从google maps的脚本里扒了段代码，没准啥时会用上。大家一块看看是怎么算的。
private const double EARTH_RADIUS = 6378.137; private static double rad(double d) { return d * Math.PI / 180.0; } public static double GetDistance(double lat1, double lng1, double lat2, double lng2) { double radLat1 = rad(lat1); double radLat2 = rad(lat2); double a = radLat1 - radLat2; double b = rad(lng1) - rad(lng2); double s = 2 * Math.Asin(Math.Sqrt(Math.Pow(Math.Sin(a/2),2) + Math.Cos(radLat1)*Math.Cos(radLat2)*Math.Pow(Math.Sin(b/2),2))); s = s * EARTH_RADIUS; s = Math.Round(s * 10000) / 10000; return s; }
看了通过两个点的经纬度计算距离这篇，据说是Google里扒来的算法，于是决定验证一下。东改西改改成了个VB的代码，通过计算发现非常正确。用30,120和31,121两个坐标点进行验证，同Mapinfo中计算的结果非常的相近。后来又把坐标系直接当成直角坐标系来计算，发现误差很大。于是仍然用直角坐标系方法来计算，给纬度加上了个0.86的参数，这样计算下来和结果相近。不上传文件了，直接贴代码了。代码如下：(五个TEXT，三个Button)
Const EARTH_RADIUS = 6378.137 Const Pi = 3.1415926535898 Function rad(ByVal d As Double) As Double rad = d * Pi / 180 End Function Function GetDistance(lat1 As Double, lng1 As Double, lat2 As Double, lng2 As Double) As Double Dim radlat1 As Double, radlat2 As Double Dim a As Double, b As Double, s As Double, Temp As Double radlat1 = rad(lat1) radlat2 = rad(lat2) a = radlat1 - radlat2 b = rad(lng1) - rad(lng2) Temp = Sqr(Sin(a / 2) ^ 2 + Cos(radlat1) * Cos(radlat2) * Sin(b / 2) ^ 2) s = 2 * Atn(Temp / Sqr(-Temp * Temp + 1)) '没有反正弦函数，网上找了个 s = s * EARTH_RADIUS GetDistance = s End Function Private Sub Command1_Click() Text5.Text = Str(GetDistance(Val(Text1.Text), Val(Text2.Text), Val(Text3.Text), Val(Text4.Text))) End Sub Private Sub Command2_Click() Text5.Text = Str((Sqr((Val(Text3.Text) - Val(Text1.Text)) ^ 2 + (Val(Text4.Text) - Val(Text2.Text)) ^ 2)) * 111.3199338) End Sub Private Sub Command3_Click() Text5.Text = Str((Sqr((Val(Text3.Text) * 0.86 - Val(Text1.Text) * 0.86) ^ 2 + (Val(Text4.Text) - Val(Text2.Text)) ^ 2)) * 111.3199338) End Sub相关资源：经纬度计算代码
