如何计算由经纬度指定的两点之间的距离?
为了澄清,我想用千米来表示距离;这些点使用WGS84系统,我想了解可用方法的相对准确性。
如何计算由经纬度指定的两点之间的距离?
为了澄清,我想用千米来表示距离;这些点使用WGS84系统,我想了解可用方法的相对准确性。
当前回答
这个脚本[在PHP中]计算两点之间的距离。
public static function getDistanceOfTwoPoints($source, $dest, $unit='K') {
$lat1 = $source[0];
$lon1 = $source[1];
$lat2 = $dest[0];
$lon2 = $dest[1];
$theta = $lon1 - $lon2;
$dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);
$miles = $dist * 60 * 1.1515;
$unit = strtoupper($unit);
if ($unit == "K") {
return ($miles * 1.609344);
}
else if ($unit == "M")
{
return ($miles * 1.609344 * 1000);
}
else if ($unit == "N") {
return ($miles * 0.8684);
}
else {
return $miles;
}
}
其他回答
这里有一个用PHP http://www.geodatasource.com/developers/php计算距离的好例子:
function distance($lat1, $lon1, $lat2, $lon2, $unit) {
$theta = $lon1 - $lon2;
$dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);
$miles = $dist * 60 * 1.1515;
$unit = strtoupper($unit);
if ($unit == "K") {
return ($miles * 1.609344);
} else if ($unit == "N") {
return ($miles * 0.8684);
} else {
return $miles;
}
}
Java实现在根据哈弗辛公式
double calculateDistance(double latPoint1, double lngPoint1,
double latPoint2, double lngPoint2) {
if(latPoint1 == latPoint2 && lngPoint1 == lngPoint2) {
return 0d;
}
final double EARTH_RADIUS = 6371.0; //km value;
//converting to radians
latPoint1 = Math.toRadians(latPoint1);
lngPoint1 = Math.toRadians(lngPoint1);
latPoint2 = Math.toRadians(latPoint2);
lngPoint2 = Math.toRadians(lngPoint2);
double distance = Math.pow(Math.sin((latPoint2 - latPoint1) / 2.0), 2)
+ Math.cos(latPoint1) * Math.cos(latPoint2)
* Math.pow(Math.sin((lngPoint2 - lngPoint1) / 2.0), 2);
distance = 2.0 * EARTH_RADIUS * Math.asin(Math.sqrt(distance));
return distance; //km value
}
function distance($lat1, $lon1, $lat2, $lon2) {
$pi80 = M_PI / 180;
$lat1 *= $pi80; $lon1 *= $pi80; $lat2 *= $pi80; $lon2 *= $pi80;
$dlat = $lat2 - $lat1;
$dlon = $lon2 - $lon1;
$a = sin($dlat / 2) * sin($dlat / 2) + cos($lat1) * cos($lat2) * sin($dlon / 2) * sin($dlon / 2);
$km = 6372.797 * 2 * atan2(sqrt($a), sqrt(1 - $a));
return $km;
}
这个脚本[在PHP中]计算两点之间的距离。
public static function getDistanceOfTwoPoints($source, $dest, $unit='K') {
$lat1 = $source[0];
$lon1 = $source[1];
$lat2 = $dest[0];
$lon2 = $dest[1];
$theta = $lon1 - $lon2;
$dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);
$miles = $dist * 60 * 1.1515;
$unit = strtoupper($unit);
if ($unit == "K") {
return ($miles * 1.609344);
}
else if ($unit == "M")
{
return ($miles * 1.609344 * 1000);
}
else if ($unit == "N") {
return ($miles * 0.8684);
}
else {
return $miles;
}
}
我已经创建了这个小Javascript LatLng对象,可能对某人有用。
var latLng1 = new LatLng(5, 3);
var latLng2 = new LatLng(6, 7);
var distance = latLng1.distanceTo(latLng2);
代码:
/**
* latLng point
* @param {Number} lat
* @param {Number} lng
* @returns {LatLng}
* @constructor
*/
function LatLng(lat,lng) {
this.lat = parseFloat(lat);
this.lng = parseFloat(lng);
this.__cache = {};
}
LatLng.prototype = {
toString: function() {
return [this.lat, this.lng].join(",");
},
/**
* calculate distance in km to another latLng, with caching
* @param {LatLng} latLng
* @returns {Number} distance in km
*/
distanceTo: function(latLng) {
var cacheKey = latLng.toString();
if(cacheKey in this.__cache) {
return this.__cache[cacheKey];
}
// the fastest way to calculate the distance, according to this jsperf test;
// http://jsperf.com/haversine-salvador/8
// http://stackoverflow.com/questions/27928
var deg2rad = 0.017453292519943295; // === Math.PI / 180
var lat1 = this.lat * deg2rad;
var lng1 = this.lng * deg2rad;
var lat2 = latLng.lat * deg2rad;
var lng2 = latLng.lng * deg2rad;
var a = (
(1 - Math.cos(lat2 - lat1)) +
(1 - Math.cos(lng2 - lng1)) * Math.cos(lat1) * Math.cos(lat2)
) / 2;
var distance = 12742 * Math.asin(Math.sqrt(a)); // Diameter of the earth in km (2 * 6371)
// cache the distance
this.__cache[cacheKey] = distance;
return distance;
}
};