我目前正在使用下面的功能，它不能正常工作。根据谷歌Maps，这些坐标之间的距离(从59.3293371,13.4877472到59.3225525,13.4619422)是2.2公里，而函数返回1.6公里。我怎样才能使这个函数返回正确的距离?
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/
使用Haversine公式，源代码:
//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
//::: :::
//::: This routine calculates the distance between two points (given the :::
//::: latitude/longitude of those points). It is being used to calculate :::
//::: the distance between two locations using GeoDataSource (TM) prodducts :::
//::: :::
//::: Definitions: :::
//::: South latitudes are negative, east longitudes are positive :::
//::: :::
//::: Passed to function: :::
//::: lat1, lon1 = Latitude and Longitude of point 1 (in decimal degrees) :::
//::: lat2, lon2 = Latitude and Longitude of point 2 (in decimal degrees) :::
//::: unit = the unit you desire for results :::
//::: where: 'M' is statute miles (default) :::
//::: 'K' is kilometers :::
//::: 'N' is nautical miles :::
//::: :::
//::: Worldwide cities and other features databases with latitude longitude :::
//::: are available at https://www.geodatasource.com :::
//::: :::
//::: For enquiries, please contact sales@geodatasource.com :::
//::: :::
//::: Official Web site: https://www.geodatasource.com :::
//::: :::
//::: GeoDataSource.com (C) All Rights Reserved 2018 :::
//::: :::
//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
function distance(lat1, lon1, lat2, lon2, unit) {
if ((lat1 == lat2) && (lon1 == lon2)) {
return 0;
}
else {
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);
if (dist > 1) {
dist = 1;
}
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;
}
}
样例代码是根据LGPLv3许可的。
Derek的解决方案对我来说很好，我只是简单地将其转换为PHP，希望它能帮助到一些人!
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;
}
试试这个。它在VB.net中，您需要将其转换为Javascript。此函数接受十进制分钟的参数。
Private Function calculateDistance(ByVal long1 As String, ByVal lat1 As String, ByVal long2 As String, ByVal lat2 As String) As Double
long1 = Double.Parse(long1)
lat1 = Double.Parse(lat1)
long2 = Double.Parse(long2)
lat2 = Double.Parse(lat2)
'conversion to radian
lat1 = (lat1 * 2.0 * Math.PI) / 60.0 / 360.0
long1 = (long1 * 2.0 * Math.PI) / 60.0 / 360.0
lat2 = (lat2 * 2.0 * Math.PI) / 60.0 / 360.0
long2 = (long2 * 2.0 * Math.PI) / 60.0 / 360.0
' use to different earth axis length
Dim a As Double = 6378137.0 ' Earth Major Axis (WGS84)
Dim b As Double = 6356752.3142 ' Minor Axis
Dim f As Double = (a - b) / a ' "Flattening"
Dim e As Double = 2.0 * f - f * f ' "Eccentricity"
Dim beta As Double = (a / Math.Sqrt(1.0 - e * Math.Sin(lat1) * Math.Sin(lat1)))
Dim cos As Double = Math.Cos(lat1)
Dim x As Double = beta * cos * Math.Cos(long1)
Dim y As Double = beta * cos * Math.Sin(long1)
Dim z As Double = beta * (1 - e) * Math.Sin(lat1)
beta = (a / Math.Sqrt(1.0 - e * Math.Sin(lat2) * Math.Sin(lat2)))
cos = Math.Cos(lat2)
x -= (beta * cos * Math.Cos(long2))
y -= (beta * cos * Math.Sin(long2))
z -= (beta * (1 - e) * Math.Sin(lat2))
Return Math.Sqrt((x * x) + (y * y) + (z * z))
End Function
编辑
javascript中的转换函数
function calculateDistance(lat1, long1, lat2, long2)
{
//radians
lat1 = (lat1 * 2.0 * Math.PI) / 60.0 / 360.0;
long1 = (long1 * 2.0 * Math.PI) / 60.0 / 360.0;
lat2 = (lat2 * 2.0 * Math.PI) / 60.0 / 360.0;
long2 = (long2 * 2.0 * Math.PI) / 60.0 / 360.0;
// use to different earth axis length
var a = 6378137.0; // Earth Major Axis (WGS84)
var b = 6356752.3142; // Minor Axis
var f = (a-b) / a; // "Flattening"
var e = 2.0*f - f*f; // "Eccentricity"
var beta = (a / Math.sqrt( 1.0 - e * Math.sin( lat1 ) * Math.sin( lat1 )));
var cos = Math.cos( lat1 );
var x = beta * cos * Math.cos( long1 );
var y = beta * cos * Math.sin( long1 );
var z = beta * ( 1 - e ) * Math.sin( lat1 );
beta = ( a / Math.sqrt( 1.0 - e * Math.sin( lat2 ) * Math.sin( lat2 )));
cos = Math.cos( lat2 );
x -= (beta * cos * Math.cos( long2 ));
y -= (beta * cos * Math.sin( long2 ));
z -= (beta * (1 - e) * Math.sin( lat2 ));
return (Math.sqrt( (x*x) + (y*y) + (z*z) )/1000);
}
谷歌的答案
https://cloud.google.com/blog/products/maps-platform/how-calculate-distances-map-maps-javascript-api
支票由三部分组成。
https://www.distancefromto.net/
static distance({ x: x1, y: y1 }, { x: x2, y: y2 }) {
function toRadians(value) {
return value * Math.PI / 180
}
var R = 6371.0710
var rlat1 = toRadians(x1) // Convert degrees to radians
var rlat2 = toRadians(x2) // Convert degrees to radians
var difflat = rlat2 - rlat1 // Radian difference (latitudes)
var difflon = toRadians(y2 - y1) // Radian difference (longitudes)
return 2 * R * Math.asin(Math.sqrt(Math.sin(difflat / 2) * Math.sin(difflat / 2) + Math.cos(rlat1) * Math.cos(rlat2) * Math.sin(difflon / 2) * Math.sin(difflon / 2)))
}
