如何计算谷歌地图V3中两个标记之间的距离?(类似于inV2的distanceFrom函数。)
谢谢. .
如何计算谷歌地图V3中两个标记之间的距离?(类似于inV2的distanceFrom函数。)
谢谢. .
当前回答
如果你想自己计算,那么你可以使用哈弗辛公式:
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
};
其他回答
/**
* Calculates the haversine distance between point A, and B.
* @param {number[]} latlngA [lat, lng] point A
* @param {number[]} latlngB [lat, lng] point B
* @param {boolean} isMiles If we are using miles, else km.
*/
function haversineDistance(latlngA, latlngB, isMiles) {
const squared = x => x * x;
const toRad = x => (x * Math.PI) / 180;
const R = 6371; // Earth’s mean radius in km
const dLat = toRad(latlngB[0] - latlngA[0]);
const dLon = toRad(latlngB[1] - latlngA[1]);
const dLatSin = squared(Math.sin(dLat / 2));
const dLonSin = squared(Math.sin(dLon / 2));
const a = dLatSin +
(Math.cos(toRad(latlngA[0])) * Math.cos(toRad(latlngB[0])) * dLonSin);
const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
let distance = R * c;
if (isMiles) distance /= 1.609344;
return distance;
}
我在网上找到了一个版本,80%是正确的,但插入了错误的参数,在使用输入时不一致,这个版本完全解决了这个问题
在我的情况下,这是最好的计算在SQL Server,因为我想采取当前位置,然后搜索所有邮政编码从当前位置一定距离内。我还有一个数据库,其中包含邮政编码及其纬度的列表。干杯
--will return the radius for a given number
create function getRad(@variable float)--function to return rad
returns float
as
begin
declare @retval float
select @retval=(@variable * PI()/180)
--print @retval
return @retval
end
go
--calc distance
--drop function dbo.getDistance
create function getDistance(@cLat float,@cLong float, @tLat float, @tLong float)
returns float
as
begin
declare @emr float
declare @dLat float
declare @dLong float
declare @a float
declare @distance float
declare @c float
set @emr = 6371--earth mean
set @dLat = dbo.getRad(@tLat - @cLat);
set @dLong = dbo.getRad(@tLong - @cLong);
set @a = sin(@dLat/2)*sin(@dLat/2)+cos(dbo.getRad(@cLat))*cos(dbo.getRad(@tLat))*sin(@dLong/2)*sin(@dLong/2);
set @c = 2*atn2(sqrt(@a),sqrt(1-@a))
set @distance = @emr*@c;
set @distance = @distance * 0.621371 -- i needed it in miles
--print @distance
return @distance;
end
go
--get all zipcodes within 2 miles, the hardcoded #'s would be passed in by C#
select *
from cityzips a where dbo.getDistance(29.76,-95.38,a.lat,a.long) <3
order by zipcode
只需将以下内容添加到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);
}
然后像这样使用函数:
var loc1 = new GLatLng(52.5773139, 1.3712427);
var loc2 = new GLatLng(52.4788314, 1.7577444);
var dist = loc2.distanceFrom(loc1);
alert(dist/1000);
//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);
}
下面是this公式的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
}
}