harder boundary value problem would have to be solved. The eikonal equation This equation is discussed in the references. The equation is highly non-linear. Given that a smooth data fitting curve n(altitude) is provided by the ITU for the refraction index n, and that the values of n differs from 1 only by something of the order 10−4, a numerical solution of the eikonal equation can be considered. Usually the equation is presented under the self-adjoint form, a more tractable equation for the ray head position vector r is given in generic parametric form: Implementations Three implementations to compute the attenuations exist: Take the ray to be a straight line. Use the optical invariant and apply the ITU recommendation. Solve the eikonal equation. The first two are only of 1st order approximation (see Orders of approximation). For the eikonal equation, many numerical schemes are available. Here only a simple second order scheme was chosen. For most standard configurations of source-target, the three methods differ little from each other. It is only in the case of rays grazing the ground that the differences are meaningful. The following was used for testing: At the latitude of 10°, when a ray starts at 5 km altitude with an elevation angle of −1° to hit a target at the same longitude but at latitude 8.84° and altitude 30 km. At 22.5 GHz, the results are: Note that 22.5 GHz is not a practical frequency but it is the most suitable for algorithms comparison. In the table, the first column gives the results in dB, the third gives the distance covered and the last gives the final altitude. Distances are in km. From the altitude 30 km up, the attenuation is negligible. The paths of the three are plotted: Note: A MATLAB version for the uplink (Telecommunications link) is available from the ITU The boundary value problem When a point S communicates with a point T, the orientation of the ray is specified by an elevation angle. In a naïve way, the angle can be given by tracing a straight line from S to T. This specification does not guarantee that the ray will reach T: the variation of refraction index bends the ray trajectory. The elevation angle has to be modified to take into account the bending effect. For the eikonal equation, this correction can be done by solving a boundary value problem. As