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。
其他回答
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
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;
}
}
如果你正在使用python; PIP安装地质
from geopy.distance import geodesic
origin = (30.172705, 31.526725) # (latitude, longitude) don't confuse
destination = (30.288281, 31.732326)
print(geodesic(origin, destination).meters) # 23576.805481751613
print(geodesic(origin, destination).kilometers) # 23.576805481751613
print(geodesic(origin, destination).miles) # 14.64994773134371
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
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。