我目前正在使用下面的功能,它不能正常工作。根据谷歌Maps,这些坐标之间的距离(从59.3293371,13.4877472到59.3225525,13.4619422)是2.2公里,而函数返回1.6公里。我怎样才能使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/


当前回答

Derek的解决方案对我来说很好,我只是简单地将其转换为PHP,希望它能帮助到一些人!

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;
}

其他回答

使用Haversine公式,源代码:

//:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
//:::                                                                         :::
//:::  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 sales@geodatasource.com                  :::
//:::                                                                         :::
//:::  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;
    }
}

样例代码是根据LGPLv3许可的。

我以前写过一个类似的方程-测试了它,也得到了1.6 km。

你的谷歌地图显示了驾驶距离。

你的函数是按照直线距离计算的。

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;
    }

大圆距离-从弦长开始

这里有一个应用策略设计模式的优雅解决方案;我希望它有足够的可读性。

TwoPointsDistanceCalculatorStrategy.js:

module.exports = () =>

class TwoPointsDistanceCalculatorStrategy {

    constructor() {}

    calculateDistance({ point1Coordinates, point2Coordinates }) {}
};

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:

module.exports = function convertDegreesToRadians({ degrees }) {

    return degrees * Math.PI / 180;
};

这是大圆距离-从弦长开始,这里有记录。

访问这个地址。 https://www.movable-type.co.uk/scripts/latlong.html 你可以使用下面的代码:

JavaScript:     

const R = 6371e3; // metres
const φ1 = lat1 * Math.PI/180; // φ, λ in radians
const φ2 = lat2 * Math.PI/180;
const Δφ = (lat2-lat1) * Math.PI/180;
const Δλ = (lon2-lon1) * Math.PI/180;

const a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
          Math.cos(φ1) * Math.cos(φ2) *
          Math.sin(Δλ/2) * Math.sin(Δλ/2);
const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));

const d = R * c; // in metres

我试着通过命名变量让代码更容易理解, 我希望这能有所帮助

function getDistanceFromLatLonInKm(point1, point2) {

  const [lat1, lon1] = point1;
  const [lat2, lon2] = point2;
  const earthRadius = 6371;
  const dLat = convertDegToRad(lat2 - lat1);
  const dLon = convertDegToRad(lon2 - lon1);
  const squarehalfChordLength =
    Math.sin(dLat / 2) * Math.sin(dLat / 2) +
    Math.cos(convertDegToRad(lat1)) * Math.cos(convertDegToRad(lat2)) *
    Math.sin(dLon / 2) * Math.sin(dLon / 2);

  const angularDistance = 2 * Math.atan2(Math.sqrt(squarehalfChordLength), Math.sqrt(1 - squarehalfChordLength));
  const distance = earthRadius * angularDistance;
  return distance;

}