' Haversine’ formula

[Orcino]

Hi, in the code below, with coord. 1 the result is 18.8 Km this is correct, but in coord. 2 the result is 31.08 Km,when the correct is 0.399 km.

Somebody can help?

Thanks

Orcino


void calcula_dist(void)

{

// float lat1=19.508016; // Coord. 1
// float lat2=19.674716; //
// float lon1=99.234144; //
// float lon2=99.234000; //

float lat1 = 18.38490; // Coord. 2
float lat2 = 18.64784; //
float lon1 = 48.11596; //
float lon2 = 48.20348; //

float distancia;

float delta_lati, delta_longe;
float a, c;

lat1 = lat1 * 0.017453; // Converte p/ radianos
lat2 = lat2 * 0.017453;
lon1 = lon1 * 0.017453;
lon2 = lon2 * 0.017453;

delta_lati = lat2 - lat1;
delta_longe= lon2 - lon1;

a = (sin(delta_lati/2) * sin(delta_lati/2)) + cos(lat1) * cos(lat2) * (sin(delta_longe/2) * sin(delta_longe/2));

c = 2 * atan2(sqrt(a),sqrt(1-a));

distancia = EARTH * c; // earth is 6371

fprintf(HOSTPC,"\nA Distancia é %f Km", distancia);


}

[sjbaxter]

You can start by explicitly defining numbers in the calculations to be 'float'. Some compilers do weird things during optimization (like converting all values in calculations to the lowest type like an int, so you'll get strange rounding errors creaping in.)

[Ttelmah]

Er.
31km, is correct for the figures you have put in for 2.
Even a quick rule of thumb test, gives a result that must be larger than the first example. If you look at the first example, there is almost no distance in longitude, and 0.1667 degrees in latitude. The second example, has 0.262884 degrees in lat, and 0.0874 degrees in long.
A quick calculation for the second figures, gives about 30.5km.
Given that 1 degree in lat, is about 112km, anywhere on the Earth, the result cannot be below 29km (0.262*112).

Best Wishes