Question

I want to find the point which is some X kms from the current point say A, Example

如果我知道中心点的纬度和长度我如何能够在环绕下获得四点的纬度。我正在搜索SO,时间相当长,但无法在 Java本或 Java找到解决办法。在我的案件中,只有几公里长的地方,才能将地球视为平地表,这样,大鼠就提出这种解决办法。

增编

Answer 1

鉴于起点、初始接合点和距离,这将计算出沿(直观距离)高圈弧行驶的目的地和最终接合点。

Java:

var lat2 = Math.asin( Math.sin(lat1)*Math.cos(d/R) + 
                      Math.cos(lat1)*Math.sin(d/R)*Math.cos(brng) );
var lon2 = lon1 + Math.atan2(Math.sin(brng)*Math.sin(d/R)*Math.cos(lat1), 
                             Math.cos(d/R)-Math.sin(lat1)*Math.sin(lat2));

Answer 2

I was crushing the code above Until I did not realized that both lat and lon are the radian Value! You have to convert them back to the degrees, and you will see real values.

这里是法典

public class cal
{         
     public static void main(String []args){
         double R = 6371.0;
         double d = 0.100;
         double lat1 = 42.873, lon1 = 74.568;
         double Radlat1 = Math.toRadians(42.873);
         double Radlon1 = Math.toRadians(74.568);
         double brng = Math.toRadians(225);
         double lat2,lon2;

        System.out.println("Hello World");
        //lat2 = Math.sin();

        lat2 = Math.asin(Math.sin(Radlat1)*Math.cos(d/R) + 
                      Math.cos(Radlat1)*Math.sin(d/R)*Math.cos(brng));



         System.out.println("Lat2degree"+"="+Math.toDegrees(lat2)+"
"+"Lat2="+lat2);

        double NewRadLat2 = Math.toRadians(Math.toDegrees(lat2));


        lon2 = Radlon1+Math.atan2(Math.sin(brng)*Math.sin(d/R)*Math.cos(Radlat1), 
                             Math.cos(d/R)-Math.sin(Radlat1)*Math.sin(lat2));

        System.out.println("Lon2"+"="+Math.toDegrees(lon2));

     }
}

友情链接