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
对于那些寻找基于WGS-84和GRS-80标准的Excel公式的人:
=ACOS(COS(RADIANS(90-Lat1))*COS(RADIANS(90-Lat2))+SIN(RADIANS(90-Lat1))*SIN(RADIANS(90-Lat2))*COS(RADIANS(Long1-Long2)))*6371
源
其他回答
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;
}
在我的项目中，我需要计算很多点之间的距离，所以我继续尝试优化我在这里找到的代码。平均而言，在不同的浏览器中，我的新实现的运行速度比获得最多好评的答案快2倍。
function distance(lat1, lon1, lat2, lon2) {
var p = 0.017453292519943295; // Math.PI / 180
var c = Math.cos;
var a = 0.5 - c((lat2 - lat1) * p)/2 +
c(lat1 * p) * c(lat2 * p) *
(1 - c((lon2 - lon1) * p))/2;
return 12742 * Math.asin(Math.sqrt(a)); // 2 * R; R = 6371 km
}
您可以在这里使用我的jsPerf并查看结果。
最近我需要在python中做同样的事情，所以这里是一个python实现:
from math import cos, asin, sqrt, pi
def distance(lat1, lon1, lat2, lon2):
p = pi/180
a = 0.5 - cos((lat2-lat1)*p)/2 + cos(lat1*p) * cos(lat2*p) * (1-cos((lon2-lon1)*p))/2
return 12742 * asin(sqrt(a)) #2*R*asin...
为了完整起见:维基百科上的Haversine。
下面是Haversine公式的java实现。
public final static double AVERAGE_RADIUS_OF_EARTH_KM = 6371;
public int calculateDistanceInKilometer(double userLat, double userLng,
double venueLat, double venueLng) {
double latDistance = Math.toRadians(userLat - venueLat);
double lngDistance = Math.toRadians(userLng - venueLng);
double a = Math.sin(latDistance / 2) * Math.sin(latDistance / 2)
+ Math.cos(Math.toRadians(userLat)) * Math.cos(Math.toRadians(venueLat))
* Math.sin(lngDistance / 2) * Math.sin(lngDistance / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return (int) (Math.round(AVERAGE_RADIUS_OF_EARTH_KM * c));
}
请注意，这里我们将答案四舍五入到最近的km。
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);
}
对于那些寻找基于WGS-84和GRS-80标准的Excel公式的人:
=ACOS(COS(RADIANS(90-Lat1))*COS(RADIANS(90-Lat2))+SIN(RADIANS(90-Lat1))*SIN(RADIANS(90-Lat2))*COS(RADIANS(Long1-Long2)))*6371
源
