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
下面是Erlang实现
lat_lng({Lat1, Lon1}=_Point1, {Lat2, Lon2}=_Point2) ->
P = math:pi() / 180,
R = 6371, % Radius of Earth in KM
A = 0.5 - math:cos((Lat2 - Lat1) * P) / 2 +
math:cos(Lat1 * P) * math:cos(Lat2 * P) * (1 - math:cos((Lon2 - Lon1) * P))/2,
R * 2 * math:asin(math:sqrt(A)).
其他回答
这个链接可能对你有帮助,因为它详细介绍了使用哈弗辛公式来计算距离。
摘录:
这个脚本计算两点之间的大圆距离 也就是说,在地球表面上的最短距离-使用 “半正矢”公式。
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)
}
下面是移植到Java的已接受的答案实现,以备任何人需要。
package com.project529.garage.util;
/**
* Mean radius.
*/
private static double EARTH_RADIUS = 6371;
/**
* Returns the distance between two sets of latitudes and longitudes in meters.
* <p/>
* Based from the following JavaScript SO answer:
* http://stackoverflow.com/questions/27928/calculate-distance-between-two-latitude-longitude-points-haversine-formula,
* which is based on https://en.wikipedia.org/wiki/Haversine_formula (error rate: ~0.55%).
*/
public double getDistanceBetween(double lat1, double lon1, double lat2, double lon2) {
double dLat = toRadians(lat2 - lat1);
double dLon = toRadians(lon2 - lon1);
double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
Math.cos(toRadians(lat1)) * Math.cos(toRadians(lat2)) *
Math.sin(dLon / 2) * Math.sin(dLon / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
double d = EARTH_RADIUS * c;
return d;
}
public double toRadians(double degrees) {
return degrees * (Math.PI / 180);
}
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;
}
下面是SQL实现,以km为单位计算距离,
SELECT UserId, ( 3959 * acos( cos( radians( your latitude here ) ) * cos( radians(latitude) ) *
cos( radians(longitude) - radians( your longitude here ) ) + sin( radians( your latitude here ) ) *
sin( radians(latitude) ) ) ) AS distance FROM user HAVING
distance < 5 ORDER BY distance LIMIT 0 , 5;
要获得通过编程语言实现的更多细节,您可以浏览这里给出的php脚本
非常感谢这一切。我在Objective-C iPhone应用程序中使用了以下代码:
const double PIx = 3.141592653589793;
const double RADIO = 6371; // Mean radius of Earth in Km
double convertToRadians(double val) {
return val * PIx / 180;
}
-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {
double dlon = convertToRadians(place2.longitude - place1.longitude);
double dlat = convertToRadians(place2.latitude - place1.latitude);
double a = ( pow(sin(dlat / 2), 2) + cos(convertToRadians(place1.latitude))) * cos(convertToRadians(place2.latitude)) * pow(sin(dlon / 2), 2);
double angle = 2 * asin(sqrt(a));
return angle * RADIO;
}
纬度和经度是十进制的。我没有在asin()调用中使用min(),因为我使用的距离非常小,以至于它们不需要min()。
它给出了错误的答案,直到我传入弧度的值-现在它几乎与从苹果地图应用程序中获得的值相同:-)
额外的更新:
如果你使用的是iOS4或更高版本,那么苹果会提供一些方法来实现相同的功能:
-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {
MKMapPoint start, finish;
start = MKMapPointForCoordinate(place1);
finish = MKMapPointForCoordinate(place2);
return MKMetersBetweenMapPoints(start, finish) / 1000;
}