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
在Mysql中使用以下函数传递参数,使用POINT(LONG,LAT)
CREATE FUNCTION `distance`(a POINT, b POINT)
RETURNS double
DETERMINISTIC
BEGIN
RETURN
GLength( LineString(( PointFromWKB(a)), (PointFromWKB(b)))) * 100000; -- To Make the distance in meters
END;
其他回答
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);
}
你可以使用CLLocationDistance中的构建来计算这个:
CLLocation *location1 = [[CLLocation alloc] initWithLatitude:latitude1 longitude:longitude1];
CLLocation *location2 = [[CLLocation alloc] initWithLatitude:latitude2 longitude:longitude2];
[self distanceInMetersFromLocation:location1 toLocation:location2]
- (int)distanceInMetersFromLocation:(CLLocation*)location1 toLocation:(CLLocation*)location2 {
CLLocationDistance distanceInMeters = [location1 distanceFromLocation:location2];
return distanceInMeters;
}
在你的例子中,如果你想要公里,只要除以1000。
正如指出的那样,精确的计算应该考虑到地球不是一个完美的球体。以下是这里提供的各种算法的一些比较:
geoDistance(50,5,58,3)
Haversine: 899 km
Maymenn: 833 km
Keerthana: 897 km
google.maps.geometry.spherical.computeDistanceBetween(): 900 km
geoDistance(50,5,-58,-3)
Haversine: 12030 km
Maymenn: 11135 km
Keerthana: 10310 km
google.maps.geometry.spherical.computeDistanceBetween(): 12044 km
geoDistance(.05,.005,.058,.003)
Haversine: 0.9169 km
Maymenn: 0.851723 km
Keerthana: 0.917964 km
google.maps.geometry.spherical.computeDistanceBetween(): 0.917964 km
geoDistance(.05,80,.058,80.3)
Haversine: 33.37 km
Maymenn: 33.34 km
Keerthana: 33.40767 km
google.maps.geometry.spherical.computeDistanceBetween(): 33.40770 km
在小范围内,Keerthana的算法似乎与谷歌Maps的算法一致。谷歌Maps似乎没有遵循任何简单的算法,这表明它可能是这里最准确的方法。
不管怎样,这里是Keerthana算法的Javascript实现:
function geoDistance(lat1, lng1, lat2, lng2){
const a = 6378.137; // equitorial radius in km
const b = 6356.752; // polar radius in km
var sq = x => (x*x);
var sqr = x => Math.sqrt(x);
var cos = x => Math.cos(x);
var sin = x => Math.sin(x);
var radius = lat => sqr((sq(a*a*cos(lat))+sq(b*b*sin(lat)))/(sq(a*cos(lat))+sq(b*sin(lat))));
lat1 = lat1 * Math.PI / 180;
lng1 = lng1 * Math.PI / 180;
lat2 = lat2 * Math.PI / 180;
lng2 = lng2 * Math.PI / 180;
var R1 = radius(lat1);
var x1 = R1*cos(lat1)*cos(lng1);
var y1 = R1*cos(lat1)*sin(lng1);
var z1 = R1*sin(lat1);
var R2 = radius(lat2);
var x2 = R2*cos(lat2)*cos(lng2);
var y2 = R2*cos(lat2)*sin(lng2);
var z2 = R2*sin(lat2);
return sqr(sq(x1-x2)+sq(y1-y2)+sq(z1-z2));
}
这是一个简单的javascript函数,从这个链接可能是有用的。不知何故相关,但我们使用谷歌地球javascript插件而不是地图
function getApproximateDistanceUnits(point1, point2) {
var xs = 0;
var ys = 0;
xs = point2.getX() - point1.getX();
xs = xs * xs;
ys = point2.getY() - point1.getY();
ys = ys * ys;
return Math.sqrt(xs + ys);
}
单位不是距离,而是相对于坐标的比率。还有其他相关的计算,你可以在这里代替getApproximateDistanceUnits函数链接
然后我使用这个函数来查看经纬度是否在半径内
function isMapPlacemarkInRadius(point1, point2, radi) {
if (point1 && point2) {
return getApproximateDistanceUnits(point1, point2) <= radi;
} else {
return 0;
}
}
点可以定义为
$$.getPoint = function(lati, longi) {
var location = {
x: 0,
y: 0,
getX: function() { return location.x; },
getY: function() { return location.y; }
};
location.x = lati;
location.y = longi;
return location;
};
然后你可以做你的事情,看看一个点是否在一个半径范围内,比如:
//put it on the map if within the range of a specified radi assuming 100,000,000 units
var iconpoint = Map.getPoint(pp.latitude, pp.longitude);
var centerpoint = Map.getPoint(Settings.CenterLatitude, Settings.CenterLongitude);
//approx ~200 units to show only half of the globe from the default center radius
if (isMapPlacemarkInRadius(centerpoint, iconpoint, 120)) {
addPlacemark(pp.latitude, pp.longitude, pp.name);
}
else {
otherSidePlacemarks.push({
latitude: pp.latitude,
longitude: pp.longitude,
name: pp.name
});
}
如果你想要驾驶距离/路线(张贴在这里,因为这是谷歌上两点之间距离的第一个结果,但对大多数人来说,驾驶距离更有用),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});
}
}
});
});
}