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by 21 2830 days ago
Intuitively this seems wrong to me. If the satellite is overhead, the error would put you 300m into the ground so to speak. I'm not sure why you project that horizontally, and especially why you take the distance to the satellite into account.

As another sanity check, if the error for 1 us is 110 km, the error for 1 ns would be 110 m, and I suspect 1 ns error is not unusual for consumer electronics:

> To reduce this error level to the order of meters would require an atomic clock. However, not only is this impracticable for consumer GPS devices, the GPS satellites are only accurate to about 10 nano seconds (in which time a signal would travel 3m)

https://wiki.openstreetmap.org/wiki/Accuracy_of_GPS_data

2 comments

> If the satellite is overhead, the error would put you 300m into the ground so to speak.

Right, I was basically calculating where that signal would just be reaching the surface at the same time it was 300m under you. This is a circle around you with a radius of ~110km (again using the approximation of the ground as a flat plane). Thinking about it more, there's not much reason to do this (GPS isn't really tied to the surface of the Earth, it gives you 3-D coordinates). I guess my point was that the 300m of distance from 1us of light propagation should not be thought of as a horizontal distance.

That would be if it were straight overhead, intersecting tangentially with another sphere. Realistically they're not overhead, but if two satellites are 30 degrees apart, the line of intersection between their spheres will move twice the distance one of the spheres moves. The magnifying factor is 1/sin(angle between the satellites from the observer).