Help Jeb fly around Kerbin!

[tryharder]

Jeb, being the badass spaceship pilot he is, wants to circumnavigate Kerbin with a plane without landing in between. However, Wernher von Kerman only designed the plane to have enough fuel for half the trip, because the space program was on a tight budget. The plane can do seriously sharp turns and transfer fuel to identical planes while flying though. Luckily for Jeb, Gene managed to rope Bob and Bill in as "willing" helpers to fly two identical copies of the plane. How can Jeb fly around Kerbin without him or his helpers crashing or landing halfway through?

Kudos to @camacju, who solved it with one method.

Edited August 2 by tryharder

[user544]

impossible unless you do some modded craft switching things that allow crafts to fly while you are in another craft + midair docking

[jost]

26 minutes ago, user544 said:

impossible unless you do some modded craft switching things that allow crafts to fly while you are in another craft + midair docking

Tryharder didn't forbid to fly to space.

Thus it should be possible to Orbit Witze the planes to space, transfer fuel and continue the circumvention.

 

Still not trivial so hats off to anybody who achieve this 

Edited July 31 by jost

[camacju]

If you can control planes at the same time and refuel the planes on the ground, a possible method would be as follows (I'm assuming this is approximately your intended restrictions).

Jeb, Bill, and Bob take off at the same time and fly 1/8 of the way around Kerbin. At this point they're all at 3/4 fuel.

Bob transfers 1/4 of a fuel tank to both Jeb and Bill, bringing them both to full fuel, and has 1/4 of a tank remaining for the trip home. Bob turns around and can fly back to KSC, landing on fumes. He refuels his plane and prepares to take off again.

Jeb and Bill fly 1/6 of the way around Kerbin. At this point they're both at 2/3 fuel, and approximately 29% of the distance around Kerbin.

Bill transfers 1/3 of a fuel tank to Jeb, bringing him to full fuel, and has 1/3 of a tank remaining for the trip home. This isn't enough, but he begins flying back toward KSC.

Bob takes off for the second time with a full load of fuel, and meets Bill 1/8 of the way around Kerbin, where he's just run out of fuel. Bob transfers 1/4 of a tank to Bill, and they both have enough fuel to get home (with Bob landing with 1/4 of a tank, so theoretically Bill could have flown further before returning, but I didn't want to calculate the exact optimum).

Meanwhile, Jeb is flying still. He can reach 75% of the way around Kerbin with some fuel margin remaining.

When Jeb reaches the halfway point, Bill and Bob take off for the second and third time respectively with full fuel tanks, heading the opposite direction.

5/6 of the way around Kerbin (1/6 from the refueler's perspective), Bob refuels Bill and returns to KSC.

3/4 of the way around Kerbin, Jeb's fuel tank is running low, and Bill has 5/6 of a tank remaining. He transfers half of it to Jeb and turns around to head back toward KSC.

Bob takes off for the fourth time heading westward with a full load of fuel.

7/8 of the way around Kerbin, Bob refuels Jeb and Bill.

All three touch down at KSC.

There's actually a bit of wiggle room in both the outbound and return trips because the refueler's return trip doesn't need to keep three planes fueled, only two. So this general scheme would work for a planet slightly larger than Kerbin. Anyone want to calculate the maximum range?

Generalizing this problem to more planes, I suspect the maximum range goes to infinity, since I feel like it would grow as the sum of 1/n (so approximately O(log(n))) where n is the number of planes.

I actually did something similar to this a couple years ago when I took the Aeris 4A stock craft to Laythe and back! But I was able to use low Kerbin orbit as a staging area, so I didn't need to mess with simultaneous control.

Spoiler

 

Edited August 2 by camacju

[tryharder]

Quote

Tryharder didn't forbid to fly to space.

Okay, you can't fly to space, only atmospheric flights allowed. And anyways its a riddle, not an actual in-game challenge.

[MioSleet]

On 8/1/2024 at 7:51 PM, camacju said:

如果您可以同时控制飞机并在地面为飞机加油，那么可能的方法如下（我假设这大致是您想要的限制）。

Jeb、Bill 和 Bob 同时起飞，飞行了 Kerbin 的 1/8 圈。此时他们的燃料都剩下 3/4。

鲍勃将 1/4 油箱的油分给杰布和比尔，这样两人的油箱都加满了，回家的路上还剩下 1/4 油箱的油。鲍勃掉头飞回肯尼迪航天中心，降落时油箱里还冒着烟。他给飞机加满油，准备再次起飞。

Jeb 和 Bill 绕 Kerbin 飞行了 1/6 圈。此时，他们的燃料都剩下 2/3，绕 Kerbin 飞行了大约 29% 的距离。

Bill 将 1/3 的油箱油量转给 Jeb，让他的油箱满油，还剩下 1/3 的油量供他返航。这还不够，但他还是开始飞回肯尼迪航天中心。

鲍勃第二次满载燃料起飞，在绕 Kerbin 飞行了 1/8 时遇到了比尔，当时他的燃料已经耗尽。鲍勃将 1/4 油箱的燃料转移给比尔，两人都有足够的燃料返回家中（鲍勃着陆时油箱里只有 1/4 的燃料，因此理论上比尔可以飞得更远再返回，但我不想计算准确的最佳值）。

与此同时，Jeb 仍在飞行。他可以飞到 Kerbin 周围 75% 的距离，并且还剩下一些燃料余量。

当杰布飞到中途时，比尔和鲍勃分别第二次和第三次满油起飞，朝着相反的方向飞行。

绕 Kerbin 飞行了 5/6 圈（从加油员的角度来看是 1/6），Bob 为 Bill 加油并返回KSC。

绕行 Kerbin 四分之三的路程，Jeb 的油箱快没油了，Bill 还剩下六分之五的油。他将一半的油转给 Jeb，然后掉头返回KSC。

鲍勃第四次起飞，满载燃料向西飞行。

在绕 Kerbin 飞行了 7/8 圈后，Bob 为 Jeb 和 Bill 补充了燃料。

三架飞机均在肯尼迪航天中心 降落。

实际上，往返行程都存在一些回旋余地，因为加油机的回程不需要为三架飞机加油，只需要两架。所以这个一般方案适用于比 Kerbin 稍大的行星。有人想计算最大航程吗？

将这个问题推广到更多的平面，我怀疑最大范围会趋于无穷大，因为我觉得它会增长为 1/n 的总和（所以大约为 O(log(n))），其中 n 是平面的数量。

几年前，当我乘坐 Aeris 4A 普通飞船往返 Laythe 时，我实际上做过类似的事情！但我能够使用低 Kerbin 轨道作为集结区，因此我不需要同时控制。

  显示隐藏内容

 

I have conducted further calculations based on your idea and arrived at more detailed conclusions. Suppose there are three airplanes, A, B, and C, each with a range of 36 units, and the total required range is 72 units (I don't want too many decimals). It also takes 72 hours for each of them to fly around Kerbin (assuming infinite fuel). Additionally, we need to assume a constant flying speed, meaning heavier planes don't fly slower.

First, let's discuss the specific refueling process and timepoints.

Stage 1: ABC take off. A refuels B and C, then returns. Assuming the distance flown at this point is x1, then 3x1 (ABC flying forward) + x1 (A returning) = 36, so x1 = 9.
The refueling timepoint is at T=9, and A lands at T=18.

Stage 2: B and C continue flying. B refuels C and returns. After returning and refueling, A takes off and meets B on its way back to refuel it. Assuming A flies a distance of x2 and B and C fly a distance of y2 forward, then
x2 (A flying forward) + 2x2 (A and B returning) = 36, so x2 = 12.
2y2 (B and C flying forward) + (y2 + 9 - x2) (B returning only until meeting A) = 36, so y2 = 13.
At T=22, B and C refuel. Subsequently, B will meet A after y2 + 9 - x2 = 10 hours, which is at T=32. A needs to take off x2 hours earlier, which is at T=20. A and B land at T=44.

Stage 3: C continues flying and exhausts its fuel at T=58, having also flown a distance of 58 units. It needs A and B to come and support it, flying in the opposite direction. The remaining distance for this final stage is 72 - 58 = 14, which means A and B take off at T=58 - 14 = 44.
There is relatively ample fuel remaining for this stage. Here's a possible scenario.

Stage 3.1: A and B take off at T=44. Assuming they fly a distance of x3 such that A can refuel B and return, exactly exhausting its fuel, then 2x3 (A and B flying in the opposite direction) + x3 (A returning) = 36, so x3 = 12. The refueling time is at T=44 + 12 = 56.

Stage 3.2: B flies in the opposite direction for 2 hours and meets C at T=58. At this time, B has remaining fuel for a range of 34 units. After refueling, C has a remaining range of 14 units, and B has a remaining range of 20 units of fuel. They both return together. Finally, they complete the flight at T=72.

Do you see the crux of the issue?
In reality, time is quite tight. A and B need to take off immediately after landing to meet C, which has just exhausted its fuel, after flying in the opposite direction. This is necessary because B and C, after separating, need to fly around opposite hemispheres of Kerbin to meet each other on the other side. This is the key requirement.

However, the refueling time of A and B at ground is unable to be 0.

Therefore, this problem is essentially unsolvable.

Quote

 

If you can simultaneously control the aircraft and refuel it on the ground, a possible approach would be as follows (I'm assuming this is roughly the constraints you want).

Jeb, Bill, and Bob take off at the same time and fly 1/8 of a circle around Kerbin, with 3/4 of their fuel remaining.

Bob gave 1/4 of the tank to Jeb and Bill, so that both of them had full tanks and had 1/4 tank left on the way home. Bob turned around and flew back to Kennedy Space Center , landing with smoke in the tank. He filled up the plane and prepared to take off again.

Jeb and Bill have flown 1/6 of the way around Kerbin. At this point, they both have 2/3 of their fuel left and have flown about 29% of the way around Kerbin.

Bill transferred 1/3 of the tank to Jeb, filling his tank and leaving 1/3 for the return flight. It wasn't enough, but he started to fly back to Kennedy Space Center .

Bob takes off a second time with a full tank of fuel and meets Bill about 1/8 of the way around Kerbin, when he has run out of fuel. Bob transfers 1/4 of his tank to Bill, and both have enough fuel to get home (Bob landed with only 1/4 of his tank, so in theory Bill could have gone further and come back, but I don't want to calculate the exact optimal value).

Meanwhile, Jeb is still flying. He can fly 75% of the way around Kerbin and still have some fuel left.

When Jeb was halfway through the flight, Bill and Bob took off with full fuel for the second and third time, respectively, flying in opposite directions.

After flying 5/6 of a circle around Kerbin (1/6 from the tanker's perspective), Bob refueled Bill and returned to KSC .

Three quarters of the way around Kerbin, Jeb's tank was running low, and Bill had five-sixths left. He transferred half of the tank to Jeb and turned back to KSC .

Bob took off for the fourth time, heading west with a full load of fuel.

After flying 7/8 of an orbit around Kerbin, Bob refueled Jeb and Bill.

All three aircraft landed at the Kennedy Space Center .

Actually, there is some wiggle room for both round trips, because the tanker doesn't need to refuel three planes on the return trip, only two. So this general scheme works for planets slightly larger than Kerbin. Anyone want to calculate the maximum range?

Generalizing this problem to more planes, I suspect the maximum range will go to infinity, as I feel it will grow as a sum of 1/n (so roughly O(log(n))), where n is the number of planes.

I actually did something similar a few years ago when I flew to Laythe and back in an Aeris 4A regular ship! But I was able to use low Kerbin orbit as a staging area so I didn't need to control both simultaneously.

 

ssss

Edited August 15 by Vanamonde
Please post in English when not using the International subforums.

[camacju]

On 8/14/2024 at 6:32 PM, MioSleet said:

This is necessary because B and C, after separating, need to fly around opposite hemispheres of Kerbin to meet each other on the other side. This is the key requirement.

However, the refueling time of A and B at ground is unable to be 0.

There's some fuel margin in the solution I gave, which should allow some time for one of the planes to slow down in flight and enable refueling on the ground. 

[user544]

uhh i cant wrap my head around this