Looking for a double-check of my thought process here...

The ISS orbit is circular but it does not orbit over the equator, it has an orbital inclination of 51.6 degrees.

That results in this ground-track:

Everything I can find says that the orbit "looks like a sine wave" but never says that it

I need to calculate the ISS

One orbit takes 90 minutes, so the Earth moves (1.5 hours / 24 hours) = about 6% of the circumference per orbit, so the error would average 3% over the course of one orbit, and I can live with that if I

Added: At the equator the answer is 51.6 degrees but that could be a northeast or southeast heading. I can figure out the correct sign by looking at the

The ISS orbit is circular but it does not orbit over the equator, it has an orbital inclination of 51.6 degrees.

That results in this ground-track:

Everything I can find says that the orbit "looks like a sine wave" but never says that it

*is*one. Because circles are involved I tend to think that it is, or close. They probably say that because the Earth rotates under the ISS, which stretches out the wave, and that if the Earth didn't rotate, the orbit*would*be a perfect sine wave.I need to calculate the ISS

*heading*at any point in its orbit, based on the latitude. Some points are easy; it crosses the equator at a heading of exactly 51.6 degrees, and at 51.6 North and South the heading is due east, and it cycles back and forth between those values. Using SIN/COS is going to produce a reasonably good approximation.One orbit takes 90 minutes, so the Earth moves (1.5 hours / 24 hours) = about 6% of the circumference per orbit, so the error would average 3% over the course of one orbit, and I can live with that if I

*have*to, but the latitude values are accurate to within 0.02% so 3% is*really*sloppy.Added: At the equator the answer is 51.6 degrees but that could be a northeast or southeast heading. I can figure out the correct sign by looking at the

*trend*of the data, so all I need is the angle.
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