[Guest]

I searched for this really hard given my last failure and didn't find it posted hear. It is an elegant pursuit puzzle.

You're on a government ship, looking for a pirate ship. You know that the pirate ship travels at a constant speed, and you know what that speed is. Your ship can travel twice as fast as the pirate ship. Moreover, you know that the pirate ship travels along a straight line, but you don't know what that line is. It's very foggy, so foggy that you see nothing. But then! All of a sudden, and for just an instant, the fog clears enough to let you determine the exact position of the pirate ship. Then, the fog closes in again and you remain (forever) in the thick fog. Although you were able to determine the position of the pirate ship during that fog-free moment, you were not able to determine its direction. How will you navigate your government ship so that you are sure to capture the pirate ship?

[Guest]

I searched for this really hard given my last failure and didn't find it posted hear. It is an elegant pursuit puzzle.

You're on a government ship, looking for a pirate ship. You know that the pirate ship travels at a constant speed, and you know what that speed is. Your ship can travel twice as fast as the pirate ship. Moreover, you know that the pirate ship travels along a straight line, but you don't know what that line is. It's very foggy, so foggy that you see nothing. But then! All of a sudden, and for just an instant, the fog clears enough to let you determine the exact position of the pirate ship. Then, the fog closes in again and you remain (forever) in the thick fog. Although you were able to determine the position of the pirate ship during that fog-free moment, you were not able to determine its direction. How will you navigate your government ship so that you are sure to capture the pirate ship?

Could you please define "capture the pirate ship" in terms of distance between the ships? also at what distance will the pirate ship be visible to the gov ship?

[Guest]

1) Gov ship travels towards the sight at which the pirate ship was seen, say point P, in a straight line at full speed

2) if the pirate ship is travelling in the exact opposite direction, the two ships are bound to meet somewhere midway. lets call this point R

3) Since pirate ship could be travelling in any direction point R is not a guaranteed rendezvous point

4) After gov ship reaches point R it changes course and follows a circular path around point P, with ever increasing radius. The rate of increase of radius should be equal to half the Gov Ship's full speed

5) this way, the gov ship is circularly scanning around point P while moving away from point P at the same speed as the pirate ship.

this should work, and if so, it was not that hard was it?

[Guest]

1) Gov ship travels towards the sight at which the pirate ship was seen, say point P, in a straight line at full speed

2) if the pirate ship is travelling in the exact opposite direction, the two ships are bound to meet somewhere midway. lets call this point R

3) Since pirate ship could be travelling in any direction point R is not a guaranteed rendezvous point

4) After gov ship reaches point R it changes course and follows a circular path around point P, with ever increasing radius. The rate of increase of radius should be equal to half the Gov Ship's full speed

5) this way, the gov ship is circularly scanning around point P while moving away from point P at the same speed as the pirate ship.

this should work, and if so, it was not that hard was it?

Very nicely done. No, not too hard once you get over the idea the pirates could be anywhere.

[Guest]

Agree with the answer that tarunark gave... Well, i took more time to work out the mathematics of the spiral to follow

So, here's the maths behind the path to follow (refer images attached):

A is the point where govt ship first sighted the pirate ship at point B

If the captain has pinpointed this location, he should be able to estimate the distance at this time

Now, move the ship in the line AB to point C such that AC = 2/3 AB

Now, the ship must follow a spiral path in order to catch the pirate ship as it may have been travelling at any angle

Defint time (t = 0) when the ship reaches point C

Now any point in time (t) when the ship is at point E, it is at angle θ with CB

If the pirate ship is not at E, it must look for it at point D, in which case the pirate ship at that time must have been at F

In time dt, while the govt ship moves from E to D, the pirate ship moves from F to D

Lets say speed of pirate ship is "s" then speed of govt ship is 2s.

So,

Now, BE = st

FD = sdt

EF = st dθ

A hypothetical point on circle with point D must have had an arc length = s(t+dt) dθ

Then ED is the average of these two lengths

ED = 1/2 (s*t*dθ + s*(t+dt)*dθ)

ED = st dθ + 1/2(s*dt*dθ)

Since dt*dθ is very small, this term can be ignored

So, ED = st dθ

Now, dt = ED/2s = FD/s

This gives,

So, t*dθ/2 = dt

or dθ = 2/t dt

integrating both sides,

θ = 2 ln(t)

or t = e^θ/2

Now, general equation of spiral is r(t) = at, where r is the radius, t is time and a is constant

In this case, a = CB as this is the starting radius

so, r = CB.e^θ/2 (considering B at origin)

Edited July 1, 2009 by DeeGee