Hal Murray skrev:
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?
OK, X, Y and Z is not arbitrary axes, they have origo at the earth mass
center, X sticks out through 0 E 0 N (0 meridian at equator), Y sticks
out through 90 E 0 N and Z sticks out through the north pole. T is in
GPS time (GPS week, Z-count, data bit in frame, C/A cycle in bit, C/A
phase and carrier phase). The sats position in X_i, Y_i, Z_i and T_i is
also known in this coordinate system as transmitted and calculated.
The actual pseudo-range to sat i is
p_i = sqrt((X-X_i)^2 + (Y-Y_i)^2 + (Z-Z_i)^2))
but this pseudo-range is skewed by c(T-T_i) where c is the speed of
light. Additional time-skew components is found from ionosphere and
troposphere among others.
The receivers time T needs to be in the neighborhood of correct, but as
soon as the first sat is being tracked, just taking the time of that sat
brings it within 100 ms at all times and the first 4-sat solution will
remove the major part of that.
That is, if I move up 100 meters along that line, how much does T change?
Notice how the above equations is per sat, but speed of light is the
conversion factor you need and it is "line of sigth" (not entierly true
as the signal goes through dispersive athmosphere) to each sat which
decides the time-skew.
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?
1 ns is 30 cm. 10 ns is 3 m. 100 ns is 30 m. 333 ns is 100 m. However,
your errors goes in different directions depending on sat, so this is
why a good constellation helps to accuratly correct errors in all
directions.
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?)
One rarely speaks of average angle, and it is highly dependent on
position on earth, so it is kind of not interesting.
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.
You can rule out those which gives too high deviation in T for the
approximated X, Y, Z and T and cancel those contributions out before
making another calculation.
Cheers,
Magnus
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