I would guess there's some rounding taking place on the site you've linked to. The same site measures the distance as 9 feet which is about 2.75 meters. Are you sure your coordinates are accurate?

//::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::://::: ::://::: 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 ::://::: 'K' is kilometers (default) ::://::: 'N' is nautical miles ::://::: ::://::: Worldwide cities and other features databases with latitude longitude ::://::: are available at http://www.geodatasource.com ::://::: ::://::: For enquiries, please contact sales@geodatasource.com ::://::: ::://::: Official Web site: http://www.geodatasource.com ::://::: ::://::: GeoDataSource.com (C) All Rights Reserved 2014 ::://::: ::://:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::function distance(lat1, lon1, lat2, lon2, unit) { var radlat1 = Math.PI * lat1/180 var radlat2 = Math.PI * lat2/180 var radlon1 = Math.PI * lon1/180 var radlon2 = Math.PI * lon2/180 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} /*53.1330940 -1.1843660TO53.1331060 -1.1843260*/var lat1=53.1330940;var lon1=-1.1843660;var lat2=53.1331060;var lon2=-1.1843260;var kilometers=distance(lat1, lon1, lat2, lon2, 'k'); meters = kilometers*1000alert(meters);

Benefit from a mission critical IT monitoring with Monitis Premium or get it FREE for your entry level monitoring needs.
-Over 200,000 users
-More than 300,000 websites monitored
-Used in 197 countries
-Recommended by 98% of users

JavaScript can be used in a browser to change parts of a webpage dynamically. It begins with the following pattern: If condition W is true, do thing X to target Y after event Z. Below are some tips and tricks to help you get started with JavaScript …

The viewer will learn the basics of jQuery, including how to invoke it on a web page.
Reference your jQuery libraries: (CODE)
Include your new external js/jQuery file: (CODE)
Write your first lines of code to setup your site for jQuery.: (CODE)

The viewer will learn the basics of jQuery including how to code hide show and toggles.
Reference your jQuery libraries:
(CODE)
Include your new external js/jQuery file:
(CODE)
Write your first lines of code to setup your site for jQuery…