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Showing results for tags 'brachistochrone'.
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Hi, What I'm trying to find out if it's possible to do a brachistochrone curve/"straight" line trajectory. Essentially, you would burn towards, say, Duna until you were halfway there, and then flip around and burn to decelerate. I've already made an engine that can burn for that long, I would mainly like to know how I could do this in-game and how I could determine the halfway point of the path (roughly speaking). Thanks! -
So, for reasons, I'm trying to get a probe to Eve as quickly as I reasonably can, for certain definitions of "reasonable" -- and decided the way to figure this out was to start with the basic kinematics equation of .5at2+vt+d and do a lot of algebra starting with the conceit that we want the acceleration portion to equal d. This resulted in just deriving the brachistochrone equation like a doof: (total trip t = 2*sqrt(d/a), total dv = 2*sqrt(da)) Throw in the min distance from Kerbin to Eve being 3,668,900km, pick a comfortable acceleration, and there's the dv and burn time. However, that's assuming crazy things like instant re-orientation, and for a = 1m/s, leads to a 22 tonne ion craft I already have around but also a pair of 16.8hr burns. I could just make everything bigger to burn harder for less time, but that could make the lower stages get a bit out of hand. So, we come to realm of having a coasting phase, which highlights the fact that the brachistochrone equation is a special case of a more general equation, and we've just set coast time to zero. As such, instead of d being just a*t2/2, which was convenient, now it's a*ta2/2 + vmaxtcoast where ttotal = ta+tcoast and vmax= roughly(a*ta). I think I can maybe constrain it with a chosen ta -- say, 10hrs (I can handle a pair of 5 hour burns over a weekend), and a vaguely acceptable acceleration rate, say, 1m/s again, for ease. So to solve for the times, we substitute a*ta for vmax and ttotal-ta for tcoast a*ta2/2 + a*ta * (ttotal-ta) = d => (d-a*ta2/2)/a*ta = ttotal-ta => d/a*ta-ta/2 = ttotal - ta => ttotal = d/a*ta + ta/2 This gives us 1m/s2*(36ks)2/2 + 36km/s*(ttotal-36ks) = 648Mm + 36km/s*(ttotal-36ks) = 3668.9Mm => 3020.9Mm/36km/s = ttotal-36ks = 83913.89ks => ttotal = not much more...that kinda gets overwhelmed by coast time...of over 23 thousand hours. I might not've done that right. I blame it being after 2:30. Any thoughts/corrections?
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Brachistochrone to Mars with Falcon 9 boosters
sevenperforce posted a topic in Science & Spaceflight
If SpaceX gets its booster return down to a science, and Falcon Heavy performs as expected, then an interesting possibility emerges. The Falcon 9 v1.1 FT Stage 1 booster is capable of SSTO on its own, though without payload or capacity for return. If a Falcon Heavy was launched without any second stage, however, you'd end up with a nearly-full first stage in orbit and two empty strap-on boosters returned safely to the ground, ready to refuel and relaunch. A single Falcon 9 launch costs $61 million, with fuel accounting for roughly $200,000 of that. Thus, Falcon Heavy would allow SpaceX to put a nearly-full Falcon 9 first stage into LEO for marginally more than the cost of a single Falcon 9 launch. With a $1 billion investment, that would be no less than fifteen nearly-full Falcon 9 first stages in LEO. Strap them together and you've got a launch stack capable of a Brachistochrone transfer to Mars for a manned mission in a minute fraction of the Hohmann transfer time. A short transfer time means your consumables budget can be much smaller, enabling an even-faster transfer. Can't think of a cheaper way of doing it.
