MioSleet

Hello everyone, I've encountered some issues with WOLF. I'm planning to use the Wolf Terminal for Crew transfer, for example from Kerbin:Shore to Kerbin:Orbit. I have sufficient Transport Credits, and I've already deployed a 10m Terminal at Shore and another terminal at orbit. Now, I deploy a new SSTO that includes a Crew Container and Transport Computer. I move it to shore biome likes the terminal, within a distance of less than 100m. However, when I attempt to "Connect to Depot"using transport computer, it still prompts me that I need to be in the same Biome as the Terminal and within a distance of less than 250m, which means I'm unable to create a Crew Transport Route. So how can I make the route to transport my crews? I've made a cargo route from Kerbin's LKO to Duna‘s LDO with ultra high ISP (~15000s) engine, witch only takes 2 transport credit and transfer 60 unit of cargo, I want to repeat that to make a crew route eventually. I hope someone familiar with Wolf can help me resolve this issue. Thank you very much.

Hello everyone, I've encountered some issues with WOLF. I'm planning to use the Wolf Terminal for Crew transfer, specifically from Kerbin:Shore to Kerbin:Orbit. I have sufficient Transport Credits, and I've already deployed a 10m Terminal at Shore. Now, I've deployed a new spacecraft that includes a Crew Container and Transport Computer. I moved it to the same Biome as the Terminal, within a distance of less than 100m. However, when I attempt to "Connect to Depot", it still prompts me that I need to be in the same Biome as the Terminal and within a distance of less than 250m, which means I'm unable to create a Crew Transport Route. On the other hand, I can normally create a Cargo Transport Route and have successfully transported 60 units of Cargo using 6 Transport Credits with nuclear air engines and high ISP nuclear liquid fuel engines from other mods. I hope someone familiar with Wolf can help me resolve this issue. Thank you very much.

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. ssss