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

下面是另一个转换为Ruby代码的代码:

include Math
#Note: from/to = [lat, long]

def get_distance_in_km(from, to)
  radians = lambda { |deg| deg * Math.PI / 180 }
  radius = 6371 # Radius of the earth in kilometer
  dLat = radians[to[0]-from[0]]
  dLon = radians[to[1]-from[1]]

  cosines_product = Math.sin(dLat/2) * Math.sin(dLat/2) + Math.cos(radians[from[0]]) * Math.cos(radians[to[1]]) * Math.sin(dLon/2) * Math.sin(dLon/2)

  c = 2 * Math.atan2(Math.sqrt(cosines_product), Math.sqrt(1-cosines_product)) 
  return radius * c # Distance in kilometer
end

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;
}

这里有一个用PHP http://www.geodatasource.com/developers/php计算距离的好例子:

 function distance($lat1, $lon1, $lat2, $lon2, $unit) {

     $theta = $lon1 - $lon2;
     $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +  cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
     $dist = acos($dist);
     $dist = rad2deg($dist);
     $miles = $dist * 60 * 1.1515;
     $unit = strtoupper($unit);

     if ($unit == "K") {
         return ($miles * 1.609344);
     } else if ($unit == "N") {
          return ($miles * 0.8684);
     } else {
          return $miles;
     }
 }

我在R中做了一个自定义函数,使用R基本包中可用的函数来计算两个空间点之间的距离(km)。

custom_hav_dist <- function(lat1, lon1, lat2, lon2) {
R <- 6371
Radian_factor <- 0.0174533
lat_1 <- (90-lat1)*Radian_factor
lat_2 <- (90-lat2)*Radian_factor
diff_long <-(lon1-lon2)*Radian_factor

distance_in_km <- 6371*acos((cos(lat_1)*cos(lat_2))+ 
                 (sin(lat_1)*sin(lat_2)*cos(diff_long)))
rm(lat1, lon1, lat2, lon2)
return(distance_in_km)
}

样例输出

custom_hav_dist(50.31,19.08,54.14,19.39)
[1] 426.3987

PS:要计算以英里为单位的距离,请将函数R(6371)替换为3958.756(海里使用3440.065)。

下面是另一个转换为Ruby代码的代码:

include Math
#Note: from/to = [lat, long]

def get_distance_in_km(from, to)
  radians = lambda { |deg| deg * Math.PI / 180 }
  radius = 6371 # Radius of the earth in kilometer
  dLat = radians[to[0]-from[0]]
  dLon = radians[to[1]-from[1]]

  cosines_product = Math.sin(dLat/2) * Math.sin(dLat/2) + Math.cos(radians[from[0]]) * Math.cos(radians[to[1]]) * Math.sin(dLon/2) * Math.sin(dLon/2)

  c = 2 * Math.atan2(Math.sqrt(cosines_product), Math.sqrt(1-cosines_product)) 
  return radius * c # Distance in kilometer
end

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
            });

        }