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。

这是“Henry Vilinskiy”为MySQL和km改编的版本:

CREATE FUNCTION `CalculateDistanceInKm`(
  fromLatitude float,
  fromLongitude float,
  toLatitude float, 
  toLongitude float
) RETURNS float
BEGIN
  declare distance float;

  select 
    6367 * ACOS(
            round(
              COS(RADIANS(90-fromLatitude)) *
                COS(RADIANS(90-toLatitude)) +
                SIN(RADIANS(90-fromLatitude)) *
                SIN(RADIANS(90-toLatitude)) *
                COS(RADIANS(fromLongitude-toLongitude))
              ,15)
            )
    into distance;

  return  round(distance,3);
END;

下面是我在Python中使用的Haversine函数:

from math import pi,sqrt,sin,cos,atan2

def haversine(pos1, pos2):
    lat1 = float(pos1['lat'])
    long1 = float(pos1['long'])
    lat2 = float(pos2['lat'])
    long2 = float(pos2['long'])

    degree_to_rad = float(pi / 180.0)

    d_lat = (lat2 - lat1) * degree_to_rad
    d_long = (long2 - long1) * degree_to_rad

    a = pow(sin(d_lat / 2), 2) + cos(lat1 * degree_to_rad) * cos(lat2 * degree_to_rad) * pow(sin(d_long / 2), 2)
    c = 2 * atan2(sqrt(a), sqrt(1 - a))
    km = 6367 * c
    mi = 3956 * c

    return {"km":km, "miles":mi}

这是我在Elixir中的实现

defmodule Geo do
  @earth_radius_km 6371
  @earth_radius_sm 3958.748
  @earth_radius_nm 3440.065
  @feet_per_sm 5280

  @d2r :math.pi / 180

  def deg_to_rad(deg), do: deg * @d2r

  def great_circle_distance(p1, p2, :km), do: haversine(p1, p2) * @earth_radius_km
  def great_circle_distance(p1, p2, :sm), do: haversine(p1, p2) * @earth_radius_sm
  def great_circle_distance(p1, p2, :nm), do: haversine(p1, p2) * @earth_radius_nm
  def great_circle_distance(p1, p2, :m), do: great_circle_distance(p1, p2, :km) * 1000
  def great_circle_distance(p1, p2, :ft), do: great_circle_distance(p1, p2, :sm) * @feet_per_sm

  @doc """
  Calculate the [Haversine](https://en.wikipedia.org/wiki/Haversine_formula)
  distance between two coordinates. Result is in radians. This result can be
  multiplied by the sphere's radius in any unit to get the distance in that unit.
  For example, multiple the result of this function by the Earth's radius in
  kilometres and you get the distance between the two given points in kilometres.
  """
  def haversine({lat1, lon1}, {lat2, lon2}) do
    dlat = deg_to_rad(lat2 - lat1)
    dlon = deg_to_rad(lon2 - lon1)

    radlat1 = deg_to_rad(lat1)
    radlat2 = deg_to_rad(lat2)

    a = :math.pow(:math.sin(dlat / 2), 2) +
        :math.pow(:math.sin(dlon / 2), 2) *
        :math.cos(radlat1) * :math.cos(radlat2)

    2 * :math.atan2(:math.sqrt(a), :math.sqrt(1 - a))
  end
end

我认为R中的一个算法版本仍然缺失:

gpsdistance<-function(lat1,lon1,lat2,lon2){

# internal function to change deg to rad

degreesToRadians<- function (degrees) {
return (degrees * pi / 180)
}

R<-6371e3  #radius of Earth in meters

phi1<-degreesToRadians(lat1) # latitude 1
phi2<-degreesToRadians(lat2) # latitude 2
lambda1<-degreesToRadians(lon1) # longitude 1
lambda2<-degreesToRadians(lon2) # longitude 2

delta_phi<-phi1-phi2 # latitude-distance
delta_lambda<-lambda1-lambda2 # longitude-distance

a<-sin(delta_phi/2)*sin(delta_phi/2)+
cos(phi1)*cos(phi2)*sin(delta_lambda/2)*
sin(delta_lambda/2)

cc<-2*atan2(sqrt(a),sqrt(1-a))

distance<- R * cc

return(distance)  # in meters
}

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。