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GPS纠偏算法,适用于google,高德体系的地图

2014-10-09 16:39 501 查看
/**

* gps纠偏算法,适用于google,高德体系的地图

* @author Administrator

*/

public class GpsCorrect {

final static double pi = 3.14159265358979324;

final static double a = 6378245.0;

final static double ee = 0.00669342162296594323;

public static void transform(double wgLat, double wgLon, double[] latlng) {

if (outOfChina(wgLat, wgLon)) {

latlng[0] = wgLat;

latlng[1] = 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 = Math.sin(radLat);

magic = 1 - ee * magic * magic;

double sqrtMagic = Math.sqrt(magic);

dLat = (dLat * 180.0) / ((a * (1 - ee)) / (magic * sqrtMagic) * pi);

dLon = (dLon * 180.0) / (a / sqrtMagic * Math.cos(radLat) * pi);

latlng[0] = wgLat + dLat;

latlng[1] = wgLon + dLon;

}

private static boolean 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;

}

private 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 * Math.sqrt(Math.abs(x));

ret += (20.0 * Math.sin(6.0 * x * pi) + 20.0 * Math.sin(2.0 * x * pi)) * 2.0 / 3.0;

ret += (20.0 * Math.sin(y * pi) + 40.0 * Math.sin(y / 3.0 * pi)) * 2.0 / 3.0;

ret += (160.0 * Math.sin(y / 12.0 * pi) + 320 * Math.sin(y * pi / 30.0)) * 2.0 / 3.0;

return ret;

}

private 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 * Math.sqrt(Math.abs(x));

ret += (20.0 * Math.sin(6.0 * x * pi) + 20.0 * Math.sin(2.0 * x * pi)) * 2.0 / 3.0;

ret += (20.0 * Math.sin(x * pi) + 40.0 * Math.sin(x / 3.0 * pi)) * 2.0 / 3.0;

ret += (150.0 * Math.sin(x / 12.0 * pi) + 300.0 * Math.sin(x / 30.0 * pi)) * 2.0 / 3.0;

return ret;

}

}
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