Intersecting Hyperbola

  1. Hello
    I am seeking advice on intersecting hyperbolae. Specifically, if six hyperbolae, which are generated from combinations of 4 focii, all intersect at a common point does this mean that the four distance measurements are always errorless?
    The reason I ask is for GPS accuracy. Many, many billions of dollars are spent each year on attaining better accuracy than the basic GPS system provides. It seems to me that in there ought to be sufficient information contained in the fifth satellite to fix errors better than they are fixed.
    Using 2D diagrams with four satellites my reasoning is as follows:
    (Note: the raw information from GPS is the time that signals are despatched. The latest despatch time is the shortest distance (an unknown, and frankly an unnecessary value). A distanceOverShortest can be calculated from the despatch times for the other satellites)
    1 – from four satellites there are six hyperbolae (detailed explanation excluded). If distanceOverShortest information is perfect the six hyperbolae will intersect at one point.
    2 - if distanceOverShortest information is imperfect there are four points where three of the hyperbolae intersect. Each of these four points is the fix point from a triplet of satellites.
    3 – When the distanceOverShortest information is imperfect adjusting one of the distanceOverShortest values leads to: a) - the distance difference between the adjusted satellite and the three other satellites varying, which b) - results in three of the hyperbolae moving, which c) - results in the movement of three of the four triplet fix points. The three hyperbolae that are NOT calculated from the distanceOverShortest for the adjusted satellite remain static, as does the fix point that is the intersection of these three hyperbolae.
    4 – the three fix points that move converge on or diverge from the fix point that does not move. The change in distanceOverShortest to the one satellite that is being adjusted can be set so that all four fix points become co-pointal, ie the six hyperbolae intersect at one point.
    5 – refer to the statement in point 1 – if the distanceOverShortest information is perfect the six hyperbolae will intersect at one point. With the variance in the distanceOverShortest for one satellite the fix points can become co-pointal, so the distanceOverShortest (with one adjustment) information is therefore perfect at this point. Isn’t it?
    6 – steps 3 – 5 can be repeated for the other three triplet fix points. The outcome is four distance variances (one for each satellite), each of which results in a point location that has perfect ‘adjusted distanceOverShortest’ information. The correct distanceOverShortest values are still unknown, however how much each of these distances has to be adjusted to give a perfect location IS known, as are the co-ordinates of these four perfect locations!
    7 – four equations can be expressed, each of which has one of the known distance variances added to one of the distanceOverShortest values, its co-pointal co-ordinates, as well as the other three unknown distanceOverShortest values. All other inputs are known values. These four equations can be simultaneously solved to provide the four correct distanceOverShortest values.
    A couple of points on the hyperbolae being used are pertinent. The information required from the hyperbolae are the gradient at the receiver location and the amount of normal movement that a ‘change in the distance difference between the distances from the receiver to the two satellites’ produces. At the scale of GPS systems, where the distance to the satellites is millions and millions of metres and the error at starting is tens of metres, the part of the hyperbola being considered is effectively linear. Also the normal movement can be expressed in terms of metresPerMetreOfDistanceDifferenceChange – for each hyperbolae this value is effectively constant at the distances in the GPS system.
    It is off course much simpler to use a lineal system, and this can be achieved if the satellites are considered to be equidistant from the receiver. This saves the step of having to identify the approximate position on the hyperbolae to identify the tangential gradient to use, and means that the metresPerMetreOfDistanceDifferenceChange is simply the shift off the X axis rather than the shift normal to the tangent.
    A spreadsheet & chart that demonstrates the above method, using equidistant satellites, is attached. The chart displays the hyperbolae and triplet fix points. Inputs of satellite angles and distanceOverShortest errors are reflected in the chart. The chart is also useful in demonstrating and explaining why relatively small range errors can produce large errors in fixes. The spreadsheet can also be downloaded from
    So – there are three premises I have used that I can see can be considered questionable. For instance I am satisfied that the metresPerMetreOfDistanceDifferenceChange value is effectively constant for each hyperbola on the GPS distances involved. I am also satisfied that the hyperbolic tangent gradient will be constant (or absolutely insignificantly changed) for the distance involved in fine GPS calculations. I have not produced any proof of these assumptions. The third point is my first question, if six hyperbolae intersect at a common point does that mean that the distance measurements are always errorless? I think it does and therefore I believe that the logic outlined is valid.
    I would really appreciate any comments.
    Thank you for your time

    Attached Files:

  2. jcsd

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