# 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:** 2
### Author:  [@Richard\_Pearse](https://boards.straightdope.com/u/Richard_Pearse)
#### Post date: [June 20, 2009, 4:40am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/21 "2009-06-20T04:40:50Z")
> [@Xema](#):
>
> No loss of distance if the course changes are “instant”, as specified in the OP.
>
> It can fairly be argued that in the real world there’s no such thing - as is the case with many of these hypotheticals.
If ship A and ship B start out traveling in a straight line with ship B trailing, _any_ change in direction by ship A with an immediate response from ship B will result in a loss of distance between A and B. Every time the direction of A is changed, B is given an opportunity to cut the corner.
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### Author:  [@SayTwo](https://boards.straightdope.com/u/SayTwo)
#### Post date: [June 20, 2009, 4:43am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/22 "2009-06-20T04:43:44Z")
> [@sailor](#):
>
> Wrong.
I guess that would depend on your definition of “instant.”
Then again…it was defined at the outset.
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### Author:  [@sailor](https://boards.straightdope.com/u/sailor)
#### Post date: [June 20, 2009, 4:51am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/23 "2009-06-20T04:51:11Z")
> [@SayTwo](#):
>
> I guess that would depend on your definition of “instant.”
>
> Then again…it was defined at the outset.
No. it does not matter. If both ships have the exact same speed then their distance is maintained when the first travels directly away from the second and diminishes in any other case. It never increases. Since it decreases at times it follows that the pursuer will catch up.
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### Author:  [@SayTwo](https://boards.straightdope.com/u/SayTwo)
#### Post date: [June 20, 2009, 4:57am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/24 "2009-06-20T04:57:57Z")
> [@sailor](#):
>
> No. it does not matter. If both ships have the exact same speed then their distance is maintained when the first travels directly away from the second and diminishes in any other case. It never increases. Since it decreases at times it follows that the pursuer will catch up.
And if it is instant, it neither increases nor decreases.
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### Author:  [@sailor](https://boards.straightdope.com/u/sailor)
#### Post date: [June 20, 2009, 5:08am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/25 "2009-06-20T05:08:24Z")
No, it does not matter if it is instant or not. A change in course of A results in B closing in. If it does not result in B closing in then it was not a change in course.
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### Author:  [@SayTwo](https://boards.straightdope.com/u/SayTwo)
#### Post date: [June 20, 2009, 5:16am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/26 "2009-06-20T05:16:05Z")
> [@sailor](#):
>
> No, it does not matter if it is instant or not. A change in course of A results in B closing in. If it does not result in B closing in then it was not a change in course.
So, you are saying that ship B gets a better “instant” than ship A does? If that’s the case, the game is rigged.
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### Author:  [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)
#### Post date: [June 20, 2009, 5:21am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/27 "2009-06-20T05:21:11Z")
> [@SayTwo](#):
>
> So, you are saying that ship B gets a better “instant” than ship A does? If that’s the case, the game is rigged.
Ship B doesn’t have to move in the same direction as ship A.
Edit: There’s [a book](http://www.amazon.com/Chases-Escapes-Mathematics-Pursuit-Evasion/dp/0691125147) that deals with this sort of problem. I haven’t read it, but the reviewers seem to think it’s decent.
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### Author:  [@sailor](https://boards.straightdope.com/u/sailor)
#### Post date: [June 20, 2009, 5:25am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/28 "2009-06-20T05:25:06Z")
> [@SayTwo](#):
>
> So, you are saying that ship B gets a better “instant” than ship A does? If that’s the case, the game is rigged.
WTF?
B travels in a straight line towards A. If A travels away then distance is maintained. The instant A changes direction distance diminishes. If A is forced to change direction then distance diminishes.
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### Author:  [@SayTwo](https://boards.straightdope.com/u/SayTwo)
#### Post date: [June 20, 2009, 5:29am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/29 "2009-06-20T05:29:12Z")
> [@sailor](#):
>
> WTF?
>
> B travels in a straight line towards A. If A travels away then distance is maintained. The instant A changes direction distance diminishes. If A is forced to change direction then distance diminishes.
Exactly. The instant A changes course, things change. If the instant B changes course things also change, then B never catches A. That’s what I said in the beginning.
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### Author:  [@sailor](https://boards.straightdope.com/u/sailor)
#### Post date: [June 20, 2009, 5:32am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/30 "2009-06-20T05:32:06Z")
> [@SayTwo](#):
>
> Exactly. The instant A changes course, things change. If the instant B changes course things also change, then B never catches A. That’s what I said in the beginning.
You are wrong.
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### Author:  [@SayTwo](https://boards.straightdope.com/u/SayTwo)
#### Post date: [June 20, 2009, 5:33am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/31 "2009-06-20T05:33:18Z")
> [@sailor](#):
>
> You are wrong.
That’s rather a compelling argument.
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### Author:  [@RaftPeople](https://boards.straightdope.com/u/RaftPeople)
#### Post date: [June 20, 2009, 5:38am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/32 "2009-06-20T05:38:26Z")
> [@Mangetout](#):
>
> I think the point is that if ship A is travelling directly away from ship B, then any change of direction turns it in a course that is not directly away from B, and puts it on a heading toward a point that B can reach more directly (A is travelling two sides of a triangle, B is travelling only one)
Doh! For some reason I was picturing 90 degree turns only, like on graph paper, who knows why.
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### Author:  [@Sage\_Rat](https://boards.straightdope.com/u/Sage_Rat)
#### Post date: [June 20, 2009, 5:44am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/33 "2009-06-20T05:44:07Z")
I find it amusing that the names of so many posters to this thread are related to the topic.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [June 20, 2009, 5:50am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/34 "2009-06-20T05:50:38Z")
Let’s write out the details of **sailor** , et al,'s correct point:
Suppose ship B is constantly moving directly towards A, at a speed of S. And A is constantly moving at a speed of S, though there may be some changes of direction involved (even instantaneous discontinuous changes of direction).
So these are the two factors playing into the rate of change of the distance between A and B. The first factor (the velocity of B) clearly contributes to the rate of change of this distance a term -S. As for the second factor (the velocity of A), we can, at any particular moment, split this into a linear combination of the vector from B to A and a perpendicular vector; the latter will contribute nothing to the (momentary) rate of change of the distance, while the former will contribute to this rate by an amount exactly equal to its magnitude, being either an increase or a decrease depending on whether it is in the direction from B to A or the reverse. But its magnitude can be at most the magnitude of A’s velocity, which is S. Thus, it contributes at most an increase of S to the distance between B and A [this only happening in the case where A is moving in the direction directly away from B].
Therefore, at any given moment, the total rate of change in the distance between A and B is -S + [something which, at most, is S]. At most, this is 0, while this drops to a negative whenever, as noted above, A is not moving in the direction directly away from B. Which is exactly **sailor** ’s point: whenever A is not moving in the direction directly away from B, B is gaining ground on A. Thus, if A is forced to make changes of direction, even instantaneous ones, then A is forced to, at some points, move in a direction not directly away from B (that is, if A was always moving in the direction directly away from B, then it would never be changing direction). And during those times when A is not moving directly away from B, B must be gaining ground on A.
That having been said, it doesn’t follow that B will necessarily eventually catch up; just that it will keep getting closer.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [June 20, 2009, 5:58am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/35 "2009-06-20T05:58:18Z")
Expanding on that last line: That having been said, it doesn’t follow that B will necessarily eventually catch up, or even that it will do so asymptotically; this just guarantees that it will keep getting closer. It seems easy enough to construct situations where B keeps getting closer but never gets closer than X% of its starting distance, for any appropriate X.
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### Author:  [@Sage\_Rat](https://boards.straightdope.com/u/Sage_Rat)
#### Post date: [June 20, 2009, 6:10am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/36 "2009-06-20T06:10:11Z")
> [@Indistinguishable](#):
>
> it doesn’t follow that B will necessarily eventually catch up; just that it will keep getting closer.
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.
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### Author:  [@Saint\_Cad](https://boards.straightdope.com/u/Saint_Cad)
#### Post date: [June 20, 2009, 6:10am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/37 "2009-06-20T06:10:40Z")
Very basic proof using triangles.
Imagine the segment between the ships as the base. The chaser wants to take a heading that would complete the third side of an isosceles triangle, therefore his heading must be the same angle to the base as the ship being chased.
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.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [June 20, 2009, 6:16am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/38 "2009-06-20T06:16:09Z")
> [@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.
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.
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### Author:  [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)
#### Post date: [June 20, 2009, 6:18am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/39 "2009-06-20T06:18:38Z")
> [@SaintCad](#):
>
> Very basic proof using triangles.
> Imagine the segment between the ships as the base. The chaser wants to take a heading that would complete the third side of an isosceles triangle, therefore his heading must be the same angle to the base as the ship being chased.
>
> 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.
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.
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### Author:  [@Richard\_Pearse](https://boards.straightdope.com/u/Richard_Pearse)
#### Post date: [June 20, 2009, 6:23am UTC](https://boards.straightdope.com/t/2-ships-theoretical-question/500351/40 "2009-06-20T06:23:19Z")
> [@SayTwo](#):
>
> Exactly. The instant A changes course, things change. If the instant B changes course things also change, then B never catches A. That’s what I said in the beginning.
I think you are missing that B changes directions so that it can travel in a shorter line to meet A.
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