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How much reserve fuel (delta-V) do you take?


paul23

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9 hours ago, paul23 said:

Escape velocity is very much exact science. And in ksp if you have escape velocity you *will* escape if you instantaneous reach that speed and don't have any aerodynamic drag. Actually a bit less since you don't even need to escape to infinity, having an elliptical orbit with apoapsis outside the sphere of influence counts as having escaped.

So the actual rocket equation is:

delta_v = v_exhaust * ln(m0/mf) - g * time_burn - integral(dragForce * time) ds 

Yes I'm aware, but the community maps don't use escape velocity or vis viva or anything else to compute them. Just learned experience by hopefully sane designs. Indeed, the most commonly used map says 3400 to LKO and 930 to Kerbin Escape. This totals 4330 Delta-V to Kerbin Escape. The actual escape velocity is 3431 m/s. And you are already at 174 m/s so the minimum delta-V a perfectly aerodynamic cannonball could need to escape Kerbin is actually 3257 m/s. The map overprovisions by 1070 m/s! This is to very crudely estimate turning, gravity, and aerodynamic losses. But these numbers are very beatable by high-TWR aerodynamically optimized designs. And likewise far too conservative for ungainly designs.

 

Laythe is smaller than Tylo with equal surface gravity. Yet requires a nominal 3790 m/s to escape from by the community map. Compared with 3320 for Tylo. This is despite its escape velocity from the surface actually being 2801 m/s to Tylo's 3063. A 262 m/s difference.

Edited by Pds314
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6 hours ago, Cavscout74 said:

Having grown up around aircraft & pilots, I follow the old pilot saying "The only time you can have too much fuel is if you're on fire"

Or trying to land with a craft with marginal TWR. Or in a BDA dogfight, or reentering (in fire).

Edited by Pds314
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