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

Java实现在根据哈弗辛公式

double calculateDistance(double latPoint1, double lngPoint1, 
                         double latPoint2, double lngPoint2) {
    if(latPoint1 == latPoint2 && lngPoint1 == lngPoint2) {
        return 0d;
    }

    final double EARTH_RADIUS = 6371.0; //km value;

    //converting to radians
    latPoint1 = Math.toRadians(latPoint1);
    lngPoint1 = Math.toRadians(lngPoint1);
    latPoint2 = Math.toRadians(latPoint2);
    lngPoint2 = Math.toRadians(lngPoint2);

    double distance = Math.pow(Math.sin((latPoint2 - latPoint1) / 2.0), 2) 
            + Math.cos(latPoint1) * Math.cos(latPoint2)
            * Math.pow(Math.sin((lngPoint2 - lngPoint1) / 2.0), 2);
    distance = 2.0 * EARTH_RADIUS * Math.asin(Math.sqrt(distance));

    return distance; //km value
}

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

我通过简化公式来简化计算。

下面是Ruby版本:

include Math
earth_radius_mi = 3959
radians = lambda { |deg| deg * PI / 180 }
coord_radians = lambda { |c| { :lat => radians[c[:lat]], :lng => radians[c[:lng]] } }

# from/to = { :lat => (latitude_in_degrees), :lng => (longitude_in_degrees) }
def haversine_distance(from, to)
  from, to = coord_radians[from], coord_radians[to]
  cosines_product = cos(to[:lat]) * cos(from[:lat]) * cos(from[:lng] - to[:lng])
  sines_product = sin(to[:lat]) * sin(from[:lat])
  return earth_radius_mi * acos(cosines_product + sines_product)
end

如果你想要驾驶距离/路线(张贴在这里,因为这是谷歌上两点之间距离的第一个结果,但对大多数人来说,驾驶距离更有用),你可以使用谷歌地图距离矩阵服务:

getDrivingDistanceBetweenTwoLatLong(origin, destination) {

 return new Observable(subscriber => {
  let service = new google.maps.DistanceMatrixService();
  service.getDistanceMatrix(
    {
      origins: [new google.maps.LatLng(origin.lat, origin.long)],
      destinations: [new google.maps.LatLng(destination.lat, destination.long)],
      travelMode: 'DRIVING'
    }, (response, status) => {
      if (status !== google.maps.DistanceMatrixStatus.OK) {
        console.log('Error:', status);
        subscriber.error({error: status, status: status});
      } else {
        console.log(response);
        try {
          let valueInMeters = response.rows[0].elements[0].distance.value;
          let valueInKms = valueInMeters / 1000;
          subscriber.next(valueInKms);
          subscriber.complete();
        }
       catch(error) {
        subscriber.error({error: error, status: status});
       }
      }
    });
});
}

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;
     }
 }

我已经创建了这个小Javascript LatLng对象,可能对某人有用。

var latLng1 = new LatLng(5, 3);
var latLng2 = new LatLng(6, 7);
var distance = latLng1.distanceTo(latLng2); 

代码:

/**
 * latLng point
 * @param {Number} lat
 * @param {Number} lng
 * @returns {LatLng}
 * @constructor
 */
function LatLng(lat,lng) {
    this.lat = parseFloat(lat);
    this.lng = parseFloat(lng);

    this.__cache = {};
}

LatLng.prototype = {
    toString: function() {
        return [this.lat, this.lng].join(",");
    },

    /**
     * calculate distance in km to another latLng, with caching
     * @param {LatLng} latLng
     * @returns {Number} distance in km
     */
    distanceTo: function(latLng) {
        var cacheKey = latLng.toString();
        if(cacheKey in this.__cache) {
            return this.__cache[cacheKey];
        }

        // the fastest way to calculate the distance, according to this jsperf test;
        // http://jsperf.com/haversine-salvador/8
        // http://stackoverflow.com/questions/27928
        var deg2rad = 0.017453292519943295; // === Math.PI / 180
        var lat1 = this.lat * deg2rad;
        var lng1 = this.lng * deg2rad;
        var lat2 = latLng.lat * deg2rad;
        var lng2 = latLng.lng * deg2rad;
        var a = (
            (1 - Math.cos(lat2 - lat1)) +
            (1 - Math.cos(lng2 - lng1)) * Math.cos(lat1) * Math.cos(lat2)
            ) / 2;
        var distance = 12742 * Math.asin(Math.sqrt(a)); // Diameter of the earth in km (2 * 6371)

        // cache the distance
        this.__cache[cacheKey] = distance;

        return distance;
    }
};