函数计算两个坐标之间的距离
函数计算两个坐标之间的距离
提问于 2013-09-19 06:50:18
我目前正在使用下面的函数，但它不能正常工作。根据谷歌地图，these coordinates (从59.3293371,13.4877472到59.3225525,13.4619422)之间的距离是2.2公里，而该函数返回1.6公里。怎样才能让这个函数返回正确的距离呢？
function getDistanceFromLatLonInKm(lat1, lon1, lat2, lon2) {
var R = 6371; // Radius of the earth in km
var dLat = deg2rad(lat2-lat1); // deg2rad below
var dLon = deg2rad(lon2-lon1);
var a =
Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon/2) * Math.sin(dLon/2)
;
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c; // Distance in km
return d;
}
function deg2rad(deg) {
return deg * (Math.PI/180)
}jsFiddle：http://jsfiddle.net/edgren/gAHJB/
回答 9
Stack Overflow用户
发布于 2013-09-19 07:10:17
我之前写过一个类似的方程式-测试过，也得到了1.6公里。
你的谷歌地图显示的是开车的距离。
你的函数是像乌鸦飞一样计算(直线距离)。
alert(calcCrow(59.3293371,13.4877472,59.3225525,13.4619422).toFixed(1));
//This function takes in latitude and longitude of two location and returns the distance between them as the crow flies (in km)
function calcCrow(lat1, lon1, lat2, lon2)
{
var R = 6371; // km
var dLat = toRad(lat2-lat1);
var dLon = toRad(lon2-lon1);
var lat1 = toRad(lat1);
var lat2 = toRad(lat2);
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
return d;
}
// Converts numeric degrees to radians
function toRad(Value)
{
return Value * Math.PI / 180;
}Stack Overflow用户
发布于 2016-07-19 19:45:12
Derek的解决方案对我来说工作得很好，我只是简单地把它转换成了PHP，希望它能对其他人有所帮助！
function calcCrow($lat1, $lon1, $lat2, $lon2){
$R = 6371; // km
$dLat = toRad($lat2-$lat1);
$dLon = toRad($lon2-$lon1);
$lat1 = toRad($lat1);
$lat2 = toRad($lat2);
$a = sin($dLat/2) * sin($dLat/2) +sin($dLon/2) * sin($dLon/2) * cos($lat1) * cos($lat2);
$c = 2 * atan2(sqrt($a), sqrt(1-$a));
$d = $R * $c;
return $d;
}
// Converts numeric degrees to radians
function toRad($Value)
{
return $Value * pi() / 180;
}Stack Overflow用户
发布于 2018-09-07 04:24:21
在javascript中计算两点之间的距离
function distance(lat1, lon1, lat2, lon2, unit) {
var radlat1 = Math.PI * lat1/180
var radlat2 = Math.PI * lat2/180
var theta = lon1-lon2
var radtheta = Math.PI * theta/180
var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
dist = Math.acos(dist)
dist = dist * 180/Math.PI
dist = dist * 60 * 1.1515
if (unit=="K") { dist = dist * 1.609344 }
if (unit=="N") { dist = dist * 0.8684 }
return dist
}有关更多详细信息，请参阅：Reference Link
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原文链接：
https://stackoverflow.com/questions/18883601
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