclude a direction if at some time $t$, $C$ is at the $\epsilon$-neighborhood of the corresponding point of $S_t$. Suppose Alice has a radar that she can direct to any point of $S_t$ and rule out the $2\epsilon$-neighborhood of it. Let Bob follow any strategy, and let Alice do this: whenever Bob's $C$ is in the closed $\epsilon$-neighborhood of $S_t$, she directs the radar to the projection of Bob's $C$ on $S_t$; in between those times, she turns her radar at constant speed. Clearly Alice will know more than Bob, yet by the claim, the angular speed of the radar will be bounded by $c/t$, and by Budd's argument, Alice will not be able to rule out everything.)

But the claim is fairly obvious: between $t_1$ and $t_2$, $C_t$ cannot get more than $(v_c-v_m)(t_2-t_1)+2\epsilon$ behind $S_t$, otherwise it wouldn't be able to catch up. If $t_2-t_1<t_1\cdot \frac{v_c-v_m}{100v_m}$, then this means that $C_t$ is always at distance at least $\frac12v_m t$ from the origin, and then the speed of the projection is bounded by $\mathrm{const}/t$ as in Budd's answer. If $t_2/t_1\geq 1+\frac{v_c-v_m}{100v_m}$, then we can just bound the distance between projections of $C_{t_{1,2}}$ by $2$, and take the constant $c$ to be $2\cdot \log\left(1+\frac{v_c-v_m}{100v_m}\right)$

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    $\begingroup$ Well, Timothy has answered that comprehensively: to cover a curve by a curve, you don't need an $\epsilon$-neighborhood at all, just that the length at your disposal is large enough. It's imposible to cover a 2d surface by a curve, you need a neighborhood. In our case, the size of the neighborhood at our disposal decreases too quickly. $\endgroup$ – Kostya_I Apr 19 '20 at 13:29
  • $\begingroup$ Yes, now I see why. Budd answered my comment just as I posted another one here so I deleted it. Together your answers are really conclusive. $\endgroup$ – Eric Apr 19 '20 at 14:09
  • $\begingroup$ Upon closer inspection, in this argument, why do we have to make the $S_t$ sphere our anchor and reference points in the first place? What about countless other plans that refers not to $S_t$ at all? "What is enough to show is the following claim..." Why is this claim enough? $\endgroup$ – Eric Apr 20 '20 at 2:14
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    $\begingroup$ @Eric, I added some explanation, hope that helps! $\endgroup$ – Kostya_I Apr 20 '20 at 9:06
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If $C$ stays on the growing sphere of possible locations of $M$ then there seems to be no hope. Suppose $v_m = 1$ and $v_c = \sqrt{2}$, $t$ is the time since launch and we are trying to get to within distance $\epsilon$ from $M$.

As soon as $C$ reaches the $t$-multiple of the unit sphere, say at time $t=t_0$, it will follow a trajectory $t\cdot x(t)$ with $\|x(t)\|=1$ on the unit sphere with velocity $\|\dot{x}(t)\|=1/t$ (by Pythagoras). The area of the positions of $M$ that $C$ can exclude by moving on the sphere is roughly $$ \int_{t_0}^t \frac{\mathrm{d}t}{t} \frac{\epsilon}{t} = \frac{\epsilon}{t_0} - \frac{\epsilon}{t} < \frac{\epsilon}{t_0}$$ So to exclude an area of order $4\pi$ we are in trouble if $\epsilon < 4\pi t_0$.

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  • $\begingroup$ Yes, that strategy is the natural one that comes to mind and was suggested by Robin in the comments. It would work if $v_c(t)=ct+k$, i.e., increasing linearly with time, for possibility 2. $\endgroup$ – Eric Apr 18 '20 at 15:54
  • $\begingroup$ Oops, somehow I missed that comment.... let me leave the answer as a (trivial) baseline. $\endgroup$ – Timothy Budd Apr 18 '20 at 15:59
  • $\begingroup$ I thought I understood your argument. But now I'm a little confused. What is $x(t)$? Can you explain a bit more in detail about "it will follow a trajectory $t\cdot x(t)$ ... (by Pythagoras)"? $\endgroup$ – Eric Apr 19 '20 at 2:11
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    $\begingroup$ $\mathbb{R}_+\to S^2\subset\mathbb{R}^3: t\to x(t)$ is the position of C projected onto the unit sphere. The speed of C in $\mathbb{R}^3$ is then $v_c^2=\|x(t)\|^2+t^2\|\dot{x}(t)\|^2=2$ if $\|\dot{x}(t)\|=1/t$. $\endgroup$ – Timothy Budd Apr 19 '20 at 6:26
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    $\begingroup$ There is no $\epsilon/t$ in the integral in the two-dimensional case. The excluded length on the unit circle is then $\int_{t_0}^t \mathrm{d}t/t = \log(t/t_0)$. $\endgroup$ – Timothy Budd Apr 19 '20 at 13:24

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