Reputation:
I'm trying to calculate the distance between Vancouver and Toronto using their latitude and longitudes. I'm using haversine formula. I'm expecting a distance of around 4390 km. Can someone tell me what mistake I'm making. Here is my code:
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body onload="getDistance()">
<script>
// Convert Degress to Radians
function Deg2Rad( deg ) {
return deg * Math.PI / 180;
}
function getDistance()
{
//Toronto Latitude 43.74 and longitude -79.37
//Vancouver Latitude 49.25 and longitude -123.12
lat1 = Deg2Rad(43.74);
lat2 = Deg2Rad(49.25);
lon1 = Deg2Rad(-79.37);
lon2 = Deg2Rad(-123.12);
latDiff = lat2-lat1;
lonDiff = lon2-lon1;
var R = 6371000; // metres
var φ1 = lat1;
var φ2 = lat2;
var Δφ = Deg2Rad(latDiff);
var Δλ = Deg2Rad(lonDiff);
var a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
Math.cos(φ1) * Math.cos(φ2) *
Math.sin(Δλ/2) * Math.sin(Δλ/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
alert('d: ' + d);
var dist = Math.acos( Math.sin(φ1)*Math.sin(φ2) + Math.cos(φ1)*Math.cos(φ2) * Math.cos(Δλ) ) * R;
alert('dist: ' + dist);
}
</script>
</body>
</html>
When I run this code, I get quite different numbers.
Upvotes: 0
Views: 5005
Reputation: 4568
I think you should expect more like 3358km. http://www.distancefromto.net/distance-from/Toronto/to/Vancouver
Here is a fiddle with the correct result: https://jsfiddle.net/Lk1du2L1/
And in that case, you'll be correct if you'd remove your deg2rad form your secondary results - you're doing it too often:
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body onload="getDistance()">
<script>
// Convert Degress to Radians
function Deg2Rad( deg ) {
return deg * Math.PI / 180;
}
function getDistance()
{
//Toronto Latitude 43.74 and longitude -79.37
//Vancouver Latitude 49.25 and longitude -123.12
lat1 = Deg2Rad(43.74);
lat2 = Deg2Rad(49.25);
lon1 = Deg2Rad(-79.37);
lon2 = Deg2Rad(-123.12);
latDiff = lat2-lat1;
lonDiff = lon2-lon1;
var R = 6371000; // metres
var φ1 = lat1;
var φ2 = lat2;
var Δφ = latDiff;
var Δλ = lonDiff;
var a = Math.sin(Δφ/2) * Math.sin(Δφ/2) +
Math.cos(φ1) * Math.cos(φ2) *
Math.sin(Δλ/2) * Math.sin(Δλ/2);
var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
var d = R * c;
alert('d: ' + d);
var dist = Math.acos( Math.sin(φ1)*Math.sin(φ2) + Math.cos(φ1)*Math.cos(φ2) * Math.cos(Δλ) ) * R;
alert('dist: ' + dist);
}
</script>
</body>
</html>
Upvotes: 1