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.

The first is 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:
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--> 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 (b /2), 2));
s = S * earth_radius;
s = Math.Round (S * 10000)/10000.0;

return s;
}

This formula is correct in most cases, only when dealing with the relative points on the sphere, there is a problem, there is a modified 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:
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-->
/* *
* 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.