我正在编写一个Java程序,以生成与我给定点相距固定距离的所有经度和纬度。距离必须精确到2000公里,而不是2000公里。
这是我的代码
public static void getLocation(double x0, double y0, int meters) {
Random random = new Random();
// Convert radius from meters to degrees
double radiusInDegrees = meters / 111000f;
double u = random.nextDouble();
double v = random.nextDouble();
double w = radiusInDegrees * Math.sqrt(u);
double t = 2 * Math.PI * v;
double x = w * Math.cos(t);
double y = w * Math.sin(t);
// Adjust the x-coordinate for the shrinking of the east-west distances
// double new_x = x / Math.cos(Math.toRadians(y0));
double foundLongitude = x + x0;
double foundLatitude = y + y0;
System.out.println("Longitude: " + foundLongitude + " Latitude: " + foundLatitude );
}
如何使所有点到地理点的距离相等,例如形成一个圆?
最佳答案
public static void generatePoint(double latitude, double longitude, double distanceInMetres, double bearing) {
Random random = new Random();
//int bear = random.nextInt(360);
double brngRad = Math.toRadians(bearing);
double latRad = Math.toRadians(latitude);
double lonRad = Math.toRadians(longitude);
int earthRadiusInMetres = 6371000;
double distFrac = distanceInMetres / earthRadiusInMetres;
double latitudeResult = Math.asin(Math.sin(latRad) * Math.cos(distFrac) + Math.cos(latRad) * Math.sin(distFrac) * Math.cos(brngRad));
double a = Math.atan2(Math.sin(brngRad) * Math.sin(distFrac) * Math.cos(latRad), Math.cos(distFrac) - Math.sin(latRad) * Math.sin(latitudeResult));
double longitudeResult = (lonRad + a + 3 * Math.PI) % (2 * Math.PI) - Math.PI;
System.out.println("bearing: "+bearing+ ", latitude: " + Math.toDegrees(latitudeResult) + ", longitude: " + Math.toDegrees(longitudeResult));
}
需要添加轴承