I'm unable to get back into the game until the weekend and this question has been bugging me, so I'm hoping some kind member can inform me before then.īasically, I'd like to know - if I lower the orbit of the station to say, 15Km, would that reduce the fuel requirements of my lander? My first instinct is that of course it'll have lower requirements if the destination orbit is lower. Or, more likely, I'm simply burning fuel really inefficiently. I would've thought 200 units of fuel would be enough, but the weight of all Jeb's snacks is just too much. However, in true Kerbal style I didn't bother to test it out and so to Jeb's chagrin, the lander has enough fuel to get down to the Mun but not back again to the station at 75Km. I've managed to get a behemoth of a space station with Science Lab and dock-able lander into a 75km circular more-or-less equatorial orbit around the Mun. Running a lower t/w will generate higher DV losses from gravity, *but* these losses may be offset by the reduced mass, structure, and fuel consumption attendant with carrying less engines.Įach engine has an ideal t/w for a given body fir high mass efficiency, and it's usually in the range of 1.2-1.6 for launches and *less than* 1 for landing.I'm thoroughly hooked on KSP now and am working my way through career mode. Using an intermediate method will yield somewhere in between, but even this is preferable to bombing in from a 2 million meter apoapsis.Īnd here's where it gets weird: Raising the t/w very quickly yields diminishing returns, and soon adds no measurable benefit. If you're using the "stop 'n' drop" method, it'll cost 578 m/sec to arrest your lateral velocity and then another 181 m/sec to arrest your freefall velocity at the surface for a total of 759 m/sec. this figure will go up with the approach method you're using. Again, this is a theoretical absolute minimum, but it's clearly preferable to 775 m/sec. Doing a zero descent rate approach from there would require 2 burns:īurn #1 to set Pe just above the surface (6.8 m/sec)īurn #2 to arrest lateral velocity (578 m/sec) If you're in a 10km circular orbit, you're orbiting at 557m/sec. This does *not* mean that more thrust is the way to go about it, though.Īs others have said, you really need to operate your lander from a low circular orbit. The absolute minimum DV to arrest to a safe landing that way would also be 775 m/sec, assuming infinite thrust and no weight penalty for generating the thrust. Mathematically, I'd expect you to hit the Munar surface at 775m/sec from that orbit without a burn. Not-a-pro tip (but still a good advice) : when planning your mission and building your rockets, use DV maps (google KSP DV MAP) to know how much dV you will need The answer is simple : gravity will make you accelerate.Save, and then dont burn, just let your lander crash and note the speed just before the crash : it is greater than 1000m/s. If i understand, before burning, you are going 800m/s, and still have 1000 m/s in fuel in your vessel, and you dont understand why it's not enough to land ? Thanks to anyone that has an idea on the problem. How is that possible, given that spending its fuel, my lander should be less affected by gravity? The way I understand things, I should be able to land, but in reality, I'm always short in fuel and hit the mun at around 230m/s. If I don't use my engines, I hit the mun at about 800m/s. Once I've modified the trajectory of the lander, its remaining deltaV is 1000m/s. I've placed an orbiter around the Mun (on a very elliptical orbit, pe = 50km, ap = 2200km), and detached the lander. I'm pretty new in KSP, and while I've read a lot of tutorials and wiki pages, there is something that I don't understand in my current mission.
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