Question

我正在寻找从上个月开始知道背后的概念，直到现在我得到了这些资源

How do I calculate distance between two latitude-longitude points?
geolib.js
Function to calculate distance between two coordinates shows wrong
Calculate distance, bearing and more between Latitude/Longitude points

但我仍然无法找出指定坐标之间的正确距离

作为一个例子，我有两个坐标

开始纬度：26.26594 开始：78.2095508 目的地纬度：21.24386 目的地长：81.611

我得到的结果为655.358公里，但实际上（我用自行车的里程表测量了它）那些距离只有3公里左右，那么为什么我得到这个答案。

我甚至检查了我在geolib.js上的实现（我在列表项中指定的第二项），它也显示了相同的结果（655.358）。

我无法弄清楚

JS

// initialising the distance calculation

function geoLocationInit() {
        $('.fd-loc-err').html(''); // empty the error shown, if there is any already
        $('.fd-dist-rest-user').html('<img src="/images/loader.gif" alt="loading"/>'); // loader
        var options = {timeout:120000};
        if (navigator.geolocation) {
            navigator.geolocation.watchPosition(getLocation, gotError, options);
        }
        else {
            locationError = 'Your Browser does not support geolocation';
            showLocationError(locationError);
        }
    }



//finding the coordinates of user

function getLocation(current) {
        var userLat = current.coords.latitude, userLong = current.coords.longitude;
        var distance = getDistance(userLat, userLong, restLatitude, restLongitude);
        $('.fd-dist-rest-user').html('~' + (distance/1000).toFixed(2) + ' km away'); // fd-dist-rest-user is the <div> tag where i have show the distance calculated
    }


    function toRad(value) {
        return value * Math.PI / 180;
    }


    // calculating distance using Vincenty Formula
    function getDistance(lat1, lon1, lat2, lon2) {
        var a = 6378137, b = 6356752.314245,  f = 1/298.257223563;
        var L = toRad(lon2-lon1);
        var U1 = Math.atan((1-f) * Math.tan(toRad(lat1)));
        var U2 = Math.atan((1-f) * Math.tan(toRad(lat2)));
        var sinU1 = Math.sin(U1), cosU1 = Math.cos(U1);
        var sinU2 = Math.sin(U2), cosU2 = Math.cos(U2);

        var lambda = L, lambdaP, iterLimit = 100;
        do 
        {
            var sinLambda = Math.sin(lambda), cosLambda = Math.cos(lambda);
            var sinSigma = Math.sqrt((cosU2*sinLambda) * (cosU2*sinLambda) + (cosU1*sinU2-sinU1*cosU2*cosLambda) * (cosU1*sinU2-sinU1*cosU2*cosLambda));
            if (sinSigma==0) return 0;

            var cosSigma = sinU1*sinU2 + cosU1*cosU2*cosLambda;
            var sigma = Math.atan2(sinSigma, cosSigma);
            var sinAlpha = cosU1 * cosU2 * sinLambda / sinSigma;
            var cosSqAlpha = 1 - sinAlpha*sinAlpha;
            var cos2SigmaM = cosSigma - 2*sinU1*sinU2/cosSqAlpha;
            if (isNaN(cos2SigmaM)) cos2SigmaM = 0;
            var C = f/16*cosSqAlpha*(4+f*(4-3*cosSqAlpha));
            lambdaP = lambda;
            lambda = L + (1-C) * f * sinAlpha * (sigma + C*sinSigma*(cos2SigmaM+C*cosSigma*(-1+2*cos2SigmaM*cos2SigmaM)));
        } while (Math.abs(lambda-lambdaP) > 1e-12 && --iterLimit>0);

        if (iterLimit==0) return NaN

        var uSq = cosSqAlpha * (a*a - b*b) / (b*b);
        var A = 1 + uSq/16384*(4096+uSq*(-768+uSq*(320-175*uSq)));
        var B = uSq/1024 * (256+uSq*(-128+uSq*(74-47*uSq)));
        var deltaSigma = B*sinSigma*(cos2SigmaM+B/4*(cosSigma*(-1+2*cos2SigmaM*cos2SigmaM)-B/6*cos2SigmaM*(-3+4*sinSigma*sinSigma)*(-3+4*cos2SigmaM*cos2SigmaM)));
        var s = b*A*(sigma-deltaSigma);
        return s;
    }


// Error handling for the geolocation

    function gotError(error) {
        switch(error.code) {
            case 1: 
                locationError = 'You need to give the permission to use your location to calculate the distances between this restaurant and you <button class="btn btn-danger fd-detect-lctn-try-agn">Please try again</button>';
                break;
            case 2:
                locationError = 'TIME OUT - You need to give permission to use your location to show the distance of this restaurant from your current position <button class="btn btn-danger fd-detect-lctn-try-agn">Please try again</button>';
                break;
            case 3:
                locationError = 'It took to long getting your position, <button class="btn btn-danger fd-detect-lctn">Please try again</button>';
                break;
            default:
                locationError = 'An unknown error has occurred, <button class="btn btn-danger fd-detect-lctn">Please try again</button>';
        }
        showLocationError(locationError);
    }

    function showLocationError(msg) {
        $('.fd-loc-err').html(msg);
        $('.fd-dist-rest-user').html('');
    }

我从数据库中获取纬度和经度，然后在按钮点击时调用geoLocationInit函数（下面是以下代码）

restLongitude = response['longitude'];
restLatitude = response['latitude'];
geoLocationInit(); /* getting location initialisation */

Answer 1

在Patrick Evans的评论的帮助下进行了一些更多的研究之后，我发现，由于桌面上缺少gps设备，我遇到了问题，因此，坐标不准确导致错误的距离计算。

根据Firefox - 位置感知浏览

精确度因地而异。在某些地方，我们的服务提供商可能能够提供几米之内的位置。但是，在其他领域，它可能远不止于此。我们的服务提供商退回的所有地点仅为估算值，我们不保证所提供地点的准确性。请勿在紧急情况下使用此信息。总是使用常识。

来源Location-Aware Browsing - How Accurate are the locations

因此，如果想要找到几乎准确的GPS坐标，那么应该使用GPS设备或移动设备中的HTML5 geolocation API，这通常会安装GPS硬件，因此可以获得精确坐标

Answer 2

我尝试修改方法的选项：'geoLocationInit'尝试访问最佳位置，但对于这个，如果是智能手机应该是有效的wifi和gps。

会是这样的：

var options = {timeout:120000, enableHighAccuracy: true};

两个GPS坐标之间的距离 - javascript

2 个答案: