给定当前点、距离和方位角，获取纬度/经度

只需要在 atan2(y,x) 函数中交换值即可，不是 atan2(x,y)！

虽然已经晚了，但对于那些可能会发现这个问题的人来说，您可以使用geographiclib库获得更准确的结果。查看大地测量问题描述和JavaScript示例，以便轻松入门如何回答主题问题以及许多其他问题。在包括Python在内的各种语言中实现。如果您关心准确性，比编写自己的代码要好得多；比之前“使用库”的建议中的VincentyDistance更好。正如文档所说：“重点是返回与舍入误差接近（约5-15纳米）的准确结果。”

这是一个基于Ed Williams航空公式的PHP版本。在PHP中，模数处理方式略有不同。这对我来说很有效。

function get_new_waypoint ( $lat, $lon, $radial, $magvar, $range )
{

   // $range in nm.
   // $radial is heading to or bearing from    
   // $magvar for local area.

   $range = $range * pi() /(180*60);
   $radial = $radial - $magvar ;

   if ( $radial < 1 )
     {
        $radial = 360 + $radial - $magvar; 
     }
   $radial =  deg2rad($radial);
   $tmp_lat = deg2rad($lat);
   $tmp_lon = deg2rad($lon);
   $new_lat = asin(sin($tmp_lat)* cos($range) + cos($tmp_lat) * sin($range) * cos($radial));
   $new_lat = rad2deg($new_lat);
   $new_lon = $tmp_lon - asin(sin($radial) * sin($range)/cos($new_lat))+ pi() % 2 * pi() -  pi();
   $new_lon = rad2deg($new_lon);

   return $new_lat." ".$new_lon;

}

对于那些对Java解决方案感兴趣的人，这是我的代码： 我注意到初始解决方案需要一些调整才能返回正确的经度值，特别是当点位于极点之一时。 此外，有时需要进行四舍五入操作，因为在0纬度/经度上的结果似乎略微偏离0。对于小距离，四舍五入将有所帮助。

private static final double EARTH_RADIUS = 6371; // average earth radius

    /**
     * Returns the coordinates of the point situated at the distance specified, in
     * the direction specified. Note that the value is an approximation, not an
     * exact result.
     *
     * @param startPointLatitude
     * @param startPointLongitude
     * @param distanceInKm
     * @param bearing:            0 means moving north, 90 moving east, 180 moving
     *                            south, 270 moving west. Max value 360 min value 0;
     * @return new point location
     */
    public static LocationDTO getPointAt(double startPointLatitude, double startPointLongitude, double distanceInKm,
            double bearing) {
        if (Math.abs(startPointLatitude) > 90) {
            throw new BadRequestException(ExceptionMessages.INVALID_LATITUDE);
        } else if (Math.abs(startPointLatitude) == 90) {
            startPointLatitude = 89.99999 * Math.signum(startPointLatitude); // we have to do this conversion else the formula doesnt return the correct longitude value 
        }
        if (Math.abs(startPointLongitude) > 180) {
            throw new BadRequestException(ExceptionMessages.INVALID_LONGITUDE);
        }
        double angularDistance = distanceInKm / EARTH_RADIUS;
        bearing = deg2rad(bearing);
        startPointLatitude = deg2rad(startPointLatitude);
        startPointLongitude = deg2rad(startPointLongitude);
        double latitude = Math.asin(Math.sin(startPointLatitude) * Math.cos(angularDistance)
                + Math.cos(startPointLatitude) * Math.sin(angularDistance) * Math.cos(bearing));
        double longitude = startPointLongitude
                + Math.atan2(Math.sin(bearing) * Math.sin(angularDistance) * Math.cos(startPointLatitude),
                        Math.cos(angularDistance) - Math.sin(startPointLatitude) * Math.sin(latitude));
        longitude = (rad2deg(longitude) + 540) % 360 - 180; // normalize longitude to be in -180 +180 interval 
        LocationDTO result = new LocationDTO();
        result.setLatitude(roundValue(rad2deg(latitude)));
        result.setLongitude(roundValue(longitude));
        return result;
    }

    private static double roundValue(double value) {
        DecimalFormat df = new DecimalFormat("#.#####");
        df.setRoundingMode(RoundingMode.CEILING);
        return Double.valueOf(df.format(value));
    }


    // This function converts decimal degrees to radians
    private static double deg2rad(double deg) {
        return (deg * Math.PI / 180.0);
    }

    // This function converts radians to decimal degrees
    private static double rad2deg(double rad) {
        return (rad * 180.0 / Math.PI);
    }

非常晚才来参加派对，但这里有一个用R语言的答案，供有兴趣的人参考。我唯一做出的改变是将半径设置为米，所以d也需要设置为米。

get_point_at_distance <- function(lon, lat, d, bearing, R = 6378137) {
  # lat: initial latitude, in degrees
  # lon: initial longitude, in degrees
  # d: target distance from initial point (in m)
  # bearing: (true) heading in degrees
  # R: mean radius of earth (in m)
  # Returns new lat/lon coordinate {d} m from initial, in degrees
  ## convert to radians
  lat1 <- lat * (pi/180)
  lon1 <- lon * (pi/180)
  a <- bearing * (pi/180)
  ## new position
  lat2 <- asin(sin(lat1) * cos(d/R) + cos(lat1) * sin(d/R) * cos(a))
  lon2 <- lon1 + atan2(
    sin(a) * sin(d/R) * cos(lat1),
    cos(d/R) - sin(lat1) * sin(lat2)
  )
  ## convert back to degrees
  lat2 <- lat2 * (180/pi)
  lon2 <- lon2 * (180/pi)
  ## return
  return(c(lon2, lat2))
}

lat = 52.20472
lon = 0.14056
distance = 15000
bearing = 90
get_point_at_distance(lon = lon, lat = lat, d = distance, bearing = bearing)
# [1]  0.3604322 52.2045157