[time-nuts] 1PPS accuracy of commercial GPS receivers
Hal Murray
hmurray at megapathdsl.net
Thu May 14 18:50:27 UTC 2009
> 2D positioning requires at least 3 sats for resolving Lat, Long, T
> (really X, Y, Z and T which a fixed relationship between X, Y and Z so
> given two the third will be given, as the heigth is assumed).
This has been discussed before, but I still don't really understand it.
I assume they take the data and solve what they can. The answer will be a
line in X,Y, Z and T. Do they just pick the point on Z=0? How much timing
error does that turn into?
That is, if I move up 100 meters along that line, how much does T change?
I assume the answer is "not much", but I don't have a good feel for the
numbers. Is it lost in the other sources of noise? Or at least not big
relative to them?
At 1ft per ns, 100 meters is 300 ns. But lots of places on Earth are much
higher elevation than 100 meters. (I just used 100 meters as an example.)
But that's the number for a satellite directly overhead. The geometry fudges
things. If the satellites are low, the change in time will be close to zero.
What's the average angle of a satellite? (or ones used in a GPS solution
when you can only get 3 of them?)
There is another layer of fudging involved in the how-many-satellites
discussion. Sometimes the geometry is degenerate. If the satellites are in
(or close) to a line, you don't get a position offset to the side of the
projection of that line on the surface of the Earth.
--
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