- 19 Apr '06 23:21Four ghostly galleons – call them E, F, G and H, – sail on a ghostly sea so foggy that visibility is nearly zero. Each pursues its course steadily, changing neither its speed nor heading. G collides with H amidships; but since they are ghostly galleons they pass through each other with no damage nor change in course. As they part, H’s captain hears G’s say “Tarnation! That’s our third collision this night!” A little while later, F runs into H amidships with the same effect (none) and H’s captain hears the same outburst from F’s. What can H’s captain do to avoid a third collision and yet reach his original destination, whatever it may be, and why will doing that succeed?

P.S.Credit to the Berkeley Riddles page.

(Here is the link to a myriad of challenging puzzles: http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml) - 21 Apr '06 06:23

They're all going the same direction at different speeds. At the beginning of the night, they were in this order:*Originally posted by Codfish***Four ghostly galleons – call them E, F, G and H, – sail on a ghostly sea so foggy that visibility is nearly zero. Each pursues its course steadily, changing neither its speed nor heading. G collides with H amidships; but since they are ghostly galleons they pass through each other with no damage nor change in course. As they part, H’s captain hears G’s say “ ...[text shortened]... he link to a myriad of challenging puzzles: http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml)**

G F E H

G is going faster than all of them and passes right through them in order: F, then E, and third H. Then, F, who has had one "collision" with G already, passes through E and then H. To avoid a third collision, H can accelerate until he is going at the same speed as F was when F collided with him. This of course assumes the ghostly captain of H knew the velocity of the other ships, especially E, and that they wouldn't change that velocity. - 24 Apr '06 04:53

This might be true. But since the collisions are happening amidships, it seems more probable that the various courses are at angles to each other.*Originally posted by AThousandYoung***They're all going the same direction at different speeds. At the beginning of the night, they were in this order:**

G F E H

G is going faster than all of them and passes right through them in order: F, then E, and third H. Then, F, who has had one "collision" with G already, passes through E and then H. To avoid a third collision, H can accelerate ...[text shortened]... the velocity of the other ships, especially E, and that they wouldn't change that velocity. - 25 Apr '06 09:38

Head on collisions aren't "amidships" either, from what I understand.*Originally posted by Yellowhotpawn***Totally dominant is right if they all are going in the same direction.**

If however G and F are going in opposite directions as E and H then H needs to make sure he decreases his velocity so he never catches up to E. - 09 May '06 16:53I think the captain of H needs to keep on going at the same speed. since the only remaining ship he can still collide with, E, has already crossed the pathes of G and F before H did, I don't think it's possible for H to catch up with E. I'm afraid I don't have mathematical proof, but I just can't seem to draw out a situation where H still collides with E...
- 11 Jun '06 12:49firstly their speeds are constant

second they never change heading this implies a straight line (circles would be constantly changing heading)

third if we assume amidships means at right angles then H is perpendicular to G

as well as F making G parallel to F since travelling at constant speeds along straight lines G can not collide with E twice this possibility leads to a contradiction

therfore assuming that amidships is any angle that is not a multiple of 180 (head on or back to front collisions are not allowed)

we draw (selected at random) the lines F:y=x and G:y=-x thus F and G collide at (0,0)

finally given any line E that crosses F and G to the left of the y-Axis and another line H that crosses F and G to the right of the y-Axis we see that the lines E and H will eventually cross unless they are parallel. Since the captain of H doesn't know E's speed he can not predict at whate time E will transverse his heading so changing his speed won't help.

he can not change his course because he doesn't know E's heading thus he does not know to what course he should align himself.

he could stay where he is and as long as he doesn't move from the point of collision with F he will avoid E but then he can't get where he is going

Thus i conclude there is NOTHING he can do to avoid a collision (unless he has a foghorn at his disposal) - 20 Jun '06 11:26This question is too vauge, or just generally doesn't make sense, unless the sraight line idea was correct. If amidships is taken to mean right angles, then this doesn't make sense, if H is headed on a straight line, then is hit by G from a right angle, then later F at a right angle, then we can conclude that both G and F are parallel to each other and could not have hit (in other words they would not have been able to have 3 colisions).

If we take amidships to mean any angle other than, 90, 180, 270, or 360 the we can say that H is traveling along a straight line, with G farthest ahead, followed by F, then E, then H. If we say that G is traveling near a 90 degree angle to H then F must be on an angle less than that, and E's angle would be smaller still. Thus H could either slow down, or speed up and presumably miss E. Of course this is assuming E and H's course will intersect with no action taken.

Basically my solution would be to barrel on heedlessly since you can't know anything about E for sure, and it doesn't matter if H and E colide anyways since nothing will happen (expept H might lose some pride). - 20 Jun '06 19:59

unless ghost ships travel really really fast and there are no large landmasses on the globe in question.*Originally posted by TheNinjunny***If amidships is taken to mean right angles, then this doesn't make sense, if H is headed on a straight line, then is hit by G from a right angle, then later F at a right angle, then we can conclude that both G and F are parallel to each other and could not have hit (in other words they would not have been able to have 3 colisions).**

(i still cant make it make sense using eliptical gyomitry, but it _is_ another avenue.)