高德地图 计算两经纬度 之间距离 lua实现

网上很多教程类似, 不是很好，精度差别很大
比如这个c#实现：

c#代码
private const double EARTH_RADIUS = 6378.137; //地球半径
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;
}

local start = {longitude=116.368904, latitude=39.923423}
local tail = {longitude=116.387271, latitude=39.922501}
算出来是：2.0451公里，
高德地图是：1571.37米，

https://lbs.amap.com/api/javascript-api/example/calcutation/calculate-distance-between-two-markers/

通过地球半径粗略算距离，单位也不统一，误差500米左右。

找到一个好一点的介绍的比较详细，写的很不错：

https://blog.youkuaiyun.com/jianggujin/article/details/72833711

他上面使用js, mysql, java实现，误差相对较小，比较精确，我参考他的实现了lua版本的：

function TestLua:CalDistance()
    local start = {longitude=116.368904, latitude=39.923423}
    local tail = {longitude=116.387271, latitude=39.922501}
    print("[gps距离]"..tostring(self:CalculateLineDistance(start, tail)))
end

function TestLua:CalculateLineDistance(start, tail)
   local d1 = 0.01745329251994329;
   local d2 = start.longitude;
   local d3 = start.latitude;
   local d4 = tail.longitude;
   local d5 = tail.latitude;
   d2 = d2*d1;
   d3 = d3*d1;
   d4 = d4*d1;
   d5 = d5*d1;
   local d6 = math.sin(d2);
   local d7 = math.sin(d3);
   local d8 = math.cos(d2);
   local d9 = math.cos(d3);
   local d10 = math.sin(d4);
   local d11 = math.sin(d5);
   local d12 = math.cos(d4);
   local d13 = math.cos(d5);
   local arrayOfDouble1 = {};
   local arrayOfDouble2 = {};
   arrayOfDouble1[0]=(d9 * d8);
   arrayOfDouble1[1]=(d9 * d6);
   arrayOfDouble1[2]=(d7);

   arrayOfDouble2[0]=(d13 * d12);
   arrayOfDouble2[1]=(d13 * d10);
   arrayOfDouble2[2]=(d11);
   local d14 = math.sqrt((arrayOfDouble1[0] - arrayOfDouble2[0]) * (arrayOfDouble1[0] - arrayOfDouble2[0]) +
       (arrayOfDouble1[1] - arrayOfDouble2[1]) * (arrayOfDouble1[1] - arrayOfDouble2[1]) +
       (arrayOfDouble1[2] - arrayOfDouble2[2]) * (arrayOfDouble1[2] - arrayOfDouble2[2]));
   return(math.asin(d14 / 2.0) * 12742001.579854401);
end