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Sun 7 Dec, 2003 12:41 pm
On Tractor 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?
Notes:
ignore extra fuel consumption as a result of acceleration, evaporation of fuel, bleeding-heart-liberal fiscal policies, etc.
All the planes have to make it back safely, so you can't give all your fuel away to another plane.
Can the planes hover/land without any consequence?
Assumimg that they can't land anywhere other than at base, I make it that you'd need the plane making the round trip plus two support planes. Three planes in all.
If anyone can get it down to less than that I'd be most interested in hearing how.
How on earth would three work?
Tet
If it is assumed that refuelling, take-off and landing, etc., take zero time, or that there is enough slack in the system to make up for these, then there is more than one 'three plane solution' that works.
Mungo
I don't see it. Are you sure that you can do it with 3 AND get all planes back to the island? I can't see how you can do it with less than 7 planes.
Is the answer 5 ?
The first 3 planes A,B and C take off simultaneously.
At the one eighth way point each plane has 3/4 of a tank so A refuels B and C, and A returns.
At the quarter way point B and C have 3/4 left again so B refuels C leaving B with 1/2 to return, and C has enough to get to the three quarter way point to be refuelled using two more planes going the other way round to meet it with the fifth plane doing a later (inverse) rendezvous at one eighth from base.
The answer COULD be 3 if we assume the last two planes are A and B with instantaneous turnround !
D'oh! Feel like an idiot now.
The key to this IMHO is geometry. If Plane 'A' flies all of the way around the circle then plane 'B' must accompany it part of the way for refuelling purposes and then fly back, subsequently meeting it with some more fuel at the 'other side'. We are told that the planes are identical so the time it takes plane 'A' to complete half of the circle going clockwise must be the same time as it takes plane 'B' to complete the other half of the circle going anticlockwise (assuming no change-over time). The rest is mere detail.
Taking another look I find that even with the change-over taking real time the puzzle as stated is still possible with three planes.
Mungo
Yes I think you are correct. My A above has plenty of time to get to the 3/4 position to meet C, and B has time to refuel and get to 7/8 to meet B+C.
The only other condition not mentioned is that one plane can refuel two others at the same time.
If refueling and redirection are instantaneous(solution1), or if varied speeds are available(solution2), it only takes two planes (including the globetrotter).
1)
Plane A is going all the way...
Plane B takes off at the same time, and 1/4 way it refuels plane B and turns around. When plane B instantaneously lands, refuels, and takes-off (now going the other direction) then plane A is 1/2 way around. Plane B meets plane A when plane A is 3/4 around the globe. Refueling takes place, plane B turns around, and both return safely.
2)
If plane A can slow down and not lose gas mileage, then refueling takes place a little after 1/4 and a little before 3/4 - then you don't have to do things instantaneously.
Plane B would crash and burn.
Plane B would crash.
It would not burn. No fuel.
It would burn if it landed in an arsonist colony
No... plane B refueled at home and took off again. Plane B never travels more than 1/2 the circumference of the globe in solution 1, and no more than about 2/3 in solution 2.
How is it that Columbus wanted to get to the West Indies? He wanted to go east? What a crack-pot.
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