Is this the correct way to manually calculate the crash site of a spacecraft?

[MedwedianPresident]

Is the following way the correct way to calculate the exact crash site of a spacecraft on an airless, non-rotating, perfectly spherical body?

It is assumed that the spacecraft is travelling on an elliptical orbit with periapsis within the periphery of the planetary sphere, e.g. under the surface. I write down the equation of the sphere (x^2+y^2+z^2=r(Planet)) and the equation of the inclined orbital ellipse, combine both into an equation system, solve it, select the solution appropriate for the orbital direction (because there are 2 crash points, depending on whether the spacecraft is orbiting clockwise or anticlockwise) and transfer it into spherical coordinates for latitude and longitude. Alternatively, I do the same but two-dimensionally (circle + ellipse equation) and break down the coordinates to account for inclination after calculating the crash site.

[Freshmeat]

I think you'll need to write your solution out in more detail, but I think the idea is correct. Why would you want to do this, BTW? I have done numerical solutions to a descending craft under thrust to get a close touchdown, but that is a whole other problem.

[Snark]

3 hours ago, MedwedianPresident said:

Is the following way the correct way to calculate the exact crash site of a spacecraft on an airless, non-rotating, perfectly spherical body?

That sounds basically reasonable.

Take your current ellipse.  Solve it to find "at what point on the trajectory does R equal planet radius."  That's the crash site.

Of course, all KSP bodies rotate, but fortunately that's easy to do, too.  They all rotate at uniform rate, so once you've calculated "at what X,Y,Z point does my ellipse intersect the surface", just figure out when that will be, multiply the time by the planet's rotation rate, offset the projected crash longitude by that much, and there you go.  So it's easy to account for planetary rotation.

The other two approximations you mention (airless and perfectly spherical) are not necessarily valid in KSP.  Several bodies have significant atmosphere, which will radically affect crash site.  And KSP planets are much bumpier (relative to their radii) than real-life planets are, so terrain elevation can really matter when working out crash site, especially if you're approaching the surface at a shallow angle.

Even though these two factors are significant, though... they're also impossibly gnarly; there's basically no way to solve them analytically with math.  So you pretty much have to ignore them (unlike planetary rotation, which you can account for easily).

One last thing:  for a general solution, you shouldn't assume an ellipse, specifically.  The ship could be moving faster than escape velocity, in which case its orbital path is a hyperbola rather than an ellipse.

[bigcalm]

Yeah, you basically need Kepler's Laws to solve.  The wiki page linked will give an answer in spherical polar coordinates (which is what you need really).  From this you can calculate the time t that the orbit intersects the body (you'll have to fudge the altitude of collision to either datum level or an approximation of what it's likely to be, or it'll get very complicated).  You can then calculate how much the body in question rotates in time t, and thus, the exact lat long coordinates of landing.  And Snark is correct above, Kepler won't help you directly with hyperbolic orbits, but that should be a relatively simple adjustment once you get it to work for ellipses.

[Vanamonde]

Actual science content detected. Thread moved to Science & Spaceflight. 

[YNM]

You can prescribe the orbit and then see where it's orbital "radius" would equate to the orbited body's radius.

There's no use of X,Y,Z - it is more efficient to use spherical coordinates.

Edited December 6, 2017 by YNM

[A_name]

I recommend the Trajectories mod for this.

[cubinator]

2 hours ago, A_name said:

I recommend the Trajectories mod for this.

Where's the fun in that when you could learn to do it by hand?

[PB666]

On 12/5/2017 at 10:27 AM, MedwedianPresident said:

Is the following way the correct way to calculate the exact crash site of a spacecraft on an airless, non-rotating, perfectly spherical body?

It is assumed that the spacecraft is travelling on an elliptical orbit with periapsis within the periphery of the planetary sphere, e.g. under the surface. I write down the equation of the sphere (x^2+y^2+z^2=r(Planet)) and the equation of the inclined orbital ellipse, combine both into an equation system, solve it, select the solution appropriate for the orbital direction (because there are 2 crash points, depending on whether the spacecraft is orbiting clockwise or anticlockwise) and transfer it into spherical coordinates for latitude and longitude. Alternatively, I do the same but two-dimensionally (circle + ellipse equation) and break down the coordinates to account for inclination after calculating the crash site.

A surface strike from below to the inside of the shell is always a strait line along the vector of travel.

Here is the reason.

https://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

The net gravity acting inside of a shell is zero (r < R_shell). Consequently, there is a direction vector from the center. The crafts position is different from the center. In this case the inside radius is r'. Current position is x, y, z from center

and velocity is dx/dt, dy/dt, dz/dt. Consequently r'= ((x₀+t_idx/dt)² + (y₀+t_idy/dt)² + (z₀+t_idz/dt)²)^1/2 where t_i is the time of impact.

Please do your math homework yourself.