My Math Forum Around the world...

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 March 2nd, 2013, 08:38 AM #1 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,145 Thanks: 1003 Around the world... For your enjoyment(?), here's one I found; in its exact original wording: On Bagshot Island, there is an airport. The airport is the homebase of an unlimited number of identical airplanes. Each airplane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. The planes have the ability to refuel in flight without loss of speed or spillage of fuel. Though the fuel is unlimited, the island is the only source of fuel. What is the fewest number of aircraft necessary to get one plane all the way around the world assuming that all of the aircraft must return safely to the airport? How did you get to your answer? Notes: (a) Each airplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground. (b) Each airplane must have enough fuel to return to airport. (c) The time and fuel consumption of refueling can be ignored. (so we can also assume that one airplane can refuel more than one airplanes in air at the same time.) (d) The amount of fuel airplanes carrying can be zero as long as the other airplane is refueling these airplanes. What is the fewest number of airplanes and number of tanks of fuel needed to accomplish this work? (we only need airplane to go around the world)
 March 3rd, 2013, 10:20 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Around the world... Fly two planes 1/6 of the way around, fuel one to full and fly the other one back. The remaining plane can go 2/3 of the way around the globe. In the meantime, two planes go 1/6 of the way in the opposite direction then fuel each other, the one going back; the two planes meet at the 2/3 mark with enough fuel to get both to the 5/6 mark. You then have (the earlier) two fueled planes meet them there and escort them back. So an upper bound is 4 planes and 6 tanks. Can this simple strategy be improved?
March 3rd, 2013, 11:14 AM   #3
Math Team

Joined: Oct 2011

Posts: 14,145
Thanks: 1003

Re: Around the world...

Quote:
 Originally Posted by CRGreathouse So an upper bound is 4 planes and 6 tanks. Can this simple strategy be improved?
3 planes are sufficient...go forth and suffer more :P

 March 4th, 2013, 06:00 PM #4 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Re: Around the world... Hi [color=#0000BF]Denis[/color], Is the following correct? T = time to fly non-stop around the world. W = distance around the world. F = full fuel tank. Planes A, B, and C take off together and go W/8, each burning F/4. Plane A refuels both B and C to F and A has F/4 so it can return to the island at T/4. Planes B and C fly to W/4 at T/4. Each now has 3F/4. Plane B refuels C to F, B has F/2 so it can return to the island at T/2. Plane C has F at W/4 so it can go to 3W/4, arriving at 3T/4. At T/2, A takes off (it could be B, but we will use A) in the other direction and meets C at 3W/4 at 3T/4. A has F/2 and fuels C so that both A and C have F/4. This will get A and C to 7W/8 at 7T/8. At 3T/4, B takes off and meets A and C at 7W/8 and 7T/8. B has 3F/4 and A and C are empty. B gives A and C F/4 each so all planes have F/4. They all travel a distance of W/8 and all make it back to the island at T.
 March 4th, 2013, 09:02 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,145 Thanks: 1003 Re: Around the world... Looks good JKS. Main "trick" is leave a plane's tank full at the 1/4 point; that plane can then reach the 3/4 point; then more or less same procedure for last 1/4 as with first 1/4. Easier to "demonstrate" by changing whole trip to a clock-like arrangement; 1 to 12 units is the trip.

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