new puzzle time

[Aharami]

.

 

[davestar]

---------- 1/4
---------- 1/4
---------- 1/4

three planes set out at once, make it 1/4 of the way around the world. each plane will have a half tank at that point. the two supporting planes, S1 and S2, should each dump half of their remaining fuel into the main plane, M1, giving it a full tank. S1 and S2 should then return. they will need to be met at the 1/8 point by a refueling plane, S3, who will give both of them enough fuel to make it the rest of the way. S3 will also have just enough fuel to make it back.

same thing should happen on the other side of the world.

so the answer is 4 (M1, S1, S2, S3)

 

[So]

Okay, even if you send two tankers, so you can refuel at 20% of the way out, the maximum the plane can cover is 70% of the distance. But you need it to cover more than 75% to get the plane home with any number of refuelers.

 

[Kyteland]

I think the answer is 4 planes total. I'm working on my proof, but I'm sure it's right.

 

[maximus maximus]

Answer is 4 planes. 2 scout planes and 2 refuling planes. Just divide the world in 2 halves.
One could do this with 2 planes as well if simultaneous scouting is not necessary.

1 scout plane and 1 refuling plane cover 1 half of the world and the other pair cover the other half.

This is how it would work.
Scout plane takes off. Once the scout plane reaches 1/2 its way, a refuling plane takes off. The scout plane goes 3/4 of its way and then turn back to reach 1/2 the way. Scout plane meets refuling plane and refuels fully.

Refuling plane heads back and scout plane continues to go ahead. Scout plane reaches the end of the first half of the world and then turns back. Refuling plane reached base and refills itself. Turns back and meets the scout plane at 1/2 way through. Scout plane refuels and returns home.

Repeat the same process on the other side of the globe.

 

[ungsunghero]

Originally posted by: davestar
---------- 1/4
---------- 1/4
---------- 1/4

three planes set out at once, make it 1/4 of the way around the world. each plane will have a half tank at that point. the two supporting planes, S1 and S2, should each dump half of their remaining fuel into the main plane, M1, giving it a full tank. S1 and S2 should then return. they will need to be met at the 1/8 point by a refueling plane, S3, who will give both of them enough fuel to make it the rest of the way. S3 will also have just enough fuel to make it back.

same thing should happen on the other side of the world.

so the answer is 4 (M1, S1, S2, S3)

I think this is the winner.

All four planes will return with exactly no fuel. At the 3/4 mark, M1 will be tapped. You can send S1 and S2 out 1/4 of the way (in the other direction), and they could each give M1 1/4 of a tank, or 1/2 total. That will get M1 back home. S1 and S2 will be able to get at the 1/8th point, and that's where S3 comes in again.

 

[So]

wow. okay, this is probably not the best solution, but:

9 planes take off, fly to 1/5 point. 6 of the planes top off the other 3 (at 1/5 distance around the earth each plane is down 2/5 and needs 2/5 to get back so each plane that continues needs two tankers). Then at the 2/5 mark, the two tankers top the third plane off. 40% there. Now the plane itself can fly all the way to the 90% mark. One more tanker flys out and tops off the plane to cover the final distance.

Total: 9. The question is # planes, not # trips. The six planes from the first part of the flight have to take off to help the two long range refuels make it back. One of the planes that came back can also handle the last 10%

 

[James Bond]

4 Planes Total. To start off with, planes one, two, and three, fly to the 1/4 mark around the world, as shown below:
Plane 1: -----------> 1/4
Plane 2: -----------> 1/4
Plane 3: -----------> 1/4
At this point, planes 2 and 3 give plane 1 one half of their fuel each. This leaves plane 1 with a full tank, and it leaves planes 2 and 3 with 1/4 tank each.
Planes 2 and 3 fly back towards base, and have enough fuel to make it to the 1/8 mark. At the 1/8 mark around the world, they are met by plane 4 who can refuel them to 1/4 tank each, leaving planes 2, 3, and 4, at 1/4 tank. Enough to make it home.
Plane 1 then flies to the 3/4 mark, where it gets low on gas:
Plane 1: ---------->---------->----------> 3/4
Plane 2: Base
Plane 3: Base
Plane 4: Base

At this point, planes 2 and 3 do the same thing as before, going the opposite direction. They meet plane 1 at the 3/4 mark, and each give him 1/4 tank in gas. This leaves plane 1 with a half tank--just enough to make it home. Plane 2 and 3 at that point turn around, and are met by plane 4 at the 7/8 mark, who refuels them to 1/4 tank each, and they all have just enough to get home.

 

[davestar]

Originally posted by: maximus maximus
Answer is 4 planes. 2 scout planes and 2 refuling planes. Just divide the world in 2 halves.
One could do this with 2 planes as well if simultaneous scouting is not necessary.

1 scout plane and 1 refuling plane cover 1 half of the world and the other pair cover the other half.

This is how it would work.
Scout plane takes off. Once the scout plane reaches 1/2 its way, a refuling plane takes off. The scout plane goes 3/4 of its way and then turn back to reach 1/2 the way. Scout plane meets refuling plane and refuels fully.

Refuling plane heads back and scout plane continues to go ahead. Scout plane reaches the end of the first half of the world and then turns back. Refuling plane reached base and refills itself. Turns back and meets the scout plane at 1/2 way through. Scout plane refuels and returns home.

Repeat the same process on the other side of the globe.

this wouldn't work for a number of reasons, the most important being that there can only be one "scout" plane.

the planes all fly at the same rate, so they either need to take off at the same time (if travelling in the same direction) or be flying in opposite directions.

thirdly, you're not taking into account the loss of fuel during the actual refueling

 

[maximus maximus]

Originally posted by: davestar

Originally posted by: maximus maximus
Answer is 4 planes. 2 scout planes and 2 refuling planes. Just divide the world in 2 halves.
One could do this with 2 planes as well if simultaneous scouting is not necessary.

1 scout plane and 1 refuling plane cover 1 half of the world and the other pair cover the other half.

This is how it would work.
Scout plane takes off. Once the scout plane reaches 1/2 its way, a refuling plane takes off. The scout plane goes 3/4 of its way and then turn back to reach 1/2 the way. Scout plane meets refuling plane and refuels fully.

Refuling plane heads back and scout plane continues to go ahead. Scout plane reaches the end of the first half of the world and then turns back. Refuling plane reached base and refills itself. Turns back and meets the scout plane at 1/2 way through. Scout plane refuels and returns home.

Repeat the same process on the other side of the globe.

this wouldn't work for a number of reasons, the most important being that there can only be one "scout" plane.

the planes all fly at the same rate, so they either need to take off at the same time (if travelling in the same direction) or be flying in opposite directions.

thirdly, you're not taking into account the loss of fuel during the actual refueling

a) like i said, it could be done with just one scout plane and one refuling plane too.

b) why would they need to take off at the same time?

c) I am assuming that this is instantaneous.

 

[Kyteland]

Originally posted by: davestar
---------- 1/4
---------- 1/4
---------- 1/4

three planes set out at once, make it 1/4 of the way around the world. each plane will have a half tank at that point. the two supporting planes, S1 and S2, should each dump half of their remaining fuel into the main plane, M1, giving it a full tank. S1 and S2 should then return. they will need to be met at the 1/8 point by a refueling plane, S3, who will give both of them enough fuel to make it the rest of the way. S3 will also have just enough fuel to make it back.

same thing should happen on the other side of the world.

so the answer is 4 (M1, S1, S2, S3)

My answer is slightly different but has the same result.

1) 4 planes fly to the 1/6 point. Each has 2/3 of a tank. Refueling takes place so that 2 planes are full and 2 are 1/3 full.
2) 1/3 planes return to base, full planes fly to 2/6 point. Planes at base a filled and other planes refuel to full and 1/3 respectively.
3) 1/3 plane flies back to 1/6 point and is met by plane from base. 2/3 full plane refuels empty plane and both have just enough to return to base.
4) full plane can fly from 2/6 point to 5/6 point. It is met by a plane from base. 2/3 full plane refuels empty plane and both return to base.

 

[LordSnailz]

T

 

[marulee]

Not in our generation. Let tomorrow generation deal with their worries. 😀

 

[ungsunghero]

Originally posted by: maximus maximus

Originally posted by: davestar

Originally posted by: maximus maximus
Answer is 4 planes. 2 scout planes and 2 refuling planes. Just divide the world in 2 halves.
One could do this with 2 planes as well if simultaneous scouting is not necessary.

1 scout plane and 1 refuling plane cover 1 half of the world and the other pair cover the other half.

This is how it would work.
Scout plane takes off. Once the scout plane reaches 1/2 its way, a refuling plane takes off. The scout plane goes 3/4 of its way and then turn back to reach 1/2 the way. Scout plane meets refuling plane and refuels fully.

Refuling plane heads back and scout plane continues to go ahead. Scout plane reaches the end of the first half of the world and then turns back. Refuling plane reached base and refills itself. Turns back and meets the scout plane at 1/2 way through. Scout plane refuels and returns home.

Repeat the same process on the other side of the globe.

this wouldn't work for a number of reasons, the most important being that there can only be one "scout" plane.

the planes all fly at the same rate, so they either need to take off at the same time (if travelling in the same direction) or be flying in opposite directions.

thirdly, you're not taking into account the loss of fuel during the actual refueling

a) like i said, it could be done with just one scout plane and one refuling plane too.

b) why would they need to take off at the same time?

c) I am assuming that this is instantaneous.

The problem is, when a refueling plane refuels the scout plane, the refueling plane has to use its fuel from its own tank.

At the 1/2 way point, for example, the refueling plane will only have half a tank of gas to transfer to the scout. It would need that entire half tank just to get back to the airfield. If it gave any amount of fuel to the scout, the refueling plane would not make it back to the base, without assistance from another plane.

 

[KMc]

- 3 planes take off at the same time. S, R1 and R2.
- At 1/8 circumference, each plane has used 1/4 of its fuel. R1 tops off both S and R2 then returns home empty.
- At 1/4 circumference, S and R2 each have used 1/4 of their fuel again. R2 refuels S and then returns home empty.
- S continues around until 3/4 circumference (now on fumes) where it is immediately met by R1 who, upon returning home, refueled and headed the other direction to meet S. R1 used 1/2 of its fuel getting there, so it gives S 1/4 of its fuel and both planes fly on together.
- At 7/8 circumference, S and R1 (now both on fumes) are met by R2, who also refueled upon returning and headed the other direction. R2 has used 1/4 of it?s fuel to get to this point, so it gives S and R1 each 1/4 tank and they all make it home together with just enough fuel to land.

So, 3 planes total.

 

[purbeast0]

good question and good solutions :thumbsup:

 

[SViper]

2 Planes.

A plane
B plane

Both planes take off. When they get to the 1/4 of the distance, plane B dumps half of his fuel into the first plane. So plane A at 1/4 of the distance has a full tank, and plane B uses his 1/4 left to go back to base and refuel.

Plane A gets to the 3/4 mark before it becomes empty, so plane B (fully refueled) flies the 1/4 of the distance to meet up with him, gives him half of his fuel, and both planes are able to make it back to base.

Edit: Lol. I confused myself on that one. It would be 3 planes.

 

[ElFenix]

Originally posted by: KMc
- 3 planes take off at the same time. S, R1 and R2.
- At 1/8 circumference, each plane has used 1/4 of its fuel. R1 tops off both S and R2 then returns home empty.
- At 1/4 circumference, S and R2 each have used 1/4 of their fuel again. R2 refuels S and then returns home empty.
- S continues around until 3/4 circumference (now on fumes) where it is immediately met by R1 who, upon returning home, refueled and headed the other direction to meet S. R1 used 1/2 of its fuel getting there, so it gives S 1/4 of its fuel and both planes fly on together.
- At 7/8 circumference, S and R1 (now both on fumes) are met by R2, who also refueled upon returning and headed the other direction. R2 has used 1/4 of it?s fuel to get to this point, so it gives S and R1 each 1/4 tank and they all make it home together with just enough fuel to land.

So, 3 planes total.

nominated for first post of the year

 

[Kyteland]

Originally posted by: KMc
- 3 planes take off at the same time. S, R1 and R2.
- At 1/8 circumference, each plane has used 1/4 of its fuel. R1 tops off both S and R2 then returns home empty.
- At 1/4 circumference, S and R2 each have used 1/4 of their fuel again. R2 refuels S and then returns home empty.
- S continues around until 3/4 circumference (now on fumes) where it is immediately met by R1 who, upon returning home, refueled and headed the other direction to meet S. R1 used 1/2 of its fuel getting there, so it gives S 1/4 of its fuel and both planes fly on together.
- At 7/8 circumference, S and R1 (now both on fumes) are met by R2, who also refueled upon returning and headed the other direction. R2 has used 1/4 of it?s fuel to get to this point, so it gives S and R1 each 1/4 tank and they all make it home together with just enough fuel to land.

So, 3 planes total.

Yeah, I just worked that out too. It looks like that's the best possible answer.

 

[James Bond]

Originally posted by: ElFenix

Originally posted by: KMc
- 3 planes take off at the same time. S, R1 and R2.
- At 1/8 circumference, each plane has used 1/4 of its fuel. R1 tops off both S and R2 then returns home empty.
- At 1/4 circumference, S and R2 each have used 1/4 of their fuel again. R2 refuels S and then returns home empty.
- S continues around until 3/4 circumference (now on fumes) where it is immediately met by R1 who, upon returning home, refueled and headed the other direction to meet S. R1 used 1/2 of its fuel getting there, so it gives S 1/4 of its fuel and both planes fly on together.
- At 7/8 circumference, S and R1 (now both on fumes) are met by R2, who also refueled upon returning and headed the other direction. R2 has used 1/4 of it?s fuel to get to this point, so it gives S and R1 each 1/4 tank and they all make it home together with just enough fuel to land.

So, 3 planes total.

nominated for first post of the year

Damn, he wins.

 

[ungsunghero]

Originally posted by: KMc
- 3 planes take off at the same time. S, R1 and R2.
- At 1/8 circumference, each plane has used 1/4 of its fuel. R1 tops off both S and R2 then returns home empty.
- At 1/4 circumference, S and R2 each have used 1/4 of their fuel again. R2 refuels S and then returns home empty.
- S continues around until 3/4 circumference (now on fumes) where it is immediately met by R1 who, upon returning home, refueled and headed the other direction to meet S. R1 used 1/2 of its fuel getting there, so it gives S 1/4 of its fuel and both planes fly on together.
- At 7/8 circumference, S and R1 (now both on fumes) are met by R2, who also refueled upon returning and headed the other direction. R2 has used 1/4 of it?s fuel to get to this point, so it gives S and R1 each 1/4 tank and they all make it home together with just enough fuel to land.

So, 3 planes total.

I stand corrected.

 

[KMc]

Originally posted by: ElFenix

Originally posted by: KMc
- 3 planes take off at the same time. S, R1 and R2.
- At 1/8 circumference, each plane has used 1/4 of its fuel. R1 tops off both S and R2 then returns home empty.
- At 1/4 circumference, S and R2 each have used 1/4 of their fuel again. R2 refuels S and then returns home empty.
- S continues around until 3/4 circumference (now on fumes) where it is immediately met by R1 who, upon returning home, refueled and headed the other direction to meet S. R1 used 1/2 of its fuel getting there, so it gives S 1/4 of its fuel and both planes fly on together.
- At 7/8 circumference, S and R1 (now both on fumes) are met by R2, who also refueled upon returning and headed the other direction. R2 has used 1/4 of it?s fuel to get to this point, so it gives S and R1 each 1/4 tank and they all make it home together with just enough fuel to land.

So, 3 planes total.

nominated for first post of the year

Well thanks. Long time listener, first time caller. Just never had anything worth mentioning until now....

...but then, history seems to indicate that hasn't deterred some people.

 

[James Bond]

Originally posted by: KMc

Originally posted by: ElFenix

Originally posted by: KMc
- 3 planes take off at the same time. S, R1 and R2.
- At 1/8 circumference, each plane has used 1/4 of its fuel. R1 tops off both S and R2 then returns home empty.
- At 1/4 circumference, S and R2 each have used 1/4 of their fuel again. R2 refuels S and then returns home empty.
- S continues around until 3/4 circumference (now on fumes) where it is immediately met by R1 who, upon returning home, refueled and headed the other direction to meet S. R1 used 1/2 of its fuel getting there, so it gives S 1/4 of its fuel and both planes fly on together.
- At 7/8 circumference, S and R1 (now both on fumes) are met by R2, who also refueled upon returning and headed the other direction. R2 has used 1/4 of it?s fuel to get to this point, so it gives S and R1 each 1/4 tank and they all make it home together with just enough fuel to land.

So, 3 planes total.

nominated for first post of the year

Well thanks. Long time listener, first time caller. Just never had anything worth mentioning until now....

...but then, history seems to indicate that hasn't deterred some people.

Lol QFT
EDIT: And, welcome to atot 🙂

 

[brikis98]

nice puzzle... gotta agree with the concensus that the answer is 3...

 

[QED]

Originally posted by: KMc
- 3 planes take off at the same time. S, R1 and R2.
- At 1/8 circumference, each plane has used 1/4 of its fuel. R1 tops off both S and R2 then returns home empty.
- At 1/4 circumference, S and R2 each have used 1/4 of their fuel again. R2 refuels S and then returns home empty.
- S continues around until 3/4 circumference (now on fumes) where it is immediately met by R1 who, upon returning home, refueled and headed the other direction to meet S. R1 used 1/2 of its fuel getting there, so it gives S 1/4 of its fuel and both planes fly on together.
- At 7/8 circumference, S and R1 (now both on fumes) are met by R2, who also refueled upon returning and headed the other direction. R2 has used 1/4 of it?s fuel to get to this point, so it gives S and R1 each 1/4 tank and they all make it home together with just enough fuel to land.

So, 3 planes total.

Good work... that's the answer.

Congratulations, YOU'RE WINNER!