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
仅限飞镖:
import 'dart:math' show cos, sqrt, asin;
double calculateDistance(LatLng l1, LatLng l2) {
const p = 0.017453292519943295;
final a = 0.5 -
cos((l2.latitude - l1.latitude) * p) / 2 +
cos(l1.latitude * p) *
cos(l2.latitude * p) *
(1 - cos((l2.longitude - l1.longitude) * p)) /
2;
return 12742 * asin(sqrt(a));
}
其他回答
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
我通过简化公式来简化计算。
下面是Ruby版本:
include Math
earth_radius_mi = 3959
radians = lambda { |deg| deg * PI / 180 }
coord_radians = lambda { |c| { :lat => radians[c[:lat]], :lng => radians[c[:lng]] } }
# from/to = { :lat => (latitude_in_degrees), :lng => (longitude_in_degrees) }
def haversine_distance(from, to)
from, to = coord_radians[from], coord_radians[to]
cosines_product = cos(to[:lat]) * cos(from[:lat]) * cos(from[:lng] - to[:lng])
sines_product = sin(to[:lat]) * sin(from[:lat])
return earth_radius_mi * acos(cosines_product + sines_product)
end
下面是一个c#实现:
static class DistanceAlgorithm
{
const double PIx = 3.141592653589793;
const double RADIUS = 6378.16;
/// <summary>
/// Convert degrees to Radians
/// </summary>
/// <param name="x">Degrees</param>
/// <returns>The equivalent in radians</returns>
public static double Radians(double x)
{
return x * PIx / 180;
}
/// <summary>
/// Calculate the distance between two places.
/// </summary>
/// <param name="lon1"></param>
/// <param name="lat1"></param>
/// <param name="lon2"></param>
/// <param name="lat2"></param>
/// <returns></returns>
public static double DistanceBetweenPlaces(
double lon1,
double lat1,
double lon2,
double lat2)
{
double dlon = Radians(lon2 - lon1);
double dlat = Radians(lat2 - lat1);
double a = (Math.Sin(dlat / 2) * Math.Sin(dlat / 2)) + Math.Cos(Radians(lat1)) * Math.Cos(Radians(lat2)) * (Math.Sin(dlon / 2) * Math.Sin(dlon / 2));
double angle = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a));
return angle * RADIUS;
}
}
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。
我在R中做了一个自定义函数,使用R基本包中可用的函数来计算两个空间点之间的距离(km)。
custom_hav_dist <- function(lat1, lon1, lat2, lon2) {
R <- 6371
Radian_factor <- 0.0174533
lat_1 <- (90-lat1)*Radian_factor
lat_2 <- (90-lat2)*Radian_factor
diff_long <-(lon1-lon2)*Radian_factor
distance_in_km <- 6371*acos((cos(lat_1)*cos(lat_2))+
(sin(lat_1)*sin(lat_2)*cos(diff_long)))
rm(lat1, lon1, lat2, lon2)
return(distance_in_km)
}
样例输出
custom_hav_dist(50.31,19.08,54.14,19.39)
[1] 426.3987
PS:要计算以英里为单位的距离,请将函数R(6371)替换为3958.756(海里使用3440.065)。