How does GPS correct for time dilation?

[doaaron]

Hi all,as I understand it, GPS satellites are offset prior to launch in order to correct for the time dilation that they experience in orbit. I am very confused about how this works in practice. I think the velocity of a GPS satellite wrt a position on Earth would depend on the position on Earth where the GPS data is to be analyzed.

I understand that all GPS satellites have the same orbital inclination, so wrt a fixed position on earth, the average velocity over time of the GPS satellites is the same for all of the satellites. So I can safely ignore the effect of the orbit on the amount of time dilation.

But what about the position on Earth? For example, the velocity of an observer at the north pole is very different from the velocity of an observer at the equator. I believe that the relative velocities of GPS satellites wrt the two positions would also be different. So how would this affect the accuracy of GPS?thanks,
Aaron

 

[mfb]

I think the velocity of a GPS satellite wrt a position on Earth would depend on the position on Earth where the GPS data is to be analyzed.

That is true due to the rotation of earth, but it does not influence the relative synchronization of the satellites with respect to a fixed observer (resting at the poles, or some arbitrary other point in space). The motion of the receiver does not matter as (typical) receivers do not use a clock on their own, and "what happens at the same time at the same place?" is independent of the reference frame.

 

[PeterDonis]

the velocity of an observer at the north pole is very different from the velocity of an observer at the equator.

Relative to an inertial frame centered on the Earth, yes. But the observer at the equator is also further from the center of the Earth. The two effects on time dilation (slower due to speed, faster due to higher altitude) cancel. An observer at rest at sea level anywhere on the rotating Earth has the same clock rate. This clock rate is the reference clock rate that the GPS satellite clocks are adjusted to.

 

[doaaron]

Thanks for the replies.

The motion of the receiver does not matter as (typical) receivers do not use a clock on their own, and "what happens at the same time at the same place?" is independent of the reference frame.

I will need to think on this. I understand that GPS receivers do not use atomic clocks, but I don't know that that is important. I think that the clock signal that the reference frame receives from the GPS satellite will still be different depending on where he is on Earth due to the difference in relative velocities.

An observer at rest at sea level anywhere on the rotating Earth has the same clock rate. This clock rate is the reference clock rate that the GPS satellite clocks are adjusted to.

I'm also not quite convinced by this argument. I have briefly read of the argument you are making before, but I think in this case, the situation is a little different. Just taking a hypothetical example, suppose we have 3 clocks, and A is taken as the rest frame, B is moving in the -x direction with speed v, and C is moving in the +x direction with speed v. Relative to A, both B and C would have the same amount of time dilation and A would say that B and C show the same time. However, B thinks that C's time is different from his and vice versa. My point is that even if two clocks appear to be synchronized (the one at the equator and the one at the pole), it doesn't mean that a third clock moving in a different frame will show the same time wrt those two clocks.

Please feel free to correct me if I'm wrong.best regards,
Aaron

 

[PeterDonis]

I think that the clock signal that the reference frame receives from the GPS satellite will still be different depending on where he is on Earth due to the difference in relative velocities.

The frequency of the radio signal received from the GPS satellite by the Earth receiver will be different, yes. (Note that this frequency is not just affected by relative speed--it's also affected by the passage of the signal through the Earth's ionosphere, which contains charged particles that do various things to radio signals passing through.) But that frequency is not the "clock signal". The "clock signal" is a digital message encoded in the radio signal; it is the same message regardless of the frequency of the signal. The only thing the frequency change affects is what bandwidth the antenna has to have in order to be sure of receiving the signal.

I'm also not quite convinced by this argument.

Then I strongly suggest that you actually look at the math instead of trying to make arguments based on your intuitive understanding.

Just taking a hypothetical example, suppose we have 3 clocks, and A is taken as the rest frame, B is moving in the -x direction with speed v, and C is moving in the +x direction with speed v.

Speed is only one thing that affects time dilation in the gravitational field of a static, massive body like the Earth; the other is altitude. You need to take that into account.

Also, you need to be careful to distinguish "clock rate" and "clock synchronization". They are not necessarily the same. See below.

My point is that even if two clocks appear to be synchronized (the one at the equator and the one at the pole), it doesn't mean that a third clock moving in a different frame will show the same time wrt those two clocks.

This is true in general; but the specific application I think you intend here is not correct.

Consider three clocks: clock A at the North Pole, clock B at the equator at 0 degrees longitude, and clock C at the equator at 180 degrees longitude. Relative to an inertial frame centered on the Earth, these three clocks all have different speeds: clock A is at rest, clock B is moving at about +450 m/s (call this speed $v$), and clock C is moving at about -450 m/s (i.e., $-v$). Also, clock A is at altitude $R$ (the polar radius of the Earth), while clocks B and C are at altitudes $R + h$ (where $h$ is the height of the Earth's equatorial bulge).

Let me first give the correct GR formula for the rates of these clocks. In an inertial frame centered on the Earth (i.e., not rotating), the rate of a clock at altitude $r$ (distance from the center of the Earth) and with speed $v$ is

$$
\frac{d\tau}{dt} = \sqrt{ 1 - \frac{2 G M}{c^2 r} - \frac{v^2}{c^2}}
$$

where a clock rate of $1$ corresponds to an observer at rest at infinity (i.e., very far away from the Earth, so the effect of the Earth's gravity is negligible). If you plug in values for the three clocks, A, B, and C, you will see that they all give the same answer for $d\tau / dt$ to a good approximation--the effect of the increase in $r$ for clocks B and C is compensated for by the effect of the nonzero $v$. (A better approximation would apply a correction for the Earth's quadrupole moment, but we won't go into that here.) So all three clocks have the same clock rate, relative to an inertial frame centered on the Earth.

The GPS system actually applies a correction to the Earth-centered inertial frame I described just now. The correction is to rescale all clock rates so that a clock rate of $1$ corresponds to the clock rate of an observer at rest on Earth at sea level. That is, clock rates are rescaled so that clocks A, B, and C all have clock rates of $1$, and thus the clock rate of an observer at rest at infinity is now greater than $1$--it is the reciprocal of the (less than $1$) rate for clocks A, B, and C calculated from the above formula. So the "GPS clock rate" of a given clock is $d\tau / dt$ for that clock, calculated from the above formula, divided by $d\tau / dt$ for the reference clock (clock A, B, or C--they're all the same). The clocks on all the GPS satellites have their rates adjusted so that they effectively tick at the same rate as the reference clock; i.e., their adjusted GPS clock rates are all $1$.

However, having the same clock rate is not sufficient for clock synchronization. Consider clocks A, B, and C again. Each of these clocks is at rest in a different inertial frame (and for clocks B and C, which inertial frame they are at rest in changes as the Earth rotates). So even if they are all running at the same rate, they don't all have the same "natural" simultaneity convention--that is, the inertial frames in which they are at rest do not agree on which events happen "at the same time". So in order to fully synchronize these clocks, we have to agree on a single simultaneity convention that they all will share (even if it is not the "natural" simultaneity convention for the inertial frame in which they are at rest). The convention that used for GPS is that of the Earth-centered inertial frame (which is the same as the "natural" simultaneity convention of clock A). All GPS clocks have their definition of what events happen "at the same time" (which basically means what time counts as "time zero" for the clock) adjusted to be the same as that of the reference clock.

So the sense in which the GPS clocks are "synchronized" is, if you want to look at it that way, "artificial"--both the rates and the simultaneity conventions of GPS clocks are changed from their "natural" ones, in order to impose a common convention on all of them. There is nothing wrong with this: it has to be done as a practical matter in order for GPS to work. But it does mean that you can't just apply your intuitions about the "natural" rates of clocks in various states of motion (or even at various altitudes) if you want to understand how timing in the GPS system works.

 

[Filip Larsen]

The two effects on time dilation (slower due to speed, faster due to higher altitude) cancel.

I think it would be more accurate to say that the two relativistic effects are opposite, but do not exactly cancel for GPS satellites. According to [1], the L1 frequency is reduced by 4.57 mHz (milli Hertz) in the space segment so that the nominal frequency of L1 at sea-level is 10.23 MHz.

Edit: just missed Peters own post where he elaborates on this.

[1] GPS, Theory and Practice, Hofmann-Wllenhof et al., Springer 1997.

 

[PeterDonis]

I think it would be more accurate to say that the two relativistic effects are opposite, but do not exactly cancel for GPS satellites.

Just to clarify, the cancellation I was referring to is for clocks at sea level on the rotating Earth.

 

[doaaron]

First off, thanks for taking the time to reply.

The frequency of the radio signal received from the GPS satellite by the Earth receiver will be different, yes. (Note that this frequency is not just affected by relative speed--it's also affected by the passage of the signal through the Earth's ionosphere, which contains charged particles that do various things to radio signals passing through.) But that frequency is not the "clock signal". The "clock signal" is a digital message encoded in the radio signal; it is the same message regardless of the frequency of the signal. The only thing the frequency change affects is what bandwidth the antenna has to have in order to be sure of receiving the signal.

I wasn't referring to frequency...

Then I strongly suggest that you actually look at the math instead of trying to make arguments based on your intuitive understanding.

Unfortunately, I think the math involved would be far too cumbersome. I think that reasoning out the problem is a more efficient approach in this case. Or maybe I'm wrong since you seem to have done the math :P

This is true in general; but the specific application I think you intend here is not correct...

I'll need some more time to digest the rest of your post. It could take some time, so thanks for now.Aaron

 

[PeterDonis]

I think the math involved would be far too cumbersome.

Not really; I wrote down the key equation in my last post. You don't need anything more complicated than algebra.

(Of course, to derive the equation I wrote down from first principles, you need a lot more, yes; but for this discussion we can just assume it, we don't need to derive it.)

 

[mfb]

Maybe that approach helps:
The satellites know their own position and the time of the observer N at the north pole (both as seen in the frame of N) with a high precision. They send it to Earth.

Observer E at the equator has a setup with two GPS devices: one is at rest for him (so it rotates with earth), one zips past with an arbitrary velocity. Exactly at the moment they meet each other, both check the satellite signals. They will get the same signals because they measure at the same point in spacetime. The velocity does not matter - it changes the frequency of the signals, but not the synchronization of them.

 

[doaaron]

Hi PeterDonis,I've been mulling over your reply. However, it still seems that you are simply taking the clock rate at different points (e.g. north pole, equator, GPS satellite), and showing that the difference in clock rates (tGPS - tNorthPole) = (tGPS - tEquator). Have I misunderstood you?

I'm not sure that this approach is valid as (tGPS wrt tEquator) ≠ (tGPS - tEquator). You need to calculate the GPS satellite speed wrt your position on Earth. This is where the mathematical difficulty comes in (for me anyway).

Anyway I think I understand the solution based on the example below. Hi mfb,

Observer E at the equator has a setup with two GPS devices: one is at rest for him (so it rotates with earth), one zips past with an arbitrary velocity.

This is a good example to work with since we don't need to consider the odd shape of the Earth and its effect on clock rates. I think I finally understand it based on a modified version of this example. I explained it below for anybody interested...

So to make it even simpler, let's assume that we have five GPS satellites in view and their positions are fixed. Device A is at a fixed point on Earth which we can call the rest frame. Device B is whizzing around the Earth at velocity v. Since we have assumed that the GPS satellites are fixed in position, they are also in the rest frame. This is equivalent to the argument that the GPS satellites are calibrated before launch to tick at the same rate as a pre-defined rest frame on Earth.

At t = 0, let's say that device B passes device A. Now let's say that you are right, and B receives the same signals from the GPS satellites as A and therefore concludes that it is in the same position as A. B then continues on its journey, and at t = 1, it completes one revolution of the Earth and passes A again. Since B has been moving while A and all GPS satellites have been at rest, A will observe that the time passed by B is less than that passed by A. So B will observe that more time has passed for the GPS satellites than for B. However, none of this matters, as the way that B finds its position is to take the distance information from four of the satellites, and the time information from the last satellite. Therefore, although his own time will not be the same as that of A, his own time is not needed to calculate his position. Furthermore, as PeterDonis pointed out, it the real case, device B will not even be out of sync due to the cancellation of GR and SR time dilations.thanks for your help,
Aaron

 

[A.T.]

Unfortunately, I think the math involved would be far too cumbersome. I think that reasoning out the problem is a more efficient approach in this case. Or maybe I'm wrong since you seem to have done the math :P

The simple reasoning is that the rotating Earth forms an equipotential surface in it's rest frame. Since the Earth is at rest in that frame, there is no kinetic time dilation between clocks resting w.r.t to the Earth. The only time dilation can come from positions at different potentials, which is not the case for clocks on the equipotential surface.

 

[jbriggs444]

Therefore, although his own time will not be the same as that of A, his own time is not needed to calculate his position. Furthermore, as PeterDonis pointed out, it the real case, device B will not even be out of sync due to the cancellation of GR and SR time dilations.

The "GR" and "SR" time effects cancel for an object at rest on the surface of the rotating earth. In your scenario, B is in motion and loops around the Earth before coming back again to A's position. B's internal clock (if any) will not read the same as A's internal clock.

But that is irrelevant. Both A and B again receive identical satellite signals, perform identical GPS calculations and compute identical positions. Their internal clocks are not needed.

 

[doaaron]

The "GR" and "SR" time effects cancel for an object at rest on the surface of the rotating earth. In your scenario, B is in motion and loops around the Earth before coming back again to A's position. B's internal clock (if any) will not read the same as A's internal clock.

But that is irrelevant. Both A and B again receive identical satellite signals, perform identical GPS calculations and compute identical positions. Their internal clocks are not needed.

When I said the real case, I was referring to the case where A is at the equator, B is at a pole, and both are resting on the surface of the Earth...

I also already mentioned that the internal clocks are not used in practice.thanks anyway,
Aaron

 

[PeterDonis]

it still seems that you are simply taking the clock rate at different points (e.g. north pole, equator, GPS satellite), and showing that the difference in clock rates (tGPS - tNorthPole) = (tGPS - tEquator).

I have actually been trying to say two things, and that is only one of them. The proof of it is simple, if we are working in an inertial frame centered on the Earth. In that frame, we know that, for clocks at sea level and at rest on the (rotating) Earth, we have tNorthPole = tEquator. Therefore, for any given GPS satellite, the equality in the above quote must hold (since tGPS is obviously the same in both cases). Obviously, this proof is only valid in an inertial frame centered on the Earth, though; in some other inertial frame, we will not, in general, have tNorthPole = tEquator.

All of the above talks only about "natural" clock rates, however. The other thing I was trying to do is to draw an important distinction between the "natural" clock rate of a given clock (the rate at which it ticks if no corrections are applied, just due to the basic physics of the clock's path through spacetime and the choice of inertial frame in which to describe the clock) and the "GPS time" clock rate, which is artificially imposed on all the clocks in the system, since almost none of them are clocks at rest at sea level on the rotating Earth (the satellite clocks are obviously not, and even most of the ground station clocks are not at sea level), by applying corrections based on their altitude and state of motion relative to an idealized reference clock that is at rest at sea level on the rotating Earth. It looks like you now understand that GPS uses this artificial clock rate, so that the variation in "natural" clock rates does not affect its operation.

 

[doaaron]

Hi PeterDonis,

I have actually been trying to say two things, and that is only one of them.

So the first point is that (tGPS - tNorthPole) = (tGPS - tEquator).

But as I argued earlier,

even if two clocks appear to be synchronized (the one at the equator and the one at the pole), it doesn't mean that a third clock moving in a different frame will show the same time wrt those two clocks.

It looks like you now understand that GPS uses this artificial clock rate

Actually I understood this from the beginning. From my original question...

as I understand it, GPS satellites are offset prior to launch in order to correct for the time dilation that they experience in orbit.

I was also vaguely aware that clocks on the surface of the Earth run at the same rate. My big problem was that even though they run at the same rate, they have different relative velocities wrt the GPS satellites. So I was concerned that that would affect their reading of GPS time.

As a last point, what I've understood is that the whole system does end up working out because a person on the ground takes his time measurement from the GPS clocks. i.e. tEquator and tNorthPole never enter the equations. Although I was aware of this, I wasn't sure whether an observers velocity wrt a GPS satellite would affect his reading of tGPS. Now it is clear to me that it won't.

It seems that there was some misunderstanding about what each of us were saying, but anyway your information on natural clock rates on Earth was very helpful.thanks,
Aaron

 

[PeterDonis]

what I've understood is that the whole system does end up working out because a person on the ground takes his time measurement from the GPS clocks. i.e. tEquator and tNorthPole never enter the equations.

Yes. But note that in order to do this, the ground observer must be able to receive signals from four GPS satellites, whereas if only position was being determined, three would be enough.

 

[doaaron]

But note that in order to do this, the ground observer must be able to receive signals from four GPS satellites, whereas if only position was being determined, three would be enough.

Are you sure about that? I remember reading the math before. For triangulation using radar, 4 signals would be needed in a 3D coordinate system. However, GPS doesn't work by triangulation, and the time information from a 5th satellite is necessary even if you are only calculating your position. If you consider the surface of the Earth a 2D coordinate system, then 3 Satellites would be needed for triangulation, but 4 are needed using GPS. If I can find the link I will post it.

regards,
Aaron

 

[A.T.]

For triangulation using radar, 4 signals would be needed in a 3D coordinate system.

Why not only 2, if each measures the direction to the object (if that’s what you mean by "triangulation")

If you consider the surface of the Earth a 2D coordinate system, then 3 Satellites would be needed for triangulation, but 4 are needed using GPS.

I would say 2 for triangulation (direction info), 4 for timing (distance info) and 3 for timing + constraint to surface of the Earth.

 

[jbriggs444]

Are you sure about that? I remember reading the math before. For triangulation using radar, 4 signals would be needed in a 3D coordinate system. However, GPS doesn't work by triangulation, and the time information from a 5th satellite is necessary even if you are only calculating your position. If you consider the surface of the Earth a 2D coordinate system, then 3 Satellites would be needed for triangulation, but 4 are needed using GPS.

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.

 

[mfb]

You need 4 satellites to determine time and position in space - in general this will have multiple solutions, but if you have a rough idea where you are (like "I am on the surface of Earth +- 50km", a very reasonable assumption most of the time) you can find the correct solution. A 5th satellite helps if you are so lost that you don't even know if you are in space or not.

3 satellites would allow a very rough estimate for the position, using the same assumption - probably enough to get out of a desert, but impractical for navigation in a town.

More satellites give more redundant measurements and increase the accuracy.

 

[doaaron]

Why not only 2, if each measures the direction to the object (if that’s what you mean by "triangulation")

I would say 2 for triangulation (direction info), 4 for timing (distance info) and 3 for timing + constraint to surface of the Earth.

The GPS satellites are simply sending out their position and time. There's no direction information involved. If we knew the direction to a GPS satellite then we would need two satellites. One merely for timing.

Everybody else seems to be saying the same thing depending on which satellite you consider to be the extra one. cheers,
Aaron

 

[A.T.]

The GPS satellites are simply sending out their position and time. There's no direction information involved.

I know. I was referring to your comments about "triangulation" which involves directional information:

http://en.wikipedia.org/wiki/Triangulation

Everybody else seems to be saying the same thing depending on which satellite you consider to be the extra one

I agree with the others. My numbers on the distance method assumed the receiver has a synched clock too. You have to add one if that is not the case.

 

[doaaron]

I know. I was referring to your comments about "triangulation" which involves directional information

No problem. Just miscommunication I guess...

One interesting take back from the discussions though is that the deliberate offset of the GPS clocks prior to launch has no effect on position calculated by a GPS receiver. It seems to be just a secondary function of the GPS system to provide accurate time information for GPS receivers. I'm also not sure how necessary it is since I believed GPS satellites are periodically calibrated with ground stations.Aaron

 

[PeterDonis]

Are you sure about that?

Yes. Others have already commented, but one thing that is worth noting is that 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.

the deliberate offset of the GPS clocks prior to launch has no effect on position calculated by a GPS receiver

Um, yes it does. The signals received from the GPS satellites only include time information; distance is calculated from time and the speed of light. If the time information is off, the distances will be off, hence the position will be off (as well as the time calculated from the received signals, of course). So the clock rate correction aboard each satellite is critical to the proper operation of GPS.

I'm also not sure how necessary it is since I believed GPS satellites are periodically calibrated with ground stations.

They are calibrated daily from ground stations, but if the clock rates were not corrected, the timing error accumulated over the course of a day would translate into about ten kilometers of error in position.

The best detailed treatment of relativistic effects in GPS is Neil Ashby's, here:

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

I strongly recommend reading it.

 

[Fantasist]

Um, yes it does. The signals received from the GPS satellites only include time information; distance is calculated from time and the speed of light. If the time information is off, the distances will be off, hence the position will be off (as well as the time calculated from the received signals, of course). So the clock rate correction aboard each satellite is critical to the proper operation of GPS.

Considering the way GPS works in practice (by using time differences between satellites rather than comparison of the satellites times to a ground clock), the distance will only be off by the same factor the satellite clocks are off. So that is pretty much negligible if the clocks run slow/fast by a relative amount 4*10^-10 (which is what Relativity predicts).

They are calibrated daily from ground stations, but if the clock rates were not corrected, the timing error accumulated over the course of a day would translate into about ten kilometers of error in position.

The best detailed treatment of relativistic effects in GPS is Neil Ashby's, here:

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

I strongly recommend reading it.

In the officcial GPS documentation ( http://www.publications.usace.army.mil/Portals/76/Publications/EngineerManuals/EM_1110-1-1003.pdf ) you can find the following information (see Chapt. 4-6 b (Clock stability)):

"The time synchronization between the GPS satellite clocks is kept to within 20 nanoseconds (ns) through the broadcast clock corrections as determined by the ground control stations"

So any theoretical discussion of differences in clock rates between the satellites e.g. due to orbital variations (which the OP was asking about) is pretty much academic, as the clocks are synchronized so frequently that they never differ by more than 20 ns (which corresponds to about 6m uncertainty in the position determination).

 

[doaaron]

Um, yes it does. The signals received from the GPS satellites only include time information; distance is calculated from time and the speed of light. If the time information is off, the distances will be off, hence the position will be off (as well as the time calculated from the received signals, of course). So the clock rate correction aboard each satellite is critical to the proper operation of GPS.

They are calibrated daily from ground stations, but if the clock rates were not corrected, the timing error accumulated over the course of a day would translate into about ten kilometers of error in position.

I don't think that's correct, and Fantasist seems to be saying the same thing above (I think).

Here is a simple calculation for position based on four GPS satellites:

https://www.courses.psu.edu/aersp/aersp055_r81/satellites/gps_details.html

As you can see, the distance info on the left hand side of the equation is

d_n = c*(t_GPSn - t_RXn + t_error)

where t_error is the clock rate mismatch between the GPS satellite and the receiver (i.e. t_RXn and t_GPSn are in sync). Call the time it takes for the GPS signal to travel from the GPS satellite to the receiver, Δt. Then

t_RXn = t_GPSn + Δt

So finally we have,

d_n = c*(t_GPSn - t_GPSn - Δt + t_error) = c*(-Δt + t_error)

the minus sign in front of the Δt is unimportant. Since we have four equations, we can solve for x, y, z, and t_error. As you can see, the absolute time of the GPS satellite does not enter into the equations for calculating position. It would only be important if you wanted to calculate absolute time.regards,
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?

 

[mfb]

Two measurements give you a hyperbolic surface, a third one should give some "ring-like" or "parabola-like" line structure (not actual rings or parabolas, but with some similarity). Intersecting this with the surface of Earth (neglecting height differences) gives some very rough estimate for the position with at least two, but possibly more intersections - a rough knowledge of the position helps to make the point unique in most cases.
A fourth measurement will reduce this line to a set of discrete points (probably at least 2, maybe more, but most of them will be far away from the surface of earth). Again, knowledge that you are not in space helps, but this time you can also calculate the height above the surface which increases accurary to a reasonable level.
A fifth measurement, in general, will resolve the remaining ambiguity.

 

[PeterDonis]

Considering the way GPS works in practice (by using time differences between satellites rather than comparison of the satellites times to a ground clock), the distance will only be off by the same factor the satellite clocks are off.

No, it won't; it will be off by the accumulated time difference due to the clock rate difference, since the last time the clocks were synchronized.

So that is pretty much negligible if the clocks run slow/fast by a relative amount 4*10-10 (which is what Relativity predicts).

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.

 

[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.