C，计算两个GPS位置之间的距离是多少？ [英] C, Calculating the distance between two GPS locations?

问题描述

如果我有两个GPS位置，说51.507222，-0.1275和48.856667，2.350833，我可以用什么公式来计算两者之间的距离是多少？我已经听说了很多关于haversine公式，但找不到任何关于它的信息，或如何将其应用到C

我已经写了下面的code，然而，这是非常innacurate。任何人都知道为什么吗？我不明白。这个问题来自于函数本身，但我不知道它是什么。

 浮动calcDistance（float一，浮动B，浮C，浮动D）
{
    浮DLAT;
    浮dLon;
    DLAT =（C  -  A）;
    dLon =（D  -  B）;
    DLAT / = 57.29577951;
    dLon / = 57.29577951;
    浮V_A;
    浮V_C;
    浮动距离;    V_A = SIN（DLAT / 2）*罪（DLAT / 2）+ COS（A）* COS（C）*罪（dLon / 2）*罪（dLon / 2）;
    V_C = 2 * ATAN2（SQRT（V_A），SQRT（1-V_A））;
    距离= R * V_C;
    返回距离;
}
 

解决方案

HTTP： //www.movable-type.co.uk/scripts/latlong.html

这有JavaScript的一个块，我敢肯定，你可以很容易地适应足够温度。

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;
}

解决方案

http://www.movable-type.co.uk/scripts/latlong.html

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

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