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?
c = clearance
L = length of rectangular portion of tanker
w = width of tanker
s = speed of tanker
2s = speed of patrol boat
Relative to the ocean, the patrol boat takes a "loopy" path with the loops being rectangles.
As the tanker is moving also, the motion of the patrol boat across the bow or stern of the tanker is not perpendicular to the motion parallel to the sides, relative to the tanker itself, but the completed "circling" figure is still a parallelogram. Since the speed of the patrol boat is twice that of the tanker, the slope of the crossing path is 1/2 as seen by the tanker.
The radius of the semicircular cap at either the bow or the stern is w/2.
What has to be cleared, however, is a rectangle of width w + 2c and length L capped by semicircles of radius 2w + c.
The distance from the center of a semicircle to the angled path of the patrol boat relative to the tanker is sqrt(5) * (2w + c). Since there are semicircles at both ends, twice this is added to each of the long sides of the parallelogram.
The short sides of the parallelogram are just (w + 2c) * sqrt(5) / 2, and so total (w + 2c) * sqrt(5). But this is irrelevant to the time consideration, as it is relative to the tanker rather than to the ocean.
The patrol boat must traverse, relative to the tanker, L + sqrt(5) * (2w + c), traveling at a relative speed of s, and also at a speed of 3s, so the time spent traveling parallel to the tanker's path is (L + sqrt(5) * (2w + c)) / s + (L + sqrt(5) * (2w + c)) / (3s). The sideways traverses are relative to the ocean and therefore take (w + 2c) * 2 / (2s) = (w + 2c) / s.
The total time is therefore (L + sqrt(5) * (2w + c)) / s + (L + sqrt(5) * (2w + c)) / (3s) + (w + 2c) / s
= (L + sqrt(5) * (2w + c)) * 4 / (3s) + (w + 2c) / s

Posted by Charlie
on 20091023 15:26:36 