javascript 计算两个坐标之间距离的函数
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Function to calculate distance between two coordinates
提问by Erik
I'm currently using the function below and it doesn't work properly. According to Google Maps, the distance between these coordinates(from 59.3293371,13.4877472to 59.3225525,13.4619422) are 2.2kilometres while the function returns 1.6kilometres. How can I make this function return the correct distance?
我目前正在使用下面的功能，但它无法正常工作。根据谷歌地图，这些坐标（从59.3293371,13.4877472到59.3225525,13.4619422）之间的距离是2.2公里，而函数返回1.6公里。我怎样才能让这个函数返回正确的距离？
function getDistanceFromLatLonInKm(lat1, lon1, lat2, lon2) {
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; // Distance in km
return d;
}
function deg2rad(deg) {
return deg * (Math.PI/180)
}
jsFiddle: http://jsfiddle.net/edgren/gAHJB/
jsFiddle：http: //jsfiddle.net/edgren/gAHJB/
回答by Ethan Brown
What you're using is called the haversine formula, which calculates the distance between two points on a sphere as the crow flies. The Google Maps link you provided shows the distance as 2.2 km because it's not a straight line.
您使用的是所谓的半正弦公式，它计算乌鸦飞行时球体上两点之间的距离。您提供的 Google 地图链接显示距离为 2.2 公里，因为它不是直线。
Wolphram Alpha is a great resource for doing geographic calculations, and also shows a distance of 1.652 km between these two points.
Wolphram Alpha 是进行地理计算的绝佳资源，并且还显示这两个点之间的距离为1.652 公里。
If you're looking for straight-line distance (as the crow files), your function is working correctly. If what you want is driving distance (or biking distance or public transportation distance or walking distance), you'll have to use a mapping API (Googleor Bingbeing the most popular) to get the appropriate route, which will include the distance.
如果您正在寻找直线距离（如乌鸦文件），则您的函数工作正常。如果您想要的是行驶距离（或骑车距离或公共交通距离或步行距离），则必须使用映射 API（Google或Bing是最受欢迎的）来获取适当的路线，其中包括距离。
Incidentally, the Google Maps API provides a packaged method for spherical distance, in its google.maps.geometry.sphericalnamespace(look for computeDistanceBetween). It's probably better than rolling your own (for starters, it uses a more precise value for the Earth's radius).
顺便说一句，Google Maps API 在其google.maps.geometry.spherical命名空间（查找computeDistanceBetween）中提供了球面距离的封装方法。它可能比滚动你自己的更好（对于初学者来说，它使用更精确的地球半径值）。
For the picky among us, when I say "straight-line distance", I'm referring to a "straight line on a sphere", which is actually a curved line (i.e. the great-circle distance), of course.
对于我们中挑剔的人，当我说“直线距离”时，我指的是“球面上的直线”，当然，它实际上是一条曲线（即大圆距离）。
回答by Derek
I have written a similar equation before - tested it and also got 1.6 km.
我之前写过一个类似的方程 - 测试它，也得到了 1.6 公里。
Your google maps was showing the DRIVING distance.
您的谷歌地图显示了行驶距离。
Your function is calculating as the crow flies (straight line distance).
您的函数正在计算乌鸦飞行（直线距离）。
alert(calcCrow(59.3293371,13.4877472,59.3225525,13.4619422).toFixed(1));
//This function takes in latitude and longitude of two location and returns the distance between them as the crow flies (in km)
function calcCrow(lat1, lon1, lat2, lon2)
{
var R = 6371; // km
var dLat = toRad(lat2-lat1);
var dLon = toRad(lon2-lon1);
var lat1 = toRad(lat1);
var lat2 = toRad(lat2);
var a = Math.sin(dLat/2) * Math.sin(dLat/2) +
Math.sin(dLon/2) * Math.sin(dLon/2) * Math.cos(lat1) * Math.cos(lat2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
return d;
}
// Converts numeric degrees to radians
function toRad(Value)
{
return Value * Math.PI / 180;
}
回答by Behzad
Derek's solution worked fine for me, and I've just simply converted it to PHP, hope it helps somebody out there !
Derek 的解决方案对我来说效果很好，我只是简单地将它转换为 PHP，希望它可以帮助那里的人！
function calcCrow($lat1, $lon1, $lat2, $lon2){
$R = 6371; // km
$dLat = toRad($lat2-$lat1);
$dLon = toRad($lon2-$lon1);
$lat1 = toRad($lat1);
$lat2 = toRad($lat2);
$a = sin($dLat/2) * sin($dLat/2) +sin($dLon/2) * sin($dLon/2) * cos($lat1) * cos($lat2);
$c = 2 * atan2(sqrt($a), sqrt(1-$a));
$d = $R * $c;
return $d;
}
// Converts numeric degrees to radians
function toRad($Value)
{
return $Value * pi() / 180;
}
回答by Noorul
Try this. It is in VB.net and you need to convert it to Javascript. This function accepts parameters in decimal minutes.
试试这个。它在 VB.net 中，您需要将其转换为 Javascript。此函数接受以十进制分钟为单位的参数。
Private Function calculateDistance(ByVal long1 As String, ByVal lat1 As String, ByVal long2 As String, ByVal lat2 As String) As Double
long1 = Double.Parse(long1)
lat1 = Double.Parse(lat1)
long2 = Double.Parse(long2)
lat2 = Double.Parse(lat2)
'conversion to radian
lat1 = (lat1 * 2.0 * Math.PI) / 60.0 / 360.0
long1 = (long1 * 2.0 * Math.PI) / 60.0 / 360.0
lat2 = (lat2 * 2.0 * Math.PI) / 60.0 / 360.0
long2 = (long2 * 2.0 * Math.PI) / 60.0 / 360.0
' use to different earth axis length
Dim a As Double = 6378137.0 ' Earth Major Axis (WGS84)
Dim b As Double = 6356752.3142 ' Minor Axis
Dim f As Double = (a - b) / a ' "Flattening"
Dim e As Double = 2.0 * f - f * f ' "Eccentricity"
Dim beta As Double = (a / Math.Sqrt(1.0 - e * Math.Sin(lat1) * Math.Sin(lat1)))
Dim cos As Double = Math.Cos(lat1)
Dim x As Double = beta * cos * Math.Cos(long1)
Dim y As Double = beta * cos * Math.Sin(long1)
Dim z As Double = beta * (1 - e) * Math.Sin(lat1)
beta = (a / Math.Sqrt(1.0 - e * Math.Sin(lat2) * Math.Sin(lat2)))
cos = Math.Cos(lat2)
x -= (beta * cos * Math.Cos(long2))
y -= (beta * cos * Math.Sin(long2))
z -= (beta * (1 - e) * Math.Sin(lat2))
Return Math.Sqrt((x * x) + (y * y) + (z * z))
End Function
EditThe converted function in javascript
在javascript中编辑转换后的函数
function calculateDistance(lat1, long1, lat2, long2)
{
//radians
lat1 = (lat1 * 2.0 * Math.PI) / 60.0 / 360.0;
long1 = (long1 * 2.0 * Math.PI) / 60.0 / 360.0;
lat2 = (lat2 * 2.0 * Math.PI) / 60.0 / 360.0;
long2 = (long2 * 2.0 * Math.PI) / 60.0 / 360.0;
// use to different earth axis length
var a = 6378137.0; // Earth Major Axis (WGS84)
var b = 6356752.3142; // Minor Axis
var f = (a-b) / a; // "Flattening"
var e = 2.0*f - f*f; // "Eccentricity"
var beta = (a / Math.sqrt( 1.0 - e * Math.sin( lat1 ) * Math.sin( lat1 )));
var cos = Math.cos( lat1 );
var x = beta * cos * Math.cos( long1 );
var y = beta * cos * Math.sin( long1 );
var z = beta * ( 1 - e ) * Math.sin( lat1 );
beta = ( a / Math.sqrt( 1.0 - e * Math.sin( lat2 ) * Math.sin( lat2 )));
cos = Math.cos( lat2 );
x -= (beta * cos * Math.cos( long2 ));
y -= (beta * cos * Math.sin( long2 ));
z -= (beta * (1 - e) * Math.sin( lat2 ));
return (Math.sqrt( (x*x) + (y*y) + (z*z) )/1000);
}
回答by Manjunath Bilwar
Calculate the Distance between Two Points in javascript
在javascript中计算两点之间的距离
function distance(lat1, lon1, lat2, lon2, unit) {
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
}
For more details refer this: Reference Link
有关更多详细信息，请参阅：参考链接
回答by Saikat
Using Haversine formula, source of the code:
使用Haversine公式，代码来源：
//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
//::: :::
//::: This routine calculates the distance between two points (given the :::
//::: latitude/longitude of those points). It is being used to calculate :::
//::: the distance between two locations using GeoDataSource (TM) prodducts :::
//::: :::
//::: Definitions: :::
//::: South latitudes are negative, east longitudes are positive :::
//::: :::
//::: Passed to function: :::
//::: lat1, lon1 = Latitude and Longitude of point 1 (in decimal degrees) :::
//::: lat2, lon2 = Latitude and Longitude of point 2 (in decimal degrees) :::
//::: unit = the unit you desire for results :::
//::: where: 'M' is statute miles (default) :::
//::: 'K' is kilometers :::
//::: 'N' is nautical miles :::
//::: :::
//::: Worldwide cities and other features databases with latitude longitude :::
//::: are available at https://www.geodatasource.com :::
//::: :::
//::: For enquiries, please contact [email protected] :::
//::: :::
//::: Official Web site: https://www.geodatasource.com :::
//::: :::
//::: GeoDataSource.com (C) All Rights Reserved 2018 :::
//::: :::
//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
function distance(lat1, lon1, lat2, lon2, unit) {
if ((lat1 == lat2) && (lon1 == lon2)) {
return 0;
}
else {
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);
if (dist > 1) {
dist = 1;
}
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;
}
}
The sample code is licensed under LGPLv3.
示例代码在 LGPLv3 下获得许可。
回答by Loukan ElKadi
Great-circle distance - From chord length
大圆距离 - 从弦长
Here's an elegant solution applying the strategy design pattern; I hope it's readable enough.
这是一个应用策略设计模式的优雅解决方案；我希望它足够可读。
TwoPointsDistanceCalculatorStrategy.js:
两点距离计算器策略.js：
module.exports = () =>
class TwoPointsDistanceCalculatorStrategy {
constructor() {}
calculateDistance({ point1Coordinates, point2Coordinates }) {}
};
GreatCircleTwoPointsDistanceCalculatorStrategy.js:
GreatCircleTwoPointsDistanceCalculatorStrategy.js：
module.exports = ({ TwoPointsDistanceCalculatorStrategy }) =>
class GreatCircleTwoPointsDistanceCalculatorStrategy extends TwoPointsDistanceCalculatorStrategy {
constructor() {
super();
}
/**
* Following the algorithm documented here:
* https://en.wikipedia.org/wiki/Great-circle_distance#Computational_formulas
*
* @param {object} inputs
* @param {array} inputs.point1Coordinates
* @param {array} inputs.point2Coordinates
*
* @returns {decimal} distance in kelometers
*/
calculateDistance({ point1Coordinates, point2Coordinates }) {
const convertDegreesToRadians = require('../convert-degrees-to-radians');
const EARTH_RADIUS = 6371; // in kelometers
const [lat1 = 0, lon1 = 0] = point1Coordinates;
const [lat2 = 0, lon2 = 0] = point2Coordinates;
const radianLat1 = convertDegreesToRadians({ degrees: lat1 });
const radianLon1 = convertDegreesToRadians({ degrees: lon1 });
const radianLat2 = convertDegreesToRadians({ degrees: lat2 });
const radianLon2 = convertDegreesToRadians({ degrees: lon2 });
const centralAngle = _computeCentralAngle({
lat1: radianLat1, lon1: radianLon1,
lat2: radianLat2, lon2: radianLon2,
});
const distance = EARTH_RADIUS * centralAngle;
return distance;
}
};
/**
*
* @param {object} inputs
* @param {decimal} inputs.lat1
* @param {decimal} inputs.lon1
* @param {decimal} inputs.lat2
* @param {decimal} inputs.lon2
*
* @returns {decimal} centralAngle
*/
function _computeCentralAngle({ lat1, lon1, lat2, lon2 }) {
const chordLength = _computeChordLength({ lat1, lon1, lat2, lon2 });
const centralAngle = 2 * Math.asin(chordLength / 2);
return centralAngle;
}
/**
*
* @param {object} inputs
* @param {decimal} inputs.lat1
* @param {decimal} inputs.lon1
* @param {decimal} inputs.lat2
* @param {decimal} inputs.lon2
*
* @returns {decimal} chordLength
*/
function _computeChordLength({ lat1, lon1, lat2, lon2 }) {
const { sin, cos, pow, sqrt } = Math;
const ΔX = cos(lat2) * cos(lon2) - cos(lat1) * cos(lon1);
const ΔY = cos(lat2) * sin(lon2) - cos(lat1) * sin(lon1);
const ΔZ = sin(lat2) - sin(lat1);
const ΔXSquare = pow(ΔX, 2);
const ΔYSquare = pow(ΔY, 2);
const ΔZSquare = pow(ΔZ, 2);
const chordLength = sqrt(ΔXSquare + ΔYSquare + ΔZSquare);
return chordLength;
}
convert-degrees-to-radians.js:
将度数转换为弧度.js：
module.exports = function convertDegreesToRadians({ degrees }) {
return degrees * Math.PI / 180;
};
This's following the Great-circle distance - From chord length, documented here.
这是遵循大圆距离 - 从弦长开始，记录在此处。
回答by Sanjay
I have written the function to find distance between two coordinates. It will return distance in meter.
我已经编写了查找两个坐标之间距离的函数。它将以米为单位返回距离。
function findDistance() {
var R = 6371e3; // R is earth's radius
var lat1 = 23.18489670753479; // starting point lat
var lat2 = 32.726601; // ending point lat
var lon1 = 72.62524545192719; // starting point lon
var lon2 = 74.857025; // ending point lon
var lat1radians = toRadians(lat1);
var lat2radians = toRadians(lat2);
var latRadians = toRadians(lat2-lat1);
var lonRadians = toRadians(lon2-lon1);
var a = Math.sin(latRadians/2) * Math.sin(latRadians/2) +
Math.cos(lat1radians) * Math.cos(lat2radians) *
Math.sin(lonRadians/2) * Math.sin(lonRadians/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
console.log(d)
}
function toRadians(val){
var PI = 3.1415926535;
return val / 180.0 * PI;
}
