Finding landing site when descending through atmosphere.

[Meithan]

dlrk: I haven't had much time to run your test but I'll get to it when I can in the next coming days.

[dlrk]

Sounds good, thanks.

[Meithan]

Your cross-sectional area of 1m^2 looks to be way too small for a 16.7 tonnes craft. This would produce an aerodynamic drag acceleration that is 134 times smaller than the one the stock game would calculate (the stock aeordynamics assume a cross-sectional area of 8 m^2 per tonne). It just doesn't look right. The cross-sectional area used by FAR must be larger. Without knowing the correct number it'll be very hard to get a prediction that applies to FAR aerodynamics.

So, what I did was running a test using the stock aerodynamics, using your Cd of 0.36. I simulated starting on a 200 km circular prograde orbit over Kerbin and lowering the periapsis to 35 km. This is how the trajectory looks like:

The blue circle is Kerbin, the green one is the atmosphere, and the simulated trajectory is shown as the black and thick magenta lines (the magenta line starts at atmosphere contact; that's the part actually simulated through numerical integration).

The black diamond is the predicted landing site position when your craft is at apoapsis (indicated by the blue dot). The landing site leads the apoapsis by about 164 degrees. This accounts for Kerbin's rotation.

The landing site and orbit are assumed to be equatorial. My program does not yet have the capability to simulate non-equatorial orbits. Fortunately, KSC is (almost) on the equator so this could be useful for landing at KSC.

In order to use this prediction, do the following:

1) Start on a 200 km prograde equatorial circular orbit.

2) Wait until your intended landing site is 164 degrees ahead of you (this can be estimated by comparing your current longitude to that of the landing site).

3) Burn retrograde until your periapsis is 35 km (as usual, you'll have to lead the exact burn point by about half your burn time to get best results).

The trajectory should then follow the one indicated in the plot, and your landing site will be below you once you reach low altitudes.

Edited October 5, 2013 by Meithan

[Specialist290]

Your cross-sectional area of 1m^2 looks to be way too small for a 16.7 tonnes craft. This would produce an aerodynamic drag acceleration that is 134 times smaller than the one the stock game would calculate (the stock aeordynamics assume a cross-sectional area of 8 m^2 per tonne). It just doesn't look right. The cross-sectional area used by FAR must be larger. Without knowing the correct number it'll be very hard to get a prediction that applies to FAR aerodynamics.

On the other hand, consider that the stock game calculates drag for every part, regardless of whether or not it's actually exposed to airflow, while FAR seems to have some sort of system* determining which parts are exposed and which aren't. A cone viewed from the top would have a cross-sectional area equal to its lateral surface (i.e. its total surface area not including the base); if you put a cylinder behind the cone's base and didn't change the angle of intersection, the cross-sectional area wouldn't change as long as the cylinder is of equal or lesser radius regardless of how long that cylinder is.

* Admittedly, not having played with FAR myself yet, I don't know the details on how that system works.