I have been reading the following article about radio navigation in general and GPS / GLONASS receiver design:

in which the author states the following about hyperbolic radionavigation:

Hyperbolic navigation systems were first implemented as ground-based navigation systems [...] Since the transmitters are located on the Earth's surface, the geometry of the problem does not allow a three-dimensional navigation. [...] To measure the altitude, one of the transmitters should be located above or below the user's receiver or at least out of the user's horizon plane.

I do not seem to be able to comprehend why this is the case. Even if the transmitters are located on the surface (e.g. Loran-C), the differences between reception times of signals from 4 transmitters would generate 3 rotational hyperboloids in 3D space, the intersection of which could be above (or below) surface.


1 Answer 1


You are correct, the time difference generates hyperboloids in 3D space and as such can be used to find the 3D position. However, if the transmitters are far apart and/or in the same plane, the geometry of the intersecting hyperboloids is such that the accuracy in the vertical axis is very poor.

When the receiver is in the same horizontal plane as the transmitters, all intersections almost form a vertical line. The slightest inaccuracy in time difference, would cause only a slight error in the horizontal position, but a significant error in the vertical position.

The concept is called Geometric Dilution of Precision (GDOP). A good start for more detail can be found on Wikipedia

  • 6
    $\begingroup$ The concept is called Geometric Dilution of Precision (GDOP). A good start for more detail can be found on Wikipedia. $\endgroup$
    – Gerry
    Jan 20, 2020 at 16:59
  • 2
    $\begingroup$ Okay, I think I understand now. It's not that it's completely impossible to calculate the intersection with the transmitters on the same plane, it's that because the third hyperboloid is oriented the same as the other two (the lines connecting the receivers and transmitters are on the same plane), and given that the measurements are ambiguous, the "intersection volume" between the three ambiguous shapes would be high on the vertical axis, yes? (However that should not be the case for example if the third shape intersected the volume from an angle) $\endgroup$
    – rkourdis
    Jan 21, 2020 at 6:20
  • 1
    $\begingroup$ @rkourdis, that's right. $\endgroup$
    – DeltaLima
    Jan 21, 2020 at 6:22
  • 1
    $\begingroup$ @Gerry, indeed. If I find the time I may expand on GDOP, HDOP and VDOP a bit. $\endgroup$
    – DeltaLima
    Jan 21, 2020 at 6:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .