Calculating Distance Between Two Latitude and Longitude Geocoordinates

Calculate distance between two latitude-longitude points? (Haversine formula)

This link might be helpful to you, as it details the use of the Haversine formula to calculate the distance.

Excerpt:

This script [in Javascript] calculates great-circle distances between the two points –
that is, the shortest distance over the earth’s surface – using the
‘Haversine’ formula.

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)
}

Calculating Distance between two Latitude and Longitude GeoCoordinates

The GeoCoordinate class (.NET Framework 4 and higher) already has GetDistanceTo method.

var sCoord = new GeoCoordinate(sLatitude, sLongitude);
var eCoord = new GeoCoordinate(eLatitude, eLongitude);

return sCoord.GetDistanceTo(eCoord);

The distance is in meters.

You need to reference System.Device.

Function to calculate distance between two coordinates

What you're using is called the haversine formula, which calculates the distance between two points on a sphere as the crow flies. The Google Maps link you provided shows the distance as 2.2 km because it's not a straight line.

Wolfram Alpha is a great resource for doing geographic calculations, and also shows a distance of 1.652 km between these two points.

If you're looking for straight-line distance (as the crow files), your function is working correctly. If what you want is driving distance (or biking distance or public transportation distance or walking distance), you'll have to use a mapping API (Google or Bing being the most popular) to get the appropriate route, which will include the distance.

Incidentally, the Google Maps API provides a packaged method for spherical distance, in its google.maps.geometry.spherical namespace (look for computeDistanceBetween). It's probably better than rolling your own (for starters, it uses a more precise value for the Earth's radius).

For the picky among us, when I say "straight-line distance", I'm referring to a "straight line on a sphere", which is actually a curved line (i.e. the great-circle distance), of course.

Calculate distance between 2 GPS coordinates

Calculate the distance between two coordinates by latitude and longitude, including a Javascript implementation.

West and South locations are negative.
Remember minutes and seconds are out of 60 so S31 30' is -31.50 degrees.

Don't forget to convert degrees to radians. Many languages have this function. Or its a simple calculation: radians = degrees * PI / 180.

function degreesToRadians(degrees) {
  return degrees * Math.PI / 180;
}

function distanceInKmBetweenEarthCoordinates(lat1, lon1, lat2, lon2) {
  var earthRadiusKm = 6371;

  var dLat = degreesToRadians(lat2-lat1);
  var dLon = degreesToRadians(lon2-lon1);

  lat1 = degreesToRadians(lat1);
  lat2 = degreesToRadians(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)); 
  return earthRadiusKm * c;
}

Here are some examples of usage:

distanceInKmBetweenEarthCoordinates(0,0,0,0)  // Distance between same 
                                              // points should be 0
0

distanceInKmBetweenEarthCoordinates(51.5, 0, 38.8, -77.1) // From London
                                                          // to Arlington
5918.185064088764

How can I quickly estimate the distance between two (latitude, longitude) points?

The answers to Haversine Formula in Python (Bearing and Distance between two GPS points) provide Python implementations that answer your question.

Using the implementation below I performed 100,000 iterations in less than 1 second on an older laptop. I think for your purposes this should be sufficient. However, you should profile anything before you optimize for performance.

from math import radians, cos, sin, asin, sqrt
def haversine(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    """
    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
    # haversine formula 
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a)) 
    # Radius of earth in kilometers is 6371
    km = 6371* c
    return km

To underestimate haversine(lat1, long1, lat2, long2) * 0.90 or whatever factor you want. I don't see how introducing error to your underestimation is useful.

Calculating distance between two points, using latitude longitude?

The Java code given by Dommer above gives slightly incorrect results but the small errors add up if you are processing say a GPS track. Here is an implementation of the Haversine method in Java which also takes into account height differences between two points.

/**
 * Calculate distance between two points in latitude and longitude taking
 * into account height difference. If you are not interested in height
 * difference pass 0.0. Uses Haversine method as its base.
 * 
 * lat1, lon1 Start point lat2, lon2 End point el1 Start altitude in meters
 * el2 End altitude in meters
 * @returns Distance in Meters
 */
public static double distance(double lat1, double lat2, double lon1,
        double lon2, double el1, double el2) {

    final int R = 6371; // Radius of the earth

    double latDistance = Math.toRadians(lat2 - lat1);
    double lonDistance = Math.toRadians(lon2 - lon1);
    double a = Math.sin(latDistance / 2) * Math.sin(latDistance / 2)
            + Math.cos(Math.toRadians(lat1)) * Math.cos(Math.toRadians(lat2))
            * Math.sin(lonDistance / 2) * Math.sin(lonDistance / 2);
    double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
    double distance = R * c * 1000; // convert to meters

    double height = el1 - el2;

    distance = Math.pow(distance, 2) + Math.pow(height, 2);

    return Math.sqrt(distance);
}

Measuring the distance between two coordinates in PHP

Not long ago I wrote an example of the haversine formula, and published it on my website:

/**
 * Calculates the great-circle distance between two points, with
 * the Haversine formula.
 * @param float $latitudeFrom Latitude of start point in [deg decimal]
 * @param float $longitudeFrom Longitude of start point in [deg decimal]
 * @param float $latitudeTo Latitude of target point in [deg decimal]
 * @param float $longitudeTo Longitude of target point in [deg decimal]
 * @param float $earthRadius Mean earth radius in [m]
 * @return float Distance between points in [m] (same as earthRadius)
 */
function haversineGreatCircleDistance(
  $latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo, $earthRadius = 6371000)
{
  // convert from degrees to radians
  $latFrom = deg2rad($latitudeFrom);
  $lonFrom = deg2rad($longitudeFrom);
  $latTo = deg2rad($latitudeTo);
  $lonTo = deg2rad($longitudeTo);

  $latDelta = $latTo - $latFrom;
  $lonDelta = $lonTo - $lonFrom;

  $angle = 2 * asin(sqrt(pow(sin($latDelta / 2), 2) +
    cos($latFrom) * cos($latTo) * pow(sin($lonDelta / 2), 2)));
  return $angle * $earthRadius;
}

➽ Note that you get the distance back in the same unit as you pass in with the parameter $earthRadius. The default value is 6371000 meters so the result will be in [m] too. To get the result in miles, you could e.g. pass 3959 miles as $earthRadius and the result would be in [mi]. In my opinion it is a good habit to stick with the SI units, if there is no particular reason to do otherwise.

Edit:

As TreyA correctly pointed out, the Haversine formula has weaknesses with antipodal points because of rounding errors (though it is stable for small distances). To get around them, you could use the Vincenty formula instead.

/**
 * Calculates the great-circle distance between two points, with
 * the Vincenty formula.
 * @param float $latitudeFrom Latitude of start point in [deg decimal]
 * @param float $longitudeFrom Longitude of start point in [deg decimal]
 * @param float $latitudeTo Latitude of target point in [deg decimal]
 * @param float $longitudeTo Longitude of target point in [deg decimal]
 * @param float $earthRadius Mean earth radius in [m]
 * @return float Distance between points in [m] (same as earthRadius)
 */
public static function vincentyGreatCircleDistance(
  $latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo, $earthRadius = 6371000)
{
  // convert from degrees to radians
  $latFrom = deg2rad($latitudeFrom);
  $lonFrom = deg2rad($longitudeFrom);
  $latTo = deg2rad($latitudeTo);
  $lonTo = deg2rad($longitudeTo);

  $lonDelta = $lonTo - $lonFrom;
  $a = pow(cos($latTo) * sin($lonDelta), 2) +
    pow(cos($latFrom) * sin($latTo) - sin($latFrom) * cos($latTo) * cos($lonDelta), 2);
  $b = sin($latFrom) * sin($latTo) + cos($latFrom) * cos($latTo) * cos($lonDelta);

  $angle = atan2(sqrt($a), $b);
  return $angle * $earthRadius;
}

Calculating distance between two points (Latitude, Longitude)

Since you're using SQL Server 2008, you have the geography data type available, which is designed for exactly this kind of data:

DECLARE @source geography = 'POINT(0 51.5)'
DECLARE @target geography = 'POINT(-3 56)'

SELECT @source.STDistance(@target)

Gives

----------------------
538404.100197555

(1 row(s) affected)

Telling us it is about 538 km from (near) London to (near) Edinburgh.

Naturally there will be an amount of learning to do first, but once you know it it's far far easier than implementing your own Haversine calculation; plus you get a LOT of functionality.

---

If you want to retain your existing data structure, you can still use STDistance, by constructing suitable geography instances using the Point method:

DECLARE @orig_lat DECIMAL(12, 9)
DECLARE @orig_lng DECIMAL(12, 9)
SET @orig_lat=53.381538 set @orig_lng=-1.463526

DECLARE @orig geography = geography::Point(@orig_lat, @orig_lng, 4326);

SELECT *,
    @orig.STDistance(geography::Point(dest.Latitude, dest.Longitude, 4326)) 
       AS distance
--INTO #includeDistances
FROM #orig dest

Find distance between two points using latitude and longitude in mysql

I think your question says you have the city values for the two cities between which you wish to compute the distance.

This query will do the job for you, yielding the distance in km. It uses the spherical cosine law formula.

Notice that you join the table to itself so you can retrieve two coordinate pairs for the computation.

SELECT a.city AS from_city, b.city AS to_city, 
   111.111 *
    DEGREES(ACOS(LEAST(1.0, COS(RADIANS(a.Latitude))
         * COS(RADIANS(b.Latitude))
         * COS(RADIANS(a.Longitude - b.Longitude))
         + SIN(RADIANS(a.Latitude))
         * SIN(RADIANS(b.Latitude))))) AS distance_in_km
  FROM city AS a
  JOIN city AS b ON a.id <> b.id
 WHERE a.city = 3 AND b.city = 7

Notice that the constant 111.1111 is the number of kilometres per degree of latitude, based on the old Napoleonic definition of the metre as one ten-thousandth of the distance from the equator to the pole. That definition is close enough for location-finder work.

If you want statute miles instead of kilometres, use 69.0 instead.

http://sqlfiddle.com/#!9/21e06/412/0

If you're looking for nearby points you may be tempted to use a clause something like this:

   HAVING distance_in_km < 10.0    /* slow ! */
    ORDER BY distance_in_km DESC

That is (as we say near Boston MA USA) wicked slow.

In that case you need to use a bounding box computation. See this writeup about how to do that. http://www.plumislandmedia.net/mysql/haversine-mysql-nearest-loc/

The formula contains a LEAST() function. Why? Because the ACOS() function throws an error if its argument is even slightly greater than 1. When the two points in question are very close together, the expression with the COS() and SIN() computations can sometimes yield a value slightly greater than 1 due to floating-point epsilon (inaccuracy). The LEAST(1.0, dirty-great-expression) call copes with that problem.

There's a better way, a formula by Thaddeus Vincenty. It uses ATAN2() rather than ACOS() so it's less susceptible to epsilon problems.

---

Edit 2022 (by Alexio Vay):
As of today the modern solution should be the following short code:

   select ST_Distance_Sphere(
    point(-87.6770458, 41.9631174),
    point(-73.9898293, 40.7628267)) 

Please check out the answer of Naresh Kumar.

How to calculate distance between two locations using their longitude and latitude value

Here getting distance in miles (mi)

private double distance(double lat1, double lon1, double lat2, double lon2) {
    double theta = lon1 - lon2;
    double dist = Math.sin(deg2rad(lat1)) 
                    * Math.sin(deg2rad(lat2))
                    + Math.cos(deg2rad(lat1))
                    * Math.cos(deg2rad(lat2))
                    * Math.cos(deg2rad(theta));
    dist = Math.acos(dist);
    dist = rad2deg(dist);
    dist = dist * 60 * 1.1515;
    return (dist);
}

private double deg2rad(double deg) {
    return (deg * Math.PI / 180.0);
}

private double rad2deg(double rad) {
    return (rad * 180.0 / Math.PI);
}

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