【转】地球坐标系 (WGS-84) 到火星坐标系 (GCJ-02) 的转换算法 C语言

最新推荐文章于 2025-05-29 14:17:29 发布

转载最新推荐文章于 2025-05-29 14:17:29 发布 · 769 阅读

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#c/c++

本文提供了一种用于将全球定位系统(GPS)坐标转换为特定地图坐标系的算法实现。该算法适用于中国范围内的坐标转换，并提供了详细的C++代码示例。通过对经纬度进行一系列数学变换，确保了坐标在中国范围内的准确性。

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Mark下吧，这是好东西啊。 
http://blog.youkuaiyun.com/coolypf/article/details/8686588


整理成C++代码：
#include


const double pi = 3.14159265358979324;


//
// Krasovsky 1940
//
// a = 6378245.0, 1/f = 298.3
// b = a * (1 - f)
// ee = (a^2 - b^2) / a^2;
const double a = 6378245.0;
const double ee = 0.00669342162296594323;


 static bool outOfChina(double lat, double lon)
{
    if (lon < 72.004 || lon > 137.8347)
        return true;
    if (lat < 0.8293 || lat > 55.8271)
        return true;
    return false;
}


static double transformLat(double x, double y)
{
    double ret = -100.0 + 2.0 * x + 3.0 * y + 0.2 * y * y + 0.1 * x * y + 0.2 * sqrt(abs(x));
    ret += (20.0 * sin(6.0 * x * pi) + 20.0 * sin(2.0 * x * pi)) * 2.0 / 3.0;
    ret += (20.0 * sin(y * pi) + 40.0 * sin(y / 3.0 * pi)) * 2.0 / 3.0;
    ret += (160.0 * sin(y / 12.0 * pi) + 320 * sin(y * pi / 30.0)) * 2.0 / 3.0;
    return ret;
}


static double transformLon(double x, double y)
{
    double ret = 300.0 + x + 2.0 * y + 0.1 * x * x + 0.1 * x * y + 0.1 * sqrt(abs(x));
    ret += (20.0 * sin(6.0 * x * pi) + 20.0 * sin(2.0 * x * pi)) * 2.0 / 3.0;
    ret += (20.0 * sin(x * pi) + 40.0 * sin(x / 3.0 * pi)) * 2.0 / 3.0;
    ret += (150.0 * sin(x / 12.0 * pi) + 300.0 * sin(x / 30.0 * pi)) * 2.0 / 3.0;
    return ret;
}


void gps_transform( double wgLat, double wgLon, double& mgLat, double& mgLon)
{
if (outOfChina(wgLat, wgLon))
        {
            mgLat = wgLat;
            mgLon = wgLon;
            return;
        }
        double dLat = transformLat(wgLon - 105.0, wgLat - 35.0);
        double dLon = transformLon(wgLon - 105.0, wgLat - 35.0);
        double radLat = wgLat / 180.0 * pi;
        double magic = sin(radLat);
        magic = 1 - ee * magic * magic;
        double sqrtMagic = sqrt(magic);
        dLat = (dLat * 180.0) / ((a * (1 - ee)) / (magic * sqrtMagic) * pi);
        dLon = (dLon * 180.0) / (a / sqrtMagic * cos(radLat) * pi);
        mgLat = wgLat + dLat;
        mgLon = wgLon + dLon;
};