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Old 21st December 2015 | 00:56
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A Squared
 
Joined: Feb 2000
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From: Alaska, PNG, etc.
Originally Posted by D driver
you do not generate the pseudo ranges after setting up the equations, in fact its the other way round. 4 linear simultaneous equations each with 4 unknown quantities i.e. X,Y,Z coordinates and T (time) are solved by iteration to give the final 3D fix and accurate time reference.
Driver, Oktas' statment is correct. To wit:

Originally Posted by Oktas8
With a current orbital equation (almanac) and an accurate time, you can generate a pseudo-range between the receiver and the satellite.
As stated, there is nothing incorrect there. Now as a practical matter, accurate time is usually not a known, and therefore is solved for in the system of equations as you pointed out.

At any rate, you have your answer, whether you like it or not. D is incorrect because the satellites do not broadcast their *position*, they broadcast a set of orbital parameters which allow the position at any time to be calculated. That isn't the same thing.

Incidentally, a satellite does not have an ecliptic. There is only one Ecliptic and it belongs to the sun. Satellites have orbits, which are elliptical, but that isn't an ecliptic. "Ecliptic" is not a word for "elliptical orbit"
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