计算谷歌地图V3中两点之间的距离

Question

如何计算Google地图V3中两个标记之间的距离？(类似于distanceFromV2中的功能.)

谢谢..

Answer 1

如果您想自己计算,那么您可以使用Haversine公式:

var rad = function(x) {
  return x * Math.PI / 180;
};

var getDistance = function(p1, p2) {
  var R = 6378137; // Earth’s mean radius in meter
  var dLat = rad(p2.lat() - p1.lat());
  var dLong = rad(p2.lng() - p1.lng());
  var a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
    Math.cos(rad(p1.lat())) * Math.cos(rad(p2.lat())) *
    Math.sin(dLong / 2) * Math.sin(dLong / 2);
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
  var d = R * c;
  return d; // returns the distance in meter
};

Answer 2

GMap3实际上似乎有一种方法.它是google.maps.geometry.spherical命名空间的静态方法.

它将两个LatLng对象作为参数,并将使用默认的地球半径6378137米,但必要时可以使用自定义值覆盖默认半径.

确保包括:

<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false&v=3&libraries=geometry"></script>

在你的头部.

电话会是:

google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);

Answer 3

使用GPS经纬度为2点的示例.

var latitude1 = 39.46;
var longitude1 = -0.36;
var latitude2 = 40.40;
var longitude2 = -3.68;

var distance = google.maps.geometry.spherical.computeDistanceBetween(new google.maps.LatLng(latitude1, longitude1), new google.maps.LatLng(latitude2, longitude2));       

Answer 4

只需将其添加到JavaScript代码的开头:

google.maps.LatLng.prototype.distanceFrom = function(latlng) {
  var lat = [this.lat(), latlng.lat()]
  var lng = [this.lng(), latlng.lng()]
  var R = 6378137;
  var dLat = (lat[1]-lat[0]) * Math.PI / 180;
  var dLng = (lng[1]-lng[0]) * Math.PI / 180;
  var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
  Math.cos(lat[0] * Math.PI / 180 ) * Math.cos(lat[1] * Math.PI / 180 ) *
  Math.sin(dLng/2) * Math.sin(dLng/2);
  var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
  var d = R * c;
  return Math.round(d);
}

然后使用这样的函数:

var loc1 = new GLatLng(52.5773139, 1.3712427);
var loc2 = new GLatLng(52.4788314, 1.7577444);
var dist = loc2.distanceFrom(loc1);
alert(dist/1000);

Answer 5

//p1 and p2 are google.maps.LatLng(x,y) objects

function calcDistance(p1, p2) {
          var d = (google.maps.geometry.spherical.computeDistanceBetween(p1, p2) / 1000).toFixed(2);
          console.log(d);              
}

Answer 6

这是本论坛的c#实现

 public class DistanceAlgorithm
{
    const double PIx = 3.141592653589793;
    const double RADIO = 6378.16;

    /// <summary>
    /// This class cannot be instantiated.
    /// </summary>
    private DistanceAlgorithm() { }

    /// <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 * RADIO) * 0.62137;//distance in miles
    }

}    

Answer 7

与谷歌,你可以用它做球的API,google.maps.geometry.spherical.computeDistanceBetween (latLngA, latLngB);.

但是,如果球形投影或半正弦解决方案的精度不够精确(例如,如果您接近极点或计算更长的距离),则应使用不同的库.

在我在维基百科上找到该主题上的信息在这里.

查看任何给定算法的精度是否足够的技巧是填写地球的最大和最小半径,看看差异是否可能导致您的用例出现问题.在本文中可以找到更多详细信息

最终谷歌api或者半身像将毫无问题地用于大多数目的.

Answer 8

使用PHP,您可以使用这个简单的函数计算距离:

// to calculate distance between two lat & lon

function calculate_distance($lat1, $lon1, $lat2, $lon2, $unit='N') 
{ 
  $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;
      }
}

// function ends here

Answer 9

离线解决方案 - Haversine算法

在Javascript中

var _eQuatorialEarthRadius = 6378.1370;
var _d2r = (Math.PI / 180.0);

function HaversineInM(lat1, long1, lat2, long2)
{
    return (1000.0 * HaversineInKM(lat1, long1, lat2, long2));
}

function HaversineInKM(lat1, long1, lat2, long2)
{
    var dlong = (long2 - long1) * _d2r;
    var dlat = (lat2 - lat1) * _d2r;
    var a = Math.pow(Math.sin(dlat / 2.0), 2.0) + Math.cos(lat1 * _d2r) * Math.cos(lat2 * _d2r) * Math.pow(Math.sin(dlong / 2.0), 2.0);
    var c = 2.0 * Math.atan2(Math.sqrt(a), Math.sqrt(1.0 - a));
    var d = _eQuatorialEarthRadius * c;

    return d;
}

var meLat = -33.922982;
var meLong = 151.083853;


var result1 = HaversineInKM(meLat, meLong, -32.236457779983745, 148.69094705162837);
var result2 = HaversineInKM(meLat, meLong, -33.609020205923713, 150.77061469270831);

C#

using System;

public class Program
{
    public static void Main()
    {
        Console.WriteLine("Hello World");

        var meLat = -33.922982;
        double meLong = 151.083853;


        var result1 = HaversineInM(meLat, meLong, -32.236457779983745, 148.69094705162837);
        var result2 = HaversineInM(meLat, meLong, -33.609020205923713, 150.77061469270831);

        Console.WriteLine(result1);
        Console.WriteLine(result2);
    }

    static double _eQuatorialEarthRadius = 6378.1370D;
    static double _d2r = (Math.PI / 180D);

    private static int HaversineInM(double lat1, double long1, double lat2, double long2)
    {
        return (int)(1000D * HaversineInKM(lat1, long1, lat2, long2));
    }

    private static  double HaversineInKM(double lat1, double long1, double lat2, double long2)
    {
        double dlong = (long2 - long1) * _d2r;
        double dlat = (lat2 - lat1) * _d2r;
        double a = Math.Pow(Math.Sin(dlat / 2D), 2D) + Math.Cos(lat1 * _d2r) * Math.Cos(lat2 * _d2r) * Math.Pow(Math.Sin(dlong / 2D), 2D);
        double c = 2D * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1D - a));
        double d = _eQuatorialEarthRadius * c;

        return d;
    }
}

参考:https: //en.wikipedia.org/wiki/Great-circle_distance