An oil tanker travels at a constant speed ( S meters per hour ), on a calm ocean, along a straight path. The shape of the tanker is a rectangle - W meters wide and L meters long capped by two semicircles at the bow and stern.

A patrol boat "circles" the tanker looking for any oil leaks. The patrol boat goes up the port side, across the bow, down the starboard side, and across the stern, and keeps cycling like this over and over. Relative to the ocean, the path of the patrol boat is always parallel to or perpendicular to the path of the tanker. During a cycle, the patrol boat makes four and only four turns. The speed of the patrol boat is twice the speed of the tanker.

Compared to the tanker, consider the patrol boat as a point; that its turns are instantaneous; and for safety it must maintain, at a minimum, C meters between itself and the tanker.

What is the shortest amount of time for the patrol boat to complete one cycle around the tanker in terms of C, L, S, and W?

(In reply to solution by Charlie)

I’m with you most of the way Charlie, but as the patrol crosses the stern it will ‘fall further behind’ the tanker, showing that it is describing an isosceles trapezium relative to the tanker, rather than a parallelogram. I hope I’ve got this right..

If so, then, following your lead, the distance between the centre of a semicircle and the mid point of a non-parallel side will be   (w/2 + c)sqrt(5)/2,

and the two parallel sides will therefore have lengths
L + 2(w/2 + c)sqrt(5)/2 + (w + 2c)/2 and L + 2(w/2 + c)sqrt(5)/2 - (w + 2c)/2

These simplify to L + (w/2 + c)(sqrt(5) + 1) and  L + (w/2 + c)(sqrt(5) - 1)
so that the total time for these parallel legs is

[L + (w/2 + c)(sqrt(5) + 1)]/s  +  [L + (w/2 + c)(sqrt(5) - 1)]/3s

= [4L + (4sqrt(5) + 2)(w/2 + c)]/3s
= [4L + (2sqrt(5) + 1)(w + 2c)]/3s

Adding on your total time for the two sideways traverses gives the overall time as

[4L + (2sqrt(5) + 1)(w + 2c)]/3s  +  (w + 2c)/s

= [4L + 2(sqrt(5) + 2)(w + 2c)]/3s

Posted by Harry on 2009-10-24 00:19:27