查找地球坐标(经纬度),距离(米)和方位(角)

Finding Earth Coordinates (latitude,longitude), Distance (meters) and Bearing (angle)

本文关键字:方位 距离 地球 经纬度 查找 坐标      更新时间:2023-10-16

我需要以各种方式处理地球坐标。在C/c++中没有一个函数可以直接做到这一点。
参考以下问题:

  1. Python:给定当前点,距离和轴承
  2. C:给定点(纬度,经度),距离和方位,如何得到新的纬度和经度

从第一个和活字脚本网站,我发现下面是公式:

查找两个坐标之间的方位(角度)

x = cos(lat1Rad)*sin(lat2Rad) - sin(lat1Rad)*cos(lat2Rad)*cos(lon2Rad-lon1Rad);
y = sin(lon2Rad-lon1Rad) * cos(lat2Rad);
bearing = atan2(y, x);  // In radians; 
// Convert to degrees and for -ve add 360

求两个坐标之间的距离(米)

PI = 3.14159265358979323846, earthDiameterMeters = 2*6371*1000;
x = sin((lat2Rad-lat1Rad) / 2);
y = sin((lon2Rad-lon1Rad) / 2);
meters = earthDiameterMeters * asin(sqrt(x*x + y*y*cos(lat1Rad)*cos(lat2Rad)));

从坐标+距离+角度查找坐标

meters *= 2 / earthDiameterMeters;
lat2Rad = asin(sin(lat1Rad)*cos(meters) + cos(lat1Rad)*sin(meters)*cos(bearing));
lon2Rad = lon1Rad + atan2(sin(bearing)*sin(meters)*cos(lat1Rad), 
                          cos(meters) - sin(lat1Rad)*sin(lat2Rad));

下面的伪代码应相互验证以上3个方程:

struct Coordinate { double lat, lon; } c1, c2;  
auto degree = FindBearing(c1, c2);
auto meters = FindDistance(c1, c2);
auto cX = FindCoordiante(c1, degree, meters);

现在实际上答案是几乎接近但不是正确的。即cX不等于c2!
0.0005的经度值总是存在差异。例如

c1 = (12.968460,77.641308)  
c2 = (12.967862,77.653130)  
angle = 92.97         ^^^
distance = 1282.74  
cX = (12.967862,77.653613)
                      ^^^

我没有太多的数学知识。但我知道的是,从fcc.gov网站上,答案总是正确的。

我做错了什么?

仅供参考的代码

虽然语法是c++的,但所有的数学函数都来自C,并且很容易移植到C中(因此标记为两者)

#include<iostream>
#include<iomanip>
#include<cmath>

// Source: http://www.movable-type.co.uk/scripts/latlong.html
static const double PI = 3.14159265358979323846, earthDiameterMeters = 6371.0 * 2 * 1000;
double degreeToRadian (const double degree) { return (degree * PI / 180); };
double radianToDegree (const double radian) { return (radian * 180 / PI); };
double CoordinatesToAngle (double latitude1,
                           const double longitude1,
                           double latitude2,
                           const double longitude2)
{
  const auto longitudeDifference = degreeToRadian(longitude2 - longitude1);
  latitude1 = degreeToRadian(latitude1);
  latitude2 = degreeToRadian(latitude2);
  using namespace std;
  const auto x = (cos(latitude1) * sin(latitude2)) -
                 (sin(latitude1) * cos(latitude2) * cos(longitudeDifference));
  const auto y = sin(longitudeDifference) * cos(latitude2);
  const auto degree = radianToDegree(atan2(y, x));
  return (degree >= 0)? degree : (degree + 360);
}
double CoordinatesToMeters (double latitude1,
                            double longitude1,
                            double latitude2,
                            double longitude2)
{
  latitude1 = degreeToRadian(latitude1);
  longitude1 = degreeToRadian(longitude1);
  latitude2 = degreeToRadian(latitude2);
  longitude2 = degreeToRadian(longitude2);
  using namespace std;
  auto x = sin((latitude2 - latitude1) / 2), y = sin((longitude2 - longitude1) / 2);
#if 1
  return earthDiameterMeters * asin(sqrt((x * x) + (cos(latitude1) * cos(latitude2) * y * y)));
#else
  auto value = (x * x) + (cos(latitude1) * cos(latitude2) * y * y);
  return earthDiameterMeters * atan2(sqrt(value), sqrt(1 - value));
#endif
}
std::pair<double,double> CoordinateToCoordinate (double latitude,
                                                 double longitude,
                                                 double angle,
                                                 double meters)
{
  latitude = degreeToRadian(latitude);
  longitude = degreeToRadian(longitude);
  angle = degreeToRadian(angle);
  meters *= 2 / earthDiameterMeters;
  using namespace std;
  pair<double,double> coordinate;
  coordinate.first = radianToDegree(asin((sin(latitude) * cos(meters))
                             + (cos(latitude) * sin(meters) * cos(angle))));
  coordinate.second = radianToDegree(longitude
                    + atan2((sin(angle) * sin(meters) * cos(latitude)),
                    cos(meters) - (sin(latitude) * sin(coordinate.first))));
  return coordinate;
}
int main ()
{
  using namespace std;
  const auto latitude1 = 12.968460, longitude1 = 77.641308,
             latitude2 = 12.967862, longitude2 = 77.653130;
  cout << std::setprecision(10);
  cout << "(" << latitude1 << "," << longitude1 << ") --- "
          "(" << latitude2 << "," << longitude2 << ")n";
  auto angle = CoordinatesToAngle(latitude1, longitude1, latitude2, longitude2);
  cout << "Angle =  " << angle << endl;
  auto meters = CoordinatesToMeters(latitude1, longitude1, latitude2, longitude2);
  cout << "Meters = " << meters << endl;
  auto coordinate = CoordinateToCoordinate(latitude1, longitude1, angle, meters);
  cout << "Destination = (" << coordinate.first << "," << coordinate.second << ")n";
}

CoordinateToCoordinate中,您使用已经以度表示的sin(coordinate.first)。使用sin(degreeToRadian(coordinate.first))

或者更干净:

... CoordinateToCoordinate (...)
{
  ...
  coordinate.first = asin((sin(latitude) * cos(meters))
                        + (cos(latitude) * sin(meters) * cos(angle)));
  coordinate.second = longitude + atan2((sin(angle) * sin(meters) * cos(latitude)), 
         cos(meters) - (sin(latitude) * sin(coordinate.first)));
  coordinate.first = radianToDegree(coordinate.first);
  coordinate.second = radianToDegree(coordinate.second);
  return coordinate;
}

这解决了这个问题。现场演示。

部分答案

将角度92.97°转换为弧度后,调用sin/cos/tan将有效地将角度更改为2.97°。在度到弧度转换后以及在三角函数调用中发生周期缩减时,仅这一步就损失了6位精度。

具有较大角度的三角函数可以提高精度。使用幸运方向正好 360.0度在一个圆。执行"模数45°",可能使用remquo(angle, 45, &octant),并在使用radian参数调用三角函数之前将然后将转换为弧度。

示例sind()


77.641308和77.653130的答案相差约6600分之一(~13位精度)。这个答案可能不能完全解释这一点,但应该有所帮助。(如果某些地方出现了float的用法,则应改为double)