Aeroplane – Fly around the world

♠ Posted by GeekyFry in , at 10:06 AM

On Timbucktoo island, there is an airport. The airport is the homebase of an unlimited number of identical airplanes. Each airplane has a fuel capacity (F) 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?

Some conditions:
(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)

Possible Solution:

Discliamer: The follwoing solution is just one of the possible solutions. Please do post your own solutions or any mistake in the given solution

# of planes: 3
Let the 3 planes be A, B & C with A the plane which will eventually travel around the world (will cover one full circle distance)

Let all the planes start clockwise with Fuel F as depicted in the above picture.
At #2, 1/6th the distance the fuel consumed will be F/3 and remaining would be 2F/3. Now at this point B transfers F/3 fuel to C and returns back to refuel and then starts towards #3 again to meet returning C (next step) mid way. The fuel in the planes would now be:
A –> F – F/3 = 2F/3
B –> F – F/3 – F/3 = F/3  (F/3 given to C)
C –> F – F/3 + F/3 = F  (F/3 received from B)
At #3, 1/4th the distance the fuel in the respective planes would be:
A –> F/2
B –> returned at #2 and started again towards #3
C –> F/2 + F/3(received from B)
At this juncture (i.e. #3), C transfers F/2 to A thereby filling A’s tank to full tank (F). It has F/3 fuel left which is not sufficient to reach back, but while it’s way back B will come back again and refill it. Once C is refilled by B, both return to base and then start (both) anticlockwise towards #6 1/4th the distance anti-clockwise. A has enough to go till #6, 3/4th the distance
At #3, post refill of A by C, the fuel tanks read:
A –> 
B –> returned at #2 and started again towards #3
C –> F/2 + F/3 – F/2 = F/3, C returns towards starting point,  gets refilled somewhere in midst by B and both return to island, refill and start to #6
At #6, 3/4th the distance clockwise and 1/4th distance anti-clockwise, A has 0 fuel, but can be refilled by C as was the case at #3 (C comes to #6 the way it came to #3, just that this time it will be anti-clockwise instead of clockwise; the same story of B and C till #3)


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