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;
}
如前所述，函数是计算到目标点的直线距离。如果你想要行车距离/路线，你可以使用谷歌地图距离矩阵服务:
getDrivingDistanceBetweenTwoLatLong(origin, destination) {
return new Observable(subscriber => {
let service = new google.maps.DistanceMatrixService();
service.getDistanceMatrix(
{
origins: [new google.maps.LatLng(origin.lat, origin.long)],
destinations: [new google.maps.LatLng(destination.lat, destination.long)],
travelMode: 'DRIVING'
}, (response, status) => {
if (status !== google.maps.DistanceMatrixStatus.OK) {
console.log('Error:', status);
subscriber.error({error: status, status: status});
} else {
console.log(response);
try {
let valueInMeters = response.rows[0].elements[0].distance.value;
let valueInKms = valueInMeters / 1000;
subscriber.next(valueInKms);
subscriber.complete();
}
catch(error) {
subscriber.error({error: error, status: status});
}
}
});
});
}
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。
我试着通过命名变量让代码更容易理解，
我希望这能有所帮助
function getDistanceFromLatLonInKm(point1, point2) {
const [lat1, lon1] = point1;
const [lat2, lon2] = point2;
const earthRadius = 6371;
const dLat = convertDegToRad(lat2 - lat1);
const dLon = convertDegToRad(lon2 - lon1);
const squarehalfChordLength =
Math.sin(dLat / 2) * Math.sin(dLat / 2) +
Math.cos(convertDegToRad(lat1)) * Math.cos(convertDegToRad(lat2)) *
Math.sin(dLon / 2) * Math.sin(dLon / 2);
const angularDistance = 2 * Math.atan2(Math.sqrt(squarehalfChordLength), Math.sqrt(1 - squarehalfChordLength));
const distance = earthRadius * angularDistance;
return distance;
}
