以下的方法可以实现CGCS2000大地坐标系XY值转化为对应经纬度信息,
注意输入的XY值与给定的值相反,则参数X为已知的Y,参数Y为已知的X。
得出的结果为 [纬度,经度],不要应用错误。
L0参数为中央子午线的经线值,应用方法前需要确定L0的值,否则得出的值会有很大的偏差。
public static double [] xytolatlon(double X, double Y ,double L0) {
double lat ,lon;
Y-=500000;
double [] result = new double[2];
double iPI = 0.0174532925199433;//pi/180
double a = 6378137.0; //长半轴 m
double b = 6356752.31414; //短半轴 m
double f = 1/298.257222101;//扁率 a-b/a
double e = 0.0818191910428; //第一偏心率 Math.sqrt(5)
double ee = Math.sqrt(a*a-b*b)/b; //第二偏心率
double bf = 0; //底点纬度
double a0 = 1+(3*e*e/4) + (45*e*e*e*e/64) + (175*e*e*e*e*e*e/256) + (11025*e*e*e*e*e*e*e*e/16384) + (43659*e*e*e*e*e*e*e*e*e*e/65536);
double b0 = X/(a*(1-e*e)*a0);
double c1 = 3*e*e/8 +3*e*e*e*e/16 + 213*e*e*e*e*e*e/2048 + 255*e*e*e*e*e*e*e*e/4096;
double c2 = 21*e*e*e*e/256 + 21*e*e*e*e*e*e/256 + 533*e*e*e*e*e*e*e*e/8192;
double c3 = 151*e*e*e*e*e*e*e*e/6144 + 151*e*e*e*e*e*e*e*e/4096;
double c4 = 1097*e*e*e*e*e*e*e*e/131072;
bf = b0 + c1*Math.sin(2*b0) + c2*Math.sin(4*b0) +c3*Math.sin(6*b0) + c4*Math.sin(8*b0); // bf =b0+c1*sin2b0 + c2*sin4b0 + c3*sin6b0 +c4*sin8b0 +...
double tf = Math.tan(bf);
double n2 = ee*ee*Math.cos(bf)*Math.cos(bf); //第二偏心率平方成bf余弦平方
double c = a*a/b;
double v=Math.sqrt(1+ ee*ee*Math.cos(bf)*Math.cos(bf));
double mf = c/(v*v*v); //子午圈半径
double nf = c/v;//卯酉圈半径
//纬度计算
lat=bf-(tf/(2*mf)*Y)*(Y/nf) * (1-1/12*(5+3*tf*tf+n2-9*n2*tf*tf)*(Y*Y/(nf*nf))+1/360*(61+90*tf*tf+45*tf*tf*tf*tf)*(Y*Y*Y*Y/(nf*nf*nf*nf)));
//经度偏差
lon=1/(nf*Math.cos(bf))*Y -(1/(6*nf*nf*nf*Math.cos(bf)))*(1+2*tf*tf +n2)*Y*Y*Y + (1/(120*nf*nf*nf*nf*nf*Math.cos(bf)))*(5+28*tf*tf+24*tf*tf*tf*tf)*Y*Y*Y*Y*Y;
result[0] =retain6(lat/iPI);
result[1] =retain6(L0+lon/iPI);
//System.out.println(result[1]+","+result[0]);
return result;
}
private static double retain6(double num) {
String result = String.format("%.6f", num);
return Double.valueOf(result);
}