C, Calculating the distance between two GPS locati

问题:

If I have two GPS locations, say 51.507222, -0.1275 and 48.856667, 2.350833, what formula could I use to calculate the distance between the two? I've heard a lot about a haversine formula, but can't find any information about it, or how to apply it to C.

I've written the following code, however, it's very innacurate. Anybody know why? I can't figure it out. The problem comes from the function itself, but I don't know what it is.

float calcDistance(float A, float B, float C, float D) 
{
    float dLat;
    float dLon;
    dLat = (C - A);
    dLon = (D - B);
    dLat /= 57.29577951;
    dLon /= 57.29577951; 
    float v_a;
    float v_c;
    float distance;

    v_a = sin(dLat/2) * sin(dLat/2) + cos(A) * cos(C) * sin(dLon/2) * sin(dLon/2);
    v_c = 2 * atan2(sqrt(v_a),sqrt(1-v_a));
    distance = r * v_c;
    return distance;
}

回答1:

This page has a block of Javascript which I'm sure you could adapt easily enough in C:

var R = 6371e3; // metres
var φ1 = lat1.toRadians();
var φ2 = lat2.toRadians();
var Δφ = (lat2-lat1).toRadians();
var Δλ = (lon2-lon1).toRadians();

var a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
        Math.cos(φ1) * Math.cos(φ2) *
        Math.sin(Δλ/2) * Math.sin(Δλ/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));

var d = R * c;

回答2:

Your code is absolutely correct.

I would rename parameters A, B, C and D to lat1, lon1, lat2 and lon2.

It returns the distance in kilometers. You need to define r as 6371, roughly the radius of the earth.

回答3:

What you're looking for is the great circle distance.

回答4:

Maybee its too late, but this is the reason of the inaccurate calculation: -> A and C (lat1 and lat2) must also be converted from degree to radians

dLat /= 57.295779513082325;
dLon /= 57.295779513082325;
lat1 /= 57.295779513082325;
lat2 /= 57.295779513082325;

regards, omeier