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
你可以用Haversine公式计算它,它是:
a = sin²(Δφ/2) + cos φ1 ⋅ cos φ2 ⋅ sin²(Δλ/2)
c = 2 ⋅ atan2( √a, √(1−a) )
d = R ⋅ c
下面给出了一个计算两点之间距离的例子
假设我要计算从新德里到伦敦的距离,那么我该如何使用这个公式:
New delhi co-ordinates= 28.7041° N, 77.1025° E
London co-ordinates= 51.5074° N, 0.1278° W
var R = 6371e3; // metres
var φ1 = 28.7041.toRadians();
var φ2 = 51.5074.toRadians();
var Δφ = (51.5074-28.7041).toRadians();
var Δλ = (0.1278-77.1025).toRadians();
var a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
Math.cos(φ1) * Math.cos(φ2) *
Math.sin(Δλ/2) * Math.sin(Δλ/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c; // metres
d = d/1000; // km
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;
}
你可以用Haversine公式计算它,它是:
a = sin²(Δφ/2) + cos φ1 ⋅ cos φ2 ⋅ sin²(Δλ/2)
c = 2 ⋅ atan2( √a, √(1−a) )
d = R ⋅ c
下面给出了一个计算两点之间距离的例子
假设我要计算从新德里到伦敦的距离,那么我该如何使用这个公式:
New delhi co-ordinates= 28.7041° N, 77.1025° E
London co-ordinates= 51.5074° N, 0.1278° W
var R = 6371e3; // metres
var φ1 = 28.7041.toRadians();
var φ2 = 51.5074.toRadians();
var Δφ = (51.5074-28.7041).toRadians();
var Δλ = (0.1278-77.1025).toRadians();
var a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
Math.cos(φ1) * Math.cos(φ2) *
Math.sin(Δλ/2) * Math.sin(Δλ/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c; // metres
d = d/1000; // km
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
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;
}
在提供的代码中有一些错误,我在下面修复了它。
以上所有答案都假定地球是一个球体。然而,更精确的近似是扁球体。
a= 6378.137#equitorial radius in km
b= 6356.752#polar radius in km
def Distance(lat1, lons1, lat2, lons2):
lat1=math.radians(lat1)
lons1=math.radians(lons1)
R1=(((((a**2)*math.cos(lat1))**2)+(((b**2)*math.sin(lat1))**2))/((a*math.cos(lat1))**2+(b*math.sin(lat1))**2))**0.5 #radius of earth at lat1
x1=R1*math.cos(lat1)*math.cos(lons1)
y1=R1*math.cos(lat1)*math.sin(lons1)
z1=R1*math.sin(lat1)
lat2=math.radians(lat2)
lons2=math.radians(lons2)
R2=(((((a**2)*math.cos(lat2))**2)+(((b**2)*math.sin(lat2))**2))/((a*math.cos(lat2))**2+(b*math.sin(lat2))**2))**0.5 #radius of earth at lat2
x2=R2*math.cos(lat2)*math.cos(lons2)
y2=R2*math.cos(lat2)*math.sin(lons2)
z2=R2*math.sin(lat2)
return ((x1-x2)**2+(y1-y2)**2+(z1-z2)**2)**0.5