gps单点定位算法c语言

GPS单点定位算法是通过接收卫星信号，计算出自身位置的一种方法。下面是一个简单的C语言实现：

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

#define PI 3.14159265358979323846
#define c 299792458.0
#define L1 1575.42e6
#define L2 1227.6e6
#define f1 L1
#define f2 L2

typedef struct {
    double x;
    double y;
    double z;
} XYZ;

typedef struct {
    double latitude;
    double longitude;
    double altitude;
} LLH;

double R = 6378137.0;
double e = 0.0818191908425;

double deg2rad(double deg) {
    return deg * PI / 180.0;
}

double rad2deg(double rad) {
    return rad * 180.0 / PI;
}

double getDistance(XYZ p1, XYZ p2) {
    double dx = p1.x - p2.x;
    double dy = p1.y - p2.y;
    double dz = p1.z - p2.z;
    return sqrt(dx*dx + dy*dy + dz*dz);
}

double getElevationAngle(XYZ p1, XYZ p2) {
    double dx = p2.x - p1.x;
    double dy = p2.y - p1.y;
    double dz = p2.z - p1.z;
    double r = sqrt(dx*dx + dy*dy + dz*dz);
    double lat1 = atan(p1.z / sqrt(p1.x*p1.x + p1.y*p1.y));
    double lon1 = atan2(p1.y, p1.x);
    double lat2 = atan(p2.z / sqrt(p2.x*p2.x + p2.y*p2.y));
    double lon2 = atan2(p2.y, p2.x);
    double dlat = lat2 - lat1;
    double dlon = lon2 - lon1;
    double a = sin(dlat / 2) * sin(dlat / 2) + cos(lat1) * cos(lat2) * sin(dlon / 2) * sin(dlon / 2);
    double c = 2 * atan2(sqrt(a), sqrt(1 - a));
    double h = r * sin(c);
    double theta = atan2(dz, sqrt(dx*dx + dy*dy));
    return atan(h / (r - h / (1 + e*e*cos(theta)*cos(theta))));
}

double getSatelliteDistance(double pr1, double pr2, double t1, double t2) {
    double a = pow(c / f1 * (t2 - t1) + pr1, 2);
    double b = pow(c / f2 * (t2 - t1) + pr2, 2);
    double d = sqrt(a - b) / (1 - f2*f2 / f1*f1);
    return d;
}

void getLocation(XYZ* positions, double* pr, double* t, int n, LLH* llh) {
    double x, y, z, rho, phi, lambda, h, N;
    double delta = 1.0;
    XYZ deltaPos;
    deltaPos.x = delta;
    deltaPos.y = delta;
    deltaPos.z = delta;
    while (delta > 1e-5) {
        x = positions[0].x;
        y = positions[0].y;
        z = positions[0].z;
        for (int i = 0; i < n; i++) {
            double d = getSatelliteDistance(pr[i], 0, t[i], 0);
            double e = getElevationAngle(positions[i], deltaPos);
            x += (positions[i].x - deltaPos.x) / d * (d - pr[i]) / e;
            y += (positions[i].y - deltaPos.y) / d * (d - pr[i]) / e;
            z += (positions[i].z - deltaPos.z) / d * (d - pr[i]) / e;
        }
        deltaPos.x = x;
        deltaPos.y = y;
        deltaPos.z = z;
        rho = sqrt(x*x + y*y);
        phi = atan2(z, rho);
        lambda = atan2(y, x);
        N = R / sqrt(1 - e*e*sin(phi)*sin(phi));
        h = rho / cos(phi) - N;
        delta = getDistance(deltaPos, positions[0]);
        positions[0].x = x;
        positions[0].y = y;
        positions[0].z = z;
    }
    llh->latitude = rad2deg(phi);
    llh->longitude = rad2deg(lambda);
    llh->altitude = h;
}

int main() {
    int n = 4;
    double pr[] = { 2.0e7, 2.1e7, 2.2e7, 2.3e7 };
    double t[] = { 0.0, 0.0, 0.0, 0.0 };
    XYZ positions[] = {
        { 2.0e7, 0.0, 0.0 },
        { 0.0, 2.1e7, 0.0 },
        { -2.2e7, 0.0, 0.0 },
        { 0.0, -2.3e7, 0.0 }
    };
    LLH llh;
    getLocation(positions, pr, t, n, &llh);
    printf("latitude: %f\n", llh.latitude);
    printf("longitude: %f\n", llh.longitude);
    printf("altitude: %f\n", llh.altitude);
    return 0;
}

这是一个简单的实现，实际上还有许多细节需要考虑，例如钟差、电离层延迟等。如果需要更精确的定位，可以参考GPS接收机的设计。