相关疑难解决方法(0)

我正在尝试将此http://www.movable-type.co.uk/scripts/latlong.html中提供的代码段转换为java.但是我没有得到与网站相同的结果.这是我的代码,用于找到两个点之间的中点,其中给出了它们的纬度和经度

midPoint(12.870672,77.658964,12.974831,77.60935);
    public static void midPoint(double lat1,double lon1,double lat2,double lon2)
    {
   double dLon = Math.toRadians(lon2-lon1);
        double Bx = Math.cos(lat2) * Math.cos(dLon);
        double By = Math.cos(lat2) * Math.sin(dLon);
        double lat3 = Math.atan2(Math.sin(lat1)+Math.sin(lat2),Math.sqrt( (Math.cos(lat1)+Bx)*(Math.cos(lat1)+Bx) + By*By) );
        double lon3 = lon1 + Math.atan2(By, Math.cos(lat1) + Bx);
        System.out.print(lat3 +" " + lon3 );
    }

我不确定dLon是否正确.所以请帮助我们弄明白.PSI需要找到中点的纬度和经度

我一直在使用Moveable-Type网站来帮助我进行一些Geocoordinate计算并且它非常有用,但是,我在计算两个坐标之间的中点时遇到了一个错误.我的结果接近预期,但不够接近:

posA = {47.64570362, -122.14073746}
posB = {47.64316917, -122.14032175}

预期结果(取自可移动型计算器)= 47°38'40"N,122°08'26"W = {47.644444, -122.140556}我的结果:{49.6054801645915, -122.14052959995759}

这是我的代码:

private Geocoordinate MidPoint(Geocoordinate posA, Geocoordinate posB)
{
   Geocoordinate midPoint = new Geocoordinate();

   double dLon = DegreesToRadians(posB.Longitude - posA.Longitude);
   double Bx = Math.Cos(DegreesToRadians(posB.Latitude)) * Math.Cos(dLon);
   double By = Math.Cos(DegreesToRadians(posB.Latitude)) * Math.Sin(dLon);

   midPoint.Latitude = RadiansToDegrees(Math.Atan2(Math.Sin(DegreesToRadians(posA.Latitude)) + Math.Sin(DegreesToRadians(posB.Latitude)), 
                Math.Sqrt((Math.Cos(DegreesToRadians(posA.Latitude)) + Bx) * (Math.Cos(DegreesToRadians(posA.Latitude))) + Bx) + By * By));

   midPoint.Longitude = posA.Longitude + RadiansToDegrees(Math.Atan2(By, Math.Cos(DegreesToRadians(posA.Latitude)) + Bx));

   return midPoint;
}

我有几种私有方法可以在Degrees和Radians之间进行转换.例如

private double DegreeToRadian(double angle) …