# 2 ships theoretical question
**URL:** https://boards.straightdope.com/t/2-ships-theoretical-question/500351
**Category:** Factual Questions
**Created:** [June 19, 2009, 11:36pm UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351 "2009-06-19T23:36:29Z")
**Posts on this page:** 20
**Page:** 3
### Author:  [@Sage\_Rat](https://boards.straightdope.com/u/Sage_Rat)
#### Post date: [June 20, 2009, 6:23am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/41 "2009-06-20T06:23:52Z")
> [@Indistinguishable](#):
>
> The distance can keep decreasing without reaching 0 or anything close to it. E.g., suppose the distance goes from 1 to 0.95 to 0.905 to 0.9005 to 0.90005 to 0.900005 to 0.9000005… This keeps decreasing, but never hits 0, or even gets below 0.9.
That’s even better then. If it’s converging on 0.9, the minimum time at which they’ll meet is distance/0.9.
I think I added a sentence to the end of my post after you responded though.
A is, by choosing the minimum angle of turn he possibly can, deciding the formula for the outside of a circle he wants to traverse. Whether he chooses to go left or right by that amount makes no difference, it merely creates a mirror image that doesn’t change anything.
B is sitting inside that circle, traveling towards the edge. B is traveling at the same speed as A and is traveling on a line that curves more sharply, by definition. And by definition, a sharper turn creates a shorter segment. Two ships traveling at the same speed, but one traveling a longer line, the ship traveling the shorter path will overtake the one traveling the longer path.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [June 20, 2009, 6:25am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/42 "2009-06-20T06:25:55Z")
As I read it, there’s no stipulation that A must travel in a circle. We’re just told that they can’t keep travelling in a straight line. They can take any other kind of path they like, as long as they travel through at a constant speed.
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### Author:  [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)
#### Post date: [June 20, 2009, 6:28am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/43 "2009-06-20T06:28:31Z")
> [@Indistinguishable](#):
>
> Your triangles are misguided; the base is the segment between the two ships, but what are the other two sides of the triangle supposed to correspond to? It looks like you want them to correspond to the velocities of these ships, since you demand that they have the same size. But why should these two culminate in the same point to serve as the third vertex? There’s no reason they should. Indeed, there’s no reason you should want to construct a triangle whose sides mix and match distances and velocities (things of different types, with different units) in the first place.
No, **SaintCad** is solving the intercept problem. One of the legs of the isosceles triangle is drawn along A’s trajectory; B chooses his trajectory along BC, with C along A’s trajectory such that AC=BC, and thus since A and B are moving at equal speeds they will reach C at the same time. (So the lengths of the legs are a velocity times a time, and the units are happy.)
Of course this is only possible when the angle BAC\<90°, as he says. If not, then B cannot choose a course that will intercept A at finite time.
So if A is forced to choose a new heading once a minute, but is allowed to choose a heading that always has this angle \>=90°, then B will never catch up to A. If A is forced to change course by at least some minimum angle, then B will approach A arbitrarily closely, however.
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### Author:  [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)
#### Post date: [June 20, 2009, 6:29am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/44 "2009-06-20T06:29:46Z")
Missed the edit window
HOWEVER, if the ship being chased is traveling 90 degrees or more to the base, the chaser will never catch it since no isosceles triangle can be formed. Therefore, as long as the chased ship is always traveling along an acute angle to the segment between the two ships despite the twists and turns, it will always be caught. Consequently, if it maintains a right angle or obtuse angle heading relative to the segment as it steers, it will never be caught.
> [@Indistinguishable](#):
>
> Your triangles are misguided; the base is the distance between the two ships, but what are the other two sides of the triangle supposed to correspond to? It looks like you want them to correspond to the velocities of these ships, since you demand that they have the same size. But you can’t make a triangle out of mixing and matching distances velocities; they’re not the same kind of thing, they don’t have the same units!
Sure you can. They’re called vectors.
The other two sides are the distances the ships have traveled so all of the units are the same unit.
You see, for one boat to catch the other one, they must be at the same point at the same time. Distance equals rate x time (customarily abbreviated as d=rt) we can rewrite it as t = d/r and since both boats have equal speeds, if they have travel equal distances, we can show they have traveled an equal amount of time.
Since they are now at the third point of the triangle and have traveled an equal distance (that is called an isosceles triangle), we conclude that they are at the same point at the same time.
See, not misguided at all.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [June 20, 2009, 6:34am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/45 "2009-06-20T06:34:12Z")
> [@Omphaloskeptic](#):
>
> No, **SaintCad** is solving the intercept problem.
Oh, thanks, I see now what they were doing. My apologies for my misinterpretation. But they’ve only shown “If A and B are both forced to remain on a straight path, can B intercept A? No, not if A’s path is at suitable angle to the vector between their starting positions.” But that’s not the relevant question. There’s nothing forcing A and B to move in a straight path until the interception point.
That is, the distance between A’s starting point and the intercept point may be different than the distance between B’s starting point and the intercept point; for example, one might move around in a circle and come back to where it started, while the other just moves in a straight line. Or myriad other possibilities. So there’s no reason to assume that the intercept point will form an isosceles triangle with the starting locations of A and B.
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### Author:  [@Sage\_Rat](https://boards.straightdope.com/u/Sage_Rat)
#### Post date: [June 20, 2009, 6:35am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/46 "2009-06-20T06:35:58Z")
> [@Indistinguishable](#):
>
> As I read it, there’s no stipulation that A must travel in a circle. We’re just told that they can’t keep travelling in a straight line. They can take any other kind of path they like, as long as they travel through at a constant speed.
Since they must constantly turn and their wisest turn is the minimum, it will be a constant rate of turn. That defines a circle. If they alternate going left-right-left-right, then they won’t make a circle if you plot their course, but so far as the math is concerned it’s still a circle.
Say I draw a circle and then draw a spiral starting at the center and rotating clockwise to meet the edge. Then I say that instead of ship A going in a circle, it goes 90 degrees around a circle and then turns around and goes back over it’s same course, repeating this back and forth. My “spiral”, matching this, will do a near about face at every 90 degree interval, creating something like a wave shape, but in the end result, the distance it travels before hitting the edge will not change.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [June 20, 2009, 6:40am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/47 "2009-06-20T06:40:33Z")
> [@Sage\_Rat](#):
>
> Since they must constantly turn and their wisest turn is the minimum, it will be a constant rate of turn. That defines a circle. If they alternate going left-right-left-right, then they won’t make a circle if you plot their course, but so far as the math is concerned it’s still a circle.
A) As I read it, they don’t have to constantly turn, just turn infinitely often. E.g., they can trace out a piecewise linear path if they like, as long as they eventually turn off of any particular direction they turn on to
B) Even if they must constantly turn, they could start by turning at an angle of 1 degree per second, then later turning at an angle of 0.1 deg/sec, then later turning at an angle of 0.01 deg/sec, and so on, this being even “wiser” than sticking to some constant rate of turn; this, of course, would not trace out a circle.
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### Author:  [@Sage\_Rat](https://boards.straightdope.com/u/Sage_Rat)
#### Post date: [June 20, 2009, 6:49am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/48 "2009-06-20T06:49:40Z")
> [@Indistinguishable](#):
>
> A) As I read it, they don’t have to constantly turn, just turn infinitely often. E.g., they can trace out a piecewise linear path if they like, as long as they eventually turn off of any particular direction they turn on to
This is true, but again doesn’t make any difference to the math. It just makes it an n-gon arranged on a circle and less easily visualized.
> [@](#):
>
> B) Even if they must constantly turn, they could start by turning at an angle of 1 degree per second, then later turning at an angle of 0.1 deg/sec, then later turning at an angle of 0.01 deg/sec, and so on, this being even “wiser” than sticking to some constant rate of turn; this, of course, would not trace out a circle.
Why do that rather than starting with the smallest most infinitesimal, indivisible angle and keeping to it? That’s the only way to maximize the distance between the two ships. If you want them to turn at 1/infinity degrees, then go for it, but I’ll wait for you to figure out exactly what angle that would be.
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### Author:  [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)
#### Post date: [June 20, 2009, 6:52am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/49 "2009-06-20T06:52:28Z")
> [@Indistinguishable](#):
>
> Oh, thanks, I see now what they were doing. My apologies for my misinterpretation. But they’ve only shown “If A and B are both forced to remain on a straight path, can B intercept A? No, not if A’s path is at suitable angle to the vector between their starting positions.” But that’s not the relevant question. There’s nothing forcing A and B to move in a straight path until the interception point.
>
> That is, the distance between A’s starting point and the intercept point may be different than the distance between B’s starting point and the intercept point; for example, one might move around in a circle and come back to where it started, while the other just moves in a straight line. Or myriad other possibilities. So there’s no reason to assume that the intercept point will form an isosceles triangle with the starting locations of A and B.
Straight path has nothing to do with it. As long as the two ships are on a heading that meet as the sides of an isosceles triangle, they are getting closer to each other and eventually they will meet. Do me a favor, draw a segment 2" long and start to make a triangle with 2 segments 1" long 60 degrees to the base. Now connect these endpoints. This segment is shorter than the base. Now make rays 70 degrees to the top base of the trapezoid. Even though they changed direction, it is still an acute angle and they will meet . . . so what does changing direction have to do with the conditions I set forth?
Now if your argument is that A might follow a curve, I would argue that at each instantaneous point, it can be treated as following a straight line along the tangent (like in calculus). As long as A’s path, as curvy as it is, never has it on a heading 90 or more degrees from the segment AB, it will be caught. Likewise, if A never curves back to have an acute heading to the segment AB, A will never be caught.
If it goes back and forth between acute, right and obtuse, the question is unanswerable unless you can describe the exact path A is taking.
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### Author:  [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)
#### Post date: [June 20, 2009, 6:53am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/50 "2009-06-20T06:53:33Z")
> [@Indistinguishable](#):
>
> Oh, thanks, I see now what they were doing. My apologies for my misinterpretation. But they’ve only shown “If A and B are both forced to remain on a straight path, can B intercept A? No, not if A’s path is at suitable angle to the vector between their starting positions.” But that’s not the relevant question. There’s nothing forcing A and B to move in a straight path until the interception point.
No, the problem as posed is underdetermined, which of course leads to lots of people solving different problems and confusing each other. It’s stated that B responds instantaneously to A’s change of heading, but it’s not specified _how_ B reacts. Two obvious choices are (1) B sets an intercept course, if possible (as implicit in **Mangetout** ’s diagram, assumed by **SaintCad** and **dracoi** , etc.) and (2) B plots a direct pursuit course, always heading straight toward A ( **sailor** assumes this).
It’s also left unstated how often and how much A’s course changes (continuously? at some constant interval? … etc.). The cases where A’s course is piecewise linear are pretty easy to deal with. Whether the requirement that A change course “every so often” is meant to imply a piecewise linear trajectory I can’t say.
The optimum strategy for B depends on A’s behavior; if B knows that A will maintain a course for at least some period of time, then it makes sense for B to set an intercept course; otherwise this may not make sense, and the optimal chioice may lie somewhere between the intercept course and the pursuit course.
> [@](#):
>
> _That is, the distance between A’s starting point and the intercept point may be different than the distance between B’s starting point and the intercept point; for example, one might move around in a circle and come back to where it started, while the other just moves in a straight line. Or myriad other possibilities. So there’s no reason to assume that the intercept point will form an isosceles triangle with the starting locations of A and B._
Well, that comes back down to assumptions. If A only changes course “every so often” then maybe B can assume A’s course to be piecewise linear, in which case setting an intercept course makes sense if A and B are sufficiently close. Of course if B can intercept A at C then B can intercept A at any point beyond C as well, but the construction given is for the fastest intercept assuming constant velocities.
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### Author:  [@Sage\_Rat](https://boards.straightdope.com/u/Sage_Rat)
#### Post date: [June 20, 2009, 7:01am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/51 "2009-06-20T07:01:35Z")
> [@Omphaloskeptic](#):
>
> No, the problem as posed is underdetermined, which of course leads to lots of people solving different problems and confusing each other. It’s stated that B responds instantaneously to A’s change of heading, but it’s not specified _how_ B reacts.
B loses ground any time it is not pointed directly at A, for the same reason that A loses ground any time it is not pointed directly away from B.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [June 20, 2009, 7:03am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/52 "2009-06-20T07:03:15Z")
> [@Sage\_Rat](#):
>
> Why do that rather than starting with the smallest most infinitesimal, indivisible angle and keeping to it?
Because there may not be a smallest, most infinitesimal, indivisible angle. Indeed, presumably, on a most straightforward mathematical idealization of the question, there isn’t.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [June 20, 2009, 7:05am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/53 "2009-06-20T07:05:20Z")
> [@SaintCad](#):
>
> Straight path has nothing to do with it. As long as the two ships are on a heading that meet as the sides of an isosceles triangle, they are getting closer to each other and eventually they will meet. Do me a favor, draw a segment 2" long and start to make a triangle with 2 segments 1" long 60 degrees to the base. Now connect these endpoints. This segment is shorter than the base. Now make rays 70 degrees to the top base of the trapezoid. Even though they changed direction, it is still an acute angle and they will meet . . . so what does changing direction have to do with the conditions I set forth?
>
> Now if your argument is that A might follow a curve, I would argue that at each instantaneous point, it can be treated as following a straight line along the tangent (like in calculus). As long as A’s path, as curvy as it is, never has it on a heading 90 or more degrees from the segment AB, it will be caught. Likewise, if A never curves back to have an acute heading to the segment AB, A will never be caught.
>
> If it goes back and forth between acute, right and obtuse, the question is unanswerable unless you can describe the exact path A is taking.
I was simply noting that reasoning about the impossibility of constructing isosceles triangles subject to certain conditions proves nothing of any relevance, until one has established that the intercept point must form an isosceles triangle with the starting positions of A and B. And there’s no reason it must. The intercept point may be further from the starting point of one ship than it is from the starting point of the other ship, even if both ships keep moving at the same speed. The total distance, in the odometer sense, covered by a ship over the entire course of its journey need not equal the actual distance from its starting point to its ending point. It’s only the odometer sense in which the two ships must cover the same distance from start to intercept; the actual distances from their starting point to the intercept may be wildly different.
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### Author:  [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)
#### Post date: [June 20, 2009, 7:07am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/54 "2009-06-20T07:07:15Z")
> [@Sage\_Rat](#):
>
> B loses ground any time it is not pointed directly at A, for the same reason that A loses ground any time it is not pointed directly away from B.
No, as I said, that depends on your assumptions. You appear to be assuming that A is traveling in a circle. **SaintCad** is assuming that A travels on a piecewise linear course. If B can reasonably assume that A’s trajectory is known for the next (say) one minute, then B can aim ahead of A to meet him at the intercept point C, as **SaintCad** has shown. It is true that for the intercept course B does not initally approach A as rapidly as B could. The point C is chosen to minimize the time to intercept, _assuming_ A maintains a constant-velocity trajectory.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [June 20, 2009, 7:11am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/55 "2009-06-20T07:11:24Z")
> [@Omphaloskeptic](#):
>
> No, the problem as posed is underdetermined, which of course leads to lots of people solving different problems and confusing each other. It’s stated that B responds instantaneously to A’s change of heading, but it’s not specified _how_ B reacts. Two obvious choices are (1) B sets an intercept course, if possible (as implicit in **Mangetout** ’s diagram, assumed by **SaintCad** and **dracoi** , etc.) and (2) B plots a direct pursuit course, always heading straight toward A ( **sailor** assumes this).
Point well taken.
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### Author:  [@magellan01](https://boards.straightdope.com/u/magellan01)
#### Post date: [June 20, 2009, 7:49am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/56 "2009-06-20T07:49:29Z")
It seems to me that the B will get increasingly closer to A, but never catch him. The first part is easy to see. For the second part, I see that every time B gets closer, A just has to change course 90° to B’s course. This creates a series of increasingly smaller triangles, but there will always be a triangle. Granted, this goes for our hypothetical ships, which have no dimension.
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### Author:  [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)
#### Post date: [June 20, 2009, 7:51am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/57 "2009-06-20T07:51:56Z")
> [@Omphaloskeptic](#):
>
> No, as I said, that depends on your assumptions. You appear to be assuming that A is traveling in a circle. **SaintCad** is assuming that A travels on a piecewise linear course. If B can reasonably assume that A’s trajectory is known for the next (say) one minute, then B can aim ahead of A to meet him at the intercept point C, as **SaintCad** has shown. It is true that for the intercept course B does not initally approach A as rapidly as B could. The point C is chosen to minimize the time to intercept, _assuming_ A maintains a constant-velocity trajectory.
No I did not. Again, it is about the ships heading and assuming it is the tangent of a curve does NOT nullify my argument in the slightest.
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### Author:  [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)
#### Post date: [June 20, 2009, 8:23am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/58 "2009-06-20T08:23:57Z")
> [@SaintCad](#):
>
> No I did not. Again, it is about the ships heading and assuming it is the tangent of a curve does NOT nullify my argument in the slightest.
Reading your post again I agree. You are not arguing about an optimal trajectory (least time to intercept) but just saying that as long as A’s heading is less than 90° from the direction to B, B can choose a course to meet A; and if A’s heading is 90° or greater, then B cannot. No argument there.
When you said that “The chaser wants to take a heading…” I understood you to be talking about finding an optimal trajectory; this depends on B’s model for A’s behavior.
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### Author:  [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)
#### Post date: [June 20, 2009, 8:35am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/59 "2009-06-20T08:35:36Z")
> [@Sage\_Rat](#):
>
> I beg to differ. Presuming that there’s a minimum value of a turn that qualifies as a turn, so that we’re not dealing with something like 1/infinity degrees, each turn is going to create a real decrease in distance. That real value might shrink each time, but as it does it approaches zero, making it an infinitesimal and so giving you a hard value: Time of Impact. If I can calculate the ToI, then most certainly the one does catch the other in all cases.
>
> As I showed in my post, in all cases, A will be traversing the outside of a circle. B will be traversing an expanding spiral on the inside of that circle. Since their speed is the same, B has a shorter distance to travel and will intersect at a specific time.
This doesn’t follow. B’s spiral could end up getting asymptotically close to the circle without B ever actually catching up, and I think that that’s what actually happens. What I find is that if A is moving around the unit circle with unit speed and B starts from the origin at time 0, then at time t\>\>1 the distance between B and the origin is approximately 1-2/t[sup]2[/sup], and the distance between A and B is approximately 2/t.
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### Author:  [@Omphaloskeptic](https://boards.straightdope.com/u/Omphaloskeptic)
#### Post date: [June 20, 2009, 8:41am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/60 "2009-06-20T08:41:49Z")
Missed the edit window. The calculations in my previous post assume that B always heads directly toward A.
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