经纬度和平面坐标系转换C代码

### 回答1：
以下是将经纬度转换为平面坐标系的C代码示例：

#include <stdio.h>
#include <math.h>

#define PI 3.14159265358979323846
#define EARTH_RADIUS 6378137.0
#define EARTH_ECCENTRICITY 0.0818191908426

double rad(double d)
{
    return d * PI / 180.0;
}

void transform(double wgLat, double wgLon, double *mgLat, double *mgLon)
{
    double mgLat0, mgLon0, mgLat1, mgLon1, mgLat2, mgLon2;
    double x, y;
    double sinMG, cosMG, sinMG0, cosMG0, sinU, cosU, tanU, tan2U;
    double cosSqAlpha, sinSigma, cos2SigmaM, sigma, sinLambda, cosLambda, sinAlpha;
    double lambda = wgLon;
    double lat = wgLat;
    double a = EARTH_RADIUS;
    double eSquared = EARTH_ECCENTRICITY * EARTH_ECCENTRICITY;
    double k0 = 0.9996;

    lat = rad(lat);
    lambda = rad(lambda);

    sinMG = sin(lambda);
    cosMG = cos(lambda);
    tanU = (1 - eSquared) * tan(lat);
    tan2U = tanU * tanU;
    cosSqAlpha = 1 / (1 + tan2U);
    sinSigma = (1 - eSquared) * sin(lat) * sin(lat) / (1 - eSquared * sin(lat) * sin(lat));
    cos2SigmaM = cosSigma - 2 * sinSigma;
    sigma = atan2(sinSigma, cosSigma);
    sinU = tanU * cosSigma;
    cosU = 1 - sinSigma * sinU;
    tanLambda = sinMG / cosMG;
    sinAlpha = sqrt(cosSqAlpha) * (tanLambda);
    cosSqAlpha = 1 - sinAlpha * sinAlpha;
    cosLambda = 1 / sqrt(1 + tanLambda * tanLambda);
    sinLambda = tanLambda * cosLambda;
    mgLat0 = a * ((1 - eSquared / 4 - 3 * eSquared * eSquared / 64 - 5 * eSquared * eSquared * eSquared / 256) * lat -
            (3 * eSquared / 8 + 3 * eSquared * eSquared / 32 + 45 * eSquared * eSquared * eSquared / 1024) * sin(2 * lat) +
            (15 * eSquared * eSquared / 256 + 45 * eSquared * eSquared * eSquared / 1024) * sin(4 * lat) -
            (35 * eSquared * eSquared * eSquared / 3072) * sin(6 * lat));
    mgLon0 = a * (cosLambda * sinAlpha * (1 + cosSqAlpha * (-1 + 2 * cosSqAlpha)) *
            (1 + eSquared * cosSqAlpha * (cosSqAlpha - 2)) +
            0.5 / cosLambda * (-sinSigma * (cosSqAlpha - sinAlpha * sinAlpha) +
            cos2SigmaM * cosLambda * (-1 + 2 * cosSqAlpha)));
    x = k0 * mgLon0 + 500000;
    y = k0 * (mgLat0 + mgLat1 + mgLat2) + 0;

    *mgLat = y;
    *mgLon = x;
  

### 回答2：
经纬度和平面坐标系之间的转换是一个常见的问题，可以使用C代码来实现。下面是一个简单的实现示例：

#include <stdio.h>
#include <math.h>

// 定义常量
#define PI 3.141592653589793238462643383279502884

// 将经度转换为弧度
double toRadians(double degree) {
return degree * (PI / 180.0);
}

// 将弧度转换为经度
double toDegrees(double radians) {
return radians * (180.0 / PI);
}

// 经纬度转平面坐标系（球面坐标转直角坐标）
void convertLatLonToXY(double lat, double lon, double refLat, double refLon, double *x, double *y) {
double earthRadius = 6371000.0; // 地球半径（单位：米）
double latRad = toRadians(lat);
double lonRad = toRadians(lon);
double refLatRad = toRadians(refLat);
double refLonRad = toRadians(refLon);
double deltaLon = lonRad - refLonRad;

*x = earthRadius * deltaLon * cos(refLatRad);
*y = earthRadius * (latRad - refLatRad);
}

// 平面坐标系转经纬度（直角坐标转球面坐标）
void convertXYToLatLon(double x, double y, double refLat, double refLon, double *lat, double *lon) {
double earthRadius = 6371000.0;
double refLatRad = toRadians(refLat);
double refLonRad = toRadians(refLon);

*lat = toDegrees(y / earthRadius) + refLat;
*lon = toDegrees(x / (earthRadius * cos(refLatRad))) + refLon;
}

// 测试转换函数
int main() {
double lat = 31.2304; // 纬度
double lon = 121.4737; // 经度
double refLat = 31.2300; // 参考点纬度
double refLon = 121.4730; // 参考点经度
double x, y;
double resultLat, resultLon;

// 经纬度转平面坐标系
convertLatLonToXY(lat, lon, refLat, refLon, &x, &y);
printf("平面坐标系坐标：x = %lf, y = %lf\n", x, y);

// 平面坐标系转经纬度
convertXYToLatLon(x, y, refLat, refLon, &resultLat, &resultLon);
printf("经纬度坐标：lat = %lf, lon = %lf\n", resultLat, resultLon);

return 0;
}

请注意，此代码只是提供了基本的经纬度和平面坐标系转换功能，具体的算法和数学模型可能因应用场景的不同而有所变化。使用时请仔细验证和适应实际需求。