Calculate the distance between two points on the earth according to the latitude and longitude degree JS implementation code _JAVASCRIPT skills

Using JS to calculate the distance between two points on the earth based on latitude and longitude
Recently used in accordance with the latitude and longitude calculation of the Earth's surface two distance between the formula, and then use JS to achieve a bit.
There are probably two ways to calculate the distance between two points on the Earth's surface.

First KindIs the default Earth is a smooth spherical surface, and then calculate the distance between any two points, this distance is called the Great Circle distance (the great Circle Distance).
The formula is as follows:
Use JS to achieve the following:

Copy Code code as follows:

var Earth_radius = 6378137.0; Unit m
var PI = Math.PI;

function Getrad (d) {
return d*pi/180.0;
}

/**
* Caculate the Great circle distance
* @param {Object} LAT1
* @param {Object} lng1
* @param {Object} lat2
* @param {Object} lng2
*/
function Getgreatcircledistance (lat1,lng1,lat2,lng2) {
var radLat1 = Getrad (LAT1);
var radLat2 = Getrad (LAT2);

var a = RADLAT1-RADLAT2;
var B = Getrad (lng1)-Getrad (LNG2);

var s = 2*math.asin (Math.sqrt (Math.pow Math.sin (A/2), 2) + Math.Cos (RADLAT1) *math.cos (RADLAT2) *math.pow (Math.sin ), 2));
s = S*earth_radius;
s = Math.Round (s*10000)/10000.0;

return s;
}

This formula is more correct in most cases, and only when dealing with relative points on the sphere, there is a problem. There's a revised formula., because there is no need to find out, can be found on the wiki.
Of course, we all know that the earth is not really a ball body, but ellipsoid, so with the following formula:

Copy Code code as follows:

/**
* Approx distance between two points on the earth ellipsoid
* @param {Object} LAT1
* @param {Object} lng1
* @param {Object} lat2
* @param {Object} lng2
*/
function Getflatterndistance (lat1,lng1,lat2,lng2) {
var f = Getrad ((lat1 + lat2)/2);
var g = Getrad ((LAT1-LAT2)/2);
var L = Getrad ((lng1-lng2)/2);

var sg = Math.sin (g);
var sl = Math.sin (l);
var sf = Math.sin (f);

var s,c,w,r,d,h1,h2;
var a = Earth_radius;
var fl = 1/298.257;

sg = SG*SG;
SL = SL*SL;
SF = SF*SF;

s = sg* (1-SL) + (1-SF) *SL;
c = (1-SG) * (1-SL) + SF*SL;

W = Math.atan (math.sqrt (s/c));
R = math.sqrt (s*c)/w;
D = 2*w*a;
H1 = (3*r-1)/2/c;
H2 = (3*r + 1)/2/s;

return d* (1 + fl* (h1*sf* (1-SG)-h2* (1-SF) *sg));
}

The formula calculates the result to be better than the first, and of course the longitude of the final result is actually dependent on the precision of the incoming coordinates.