# How does the ISS Transit Finder website get the position of the ISS so accurately?

The recent Petapixel article How to Shoot the ISS Flying Across the Face of the Moon gives a lot of detailed, helpful advice for photographing transits of the ISS across the disk of the Moon.

For transit predictions, it recommends the web site https://transit-finder.com/

The ISS orbits at an altitude of roughly 400 km, so the ISS could be as close as 400 km to the observer. The angular radius of the moon is 0.25 degrees, or about 4.4 mrad, which at 400 km is only 1.7 km.

The regular publicly available TLEs are known to have some error, I've heard values like 5 or 7 km but of course it will depend on the details. Along-track errors would result in timing shifts of the order of 1 second, but cross-track errors would easily cause a complete miss, which really means you end up sitting several kilometers away from where you should have been to photograph the transit.

However, the example photograph shows a cross-track error of roughly a few hundred meters at most! The path of the visible transit across the UK is shown in a map in the article, where you can see it is only a few km wide.

Question: Is it known if this web site just blindly uses whatever the most recent TLE is from CelesTrak, or does it do a more intelligent job of interpreting several of the most recent TLEs along with some fitting or custom propagation, or perhaps uses some other publicly available ISS finding software?

below: From here, photo credit should be Mathew Browne (author). "The photo you see here is a composite of 22 consecutive frames illustrating the movement of the ISS across the face of the moon."

A typical TLE for a maintained satellite (as opposed to debris tracked via ground assets) has an error of only around 1 km at epoch (Vallado, et al, Appendix B), which grows at a rate of around 1-3 km per day. Worst case, a 1 km cross-track deviation from a distance of 400 km (approximate distance to ISS for a zenith pass) gives a deviation of only 0.143 degrees from the nominal path. This is only 30% of the lunar diameter. In reality though, the 1 km error is the total error in all directions, so a more realistic expectation of the cross track error would be $$\sqrt{\frac{1}{3}} \approx 0.577$$ km, for an angular deviation of about 0.083 degrees.