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This question and its answers explain the basic theory behind GPS. I particularly like the second answer by @jah138. Using jah138's notation, GPS satellites provide extremely accurate $x_i$, $y_i$, and $z_i$. Meanwhile, $c$ and $\rho_i$ are both known with extreme accuracy. $\Delta t$ is known from the cell phone. With multiple satellites, you end up with a system of equations, and can solve for $x_u$, $y_u$, and $z_u$, make relativistic corrections, blah blah blah, you have your answer.

Where I get held up on is that $\Delta t$ is not known with extreme accuracy since (as I understand, and this is where I think I'm wrong) it comes from my relatively cheap and inaccurate cell phone. GPS obviously works which I believe means it requires a very accurate knowledge of $\Delta t$. How does a cell phone get extremely accurate $\Delta t$ without use of an on board atomic clock?

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  • $\begingroup$ @ForgeMonkey, I think I follow you. Does this mean the cell phone's internally kept time has nothing to do with the GPS calculation? I think I misunderstood the $\rho_i$ term. EDIT: ForgeMonkey's comment disappeared, in brief, it said that the $\Delta t$ term is between satellites and not between a satellite and my phone. $\endgroup$ – chessofnerd Oct 6 '16 at 22:52
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    $\begingroup$ It's mentioned in the cited answer: $\Delta t$ is the fourth unknown quantity in that equation. So, it's not known, it's solved for. $\endgroup$ – Chris Oct 6 '16 at 23:11
  • $\begingroup$ @Chris, you're right. I think I got confused because I always assumed the $\Delta t$ was a computed quantity. So much so that I missed it in the answer. What I think is interesting here is that it gets solved for and then completely ignored because the cell phone only cares about $x_u$, $y_u$, and $z_u$. $\endgroup$ – chessofnerd Oct 7 '16 at 0:17
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    $\begingroup$ This is still worth leaving open (IMO), since there is still the possibility of an interesting expanded answer. $\endgroup$ – Chris Oct 7 '16 at 0:19
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    $\begingroup$ I think this highlights the GPS misconception (which I held until today) that GPS triangulates the location which requires one to know the distance to the satellites before the position can be calculated. That would require my cell phone to have an on board atomic clock. $\endgroup$ – chessofnerd Oct 7 '16 at 0:30
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Actually, the accuracy of a quartz clock is sufficient, since only very short periods are measured by the GPS receiver.

The following is a simple sketch on how GPS works: Three satellites A, B, C send out a signal at the same time, containing their position.

So, the receiver knows the position of the satellites, but since it does not have an atomic clock, it does not know the time of flight, i.e. the distance to each satellite. Instead, it measures the delay between the signals, and so the difference in distance to the satellites.
Mathematically, the receiver must be on a hyperbola, the blue one from the delay between A and B and green from the delay between A and C. (Usually, a hyperbola consist of two curves, but since the signal from A came later than that of B, the receiver must be on the curve closer to B. I didn't draw the other curve.)

enter image description here

Now, a quartz clock isn't as precise as an atomic clock. It is said that it can drift up to 30s per month, i.e. the clock is too slow/fast by a factor of 0.00001. If the delay between the signals of two satellites is about 1 second, this gives an error of 300000km/s*0.00001s = 3km.
But GPS satellites don't send just a single "ping" each second, they continuously send a data stream including time stamp. From the time stamp, a coarse delay between the signals is calculated. Looking at the waveform of the signals, the single bits have a duration of about 1ms, so it is possible to determine the delay between two signals with a precision of better than 1ms.

If the quartz clock has to measure time delays of not more than 1ms, the drift error corresponds to not more than 3 meters!

Since there are many other uncertainties e.g. due to weather, the receiver uses more satellites than necessary to calculate the most probable position. In my sketch, only timing A<->B and A<->C was used, if we use B<->C, too, we get three possible positions, and the truth is somewhere in the middle:

enter image description here

(The three curves would cross on one point if there were no uncertainties)

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  • $\begingroup$ The internal oscillator of the receiver should not play any role in the accuracy: The receiver has a PLL that locks on the signal received and because of that has exactly the correct frequency. They usually don't have the absolute time available, but at least frequency is fixed. Additionally, there are ways to determine the relative phase of two signals without any precise oscillator at all. $\endgroup$ – asdfex Oct 7 '16 at 11:39
  • $\begingroup$ So what it comes down to is that there are sneaky ways to get the exact time delay without a fancy atomic clock in board the receiver? $\endgroup$ – chessofnerd Oct 7 '16 at 14:54
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    $\begingroup$ @chessofnerd: Yes. asdfex is right, there are even other methods than using a quartz clock. But finally, even a quartz clock is accurate enough on small scale for a reasonable GPS accuracy. You don't need an absolute time, just time delays. $\endgroup$ – sweber Oct 7 '16 at 15:41
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    $\begingroup$ You need four satellites: the two dimensional hyperbola you mention is really a three dimensional hyperboloid of revolution $\endgroup$ – DJohnM Oct 8 '16 at 5:47
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As you said yourself, you get a system of equations from which the location can be calculated. Here the Δt consists of the difference between the time of the signal being sent at the satellite and the time of arrival. Now it is important that if we have 2 satellites next to each other in space, the Δt should be the same.

This requires that the 2 clocks in the satellites should be equal, hence the atomic clocks which are synchronized from earth. But the clock in the receiver is in both cases the same, so even if your phone is of by 2 hours, the Δt will still be the same in both cases.

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  • $\begingroup$ Sure. But this misses the crux of my question. How is the $\Delta t$ measured without an extremely accurate clock? $\endgroup$ – chessofnerd Oct 7 '16 at 14:46

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