[zwegner]

> A 1 microsecond error is 300 meters.

While the speed-of-light propagation is about 300 meters in a microsecond, isn't the final position error possibly much greater? For calculating position on Earth, you can think about a sphere expanding at the speed of light from each satellite. The 1 microsecond error here corresponds to a radius 300m bigger or smaller, which only corresponds to 300m horizontal distance on the ground if the satellite is on the horizon (assuming that Earth is locally a flat plane for simplicity here). For a satellite directly overhead, the 300m error is a vertical distance. Calculating the difference in horizontal position from this error is then finding the length of a leg of a right triangle with other leg length D and hypotenuse length D+300m, where D is the orbital distance from the satellite (according to Wikipedia, 20180km). The final horizontal distance error is then sqrt((D+300)^2 - D^2), or about 110km.

Of course, this is just the effect of a 1us error in a single satellite, I'm sure there's ways to detect and compensate for these errors.

[21]

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

[zwegner]

> 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.

[SamReidHughes]

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).