这段Lua代码改编自维基百科和Robert Lipe的GPSbabel工具:

local EARTH_RAD = 6378137.0 
  -- earth's radius in meters (official geoid datum, not 20,000km / pi)

local radmiles = EARTH_RAD*100.0/2.54/12.0/5280.0;
  -- earth's radius in miles

local multipliers = {
  radians = 1, miles = radmiles, mi = radmiles, feet = radmiles * 5280,
  meters = EARTH_RAD, m = EARTH_RAD, km = EARTH_RAD / 1000, 
  degrees = 360 / (2 * math.pi), min = 60 * 360 / (2 * math.pi)
}

function gcdist(pt1, pt2, units) -- return distance in radians or given units
  --- this formula works best for points close together or antipodal
  --- rounding error strikes when distance is one-quarter Earth's circumference
  --- (ref: wikipedia Great-circle distance)
  if not pt1.radians then pt1 = rad(pt1) end
  if not pt2.radians then pt2 = rad(pt2) end
  local sdlat = sin((pt1.lat - pt2.lat) / 2.0);
  local sdlon = sin((pt1.lon - pt2.lon) / 2.0);
  local res = sqrt(sdlat * sdlat + cos(pt1.lat) * cos(pt2.lat) * sdlon * sdlon);
  res = res > 1 and 1 or res < -1 and -1 or res
  res = 2 * asin(res);
  if units then return res * assert(multipliers[units])
  else return res
  end
end

我猜你想让它沿着地球的曲率运动。你的两点和地心在一个平面上。地球的中心是这个平面上的圆心，这两个点(大致)在这个圆的周长上。由此你可以通过求一点到另一点的角度来计算距离。

如果点的高度不一样，或者如果你需要考虑地球不是一个完美的球体，这就有点困难了。

在SQL Server 2008中使用地理类型非常容易做到这一点。

SELECT geography::Point(lat1, lon1, 4326).STDistance(geography::Point(lat2, lon2, 4326))
-- computes distance in meters using eliptical model, accurate to the mm

4326是WGS84椭球地球模型的SRID

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;
}

如果你使用的是。net，不要重新启动轮子。看到System.Device.Location。在另一个答案的评论中赞扬fnx。

using System.Device.Location;

double lat1 = 45.421527862548828D;
double long1 = -75.697189331054688D;
double lat2 = 53.64135D;
double long2 = -113.59273D;

GeoCoordinate geo1 = new GeoCoordinate(lat1, long1);
GeoCoordinate geo2 = new GeoCoordinate(lat2, long2);

double distance = geo1.GetDistanceTo(geo2);

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