I came across this question on StackOverflow about Satellite Trajectory plot. In the code mentioned in the question, the user takes only the Azimuth & Elevation angles (radians) & converts them into cartesian (x, y) coordinates & plots the Satellite trajectory onto a polar plot.

However, to convert from Spherical to Cartesian coordinates, one needs Range, Azimuth & Elevation. But the formula used in the code is
$$ x = \frac{(\pi/2) - Elevation} {(\pi/2)*cos(Azimuth - (\pi/2))} $$
$$ y = \frac{Elevation - (\pi/2)} {(\pi/2)*sin(Azimuth - (\pi/2))} $$
How was the OP able to draw a satellite ground track using only Azimuth & Elevation? How to arrive at the above equations from
$$ x = r*cos(Elevation)*cos(Azimuth) $$
$$ y = r*cos(Elevation)*sin(Azimuth) $$
$$ z = r*sin(Elevation) $$

  • $\begingroup$ You should ask questions about that plot there, although the question is from 2012. It's not a ground-track plot at all as far as I can tell, it's just a plot of altitude and azimuth of several GPS satellites, shown only when they are above the horizon for that viewing location. $\endgroup$
    – uhoh
    Jun 15, 2018 at 13:04

2 Answers 2


Yes. This is a classical astrodynamics problem of orbit determination.

The technique you would use is called Gauss' method. It allows you to determine an approximate orbit from three timed observations of azimuth and elevation. The details are well-described in the link, but are too lengthy to reasonably list here.

  • $\begingroup$ Nitpicking: in the process, you actually do first determine the missing range (encoded as magnitude of [the satellite position vector minus your position vector]) - then you determine the ground track using it. So 'no, you need the range, but you can find it using this method.' $\endgroup$
    – SF.
    Jun 15, 2018 at 21:25
  • 1
    $\begingroup$ @SF. a process that determines orbit will of course yield range in the process, but range is not a given; unhelpful nit in my opinion. $\endgroup$
    – uhoh
    Jun 16, 2018 at 1:45
  • $\begingroup$ @Tristan I followed your link to the Wikipedia article, which is difficult to understand so instead read the derivation as given by Howard Curtis in his Orbital Mechanics for Engineering Students. I understood that derivation. The equation is for geocentric position & velocity vector. But how does that 8th order polynomial equation reduce to the equation mentioned in the question? Can you explain that $\endgroup$ Jun 20, 2018 at 9:59

As far as I can tell, in the question by the link, the OP is not trying to plot satellites' ground tracks in any way. He simply plots their elevations and azimuths, that's all, so he doesn't need to know the ranges. The $x$ and $y$ his program computes have nothing to do with satellite's Cartesian coordinates; they are just the coordinates of the point which represent the given elevation and azimuth on the plot.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.