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
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)
}
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)
}
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;
}
下面是VB的实现。NET,这个实现将根据您传递的Enum值以KM或Miles为单位给您结果。
Public Enum DistanceType
Miles
KiloMeters
End Enum
Public Structure Position
Public Latitude As Double
Public Longitude As Double
End Structure
Public Class Haversine
Public Function Distance(Pos1 As Position,
Pos2 As Position,
DistType As DistanceType) As Double
Dim R As Double = If((DistType = DistanceType.Miles), 3960, 6371)
Dim dLat As Double = Me.toRadian(Pos2.Latitude - Pos1.Latitude)
Dim dLon As Double = Me.toRadian(Pos2.Longitude - Pos1.Longitude)
Dim a As Double = Math.Sin(dLat / 2) * Math.Sin(dLat / 2) + Math.Cos(Me.toRadian(Pos1.Latitude)) * Math.Cos(Me.toRadian(Pos2.Latitude)) * Math.Sin(dLon / 2) * Math.Sin(dLon / 2)
Dim c As Double = 2 * Math.Asin(Math.Min(1, Math.Sqrt(a)))
Dim result As Double = R * c
Return result
End Function
Private Function toRadian(val As Double) As Double
Return (Math.PI / 180) * val
End Function
End Class
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;
计算距离——尤其是大距离——的主要挑战之一是解释地球的曲率。如果地球是平的,计算两点之间的距离就会像计算直线一样简单!哈弗辛公式包括一个常数(下面是R变量),它表示地球的半径。根据你是用英里还是公里来测量,它分别等于3956英里或6367公里。 基本公式是:
Dlon = lon2 - lon1 dat = lat2 - lat1 = (sin (dlat / 2)) ^ 2 + cos (lat1) * cos (lat2) * (sin (dlon / 2)) ^ 2 C = 2 * atan2(√(a),√(1-a)) distance = R * c(其中R为地球半径) R = 6367公里OR 3956英里
lat1, lon1: The Latitude and Longitude of point 1 (in decimal degrees)
lat2, lon2: The Latitude and Longitude of point 2 (in decimal degrees)
unit: The unit of measurement in which to calculate the results where:
'M' is statute miles (default)
'K' is kilometers
'N' is nautical miles
样本
function distance(lat1, lon1, lat2, lon2, unit) {
try {
var radlat1 = Math.PI * lat1 / 180
var radlat2 = Math.PI * lat2 / 180
var theta = lon1 - lon2
var radtheta = Math.PI * theta / 180
var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
dist = Math.acos(dist)
dist = dist * 180 / Math.PI
dist = dist * 60 * 1.1515
if (unit == "K") {
dist = dist * 1.609344
}
if (unit == "N") {
dist = dist * 0.8684
}
return dist
} catch (err) {
console.log(err);
}
}