How to track the visual path of a LEO satellite as seen from the ground

Question

I have been struggling with this problem for a while so I decided to ask. I'm new here and I'm not sure where this type of question belongs, so forgive me if this isn't the right section.

I am working on a satellite tracking program. I have it using TLE data from Celestrak to track a variety of satellites. I have the basic information for a pass, including the AOS azimuth, LOS azimuth, and Max Elevation, as well as the total time it takes to cross the sky as seen from the observer on the ground. I need some way to translate this info into a list of azimuth and elevation values that represent the path the satellite will take across the sky. I've searched quite a bit and haven't been able to solve it yet.

In other words, given the starting azimuth, ending azimuth, max elevation, and total time, how can I create a function to tell me the azimuth and elevation at any particular moment in time?

Answer 1

All the derivations below are with respect to this diagram of observer's celestial sphere

Red-Path of Satellite, Red/Green Intersection - Position of Satellite at a given time.

$$ T, \ Total \ Time \ in \ Sky \\ i, \ inclination \ of \ satellite \ at \ rise/set(ACB) \\ e, \ maximum \ elevation(ND) \\ A_r, \ Aziumth \ at \ rise \\ A_s, \ Aziumth \ at\ set(NSC) \\ l, \ Half \ Length \ of \ Path \ in \ sky(CD) $$

Using spherical trigonometry, $$ \cos(l) = \cos(\frac{A_r+A_s}{2}).\cos(e) \ (1) \\ Since, \ DNB = 90^°, \ \sin(l) = \frac{\sin(e)}{\sin(i)} \ (2) $$

That's one step. Again using spherical trigonometry,

$$ AC = \frac{2lt}{T} \ (3)\, \ for \ some \ given \ time \ t \\ \cos(\frac{2lt}{T}) = \cos(a).\cos(A), \ where \ a \ and \ A \ are \ altitude \ and \ aziumth \ to \ be \ found \\ \sin(\frac{2lt}{T}) = \frac{\sin(a)}{\sin(e)} \ (4) $$

Now, you simply have to solve $(1),(2),(3),(4)$ to get $a$ as a function of $A$.

You're welcome! :)