GIS中通過兩點經緯度確定方位角與方位
阿新 • • 發佈:2019-01-22
確定逆向地理編碼時服務商對地址的方位沒有清楚的描述,導致偏遠的地區沒有明確的描述,此演算法通過兩個座標的相對位置計算出了方位角得到方位,可以清楚的描述 A地址距離B地址南北方向5000米,類似這樣的說明。
此程式碼為後端C#程式碼,也是就是這個,靈魂已有,各自實施。
public static class LatLonGetDirection { /// <summary> /// 根據經緯度計算角度 /// </summary> /// <param name="lat1"></param> /// <param name="lng1"></param> /// <param name="lat2"></param> /// <param name="lng2"></param> /// <returns></returns> public static double GetJiaoDu(location ll, location ll2) { double x1 = Convert.ToDouble(ll.lng); double y1 = Convert.ToDouble(ll.lat); double x2 = Convert.ToDouble(ll2.lng); double y2 = Convert.ToDouble(ll2.lat); double pi = Math.PI; double w1 = y1 / 180 * pi; double j1 = x1 / 180 * pi; double w2 = y2 / 180 * pi; double j2 = x2 / 180 * pi; double ret; if (j1 == j2) { if (w1 > w2) return 270; //北半球的情況,南半球忽略 else if (w1 < w2) return 90; else return -1;//位置完全相同 } ret = 4 * Math.Pow(Math.Sin((w1 - w2) / 2), 2) - Math.Pow(Math.Sin((j1 - j2) / 2) * (Math.Cos(w1) - Math.Cos(w2)), 2); ret = Math.Sqrt(ret); double temp = (Math.Sin(Math.Abs(j1 - j2) / 2) * (Math.Cos(w1) + Math.Cos(w2))); ret = ret / temp; ret = Math.Atan(ret) / pi * 180; if (j1 > j2) // 1為參考點座標 { if (w1 > w2) ret += 180; else ret = 180 - ret; } else if (w1 > w2) ret = 360 - ret; return ret; } /// <summary> /// 計算方位 按360計算 /// </summary> /// <param name="lat1">參照物緯度</param> /// <param name="lng1">參照物經度</param> /// <param name="lat2">目標緯度</param> /// <param name="lng2">目標經度</param> /// <returns></returns> public static string GetDirection(location ll, location ll2) { double jiaodu = GetJiaoDu(ll, ll2); if ((jiaodu <= 10) || (jiaodu > 350)) return "東"; if ((jiaodu > 10) && (jiaodu <= 80)) return "東北"; if ((jiaodu > 80) && (jiaodu <= 100)) return "北"; if ((jiaodu > 100) && (jiaodu <= 170)) return "西北"; if ((jiaodu > 170) && (jiaodu <= 190)) return "西"; if ((jiaodu > 190) && (jiaodu <= 260)) return "西南"; if ((jiaodu > 260) && (jiaodu <= 280)) return "南"; if ((jiaodu > 280) && (jiaodu <= 350)) return "東南"; return string.Empty; } }