兩個GPS座標之間的距離 - 的JavaScript

Question

我正在尋找了解自上個月這背後的理念，到現在我得到了這些資源

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

但我i st生病無法找出正確的距離指定座標之間

舉一個例子，我有兩個座標

開始緯度：26.26594 開始長：78.2095508 目的地緯度：21.24386 目的地長：81.611

我得到的結果爲655.358公里，但實際上（我用我的自行車的里程錶測量過）這些距離只有3公里，那麼爲什麼我得到這個答案。

我甚至檢查過我在geolib.js上的實現（我在列表項中指定了第二項），它也顯示了相同的結果（655.358）。

我無法弄清楚

JS

// inititalising 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(''); 
    } 

我正在從我的數據庫，然後調用上的按鈕，點擊geoLocationInit功能的緯度和經度（下面是下面的代碼）

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

在此先感謝誰可以解決我的萬阿英，蔣達清

Answer 1

在經過Patrick Evans評論的幫助後進行了一些更多的研究，我發現，由於缺少臺式機上的gps設備，我得到了問題，因此，座標不準確，導致計算錯誤距離。

據火狐 - 位置感知瀏覽

精度差異很大，從位置到位置。在某些地方，我們的服務提供商可能能夠提供幾米之內的位置。但是，在其他領域，這可能遠不止於此。我們的服務提供商返回的所有地點僅爲估算值，我們不保證所提供地點的準確性。請不要將這些信息用於緊急情況。始終使用常識。

來源，如果一個人想找到幾乎精確的GPS座標，那麼，應該在移動設備一般自帶GPS硬件請使用GPS裝置或HTML5 geolocation API安裝，所以一會就搞定Location-Aware Browsing - How Accurate are the locations

所以確切的座標

Answer 2

我嘗試修改方法的選項：'geoLocationInit'試圖訪問最佳位置，但對於這個如果它是智能手機應該是活躍的WiFi和GPS。

會是這樣的：

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