已知一个点的经纬度、方位角、距离,求另一点经纬度

已知一个点的经纬度、方位角、距离,求另一点经纬度
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大家好,我是架构君,一个会写代码吟诗的架构师。今天说一说已知一个点的经纬度、方位角、距离,求另一点经纬度,希望能够帮助大家进步!!!

 /** * 长半径a=6378137 米 */ public static double EARTH_RADIUS = 6378137; /** * 短半径b=6356752.3142 */ public static double b = 6356752.3142; /** * 扁率f=1/298.2572236 */ public static double f = 1 / 298.2572236; /** * 已知一点经纬度,方位角,距离,求另一点经纬度 * 通过三角函数求终点坐标-球面坐标系 * </summary> * <param name="angle">角度</param> *<param name="startPoint">起点</param> * <param name="distance">距离(米)</param> * <returns>终点坐标</returns> */ public static double[] getEndPointByTrigonometric(double angle, double[] startPoint, double distance) { double lon=startPoint[0]; double lat=startPoint[1]; double alpha1 = rad(angle); double sinAlpha1 = Math.sin(alpha1); double cosAlpha1 = Math.cos(alpha1); double tanU1 = (1 - f) * Math.tan(rad(lat)); double cosU1 = 1 / Math.sqrt((1 + tanU1 * tanU1)); double sinU1 = tanU1 * cosU1; double sigma1 = Math.atan2(tanU1, cosAlpha1); double sinAlpha = cosU1 * sinAlpha1; double cosSqAlpha = 1 - sinAlpha * sinAlpha; double uSq = cosSqAlpha * (EARTH_RADIUS * EARTH_RADIUS - b * b) / (b * b); double A = 1 + uSq / 16384 * (4096 + uSq * (-768 + uSq * (320 - 175 * uSq))); double B = uSq / 1024 * (256 + uSq * (-128 + uSq * (74 - 47 * uSq))); double cos2SigmaM = 0; double sinSigma = 0; double cosSigma = 0; double sigma = distance / (b * A), sigmaP = 2 * Math.PI; while (Math.abs(sigma - sigmaP) > 1e-12) { cos2SigmaM = Math.cos(2 * sigma1 + sigma); sinSigma = Math.sin(sigma); cosSigma = Math.cos(sigma); double deltaSigma = B * sinSigma * (cos2SigmaM + B / 4 * (cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM) - B / 6 * cos2SigmaM * (-3 + 4 * sinSigma * sinSigma) * (-3 + 4 * cos2SigmaM * cos2SigmaM))); sigmaP = sigma; sigma = distance / (b * A) + deltaSigma; } double tmp = sinU1 * sinSigma - cosU1 * cosSigma * cosAlpha1; double lat2 = Math.atan2(sinU1 * cosSigma + cosU1 * sinSigma * cosAlpha1, (1 - f) * Math.sqrt(sinAlpha * sinAlpha + tmp * tmp)); double lambda = Math.atan2(sinSigma * sinAlpha1, cosU1 * cosSigma - sinU1 * sinSigma * cosAlpha1); double C = f / 16 * cosSqAlpha * (4 + f * (4 - 3 * cosSqAlpha)); double L = lambda - (1 - C) * f * sinAlpha * (sigma + C * sinSigma * (cos2SigmaM + C * cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM))); double revAz = Math.atan2(sinAlpha, -tmp); // final bearing
// System.out.println(revAz);
// System.out.println(lon + deg(L) + "," + deg(lat2)); double[] endPoint=new double[2]; endPoint[0]=lon + deg(L); endPoint[1]= deg(lat2); return endPoint; } /** * 度换成弧度 * @param d 度 * @return 弧度 */ public static double rad(double d) { return d * Math.PI / 180.0; } /** * 弧度换成度 * @param x 弧度 * @return 度 */ public static double deg(double x) { return x * 180 / Math.PI; }

本文来源huayang183,由架构君转载发布,观点不代表Java架构师必看的立场,转载请标明来源出处:https://javajgs.com/archives/17953

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