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I was looking for the equations that are used by the GPS receiver to calculate its position and time correction (bias). By looking at the Navigation Equations on Wikipedia, I don't see any account for special or general relativity in the equations. Where and when does that correction occur?

Hopefully this is not a duplicate of "Why GPS depends on relativity". I understand the why. But what I don't get is why does not the time correction (i.e. bias) that is calculated from the navigation equations already account for the relativistic difference between the clocks?

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    $\begingroup$ Possible duplicate of Why does GPS depend on relativity? $\endgroup$ – pentane Mar 6 '18 at 3:46
  • $\begingroup$ pentane, the question you referred to explains in general the why. My question is a bit more specific. $\endgroup$ – mhmhsh Mar 6 '18 at 4:53
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The correction is just making the clock multiplier in the on-board computer slightly different so that the computer's knowledge of where it is in its orbit is correct.

See also That 10km/day error predicted if GPS satellite clocks not corrected for relativity

ps Rather than retypeset them here, the equations for the SR/GR corrections and their derivations are described very well by Neil Ashby of U Colorado

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  • $\begingroup$ Thanks Martin! From Nail's paper: " In the older satellites these terms were compensated by setting the atomic clock frequencies down by this amount before launch–the so-called “factory frequency offset.” Atomic clocks that have been recently launched are based on Rubidium atoms. These clock frequencies may be bumped during launch so they are measured after orbit insertion and the necessary frequency corrections are transmitted to the receivers in the navigation message." ---> if the correction is transmitted to the receiver, why cannot the bias parameter already account for that? $\endgroup$ – mhmhsh Mar 6 '18 at 4:29
  • $\begingroup$ Perhaps the new clocks are so accurate that it is difficult to calculate the shift precisely enough? It would depend on the precise gravitational field at that altitude and the exact speed. $\endgroup$ – Martin Beckett Mar 7 '18 at 2:56

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