Actually, the assumption I made is that the poster is acting in good faith and not trying to waste our time. If it is some kind of trick question, then I’m not particularly interested in it.
The short answer is that Ship B cannot catch Ship A.
Here’s the slightly longer answer:
Assume that the ships travel at unit speed.
Let’s give the captains names: I propose Ahab and (Black) Bart.
Let’s ignore the bit about changing directions for a moment. There seems to be a notion running through this thread that Ahab can only avoid Bart if it sails always directly away from Bart. This is not true, as it is easy to check. Start Ahab at the origin and Bart at the point (1,0). Suppose that Ahab sails in the y direction. Bart can never catch Ahab if he maintains this course. Suppose that Bart did intercept Ahab. We know that Ahab is traveling in the y direction, so the intercept point must be of the form (0,t) where t is the time elapsed since the chase began.
This supposed interception point is distance t from Ahab’s starting point and distance (t[sup]2[/sup] + 1)[sup]1/2[/sup] from Bart’s starting point. Because Bart travels at unit speed, he cannot reach this point. (…in space and time. He can reach that point in space, but by the time he gets there, Ahab has vamoosed) So if Ahab need not change course, he cannot be caught. All the captain needs to do is pick a heading and stick with it. (As long as the horizontal component of Ahab’s travel is negative, Bart will not be able to catch him.)
Fine, but kanicbird said that Ahab is required to periodically change course. Anytime Queequeg informs Ahab that he need to change course, he should adjust their heading so that they are sailing perpendicularly to the direction of Bart. That is to say: draw a line connecting Ship A and Ship B; Ahab should steer so that it is moving perpendicularly to this line. Thus any time the course is adjusted, we simply restart the clock in the previous situation. In this way Ahab can evade Bart, even while making periodic course adjustments.
If we strengthen the meaning of “every once in a while” to continuously, we see that more or less the same situation holds. If Ahab must make continuous course adjustments, he can simply steer to be moving perpendicularly to the apparent direction of Bart.
Now, what if Bart decides to live up to his piratical name and arranges to steal Ahab’s sailing plans? Now Ahab needs to plot a course so that, after time t has elapsed, he is at least distance t from Bart’s starting point. Again, such a course can be constructed. If course changes are not required, simply choose any straight line course which avoids interception. If course changes are required, it is a little more difficult to describe a course, but it can be done. (If you don’t believe me, I can post such a curve tomorrow. It can be done with continuously changing heading, or as a piecewise linear curve, whichever you like. At the moment, it’s quite late so I’m going to go to bed.)
As the last part of my post demonstrates, it is possible for Ahab to chart a course on which it is impossible for Bart to catch him. It doesn’t matter whether Bart is attempting to chart interception courses or is following in pursuit, Ahab can remain uncaught if he so chooses. (A pursuit in the mathematical sense that Bart is always sailing directly toward Ahab.)
In your example, although Bart hasn’t caught Ahab, he has got closer. In fact, every time Ahab changes direction, Bart is given an opportunity to get closer because he can cut the corner and track straight to Ahab without taking the longer route to the same point that Ahab has taken. So Bart gets closer and closer with each change of direction. I’ll leave it to someone who knows the maths to prove that Bart actually reaches Ahab in theory, however, in reality, once the distance gets down to microscopic levels the ships will be what we earthlings call “touching” and they will be able to get no closer.
So that’s the reality, they’ll eventually get close enough to be physically touching in real world terms, but it’s more interesting, in my opinion, to mathematically prove that they will touch, but that is out of my league.
I believe I did a few posts back.
Let’s assume that one or both ship have a finite length or beam. Chose one point within the ships which we will use as a reference point for speed and direction. Since those points are a finite distance from the ends and sides of the ship, the ships must eventually touch as the reference points within them become asymptotically close.
Welcome to the SDMB.
OK, so I’m only at post 78, but I’ve got to run so I want to post this before I leave. Sailor keeps saying that B gains ground when A diverts course. This is slightly untrue. More accurate is to say that B begins to gain ground. He has to travel over some time interval to actually get closer. There is no argument from me that that is exactly what happens. What I will argue, though, is that the time needed for d=0 is infinity, for the reasons SaintCal points out.
Consider this: If they do meet, what was the position of the boats at a given T previous to them meeting? In other words, imagine the end scenario and back up in time as far as you like. I would require that A was pointed at an acute angle WRT the segment connecting the boats. We’ve already shown that this can’t happen with any strategy by A. Only though random folly could this course be selected.
BTW, has anyone pointed out yet that we don’t even know if the boats can see each other?
I think you’re misunderstanding. When Ahab makes his changes in course, he is actually moving away from Bart. It is true that, so long as Ahab does not turn directly away from Bart, the distance between them is decreasing, but when Ahab turns he decreases the rate of approach.
In fact the bolded part is not true. Ahab can plot a course in such a way that Bart can never catch him. That is, Ship A and Ship B will never meet, in the mathematical sense. Depending on how Bart tries to catch Ahab, he may be able to approach Ahab asymptotically, but he may not.
If Bart knows what Ahab’s course will be ahead of time, he can plot a course such that, after some time t[sub]0[/sub] the distance between Ship A and Ship B is less than any constant k. So if Ahab and Bart are sailing physical boats on the infinite sea, the boats will eventually touch.
I actually demonstrated that in mathematical terms that the two ships won’t touch, if Ahab is suitably wily.
Now, there is some question of how Bart is going to follow Ahab. The argument that I was explaining actually shows that, even if Bart has perfect information of all of Ahab’s future movements, Bart cannot catch Ahab. However, if we assume that Bart only knows where Ahab is at this moment, and what direction Ahab is traveling in, then Bart’s task becomes much more difficult. Because he does not know where Ahab is going to go next, his best strategy is to head for where Ahab is now. That is, he always steers straight toward Ahab. In this way he is always decreasing the distance. (Unless Ahab is sailing directly away from Bart, which he cannot do indefinitely because Ahab has been tasked to change course “every once in a while”.) However, it may be that he does not approach Ahab asymptotically. I will check later this afternoon when I have some paper and can compute the limit, but I believe that Ahab will actually remain some positive distance, bounded away from zero, ahead of Bart.
That is to say, he will never be caught even in the sense of “the boats are so close I can step from one to the other.” The equations for the pursuit curve are derived here if you want to look at them yourself.
Cheers.
[quote=“Omphaloskeptic, post:80, topic:500351”]
[ol][li]B can catch A in finite time;[]B asymptotically approaches A, with the distance between A and B approaching zero but never mathematically equal to zero at finite time;[]The distance between A and B asymptotically approaches some nonzero value.[/ol][/li][/QUOTE]
After looking at Pursuit Curve -- from Wolfram MathWorld I think it is clear to me that the answer is (2) with the distance between A and B approaching zero but never mathematically equal to zero at finite time. In real world terms the answer is that the pursuer does catch up but not in mathematical terms.
As I said, if A is heading directly away from B then the distance between them remains the same. If B points in any other direction then the distance diminishes.
But because it diminishes in proportion to the distance which separates them, it can never quite reach zero. It is similar to the charging of a capacitor through a resistor. Mathematically it can never reach the voltage of the supply. In practice it reaches it very fast.
Wow, When I posted this I got a error message and thought it didn’t post, now that I see it did and is 3 pages long, never expect that, and now the 2 ships have names too.
It appears as long as Ahab doesn’t do anything stupid, like turning towards Bart, Ahab will run forever with Bart closing if they have infinitely small ships. In real terms it appears that they will eventually contact each other.
Exacly, and it doesn’t have to be an infinite sea; any circular course will do.
<OBLIGATORY> Are the ships on a treadmill? </OBLIGATORY>
Aha, yes! Finally! The mathematicaly treatment of the Bart strategy of always moving towards Ahab. I was about to do it myself, using a mess of integrals (just to see if i could still remember how to…) but I love the matrix approach used by Wolfram. I have to figure out how it is done. Anyway, as can be seen, this approach is dumb and mostly useless. The other proposed strategy is “interception.” Unfortunately, it can’t so easily be defined.
Btw, I’d like to ammend my earlier post. If Ahab uses even the simple and transparent strategy of bumming around randomly, that won’t give Bart a huge advantage. Sure, Ahab won’t really get anywhere with it, and Bart can move to whereever Ahab is aimlessly buzzing. But the infinitessimal sizes of Ahab and Bart put a nail into any sort of strategy of using chance to snare the victim. If you can’t use chance, you can’t do shit, and there’s little room for “intelligent” strategies. But, I’m allowing, that there is some trivial strategy that might work.
I’d love someone to plug in a random function into the Wolfram math for the Pursuit Curve. (Although… is that even possible?)