最近在做一个地图项目，关于GPS转火星高德坐标的算法如下

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希望对各位开发者有用，测试误差不大

#pragma mark - 高德地图GPS转坐标
// location lati， longi
#define pi 3.14159265358979324
NSArray * TransformGPStoMAMAP(NSString * location) {
    NSArray * result = nil;
    // 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;
    
    NSArray * tmpArr = [location componentsSeparatedByString:@","];
    double wgLat = [[tmpArr firstObject] doubleValue];
    double wgLon = [[tmpArr lastObject] doubleValue];
    
    //
    // World Geodetic System ==> Mars Geodetic System
    if (outOfChina(wgLat, wgLon))
    {
        result = @[[NSNumber numberWithDouble:wgLat], [NSNumber numberWithDouble:wgLon]];
        
        return result;
    }
             
    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);
    wgLat = wgLat + dLat;
    wgLon = wgLon + dLon;
    
    result = @[[NSNumber numberWithDouble:wgLat], [NSNumber numberWithDouble:wgLon]];
    
    return result;
}

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;
}

最近在做一个地图项目，关于GPS转火星高德坐标的算法如下

标签：

原文地址：http://blog.csdn.net/neanoher/article/details/45024813