我需要以各种方式处理地球坐标。C/C++ 中没有可以立即执行此操作的函数。
参考以下问题：
从第一个和活字脚本网站，我发现以下是公式：
查找 2 个坐标之间的方位角（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
查找 2 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+角度求坐标
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));
下面的伪代码应该相互验证以上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);
现在实际上答案几乎接近但不正确。即cX不等于c2！经度值的差异
总是存在差异的。0.0005例如
c1 = (12.968460,77.641308)
c2 = (12.967862,77.653130)
angle = 92.97 ^^^
distance = 1282.74
cX = (12.967862,77.653613)
^^^
我对数学的知识并不多。 Havesine Forumla。但我所知道的是，从fcc.gov 网站上，答案总是正确的。
我究竟做错了什么？
代码仅供参考
虽然语法是 C++，但所有数学函数都来自 C，并且在 C 中也很容易移植（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";
}