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<script type="text/javascript" src="http://apps.bdimg.com/libs/jquery/1.6.4/jquery.js"></script>
<script type="text/javascript">
var EARTH_RADIUS = 6378137.0; //单位M
var PI = Math.PI;
function getRad(d) {
return d * PI / 180.0;
}
/**
* caculate the great circle distance
* @param {Object} lat1(纬度1)
* @param {Object} lng1(经度1)
* @param {Object} lat2(纬度2)
* @param {Object} lng2(经度2)
*/
//第一种方法:这种算法是把地球当作规则的球面来计算的咯,这种方法不是很精准咯,这个还要取决你定位的精准度咯
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 = * Math.asin(Math.sqrt(Math.pow(Math.sin(a / ), ) + Math.cos(radLat1) * Math.cos(radLat2) * Math.pow(Math.sin(b / ), )));
s = s * EARTH_RADIUS;
s = Math.round(s * ) / 10000.0;
alert(s);
}
//第一种方法:地球是椭圆的,所以会有这种算法
function getFlatternDistance(lat1, lng1, lat2, lng2) {
var f = getRad((lat1 + lat2) / );
var g = getRad((lat1 - lat2) / );
var l = getRad((lng1 - lng2) / );
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 = / 298.257;
sg = sg * sg;
sl = sl * sl;
sf = sf * sf;
s = sg * ( - sl) + ( - sf) * sl;
c = ( - sg) * ( - sl) + sf * sl;
w = Math.atan(Math.sqrt(s / c));
r = Math.sqrt(s * c) / w;
d = * w * a;
h1 = ( * r - ) / / c;
h2 = ( * r + ) / / s;
alert( d * ( + fl * (h1 * sf * ( - sg) - h2 * ( - sf) * sg)));
}
$(function () {
getFlatternDistance(28.89596, 105.443985, 28.896462, 105.444291);
});
</script>
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63.3422
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