Question

考虑点P1（60°N，20°E，0）和P2（60°N，22°E，0）地球表面

点P1和P2之间的最短距离是什么时候的形状地球是用WGS-84椭圆体建模的吗？

Answer 1

不幸的是，Vincenty的算法无法收敛某些输入。 GeographicLib提供了一种始终收敛的替代方案（和也更准确）。 C ++，C，Fortran，Javascript中的实现，提供了Python，Java和Matlab。例如，使用 Matlab package：

format long;
geoddistance(60,20,60,22)
->
111595.753650629

Answer 2

正如您对问题的评论中所指出的，您应该使用Vincenty的公式来解决逆问题。

您的问题的答案是：111595.75米（或60.257海里）。

从http://jsperf.com/vincenty-vs-haversine-distance-calculations复制的Vincenty反演公式的Javascript实现：

/**
 * Calculates geodetic distance between two points specified by latitude/longitude using 
 * Vincenty inverse formula for ellipsoids
 *
 * @param   {Number} lat1, lon1: first point in decimal degrees
 * @param   {Number} lat2, lon2: second point in decimal degrees
 * @returns (Number} distance in metres between points
 */

function distVincenty(lat1, lon1, lat2, lon2) {
  var a = 6378137,
      b = 6356752.314245,
      f = 1 / 298.257223563; // WGS-84 ellipsoid params
  var L = (lon2 - lon1).toRad();
  var U1 = Math.atan((1 - f) * Math.tan(lat1.toRad()));
  var U2 = Math.atan((1 - f) * Math.tan(lat2.toRad()));
  var sinU1 = Math.sin(U1),
      cosU1 = Math.cos(U1);
  var sinU2 = Math.sin(U2),
      cosU2 = Math.cos(U2);

  var lambda = L,
      lambdaP, iterLimit = 100;
  do {
    var sinLambda = Math.sin(lambda),
        cosLambda = Math.cos(lambda);
    var sinSigma = Math.sqrt((cosU2 * sinLambda) * (cosU2 * sinLambda) + (cosU1 * sinU2 - sinU1 * cosU2 * cosLambda) * (cosU1 * sinU2 - sinU1 * cosU2 * cosLambda));
    if (sinSigma == 0) return 0; // co-incident points
    var cosSigma = sinU1 * sinU2 + cosU1 * cosU2 * cosLambda;
    var sigma = Math.atan2(sinSigma, cosSigma);
    var sinAlpha = cosU1 * cosU2 * sinLambda / sinSigma;
    var cosSqAlpha = 1 - sinAlpha * sinAlpha;
    var cos2SigmaM = cosSigma - 2 * sinU1 * sinU2 / cosSqAlpha;
    if (isNaN(cos2SigmaM)) cos2SigmaM = 0; // equatorial line: cosSqAlpha=0 (§6)
    var C = f / 16 * cosSqAlpha * (4 + f * (4 - 3 * cosSqAlpha));
    lambdaP = lambda;
    lambda = L + (1 - C) * f * sinAlpha * (sigma + C * sinSigma * (cos2SigmaM + C * cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM)));
  } while (Math.abs(lambda - lambdaP) > 1e-12 && --iterLimit > 0);

  if (iterLimit == 0) return NaN // formula failed to converge
  var uSq = cosSqAlpha * (a * a - b * b) / (b * b);
  var A = 1 + uSq / 16384 * (4096 + uSq * (-768 + uSq * (320 - 175 * uSq)));
  var B = uSq / 1024 * (256 + uSq * (-128 + uSq * (74 - 47 * uSq)));
  var deltaSigma = B * sinSigma * (cos2SigmaM + B / 4 * (cosSigma * (-1 + 2 * cos2SigmaM * cos2SigmaM) - B / 6 * cos2SigmaM * (-3 + 4 * sinSigma * sinSigma) * (-3 + 4 * cos2SigmaM * cos2SigmaM)));
  var s = b * A * (sigma - deltaSigma);

  s = s.toFixed(3); // round to 1mm precision
  return s;
}

Answer 3

Haversine Formula是常用的（错误<0,5％）

使用WGS84-椭圆体的距离

3 个答案: