54 lines
2.0 KiB
C#
54 lines
2.0 KiB
C#
using System;
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namespace 电子展板.Utility.Other
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{
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public class CalcUtils
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{
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/// <summary>
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/// 计算两个坐标的距离
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/// </summary>
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/// <param name="long1">经度1</param>
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/// <param name="lat1">纬度1</param>
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/// <param name="long2">经度2</param>
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/// <param name="lat2">纬度2</param>
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/// <returns></returns>
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public static double getDistance(double long1, double lat1, double long2, double lat2)
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{
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double a, b, R;
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R = 6371393; // 地球半径
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lat1 = lat1 * Math.PI / 180.0;
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lat2 = lat2 * Math.PI / 180.0;
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a = lat1 - lat2;
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b = (long1 - long2) * Math.PI / 180.0;
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double d;
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double sa2, sb2;
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sa2 = Math.Sin(a / 2.0);
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sb2 = Math.Sin(b / 2.0);
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d = 2 * R * Math.Asin(Math.Sqrt(sa2 * sa2 + Math.Cos(lat1) * Math.Cos(lat2) * sb2 * sb2));
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return d;
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}
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public static double algorithm(double longitude1, double latitude1, double longitude2, double latitude2)
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{
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double Lat1 = rad(latitude1); // 纬度
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double Lat2 = rad(latitude2);
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double a = Lat1 - Lat2;//两点纬度之差
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double b = rad(longitude1) - rad(longitude2); //经度之差
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double s = 2 * Math.Asin(Math.Sqrt(Math.Pow(Math.Sin(a / 2), 2) + Math.Cos(Lat1) * Math.Cos(Lat2) * Math.Pow(Math.Sin(b / 2), 2)));//计算两点距离的公式
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s = s * 6378137.0;//弧长乘地球半径（半径为米）
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s = Math.Round(s * 10000d) / 10000d;//精确距离的数值
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return s;
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}
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private static double rad(double d)
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{
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return d * Math.PI / 180.00; //角度转换成弧度
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}
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public static double getVectorAngle(double x1, double y1, double x2, double y2)
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{
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return Math.Acos((x1 * x2 + y1 * y2) / (Math.Sqrt(x1 * x1 + y1 * y1) * Math.Sqrt(x2 * x2 + y2 * y2))) * 180.0 / Math.PI;
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}
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}
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}
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