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Homework Help: GPS - the fourth satellite?

  1. Jul 7, 2010 #1
    Hey all,
    im doing a individual info search on GPS. i can understand everything (well its just a basic overview of GPS) but the one thing i cant get my head around is why the need for a fourth satellite? i know it corrects the timing as the receiver does not have an atomic clock. ive read 4 or 5 different explanations about it but i just dont get it..
    here is an excerpt from wikipedia (i know its unreliable..), if someone can, can you elaborate on it and how they derive the equation? i can tell its REALLY simple, but its just not going through to me.

    "It is likely that the surfaces of the three spheres intersect, since the circle of intersection of the first two spheres is normally quite large, and thus the third sphere surface is likely to intersect this large circle. It is very unlikely that the surface of the sphere corresponding to the fourth satellite will intersect either of the two points of intersection of the first three, since any clock error could cause it to miss intersecting a point. However, the distance from the valid estimate of GPS receiver position to the surface of the sphere corresponding to the fourth satellite can be used to compute a clock correction. Let denote the distance from the valid estimate of GPS receiver position to the fourth satellite and let denote the pseudorange of the fourth satellite. Let . is the distance from the computed GPS receiver position to the surface of the sphere corresponding to the fourth satellite. Thus the quotient, , provides an estimate of
    (correct time) − (time indicated by the receiver's on-board clock),
    and the GPS receiver clock can be advanced if is positive or delayed if is negative. However, it should be kept in mind that a less simple function of may be needed to estimate the time error in an iterative algorithm as discussed in the Navigation section."


    edit: also theres an image on wiki that is explaining it. hope it helps you help me with it :)
  2. jcsd
  3. Jul 8, 2010 #2
    As the wikipedia says, if the receiver's GPS clock was perfect, the four sphere's would intersect in one point. However, because it isn't, they don't. Think about how precise the clock would have to be to have all those giant sphere's intersect in one really small dot...
    But the GPS knows that they should be intersecting at on point, which means p4 should be equal to r4 (but it isn't due to the time problem). What that formula does is calculates how much time the GPS receiver is off by, so that p4 would in fact be equal to r4. See in the sample provided, p4 is less than r4, the real distance from the Satellite 4, which means that the GPS thinks the signal took less time to arrive that it actually did. This would mean that the GPS's clock is slow; and if you advance it by that amount provided, it'll be good again.
    I hope this was clear, but if it isn't, ask again:)

  4. Jul 8, 2010 #3
    thanks heaps Tusike!
    it is pretty clear, thanks for referrin to that image.
    im still a little baffled..
    so the circle around fourth satellite(with r = p4) is the distance the reciever thinks it is from the satellite, but the real distance is r4? i understand now the correct time - receivers estimated time = margin of error thing. i dont know how that slipped past me lol.

    ie. p4 = right
    r4= wrong
    and p4 - r4 = error

    i hope thats right, because it sounds simple now :D

    another question which is relataively easy lol (this stuff baffles me for some odd reason):
    how do the receivers get corrected? does it correct itself after the 4th satellite 'tells' it how much its off by?
  5. Jul 8, 2010 #4
    I think it's exactly the opposite way:)
    r4 = right
    p4 = wrong

    I myself had to think about which was is right or wrong, but r4 is the actual distance, no matter what; and p4 is the distance calculated by the GPS using the formula p4=c*(time the gps think passed between the sending and the receiving of the signal).

    As for your other question, I thought about it too, since if the 4th satellite's timing was off by a bit, the other three's were as well. I think what the Wiki says, "it should be kept in mind that a less simple function of da may be needed to estimate the time error in an iterative algorithm" is exactly what you need. If you scroll down to the Navigation section, you will find a bunch of formulas and all that tell you how the correction is done.

    If I had more time, I'd look into those formula's, but I doubt they can be explained really simply because of the fact that they're not... Even I may not understand them. I love physics and everything, the only thing that really bothers me is that I just don't understand the math for those more advanced things, being only in 11th grade and not following any special math courses... So, for the correction, see those formulas, and good luck:)
  6. Jul 8, 2010 #5
    ahh great haha.
    i google defined pseudorange (the term used in wiki article) and states;
    "A distance measurement based on the correlation of a satellite transmitted code and the local receiver's reference code, that has not been corrected for errors in synchronization between the transmitter's clock and the receiver's clock."

    so to me it seems as though that is the incorrect one. which is exactly what you said. i think i was supposed to say p4 was wrong in the first place tbh but i got it mixed up LOL.

    ill have a squizz at the formulas is navigation when its not 12:10am lol

    PS: im in year 12..but by the sounds of it youve been doing physics since year 9-10? my physics coruse started yr 11 (and physics teacher was horrible lol). this year im pretty much learning what i was emant to learn last year and the new stuff o_O

    thanks again tusike
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