Blake said:
The answer may be that it is really hard, and that is why NASA pays out the big bucks to math majors. If that's the case I'll spend some quality time with the LaunchMFD code (which I will likely do anyway).
I wanted to take a crack at this without reading the links, to see if I can get it right. Turns out that it's not very hard to do. A little bit of spherical trigonometry and you are set.
First we have to find what the launch window's "conditions" are. We have three separate cases.
1st case: The target orbit has a lower inclination than the latitude of the launch site. [math]T.Inc < [/math]
In this case launching in plane with the target orbit is impossible. The best we can do is launch at a 90° heading when the LAN of the target orbit is the same as the LAN of the of the spacecraft (as shown in OrbitMFD from the EQU frame). The resulting relative inclination to the target orbit will be:[math] R.Inc = L.Lat -T.Inc [/math]
2nd case: The target orbit has the same inclination as the latitude of the launch site. [math]T.Inc = L.lat[/math]
This is a best case scenario. We launch at a 90° heading when the LAN of the target orbit is the same as the LAN of the spacecraft (as shown in OrbitMFD from the EQU frame). The resulting relative inclination to the target orbit will be zero.
3rd case: The target orbit has a higher inclination than the latitude of the launch site. [math]T.Inc > L.lat[/math]. In this case we want to launch with a specific launch azimuth when the LAN of the target orbit and the LAN of the spacecraft have a specific difference. Have a look at this pic:
Here we have the orbital plane of the ISS and the landed spacecraft near the time of launch. (the plane of the ISS is about to cross the launch site).
That "shark-fin" in MapMFD is a spherical triangle projected in 2D. In the blue window below you can see a "zoom" of it.
The points A B and C are:
A = Launch site
B = LAN of the target orbit (equatorial)
C = LAN of the landed spacecraft's "orbit" (equatorial)
The angles α β and γ are:
α = Launch latitude [math]L.lat[/math]
β = 180° minus the target inclination [math]180^{o}-T.Inc[/math]
γ = "Rough" Launch Azimuth = [math]acos \left( \frac{cos(T.Inc)}{cos(L.lat)}\right)[/math]
a b and c are also angles, this time one point is at the center of the Earth. If we name the center of the Earth "D" then the angle [math]A \hat{D} B[/math] is the angle a and so on.
What we are interested in is the angle c, which is the difference in LAN at the time of launch. Since we know the angles α β and γ, we can use the spherical law of cosines and solve for c
[math] c = acos \left ( \frac{cos\gamma + cos \alpha cos\beta}{sin\alpha sin\beta} \right)[/math]
Let's try an example with the orbits shown in the pic above. We have a launch latitude of 28.63° and the target inclination is 51.52°
α = 28.63°
β = 180° - 51.52° = 128.48°
γ = [math] acos \left( \frac{cos(51.52^{o})}{cos(28.63^{o})}\right)= 44.853^{o}[/math]
[math] c = acos \left ( \frac{cos(44.853^{o}) + cos(28.63^{o}) cos(128.48^{o})}{sin(28.63^{o}) sin(128.48^{o})} \right)=64.28^{o}
[/math]
So we need to launch when the LAN of the target orbit and the LAN of the spacecraft have difference of ±64.28°.
To find the accurate launch azimuth, you can use the equations from the link from OrbiterWiki.
The rate of the rotation of the landed spacecraft's LAN is 360°/siderial day ~ 0.25° per minute.
There are more things to take into consideration, such as the time the spacecraft needs to reach orbital velocity and also the rate of the nodal regression of the target orbit (if you are flying with nonspherical gravity switched on), but I hope this will do for now.
