一、关于“面包屑”方法

地球半径在不同的纬度上是不同的。在Haversine算法中必须考虑到这一点。 考虑轴承的变化，它将直线变成拱门(更长的) 考虑到速度变化将把拱门变成螺旋(比拱门更长或更短) 高度变化将使平面螺旋变成3D螺旋(再次变长)。这对丘陵地区非常重要。

下面是考虑#1和#2的C语言函数:

double   calcDistanceByHaversine(double rLat1, double rLon1, double rHeading1,
       double rLat2, double rLon2, double rHeading2){
  double rDLatRad = 0.0;
  double rDLonRad = 0.0;
  double rLat1Rad = 0.0;
  double rLat2Rad = 0.0;
  double a = 0.0;
  double c = 0.0;
  double rResult = 0.0;
  double rEarthRadius = 0.0;
  double rDHeading = 0.0;
  double rDHeadingRad = 0.0;

  if ((rLat1 < -90.0) || (rLat1 > 90.0) || (rLat2 < -90.0) || (rLat2 > 90.0)
              || (rLon1 < -180.0) || (rLon1 > 180.0) || (rLon2 < -180.0)
              || (rLon2 > 180.0)) {
        return -1;
  };

  rDLatRad = (rLat2 - rLat1) * DEGREE_TO_RADIANS;
  rDLonRad = (rLon2 - rLon1) * DEGREE_TO_RADIANS;
  rLat1Rad = rLat1 * DEGREE_TO_RADIANS;
  rLat2Rad = rLat2 * DEGREE_TO_RADIANS;

  a = sin(rDLatRad / 2) * sin(rDLatRad / 2) + sin(rDLonRad / 2) * sin(
              rDLonRad / 2) * cos(rLat1Rad) * cos(rLat2Rad);

  if (a == 0.0) {
        return 0.0;
  }

  c = 2 * atan2(sqrt(a), sqrt(1 - a));
  rEarthRadius = 6378.1370 - (21.3847 * 90.0 / ((fabs(rLat1) + fabs(rLat2))
              / 2.0));
  rResult = rEarthRadius * c;

  // Chord to Arc Correction based on Heading changes. Important for routes with many turns and U-turns

  if ((rHeading1 >= 0.0) && (rHeading1 < 360.0) && (rHeading2 >= 0.0)
              && (rHeading2 < 360.0)) {
        rDHeading = fabs(rHeading1 - rHeading2);
        if (rDHeading > 180.0) {
              rDHeading -= 180.0;
        }
        rDHeadingRad = rDHeading * DEGREE_TO_RADIANS;
        if (rDHeading > 5.0) {
              rResult = rResult * (rDHeadingRad / (2.0 * sin(rDHeadingRad / 2)));
        } else {
              rResult = rResult / cos(rDHeadingRad);
        }
  }
  return rResult;
}

2有一种更简单的方法，效果很好。

按平均速度。

Trip_distance = Trip_average_speed * Trip_time

由于GPS速度是由多普勒效应检测的，与[Lon,Lat]没有直接关系，如果不是主要的距离计算方法，至少可以考虑作为次要的(备份或校正)。

下面是c#语言(用纬度和弧度表示):

double CalculateGreatCircleDistance(double lat1, double long1, double lat2, double long2, double radius)
{
    return radius * Math.Acos(
        Math.Sin(lat1) * Math.Sin(lat2)
        + Math.Cos(lat1) * Math.Cos(lat2) * Math.Cos(long2 - long1));
}

如果你的纬度和长度是用角度表示的，那么除以180/PI就可以转换成弧度。

这取决于你需要它有多准确。如果你需要精确到毫米的精度，最好看看使用椭球的算法，而不是球体，比如Vincenty的算法。

下面是答案中的Swift实现

func degreesToRadians(degrees: Double) -> Double {
    return degrees * Double.pi / 180
}

func distanceInKmBetweenEarthCoordinates(lat1: Double, lon1: Double, lat2: Double, lon2: Double) -> Double {

    let earthRadiusKm: Double = 6371

    let dLat = degreesToRadians(degrees: lat2 - lat1)
    let dLon = degreesToRadians(degrees: lon2 - lon1)

    let lat1 = degreesToRadians(degrees: lat1)
    let lat2 = degreesToRadians(degrees: lat2)

    let a = sin(dLat/2) * sin(dLat/2) +
    sin(dLon/2) * sin(dLon/2) * cos(lat1) * cos(lat2)
    let c = 2 * atan2(sqrt(a), sqrt(1 - a))
    return earthRadiusKm * c
}

Unity版本c#

Haversine Algorithm。

public float Distance(float lat1, float lon1, float lat2, float lon2)
{
    var earthRadiusKm = 6371;

    var dLat = (lat2 - lat1) * Mathf.Rad2Deg;
    var dLon = (lon2 - lon1) * Mathf.Rad2Deg;

    var a = Mathf.Sin(dLat / 2) * Mathf.Sin(dLat / 2) +
            Mathf.Sin(dLon / 2) * Mathf.Sin(dLon / 2) * 
            Mathf.Cos(lat1 * Mathf.Rad2Deg) * Mathf.Cos(lat2 * Mathf.Rad2Deg);

    var c = 2 * Mathf.Atan2(Mathf.Sqrt(a), Mathf.Sqrt(1 - a));
    return earthRadiusKm * c;
}

我需要在PowerShell中实现这个，希望它可以帮助其他人。 关于这种方法的一些注意事项

Don't split any of the lines or the calculation will be wrong To calculate in KM remove the * 1000 in the calculation of $distance Change $earthsRadius = 3963.19059 and remove * 1000 in the calculation of $distance the to calulate the distance in miles I'm using Haversine, as other posts have pointed out Vincenty's formulae is much more accurate Function MetresDistanceBetweenTwoGPSCoordinates($latitude1, $longitude1, $latitude2, $longitude2) { $Rad = ([math]::PI / 180); $earthsRadius = 6378.1370 # Earth's Radius in KM $dLat = ($latitude2 - $latitude1) * $Rad $dLon = ($longitude2 - $longitude1) * $Rad $latitude1 = $latitude1 * $Rad $latitude2 = $latitude2 * $Rad $a = [math]::Sin($dLat / 2) * [math]::Sin($dLat / 2) + [math]::Sin($dLon / 2) * [math]::Sin($dLon / 2) * [math]::Cos($latitude1) * [math]::Cos($latitude2) $c = 2 * [math]::ATan2([math]::Sqrt($a), [math]::Sqrt(1-$a)) $distance = [math]::Round($earthsRadius * $c * 1000, 0) #Multiple by 1000 to get metres Return $distance }