c# – 计算给定距离,轴承,起点的端点

概述我试图找到目的地点,给定起点lat / long,轴承&距离.下面这个网站的计算器给了我所需的结果. http://www.movable-type.co.uk/scripts/latlong.html 当我尝试通过代码实现相同的时候,我没有得到正确的结果. 以下是我的代码 – private GLatLng pointRadialDistance(double lat1, double lon 我试图找到目的地点,给定起点lat / long,轴承&距离.下面这个网站的计算器给了我所需的结果.

http://www.movable-type.co.uk/scripts/latlong.html

当我尝试通过代码实现相同的时候,我没有得到正确的结果.

以下是我的代码 –

private  GLatLng poinTradialdistance(double lat1,double lon1,double radianbearing,double radialdistance)    {        double rEarth = 6371.01;        lat1 = DegreetoRadian(lat1);        lon1 = DegreetoRadian(lon1);        radianbearing = DegreetoRadian(radianbearing);        radialdistance = radialdistance / rEarth;        double lat = Math.Asin(Math.Sin(lat1) * Math.Cos(radialdistance) + Math.Cos(lat1)                        * Math.Sin(radialdistance) * Math.Cos(radianbearing));        double lon;        if (Math.Cos(lat) == 0)        {  // Endpoint a pole             lon = lon1;        }        else        {            lon = ((lon1 - Math.Asin(Math.Sin(radianbearing) * Math.Sin(radialdistance) / Math.Cos(lat))                            + Math.PI) % (2 * Math.PI)) - Math.PI;        }        lat = RadianToDegree(lat);        lon = RadianToDegree(lon);        GLatLng newLatLng = new GLatLng(lat,lon);        return newLatLng;    }    public  double bearing(double lat1,double long1,double lat2,double long2)    {        //Convert input values to radians           lat1 = DegreetoRadian(lat1);        long1 = DegreetoRadian(long1);        lat2 = DegreetoRadian(lat2);        long2 = DegreetoRadian(long2);        double deltaLong = long2 - long1;        double y = Math.Sin(deltaLong) * Math.Cos(lat2);        double x = Math.Cos(lat1) * Math.Sin(lat2) -                Math.Sin(lat1) * Math.Cos(lat2) * Math.Cos(deltaLong);        double bearing = Math.atan2(y,x);        return bearing;    }      public double DegreetoRadian(double angle)    {    return Math.PI * angle / 180.0;    }    public double RadianToDegree(double angle)    {        return 180.0 * angle / Math.PI;    }

从主程序,我调用子程序如下 –

double bearing = bearing(-41.294444,174.814444,-40.90521,175.6604);GLatLng endLatLng = poinTradialdistance(-41.294444,bearing,80);

我得到以下结果 –

bearing=1.02749621782165endLatLng=-40.5751022737927,174.797458881699

我期望的答案是-40.939722,175.646389(从上面的网站链接).

任何人都可以建议我在代码中出现什么错误？

@H_404_22@解决方法 这是一些实现你想要做的代码.

public static Geolocation FindPointAtdistanceFrom(Geolocation startPoint,double initialbearingradians,double distanceKilometres){    const double radiusEarthKilometres = 6371.01;    var distRatio = distanceKilometres / radiusEarthKilometres;    var distRatioSine = Math.Sin(distRatio);    var distRatioCosine = Math.Cos(distRatio);    var startLaTrad = degreesToradians(startPoint.Latitude);    var startLonRad = degreesToradians(startPoint.Longitude);    var startLatCos = Math.Cos(startLaTrad);    var startLatSin = Math.Sin(startLaTrad);    var endLaTrads = Math.Asin((startLatSin * distRatioCosine) + (startLatCos * distRatioSine * Math.Cos(initialbearingradians)));    var endLonRads = startLonRad        + Math.atan2(            Math.Sin(initialbearingradians) * distRatioSine * startLatCos,distRatioCosine - startLatSin * Math.Sin(endLaTrads));    return new Geolocation    {        Latitude = radiansTodegrees(endLaTrads),Longitude = radiansTodegrees(endLonRads)    };}public struct Geolocation{    public double Latitude { get; set; }    public double Longitude { get; set; }}public static double degreesToradians(double degrees){    const double degToRadFactor = Math.PI / 180;    return degrees * degToRadFactor;}public static double radiansTodegrees(double radians){    const double radToDegFactor = 180 / Math.PI;    return radians * radToDegFactor;}

@H_404_22@ @H_404_22@ 总结

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