perplexus.info :: Just Math : Oil Tanker

Oil Tanker (Posted on 2009-10-23)

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?

Submitted by Bractals

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Solution: (Hide)

The path of the patrol boat has a rectangular shape with respect to the ocean, but calculating the lengths is difficult. So we will look at things with respect to the tanker. This is done by substracting the velocity of the tanker from the velocity of the patrol boat in each path segment.

First, let i and j be unit vectors perpendicular and parallel to the path of the tanker respectively. With i in the direction port to starboard and j in the direction stern to bow. With these we have
Velocity in path segments with respect to Segment Ocean Tanker ------------- --------- ------------ Port 2Sj Sj Bow 2Si S(2i-j) Starboard -2Sj -3Sj Stern -2Si -S(2i+i)
From these velocities with respect to the tanker, it is easy to see that the path of the patrol boat with respect to the tanker is an isosceles trapezoid. The parallel sides of the trapezoid are parallel to the sides of the tanker and C distance from them. The shorter side is to starboard. The slant sides of the trapezoid are parallel to the velocities for the bow and stern and C distance from the bow and stern.

It is fairly easy to calculate the lengths of the path segments and the speed of the patrol boat in each of them.
Segment Length Speed ------------- -------------------------- --------- Port L + (C + W/2)*(√5 + 1) S Bow (C + W/2)*√5 S*√5 Starboard L + (C + W/2)*(√5 - 1) 3S Stern (C + W/2)*√5 S*√5
Summing the Length/Speed for the four segments we get the time for the cycle:
2[2L + (W + 2C)*(√5 + 2)] ----------------------------- 3S

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Answer	K Sengupta	2023-11-16 08:01:38
No Subject	CodyDunning	2023-11-16 03:18:16
re(4): solution - relativity	brianjn	2009-10-25 08:06:40
re(3): solution - relativity	Charlie	2009-10-24 11:48:21
re(2): solution - relativity	brianjn	2009-10-24 00:39:20
re: solution	Harry	2009-10-24 00:19:27
solution	Charlie	2009-10-23 15:26:36

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