Calculating GPS Locations: The Impact of Special Relativity

[russ_watters]

But how does it determine positions from just time? I may seem stupid but I can't make the connection.

What the code time signals tell you is that one satellite is some distance further from you than another. Since the receiver knows how far apart the satellites are, that is enough information to triangulate your distance from them and your position on Earth (giving a two-position solution).

 

[cesiumfrog]

A differential receiver uses both real satellites and Earth-bound "satellites" to determine its position, so there are GR timing differences between two "satellites" due to their much different r coordinates.

It seems to me like the maximum positional error caused by relativistic effects will be (38\mus/day)*(velocity of satellite, not of light). What do you think?

 

[mgb_phys]

Roughly correct.
The position fix relies on knowing where the satelite was when it broadcast the data block, this is based on it's onboard clock - so you would be out by the distance the satelite travels in 38us ( actually half that since it makes two orbits/day)

In practice I don't think it changes your actual position very much, if you have good geometry with satelites both east and west of you then the errors due to clock difference will apply in both directions, you will get a poorer fit but will be in roughly the same place.

( Except that the designers of the satelites knew about relativity and set the clocks to run fast ;-)

 

[cesiumfrog]

And that's for differential GPS. For normal GPS I'd expect the relativistic error to be smaller again, since the satellites have less time dilation relative to each other than to ground stations (presuming that the satellite positions are determined by triangulation from ground stations, as would seem obvious and consistent, rather than by some kind of comparison between satellite and ground station clocks).

 

[mgb_phys]

The satelite position is calcualted in the receiver based on the tranmsitted empheris.
The onboard empheris (e the satelites knowledge of it's orbit) is recalculated (and uploaded) every few days when the satelite passes over one of the ground stations - not sure if they fix the satelite position optically with a transit telescope (which would be most accurate) or a radio fix.
The satelites store and transmit an almanac giving each others approximate positions - which helps single channel receivers find other sats - but they don't send signals between each other (actaully they do but this is nothing to do with the GPS position)

Differential GPs just receives the GPS signal at a fixed ground station and retransmits it to the reciever (either via satelite WAAS/EGNOS or by radio starFire) it takes out large scale ionospheric effects.

Ironically the simplest GPS to understand is the highest accuracy RTK mode - where you have two receivers close together looking at the same satelites - this just uses the pure time difference of the high frequency carrier wave to get a distance between the two receivers directly to an accuracy of cm. It then uses the low accuracy GPS position to work out where it is to a much poorer accuracy.

 

[p4h]

Kinda afraid to post after the verbal spanking Randall gave me, but got another question:

contraction = sqrt( 1 - v^2/c^2 ) = sqrt ( 1 - 1.6e-10 ) = 0.99999999991699995
times 24hours = 7 microsecs

I'm sorry to seem stupid, but I've tried two hours now to get 0.99999999991699995
times 24hours = 7 microsecs.. I just don't see how?

 

[mgb_phys]

I'm sorry to seem stupid, but I've tried two hours now to get 0.99999999991699995
times 24hours = 7 microsecs.. I just don't see how?

Remember it's the fraction of a day so,
1 - 0.99999999991699995 = 8.3 E-11 of a day

And 24 * 60 * 60 * 8.3E-11 = 7.2 E-6 seconds

 

[p4h]

Ah! Thanks again..

If I do:

\sqrt{1-\frac{3900}{299792458}} = .9999934955

So we don't get the same. >.<

Where do you get the 1 - .9999934955 from?

 

[mgb_phys]

The result of the sqrt is the contraction - this is the fraction of a day that the satelite expereinces, so a satelite's view of a day is 0.9999934955 of an Earth's view of a day.
The difference between these is simply 1 - 0.9999934955
or if you prefer, satelite day = 0.9999934955 * 24 * 3600 seconds, Earth day = 1 * 24 * 3600 seconds.
The difference is 7.2 us.

edit sorry just realized you are asking about the actual numerical value
You forgot to square the v and c, it should be sqrt( 1 - (3900^2)/(3E8^2))

 

[p4h]

The result of the sqrt is the contraction - this is the fraction of a day that the satelite expereinces, so a satelite's view of a day is 0.9999934955 of an Earth's view of a day.
The difference between these is simply 1 - 0.9999934955
or if you prefer, satelite day = 0.9999934955 * 24 * 3600 seconds, Earth day = 1 * 24 * 3600 seconds.
The difference is 7.2 us.

edit sorry just realized you are asking about the actual numerical value
You forgot to square the v and c, it should be sqrt( 1 - (3900^2)/(3E8^2))

Yeah thanks it was the top part I wanted..

And ye forgot to square them, but that only got my answer closer to 1

But thanks yet again!

 

[RandallB]

Kinda afraid to post after the verbal spanking Randall gave me, but got another question:

No reason to fear posting – often newbes accidentally work on homework in the discussion forums instead of the homework forum. If someone doesn’t tell you, you won’t learn how to best use PF.

By the way if a Mentor has not already (I am not one) welcome to the PF forums.

PS: Also, no need to abuse your teacher behind their back (post #28), at least be sure what they want you to learn from doing a report. Remember your writing a report not a textbook.

 

[p4h]

Well I just got alittle frustrated because I couldn't see my way out of the project, since I also have this graph I can't figure out.

 

[p4h]

Okay I got two more things left before I stop reviving this thread, as I figure I might as well use this one, instead of starting a new one.

1. When thinking of what impact the theory of relativity (both SR and GR) have on things around us, I couldn't think of anything except Redshift / Doppler effect. I'm sure there are a lot more areas but I can't think of any?

2. The image attached is a graph I'm supposed to be able to figure out. I'm supposed to be able to figure out the distance between a Satelite and a GPS reciever.
X axis: Time / seconds
Y axis: Signal strength / no unit
There is no legend to the graph, so I don't know which curve represents the satellite and which represents the reciever.

I'd appreciate it if only hints is dropped, so I can fiddle with it myself, but I may ask again if I'm lost :)

PS: I realize this should be in the homework section, but as I already have a thread with this subject, I might as well use it.

 

[RandallB]

Cute problem.
I assume you can translate the fairly intuitive answer of about “0.065 tid/s” into the appropriate units of distance.
The correct value answer of course is not the important part.
I’m sure they expect you to report on what detailed information has to be known at the receiver to use in observing this one satellite.
And why using SR & GR in so important to maintaining that information.

A good problem in that it is easy to answer, once you understand the issues of running a GPS.
Hint: think simple

Mentors can move the thread if they feel it is needed, interesting approch at least.

 

[p4h]

I assume I have to make a connection between length and signal strength, but to me, when I think of signal strength, I just think that it has to vary a lot due to different factors, being it the ionozation in the Ionosphere, or being it the humidity in the Troposphere? I may be overthinking it.

 

[RandallB]

Signal strength has nothing to do with it.
You are making it harder than it is.
Radio photons are no different than light photons.
Think simple & how do you measure distance with them.