How does GPS correct for time dilation?

[PeterDonis]

terror is the clock rate mismatch between the GPS satellite and the receiver

No, it isn't; it is the amount that the receiver's clock needs to be changed by to be in sync with the GPS satellite's clock. It has nothing whatsoever to do with the correction of the GPS satellite's clock rate relative to the master clocks on the ground that are the standards for GPS time, which is the subject under discussion.

 

[PeterDonis]

Four measurements assumes that you are on the surface, which reduces the position to two unknowns, doesn't it?

If you are exactly at sea level, yes. But if you're not, even if the GPS receiver is only telling you latitude and longitude, it can't obtain a correct solution for those two unknowns without also solving for altitude, which is the third unknown for position. If it assumes it is at sea level when it isn't, it will give incorrect latitude and longitude.

 

[doaaron]

No, it isn't; it is the amount that the receiver's clock needs to be changed by to be in sync with the GPS satellite's clock.

that's exactly what I said ( terror is the clock rate mismatch between the GPS satellite and the receiver). In my sentence, the GPS receiver would be your Tom Tom or whatever, and the GPS satellite would be the time at the satellite. The final equation does not need the correct ground reference time to calculate position. It only needs the correct ground reference time if your Tom Tom also wants to display accurate time.

Based on 38 us error accumulated over 1 day, and a GPS speed of 3900m/s (from google), the location error of any particular satellite would be about 15cm/day. And the time error sent by the GPS satellite would be 38us/day.

A fifth measurement, in general, will resolve the remaining ambiguity

this is my understanding too.thanks,
Aaron

 

[A.T.]

the basic reason you need four signals is that you have four unknowns (three position coordinates plus time), so you need four measurements to determine them all.

Four measurements assumes that you are on the surface, which reduces the position to two unknowns, doesn't it?

If you are exactly at sea level, yes.

But then your general rule, equating the number of unknowns to the number of required measurements doesn't work, does it? It rather seems that you need one measurement more than unknowns.

 

[doaaron]

But then your general rule, equating the number of unknowns to the number of required measurements doesn't work, does it? It rather seems that you need one measurement more than unknowns.

I noticed that too, but I guess it only works on linear equations. For example, y = x². Given y, solve for x. There are two solutions...Aaron

 

[mfb]

Not if the GPS satellite clock runs fast for a whole day. That gives an accumulated time difference of $4 \times 10^{-10}$ times $86,400$ seconds, or about 38 microseconds. That corresponds to an error in position of about 10 km, as I said in my previous post.

Only if you rely on a clock in your receiver. Typical distance differences between two satellites are of the order of 10000km. If both satellite clocks are wrong in the same way, then the difference is only $4 \times 10^{-10}$ times the distance difference, a few millimeters. The incorrect data about the satellite position (because keeping track of the orbit will be wrong) is more relevant.

I noticed that too, but I guess it only works on linear equations. For example, y = x2. Given y, solve for x. There are two solutions...

That is exactly the point. A second equation like z = -x^2 + 2 x allows to get a unique value of x if you know y and z.

 

[PeterDonis]

that's exactly what I said ( terror is the clock rate mismatch between the GPS satellite and the receiver)

38 microseconds per day is not the clock rate mismatch; that's $4 \times 10^{-10}$, which is a pure number. The 38 microseconds per day is the accumulated difference in time over 86,400 seconds, i.e., it's the clock rate mismatch (a pure number) multiplied by 86,400 (seconds) to get an accumulated time difference (in seconds) between the two clock readings.

Based on 38 us error accumulated over 1 day, and a GPS speed of 3900m/s (from google

No, that's not how you calculate the location error; the speed of the GPS satellite is not even a factor in that calculation. The location error is caused by the error in the time stamp sent by the GPS satellite in the signal that is picked up by the receiver. That time stamp is what is used by the receiver to calculate its own position; the position error is the time stamp error, 38 microseconds, times the speed of light, $3 \times 10^{8}$ meters per second, or about 10 km.

 

[PeterDonis]

But then your general rule, equating the number of unknowns to the number of required measurements doesn't work, does it?

I don't understand. If you are exactly at sea level, then your altitude is known, so the number of position unknowns is two, not three. That means one less measurement is required (3 instead of 4, since you now have three unknowns--2 position + time--instead of four). But just "being on the surface of the Earth" does not mean you're exactly at sea level; you could be at any altitude within a range of several km at least. So your altitude is not known, and must be solved for.

 

[doaaron]

38 microseconds per day is not the clock rate mismatch; that's 4×10−10, which is a pure number. The 38 microseconds per day is the accumulated difference in time over 86,400 seconds, i.e., it's the clock rate mismatch (a pure number) multiplied by 86,400 (seconds) to get an accumulated time difference (in seconds) between the two clock readings

The equation seems quite clear to me. Assume that both the GPS receiver and the GPS satellite are running at the same rate, then

dn = c*(tGPSn - tRXn) = c*(Δt)

where Δt is the time it takes for the signal to travel from the GPS satellite to the GPS receiver. In this equation there is no error, and the distance is calculated precisely. Now assume an error of 38us per day accumulates, so that after one day,

dn = c*(tGPSn - tRXn + 38us) = c*(Δt + 38us)

So you are claiming that the distance to this satellite will be off by 38us*c. In this case, 38us is t_error. What I am claiming is that the addition of a fourth satellite allows me to calculate the value "38us" an de-embed it. i.e. the 10km error which you keep referring to is a myth.

No, that's not how you calculate the location error; the speed of the GPS satellite is not even a factor in that calculation. The location error is caused by the error in the time stamp sent by the GPS satellite in the signal that is picked up by the receiver.

As others have pointed out, the more important source of error comes from the right hand side of the original equation,

dn = √{(x₁ - x)² + (y₁ - y)² + (z₁ - z)²}

If (x₁, y₁, z₁) is precisely known, then we can calculate dn precisely. However, if we assume the 38us per day timing error, then the GPS satellite will calculate its own position (x₁, y₁, z₁) wrongly based on its orbit by 38us * v which is much less than 38us * c. Its only 15cm/day.

The equations are very simple, so if you think I am wrong, please show me where in these equations I have gone wrong. regards,
Aaron

 

[PeterDonis]

The equation seems quite clear to me.

Rather than continue to try to explain this myself, I'll point you to this excellent reference:

http://relativity.livingreviews.org/Articles/lrr-2003-1/

Note the statement in the introduction: "Timing errors of one ns will lead to positioning errors of the order of 30 cm." A timing error of 38 us = 38,000 ns would therefore lead to a positioning error of 38,000 * 30 cm, or approximately 10 km. Reading through the reference in detail will show you why this is true. What I've been saying is consistent with what's there. What you have been saying is not: the calculations you are making are not the ones that actually determine the position error due to a 38 us accumulated time difference.

 

[PeterDonis]

If both satellite clocks are wrong in the same way, then the difference is only $4 \times 10^{-10}$ times the distance difference, a few millimeters.

This is not the error I'm talking about.

The incorrect data about the satellite position (because keeping track of the orbit will be wrong) is more relevant.

If by this you mean that the time at which the satellite thinks it is at a given position is wrong, yes, that's the error I'm talking about. Equation (1) in the introduction of the reference I linked to shows that clearly: any error in $t_j$ (the time that satellite $j$ thinks it is at position $r_j$) is multiplied by $c$ in the solution of the 4 simultaneous equations to get the coordinates of the reception event.

 

[A.T.]

I don't understand. If you are exactly at sea level, then your altitude is known, so the number of position unknowns is two, not three. That means one less measurement is required (3 instead of 4, since you now have three unknowns--2 position + time--instead of four).

See this explanation by jbriggs444, which suggests that you do need 4 satellites, if at sea level with no clock, and 5 satelites if at any postion with no clock:

If I remember correctly, the additional satellite is needed is to disambiguate between multiple possible solutions. Take the simple case of two satellites and a receiver that already has a properly synchronized time source. With a signal from one satellite you can solve for distance from the satellite. The solution set is a spherical shell. With two satellites you can solve for distance from both. The solution set is a circle. With the additional constraint that you are on the surface of the earth, the solution set is reduced to two points. So you need a third satellite to figure out which is the correct point.

If you receiver has no clock source, you need an additional satellite for that. If you are not figuring for a receiver tied for the surface of the Earth you need a satellite for that. Total of five satellites.

Where is the error in jbriggs444 explanation? Is it that one of the two possible solution points is always inside the Earth (for most practical altitudes)?

 

[PeterDonis]

Is it that one of the two possible solution points is always inside the Earth (for most practical altitudes)?

Yes. It is true that there are multiple discrete points that solve the simultaneous equations; they don't have a unique solution. But for an earth-bound receiver, all but one of the solutions will be either deep inside the Earth or too far above the surface to be possible. (I think mfb mentioned this earlier in the thread as well.)

 

[A.T.]

Yes. It is true that there are multiple discrete points that solve the simultaneous equations; they don't have a unique solution. But for an earth-bound receiver, all but one of the solutions will be either deep inside the Earth or too far above the surface to be possible. (I think mfb mentioned this earlier in the thread as well.)

Okay, so we have to use that extra constraint to have #measruments = #variables. It's not a general rule for this kind of system.

 

[jbriggs444]

See this explanation by jbriggs444, which suggests that you do need 4 satellites, if at sea level with no clock, and 5 satelites if at any postion with no clock:

Where is the error in jbriggs444 explanation? Is it that one of the two possible solution points is always inside the Earth (for most practical altitudes)?

The explanation I gave explicitly assumes (contrary to fact) that the receiver has a correctly synchronized clock. In this case, measurements from two satellites immediately give you the distance to those satellites. Two measurements constrain your position to a circle. That circle will intersect with the surface of the Earth at two points. Neither point will be sub-surface. By construction, they are both exactly at the surface.

The error in the explanation is, I believe, in the counter-factual assumption that the receivers have a clock. It was intended not so much as a correct explanation as an easy to follow hand-wave.

 

[A.T.]

The explanation I gave explicitly assumes (contrary to fact) that the receiver has a correctly synchronized clock. In this case, measurements from two satellites immediately give you the distance to those satellites. Two measurements constrain your position to a circle. That circle will intersect with the surface of the Earth at two points. Neither point will be sub-surface. By construction, they are both exactly at the surface.

So you need a 3rd satellite to choose one of the two points, while the number of variables is 2 (position on the surface). If you don't have a clock you need a 4th satellite. And if you are not on the surface a 5th satellite.

Do you agree @PeterDonis ?

 

[PeterDonis]

So you need a 3rd satellite to choose one of the two points

You do if you have no other knowledge at all about your position, so you have no basis on which to choose one of the muliple discrete points that are possible solutions. But if you have enough other knowledge about your (approximate) position (for example, that you are within, say, 10 km of the geoid), you may be able to rule out all but one of the possible solutions on that basis.

In practice, I think what a given GPS receiver will actually do would depend on the receiver. The system is designed so that at least four satellites are always in view, but there may not be five, so I would expect a receiver to have some mechanism for applying prior knowledge of approximate position as I described above. But if five satellites happened to be in view, I think some receivers might use the additional signal as a further check, while others might not care (the latter would probably be the cheaper models).

 

[doaaron]

Rather than continue to try to explain this myself, I'll point you to this excellent reference:

http://relativity.livingreviews.org/Articles/lrr-2003-1/

Note the statement in the introduction: "Timing errors of one ns will lead to positioning errors of the order of 30 cm." A timing error of 38 us = 38,000 ns would therefore lead to a positioning error of 38,000 * 30 cm, or approximately 10 km. Reading through the reference in detail will show you why this is true. What I've been saying is consistent with what's there. What you have been saying is not: the equations you are using are not the ones that determine the position error due to a 38 us accumulated time difference.

I read that part of the document. The equation he uses is essentially the same as the one I used.

c²(tj - t)² = (r - rj)²

He is saying that with four equations he can solve for t, and r. And then he says that timing errors of 1 ns result in 30cm of error. This is obviously just c*t_error. I am going to give Neil Ashby the benefit of the doubt here and suggest that by timing errors he is referring to mismatch between the satellites or circuit delays, or something to that effect. He doesn't specify that he is talking about the absolute clock rates of the GPS satellites. Furthermore, any time, t, calculated from the above equations must be relative to tj, and since all tj are in sync, the timing error will not be 38us per day.Aaron

 

[Fantasist]

Not if the GPS satellite clock runs fast for a whole day. That gives an accumulated time difference of $4 \times 10^{-10}$ times $86,400$ seconds, or about 38 microseconds. That corresponds to an error in position of about 10 km, as I said in my previous post.

This only holds if you compare the (off-frequency) satellite clock to the ground clock. But this is not how GPS works in practice, which is by taking differences between different satellite time signals, and if those are all subject to the same time offset, there won't be any accumulation of the error.

This is easy to see by considering a one-dimensional example: if you have s straight line with transmitters at both ends, the time signals of both will always meet at the same point if the signals from both are emitted at the same rate, regardless of what that rate is. You still will be able to accurately determine your relative position on this line, it is only that the absolute values are contracted or expanded. The position offset from the meeting-(zero-) point is

x1-x2 = c*(1+δ)*(t1-t2)

where δ is the common relative tick rate of both clocks. So if δ=4*10^-10 , then the position also has a relative offset of 4*10^-10. In this case this would only amount to a constant offset of about 1 cm (assuming x1-x2 = 20000 km for δ=0 ).

On the other hand, if the clock rates of the two clocks are different, it is obvious that the meeting (zero) point will drift off to one side (with a speed δ*c if δ=0 for one of the clocks). But as I said, this is not the case we are considering here, as the same clock rate offset δ is shared by all satellites.

 

[PeterDonis]

this is not how GPS works in practice, which is by taking differences between different satellite time signals, and if those are all subject to the same time offset, there won't be any accumulation of the error.

The bolded phrase is critical, and I think it is where the issue is. Under the actual regime we have, where GPS satellite clocks have effective clock rates the same as ground clocks, to within some small error, we can assume that all satellite clocks have the same offset, again to within some small error. (Actually, what we are assuming is that our receiver's clock bias is the same compared to all the satellite clocks, but it amounts to the same thing.)

But suppose the GPS satellite clock rates were uncorrected (i.e., they all gained 38 us/day on ground clocks), and they got re-synchronized once a day. And suppose the re-synchronizations didn't all happen at the same time. Then you could be receiving signals from different satellites with up to 38 us difference in their offsets (in the worst case), because some had just been updated and some had gone a day without being updated yet.

This scenario may be what Ashby's statement was referring to; I'm not sure.

 

[A.T.]

So you need a 3rd satellite to choose one of the two points,

You do if you have no other knowledge at all about your position,

No, the knowledge about being at sea levels is already included there. If you do have a clock, two satellites will give you a solution circle, which intersects the surface at two points. So you need a 3rd satellite to choose one of the two surface points. That's 3 satellites for 2 variables (position on surface).

 

[Fantasist]

Only if you rely on a clock in your receiver. Typical distance differences between two satellites are of the order of 10000km. If both satellite clocks are wrong in the same way, then the difference is only $4 \times 10^{-10}$ times the distance difference, a few millimeters. The incorrect data about the satellite position (because keeping track of the orbit will be wrong) is more relevant..

They are potentially more relevant, but they are still insignificant here: in 38 μs (the daily relativistic clock drift) a satellite moves by merely about 15 cm. In fact, as stated in the GPS documentation I referenced in post #26 already, the satellite clocks are allowed to drift up to 1 ms relatively to the ground clock before they are corrected. So even without a correction for the overall relativistic clock drift, the satellite positions would be off by only about 5 m in one month (before the clocks are corrected).

 

[Fantasist]

But suppose the GPS satellite clock rates were uncorrected (i.e., they all gained 38 us/day on ground clocks), and they got re-synchronized once a day. And suppose the re-synchronizations didn't all happen at the same time. Then you could be receiving signals from different satellites with up to 38 us difference in their offsets (in the worst case), because some had just been updated and some had gone a day without being updated yet.
This scenario may be what Ashby's statement was referring to; I'm not sure.

My understanding from the GPS documentation is that no two satellite clocks differ by more than 20 ns at any time.

 

[doaaron]

Then you could be receiving signals from different satellites with up to 38 us difference in their offsets (in the worst case), because some had just been updated and some had gone a day without being updated yet.

This makes some sense. But you're suggesting that the GPS satellites are out of sync with each other which no reference I've seen makes any mention of...anyway the document referenced by Fantasist looks quite legitimate and does say that the satellites are in sync to within 20ns Aaron

 

[PeterDonis]

If you do have a clock, two satellites will give you a solution circle, which intersects the surface at two points. So you need a 3rd satellite to choose one of the two surface points.

Once again: only if you have no other prior knowledge about your position. But suppose the two surface points are 1000 km apart, and you know your approximate position to within 100 km. Then you don't need a third satellite to pick which surface point is the correct one; only one of them will be within the 100 km-wide approximate area you already know.

The key difference here is that, if you only have one satellite (in this idealized case), you have a continuous range of possible positions. A second satellite reduces the possibilities to a discrete set. Approximate knowledge of your position does not pick out a particular point from a continuous range--if you know your approximate position to within 100 km, knowing that it's also somewhere on a circle (the intersection of one satellite sphere with the surface of the Earth) does not pick out a single point; it only picks out the arc of the circle which is within the 100 km approximate range. But approximate knowledge can pick out a unique point from a discrete set: adding the second satellite, by eliminating all but two of the points on the whole circle that the first satellite gave you, eliminates all but one of the points on that circular arc (the other possible point is outside the 100-km area so it's not a possible solution given your prior knowledge).

 

[PeterDonis]

you're suggesting that the GPS satellites are out of sync with each other

I'm suggesting no such thing. I'm saying that in the hypothetical situation where the GPS satellite clock rates were not corrected, and they were only re-synchronized with ground clocks once a day, there could be as much as 38 us difference between the offset of the clocks in two different satellites (more precisely, the bias of a given receiver's clock relative to the clocks on two different satellites could differ by as much as 38 us).

Of course this hypothetical situation is not the actual situation, because on the actual GPS satellites the clock rates are corrected to be the same as the rate of clocks on the geoid. (I said this explicitly in post #50.) But thinking about what would have happened in that hypothetical situation may help to explain why physicists say that the correction of the clock rates of the GPS satellites is in fact a critical feature in making sure position fixes are accurate.

 

[doaaron]

I'm suggesting no such thing.

What I meant was that you are suggesting that the GPS satellites would be out of sync with each other without the offset. So far, I haven't seen any reference mentioning that the GPS satellites would be out of sync with each other, they all seem to talk about how they would be out of sync with the receiver clock.

Anyway I think you have a good point. I'll do a bit more reading to see if I can confirm your intuition.

BTW, just as an aside, I think that when the GPS clocks are periodically calibrated to a ground station, they should be able to calibrate not just their offset, but their rate too.regards,
Aaron

 

[mfb]

I'm suggesting no such thing. I'm saying that in the hypothetical situation where the GPS satellite clock rates were not corrected, and they were only re-synchronized with ground clocks once a day, there could be as much as 38 us difference between the offset of the clocks in two different satellites (more precisely, the bias of a given receiver's clock relative to the clocks on two different satellites could differ by as much as 38 us).

Of course this hypothetical situation is not the actual situation, because on the actual GPS satellites the clock rates are corrected to be the same as the rate of clocks on the geoid. (I said this explicitly in post #50.) But thinking about what would have happened in that hypothetical situation may help to explain why physicists say that the correction of the clock rates of the GPS satellites is in fact a critical feature in making sure position fixes are accurate.

Even in this hypothetical situation (it was not clear to me that you were considering this), you could avoid these errors - you could correct for it in satelitte software, or at least include data about the last synchronization event (so the receiver can correct for it).

 

[Chronos]

I agree with Peter: GPS clocks are calibrated to Earth based clocks because they are the clocks most relevant to a location on earth. If you were intent on locating a position on the moon, a moon based clock would be best. The target is distinct from the gun.

 

[PeterDonis]

I think that when the GPS clocks are periodically calibrated to a ground station, they should be able to calibrate not just their offset, but their rate too.

I'm not sure that's possible with the current satellite design; AFAIK the clock rate correction is fixed when the satellite is built. However, I don't see any reason in principle why you couldn't design a satellite clock whose rate was adjustable.