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
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
function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2,units) {
var R = 6371; // Radius of the earth in km
var dLat = deg2rad(lat2-lat1); // deg2rad below
var dLon = deg2rad(lon2-lon1);
var a =
Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) *
Math.sin(dLon/2) * Math.sin(dLon/2)
;
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
var miles = d / 1.609344;
if ( units == 'km' ) {
return d;
} else {
return miles;
}}
查克的解决方案,也适用于英里。
//JAVA
public Double getDistanceBetweenTwoPoints(Double latitude1, Double longitude1, Double latitude2, Double longitude2) {
final int RADIUS_EARTH = 6371;
double dLat = getRad(latitude2 - latitude1);
double dLong = getRad(longitude2 - longitude1);
double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.cos(getRad(latitude1)) * Math.cos(getRad(latitude2)) * Math.sin(dLong / 2) * Math.sin(dLong / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return (RADIUS_EARTH * c) * 1000;
}
private Double getRad(Double x) {
return x * Math.PI / 180;
}
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
});
}
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;
}
};
计算距离——尤其是大距离——的主要挑战之一是解释地球的曲率。如果地球是平的,计算两点之间的距离就会像计算直线一样简单!哈弗辛公式包括一个常数(下面是R变量),它表示地球的半径。根据你是用英里还是公里来测量,它分别等于3956英里或6367公里。 基本公式是:
Dlon = lon2 - lon1 dat = lat2 - lat1 = (sin (dlat / 2)) ^ 2 + cos (lat1) * cos (lat2) * (sin (dlon / 2)) ^ 2 C = 2 * atan2(√(a),√(1-a)) distance = R * c(其中R为地球半径) R = 6367公里OR 3956英里
lat1, lon1: The Latitude and Longitude of point 1 (in decimal degrees)
lat2, lon2: The Latitude and Longitude of point 2 (in decimal degrees)
unit: The unit of measurement in which to calculate the results where:
'M' is statute miles (default)
'K' is kilometers
'N' is nautical miles
样本
function distance(lat1, lon1, lat2, lon2, unit) {
try {
var radlat1 = Math.PI * lat1 / 180
var radlat2 = Math.PI * lat2 / 180
var theta = lon1 - lon2
var radtheta = Math.PI * theta / 180
var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta);
dist = Math.acos(dist)
dist = dist * 180 / Math.PI
dist = dist * 60 * 1.1515
if (unit == "K") {
dist = dist * 1.609344
}
if (unit == "N") {
dist = dist * 0.8684
}
return dist
} catch (err) {
console.log(err);
}
}