c++ - 为什么 boost::geometry 地理 Vincenty 距离在赤道附近不准确？

Question

我需要一个函数来计算一对 WGS 84 之间的距离定位到高精度，我计划使用 boost geometry 中的 geographic 函数.

boost geometry Design Rational状态:

There is the Andoyer method, fast and precise, and there is the Vincenty method, slower and more precise..

但是，当使用 Andoyer 和 Vincenty 策略测试 boost::geometry::distance 函数时，我得到了以下结果:

WGS 84 values (metres)
    Semimajor axis:         6378137.000000
    Flattening:             0.003353
    Semiminor axis:         6356752.314245

    Semimajor distance:     20037508.342789
    Semiminor distance:     19970326.371123

Boost geometry near poles
Andoyer function:
    Semimajor distance:     20037508.151445
    Semiminor distance:     20003917.164970
Vincenty function:
    Semimajor distance:     **19970326.180419**
    Semiminor distance:     20003931.266635

Boost geometry at poles
Andoyer function:
    Semimajor distance:     0.000000
    Semiminor distance:     0.000000
Vincenty function:
    Semimajor distance:     **19970326.371122**
    Semiminor distance:     20003931.458623

Vincenty 沿半长轴(即赤道周围)的距离小于南北极之间的半短轴周围的距离。这不可能是正确的。

Semiminor 和 Andoyer 距离看起来很合理。除非点位于地球的另一侧，当 boost Andoyer 函数返回零时!

问题出在:Vincenty 算法、它的boost geometry 实现，还是我的测试代码？

测试代码:

/// boost geometry WGS84 distance issue

// Note: M_PI is not part of the C or C++ standards, _USE_MATH_DEFINES enables it
#define _USE_MATH_DEFINES
#include <boost/geometry.hpp>
#include <cmath>
#include <iostream>
#include <ios>

// WGS 84 parameters from: Eurocontrol WGS 84 Implementation Manual
// Version 2.4 Chapter 3, page 14

/// The Semimajor axis measured in metres.
/// This is the radius at the equator.
constexpr double a = 6378137.0;

/// Flattening, a ratio.
/// This is the flattening of the ellipse at the poles
constexpr double f = 1.0/298.257223563;

/// The Semiminor axis measured in metres.
/// This is the radius at the poles.
/// Note: this is derived from the Semimajor axis and the flattening.
/// See WGS 84 Implementation Manual equation B-2, page 69.
constexpr double b = a * (1.0 - f);

int main(int /*argc*/, char ** /*argv*/)
{
  std::cout.setf(std::ios::fixed);

  std::cout << "WGS 84 values (metres)\n";
  std::cout << "\tSemimajor axis:\t\t"   << a << "\n";
  std::cout << "\tFlattening:\t\t"       << f << "\n";
  std::cout << "\tSemiminor axis:\t\t"   << b << "\n\n";

  std::cout << "\tSemimajor distance:\t" << M_PI * a << "\n";
  std::cout << "\tSemiminor distance:\t" << M_PI * b << "\n";
  std::cout << std::endl;

  // Min value for delta. 0.000000014 causes Andoyer to fail.
  const double DELTA(0.000000015);

  // For boost::geometry:
  typedef boost::geometry::cs::geographic<boost::geometry::radian> Wgs84Coords;
  typedef boost::geometry::model::point<double, 2, Wgs84Coords> GeographicPoint;
  // Note boost points are Long & Lat NOT Lat & Long
  GeographicPoint near_north_pole   (0.0,  M_PI_2 - DELTA);
  GeographicPoint near_south_pole   (0.0, -M_PI_2 + DELTA);

  GeographicPoint near_equator_east ( M_PI_2 - DELTA, 0.0);
  GeographicPoint near_equator_west (-M_PI_2 + DELTA, 0.0);

  // Note: the default boost geometry spheroid is WGS84
  // #include <boost/geometry/core/srs.hpp>
  typedef boost::geometry::srs::spheroid<double> SpheroidType;
  SpheroidType spheriod;

  //#include <boost/geometry/strategies/geographic/distance_andoyer.hpp>
  typedef boost::geometry::strategy::distance::andoyer<SpheroidType>
                                                               AndoyerStrategy;
  AndoyerStrategy andoyer(spheriod);

  std::cout << "Boost geometry near poles\n";
  std::cout << "Andoyer function:\n";
  double andoyer_major(boost::geometry::distance(near_equator_east, near_equator_west, andoyer));
  std::cout << "\tSemimajor distance:\t" << andoyer_major << "\n";
  double andoyer_minor(boost::geometry::distance(near_north_pole, near_south_pole, andoyer));
  std::cout << "\tSemiminor distance:\t" << andoyer_minor << "\n";

  //#include <boost/geometry/strategies/geographic/distance_vincenty.hpp>
  typedef boost::geometry::strategy::distance::vincenty<SpheroidType>
                                                               VincentyStrategy;
  VincentyStrategy vincenty(spheriod);

  std::cout << "Vincenty function:\n";
  double vincenty_major(boost::geometry::distance(near_equator_east, near_equator_west, vincenty));
  std::cout << "\tSemimajor distance:\t" << vincenty_major << "\n";
  double vincenty_minor(boost::geometry::distance(near_north_pole, near_south_pole, vincenty));
  std::cout << "\tSemiminor distance:\t" << vincenty_minor << "\n\n";

  // Note boost points are Long & Lat NOT Lat & Long
  GeographicPoint north_pole   (0.0,  M_PI_2);
  GeographicPoint south_pole   (0.0, -M_PI_2);

  GeographicPoint equator_east ( M_PI_2, 0.0);
  GeographicPoint equator_west (-M_PI_2, 0.0);

  std::cout << "Boost geometry at poles\n";
  std::cout << "Andoyer function:\n";
  andoyer_major = boost::geometry::distance(equator_east, equator_west, andoyer);
  std::cout << "\tSemimajor distance:\t" << andoyer_major << "\n";
  andoyer_minor = boost::geometry::distance(north_pole, south_pole, andoyer);
  std::cout << "\tSemiminor distance:\t" << andoyer_minor << "\n";

  std::cout << "Vincenty function:\n";
  vincenty_major = boost::geometry::distance(equator_east, equator_west, vincenty);
  std::cout << "\tSemimajor distance:\t" << vincenty_major << "\n";
  vincenty_minor = boost::geometry::distance(north_pole, south_pole, vincenty);
  std::cout << "\tSemiminor distance:\t" << vincenty_minor << "\n";

  return 0;
}

Answer 1

我按照@jwd630 的建议查看了geographiclib .
以下是结果:

WGS 84 values (metres)
    Semimajor distance:    20037508.342789
    Semiminor distance:    19970326.371123

GeographicLib near antipodal
    Semimajor distance:    20003931.458625
    Semiminor distance:    20003931.455275

GeographicLib antipodal
    Semimajor distance:    20003931.458625
    Semiminor distance:    20003931.458625

GeographicLib verify
    JFK to LHR distance:   5551759.400319

即对于极点之间的半短距离(至 5dp)，它提供与 Vincenty 相同的距离，并且它为赤道上的对映点计算相同的距离。

这是正确的，因为赤道对映点之间的最短距离是通过其中一个极点，而不是像默认的 boost Andoyer 算法计算的那样围绕赤道。

所以@jwd630 上面的答案是正确的，在三个算法中，geographiclib是唯一一个计算整个 WGS84 大地水准面的正确距离的方法。

测试代码如下:

/// GeographicLib  WGS84 distance

// Note: M_PI is not part of the C or C++ standards, _USE_MATH_DEFINES enables it
#define _USE_MATH_DEFINES
#include <GeographicLib/Geodesic.hpp>
#include <cmath>
#include <iostream>
#include <ios>

// WGS 84 parameters from: Eurocontrol WGS 84 Implementation Manual
// Version 2.4 Chapter 3, page 14

/// The Semimajor axis measured in metres.
/// This is the radius at the equator.
constexpr double a = 6378137.0;

/// Flattening, a ratio.
/// This is the flattening of the ellipse at the poles
constexpr double f = 1.0/298.257223563;

/// The Semiminor axis measured in metres.
/// This is the radius at the poles.
/// Note: this is derived from the Semimajor axis and the flattening.
/// See WGS 84 Implementation Manual equation B-2, page 69.
constexpr double b = a * (1.0 - f);

int main(int /*argc*/, char ** /*argv*/)
{
  const GeographicLib::Geodesic& geod(GeographicLib::Geodesic::WGS84());

  std::cout.setf(std::ios::fixed);

  std::cout << "WGS 84 values (metres)\n";
  std::cout << "\tSemimajor axis:\t\t"   << a << "\n";
  std::cout << "\tFlattening:\t\t"       << f << "\n";
  std::cout << "\tSemiminor axis:\t\t"   << b << "\n\n";

  std::cout << "\tSemimajor distance:\t" << M_PI * a << "\n";
  std::cout << "\tSemiminor distance:\t" << M_PI * b << "\n";
  std::cout << std::endl;

  // Min value for delta. 0.000000014 causes boost Andoyer to fail.
  const double DELTA(0.000000015);

  std::pair<double, double> near_equator_east (0.0, 90.0 - DELTA);
  std::pair<double, double> near_equator_west (0.0, -90.0 + DELTA);

  std::cout << "GeographicLib near antipodal\n";
  double distance_metres(0.0);
  geod.Inverse(near_equator_east.first, near_equator_east.second,
               near_equator_west.first, near_equator_west.second, distance_metres);
  std::cout << "\tSemimajor distance:\t" << distance_metres << "\n";

  std::pair<double, double> near_north_pole   (90.0 - DELTA, 0.0);
  std::pair<double, double> near_south_pole   (-90.0 + DELTA, 0.0);

  geod.Inverse(near_north_pole.first, near_north_pole.second,
               near_south_pole.first, near_south_pole.second, distance_metres);
  std::cout << "\tSemiminor distance:\t" << distance_metres << "\n\n";

  std::pair<double, double> equator_east (0.0, 90.0);
  std::pair<double, double> equator_west (0.0, -90.0);

  std::cout << "GeographicLib antipodal\n";
  geod.Inverse(equator_east.first, equator_east.second,
               equator_west.first, equator_west.second, distance_metres);
  std::cout << "\tSemimajor distance:\t" << distance_metres << "\n";

  std::pair<double, double> north_pole   (90.0, 0.0);
  std::pair<double, double> south_pole   (-90.0, 0.0);

  geod.Inverse(north_pole.first, north_pole.second,
               south_pole.first, south_pole.second, distance_metres);
  std::cout << "\tSemiminor distance:\t" << distance_metres << "\n\n";

  std::pair<double, double> JFK   (40.6, -73.8);
  std::pair<double, double> LHR   (51.6, -0.5);

  std::cout << "GeographicLib verify distance\n";
  geod.Inverse(JFK.first, JFK.second,
               LHR.first, LHR.second, distance_metres);
  std::cout << "\tJFK to LHR distance:\t" << distance_metres << std::endl;

  return 0;
}

在他的论文中Algorithms for geodesics ,Charles F. F. Karney 指出，“Vincenty 的方法无法收敛于几乎对映的点”。这可能会回答我原来的问题，即 Vincenty 算法不适合对映点。

注意:我提出了 boost 票 #11817描述问题所在Andoyer 算法为对映点返回零，并向 boost 发送了一个拉取请求并修复了它。

然而，对于不正确距离的唯一正确修复是使用正确的算法，即:geographiclib

非常感谢 Charles F. F. Karney (@cffk) 礼貌地指出我的愚蠢错误!