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;
}
}
其他回答
这是我的java实现计算距离经过一些搜索。我用的是世界平均半径(来自维基百科),单位是千米。İf你想要的结果英里,然后使用世界半径英里。
public static double distanceLatLong2(double lat1, double lng1, double lat2, double lng2)
{
double earthRadius = 6371.0d; // KM: use mile here if you want mile result
double dLat = toRadian(lat2 - lat1);
double dLng = toRadian(lng2 - lng1);
double a = Math.pow(Math.sin(dLat/2), 2) +
Math.cos(toRadian(lat1)) * Math.cos(toRadian(lat2)) *
Math.pow(Math.sin(dLng/2), 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
return earthRadius * c; // returns result kilometers
}
public static double toRadian(double degrees)
{
return (degrees * Math.PI) / 180.0d;
}
function getDistanceFromLatLonInKm(position1, position2) {
"use strict";
var deg2rad = function (deg) { return deg * (Math.PI / 180); },
R = 6371,
dLat = deg2rad(position2.lat - position1.lat),
dLng = deg2rad(position2.lng - position1.lng),
a = Math.sin(dLat / 2) * Math.sin(dLat / 2)
+ Math.cos(deg2rad(position1.lat))
* Math.cos(deg2rad(position2.lat))
* Math.sin(dLng / 2) * Math.sin(dLng / 2),
c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return R * c;
}
console.log(getDistanceFromLatLonInKm(
{lat: 48.7931459, lng: 1.9483572},
{lat: 48.827167, lng: 2.2459745}
));
我不喜欢添加另一个答案,但谷歌地图API v.3具有球形几何(以及更多)。在将你的WGS84转换为十进制度后,你可以这样做:
<script src="http://maps.google.com/maps/api/js?sensor=false&libraries=geometry" type="text/javascript"></script>
distance = google.maps.geometry.spherical.computeDistanceBetween(
new google.maps.LatLng(fromLat, fromLng),
new google.maps.LatLng(toLat, toLng));
关于谷歌的计算有多精确,甚至使用了什么模型都没有任何消息(尽管它说的是“球面”而不是“大地水准面”。顺便说一下,“直线”距离显然不同于一个人在地球表面旅行的距离,而这似乎是每个人都在假设的。
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;
}
下面是一个Scala实现:
def calculateHaversineDistance(lat1: Double, lon1: Double, lat2: Double, lon2: Double): Double = {
val long2 = lon2 * math.Pi / 180
val lat2 = lat2 * math.Pi / 180
val long1 = lon1 * math.Pi / 180
val lat1 = lat1 * math.Pi / 180
val dlon = long2 - long1
val dlat = lat2 - lat1
val a = math.pow(math.sin(dlat / 2), 2) + math.cos(lat1) * math.cos(lat2) * math.pow(math.sin(dlon / 2), 2)
val c = 2 * math.atan2(Math.sqrt(a), math.sqrt(1 - a))
val haversineDistance = 3961 * c // 3961 = radius of earth in miles
haversineDistance
}