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Distance calculation gives strange output javascript

I need to calculate the distance between two latitude and longitude points. I found this javascript code which I suppose I want.
Here comes the problem. I add the two positions lat and lng values in, and sometimes it just gives random output. What happens is two points literally next to other are sometimes like 8000 meters away, but two other much furthest points return only 1500 meters for example.
function degreesToRadians(degrees) {
return degrees * Math.PI / 180;
}
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)
}
const distanceR = getDistanceFromLatLonInKm(userLat, userLng, solLat, solLng)
Lat: 68.00757101804007, lng: -49.306640625
to
lat: 73.26312194058698 lng: -23.535461425781254
is 1143 kilometres, but these two points are next to each other.
lat: 66.75724984139227, lng: -16.259765625000004
to
lat: 71.99597405683693 lng:-42.31933593750001
is 1161 metres and the points are much farther then the previus one.
Here I think it calculates fine, unlike the two previous lat and lng points.

I've tested the examples you provided and did a few on my own and I believe your implementation is working fine. I did however get different results using your examples and code.
The first example returns approx. 1102km, which seems close to the distance using a visualizer.
The second example returns 1161 kilometres which visualized again seems about right.
Please note: the images in the links were constructed using gpsvisualizer.com which uses a Vincenty formula to calculate distance, hence the slight variation in distance numbers.
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)
}
const e1 = {
lat1: 68.00757101804007,
lon1: -49.306640625,
lat2: 73.26312194058698,
lon2: -23.535461425781254
}
const e2 = {
lat1: 66.75724984139227,
lon1: -16.259765625000004,
lat2: 71.99597405683693,
lon2: -42.31933593750001
}
const dist1 = getDistanceFromLatLonInKm(e1.lat1, e1.lon1, e1.lat2, e1.lon2);
const dist2 = getDistanceFromLatLonInKm(e2.lat1, e2.lon1, e2.lat2, e2.lon2);
console.log(`Example 1 distance: ${dist1}km. Example 2 distance: ${dist2}km`);
I can not reproduce the issue you are having so I believe your error lies in your visualization of coordinates.

Geometric rotations of a group of objects

Referring to the first diagram, I am trying to copy the three objects, looking at them from an arbitrary angle(A1). The distance between where I'm and the first object does not matter just the relative location of the object to one another.
In the second diagram, I select a point to copy these objects, facing another arbitrary angle(B1).
Angle (C1) shows the approximate position of -90 degrees.
I can get this to work if A1 = 0,90,180,270 and even 45,135 etc but the equations I come up with only work for 0 and 180 or 90 and 270. I have to modify them to work in those directions by changing a hardcoded offset angle and putting/removing a negative sign before the offset.
I'm doing this is javascript (and its Minecraft) usually I can figure out this but I have been working on it for weeks.
Here is some pseudo-code that works some of the time in certain right-angle directions. I have updated this to be more accurate, the 1x and 2x are the blocks x coordinate, etc. - everything is relative from the (1) block.
Minecraft's coordinate system is a little different from normal - 0 is south, +90 is west, 180 is north, 270 is east.
the only difference is that I am making negative az, ax.
// works for north/south looking - A1 is either 180/0 , B1 can be anything
var x = 1x - 2x;
var z = 1z - 2z;
var direction = Math.atan2(z1, x1);
var L1 = Math.sqrt(Math.pow(x1, 2) + Math.pow(z1, 2));
var az = Math.round(L1 * Math.sin((B1 + A1 + (direction * 180 / Math.PI)) * Math.PI / 180));
var ax = Math.round(L1 * Math.cos((B1 + A1 + (direction * 180 / Math.PI)) * Math.PI / 180));
// works for east/west looking - A1 is either 90/270 , B1 can be anything
var x = 1x - 2x;
var z = 1z - 2z;
var direction = Math.atan2(z1, x1);
var L1 = Math.sqrt(Math.pow(x1, 2) + Math.pow(z1, 2));
var az = -Math.round(L1 * Math.sin((B1 + A1 + (direction * 180 / Math.PI)) * Math.PI / 180));
var ax = -Math.round(L1 * Math.cos((B1 + A1 + (direction * 180 / Math.PI)) * Math.PI / 180));
First diagram
Second diagram

Not able to calculate distance(between two coordinates, dont consider the Travelmode like cycling,driving) in Map v3 API [duplicate]

How do I calculate the distance between two points specified by latitude and longitude?
For clarification, I'd like the distance in kilometers; the points use the WGS84 system and I'd like to understand the relative accuracies of the approaches available.

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)
}

I needed to calculate a lot of distances between the points for my project, so I went ahead and tried to optimize the code, I have found here. On average in different browsers my new implementation runs 2 times faster than the most upvoted answer.
function distance(lat1, lon1, lat2, lon2) {
var p = 0.017453292519943295; // Math.PI / 180
var c = Math.cos;
var a = 0.5 - c((lat2 - lat1) * p)/2 +
c(lat1 * p) * c(lat2 * p) *
(1 - c((lon2 - lon1) * p))/2;
return 12742 * Math.asin(Math.sqrt(a)); // 2 * R; R = 6371 km
}
You can play with my jsPerf and see the results here.
Recently I needed to do the same in python, so here is a python implementation:
from math import cos, asin, sqrt, pi
def distance(lat1, lon1, lat2, lon2):
p = pi/180
a = 0.5 - cos((lat2-lat1)*p)/2 + cos(lat1*p) * cos(lat2*p) * (1-cos((lon2-lon1)*p))/2
return 12742 * asin(sqrt(a)) #2*R*asin...
And for the sake of completeness: Haversine on Wikipedia.

Here is a C# Implementation:
static class DistanceAlgorithm
{
const double PIx = 3.141592653589793;
const double RADIUS = 6378.16;
/// <summary>
/// Convert degrees to Radians
/// </summary>
/// <param name="x">Degrees</param>
/// <returns>The equivalent in radians</returns>
public static double Radians(double x)
{
return x * PIx / 180;
}
/// <summary>
/// Calculate the distance between two places.
/// </summary>
/// <param name="lon1"></param>
/// <param name="lat1"></param>
/// <param name="lon2"></param>
/// <param name="lat2"></param>
/// <returns></returns>
public static double DistanceBetweenPlaces(
double lon1,
double lat1,
double lon2,
double lat2)
{
double dlon = Radians(lon2 - lon1);
double dlat = Radians(lat2 - lat1);
double a = (Math.Sin(dlat / 2) * Math.Sin(dlat / 2)) + Math.Cos(Radians(lat1)) * Math.Cos(Radians(lat2)) * (Math.Sin(dlon / 2) * Math.Sin(dlon / 2));
double angle = 2 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1 - a));
return angle * RADIUS;
}
}

Here is a java implementation of the Haversine formula.
public final static double AVERAGE_RADIUS_OF_EARTH_KM = 6371;
public int calculateDistanceInKilometer(double userLat, double userLng,
double venueLat, double venueLng) {
double latDistance = Math.toRadians(userLat - venueLat);
double lngDistance = Math.toRadians(userLng - venueLng);
double a = Math.sin(latDistance / 2) * Math.sin(latDistance / 2)
+ Math.cos(Math.toRadians(userLat)) * Math.cos(Math.toRadians(venueLat))
* Math.sin(lngDistance / 2) * Math.sin(lngDistance / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return (int) (Math.round(AVERAGE_RADIUS_OF_EARTH_KM * c));
}
Note that here we are rounding the answer to the nearest km.

Thanks very much for all this. I used the following code in my Objective-C iPhone app:
const double PIx = 3.141592653589793;
const double RADIO = 6371; // Mean radius of Earth in Km
double convertToRadians(double val) {
return val * PIx / 180;
}
-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {
double dlon = convertToRadians(place2.longitude - place1.longitude);
double dlat = convertToRadians(place2.latitude - place1.latitude);
double a = ( pow(sin(dlat / 2), 2) + cos(convertToRadians(place1.latitude))) * cos(convertToRadians(place2.latitude)) * pow(sin(dlon / 2), 2);
double angle = 2 * asin(sqrt(a));
return angle * RADIO;
}
Latitude and Longitude are in decimal. I didn't use min() for the asin() call as the distances that I'm using are so small that they don't require it.
It gave incorrect answers until I passed in the values in Radians - now it's pretty much the same as the values obtained from Apple's Map app :-)
Extra update:
If you are using iOS4 or later then Apple provide some methods to do this so the same functionality would be achieved with:
-(double)kilometresBetweenPlace1:(CLLocationCoordinate2D) place1 andPlace2:(CLLocationCoordinate2D) place2 {
MKMapPoint start, finish;
start = MKMapPointForCoordinate(place1);
finish = MKMapPointForCoordinate(place2);
return MKMetersBetweenMapPoints(start, finish) / 1000;
}

This is a simple PHP function that will give a very reasonable approximation (under +/-1% error margin).
<?php
function distance($lat1, $lon1, $lat2, $lon2) {
$pi80 = M_PI / 180;
$lat1 *= $pi80;
$lon1 *= $pi80;
$lat2 *= $pi80;
$lon2 *= $pi80;
$r = 6372.797; // mean radius of Earth in km
$dlat = $lat2 - $lat1;
$dlon = $lon2 - $lon1;
$a = sin($dlat / 2) * sin($dlat / 2) + cos($lat1) * cos($lat2) * sin($dlon / 2) * sin($dlon / 2);
$c = 2 * atan2(sqrt($a), sqrt(1 - $a));
$km = $r * $c;
//echo '<br/>'.$km;
return $km;
}
?>
As said before; the earth is NOT a sphere. It is like an old, old baseball that Mark McGwire decided to practice with - it is full of dents and bumps. The simpler calculations (like this) treat it like a sphere.
Different methods may be more or less precise according to where you are on this irregular ovoid AND how far apart your points are (the closer they are the smaller the absolute error margin). The more precise your expectation, the more complex the math.
For more info: wikipedia geographic distance

I post here my working example.
List all points in table having distance between a designated point (we use a random point - lat:45.20327, long:23.7806) less than 50 KM, with latitude & longitude, in MySQL (the table fields are coord_lat and coord_long):
List all having DISTANCE<50, in Kilometres (considered Earth radius 6371 KM):
SELECT denumire, (6371 * acos( cos( radians(45.20327) ) * cos( radians( coord_lat ) ) * cos( radians( 23.7806 ) - radians(coord_long) ) + sin( radians(45.20327) ) * sin( radians(coord_lat) ) )) AS distanta
FROM obiective
WHERE coord_lat<>''
AND coord_long<>''
HAVING distanta<50
ORDER BY distanta desc
The above example was tested in MySQL 5.0.95 and 5.5.16 (Linux).

In the other answers an implementation in r is missing.
Calculating the distance between two point is quite straightforward with the distm function from the geosphere package:
distm(p1, p2, fun = distHaversine)
where:
p1 = longitude/latitude for point(s)
p2 = longitude/latitude for point(s)
# type of distance calculation
fun = distCosine / distHaversine / distVincentySphere / distVincentyEllipsoid
As the earth is not perfectly spherical, the Vincenty formula for ellipsoids is probably the best way to calculate distances. Thus in the geosphere package you use then:
distm(p1, p2, fun = distVincentyEllipsoid)
Off course you don't necessarily have to use geosphere package, you can also calculate the distance in base R with a function:
hav.dist <- function(long1, lat1, long2, lat2) {
R <- 6371
diff.long <- (long2 - long1)
diff.lat <- (lat2 - lat1)
a <- sin(diff.lat/2)^2 + cos(lat1) * cos(lat2) * sin(diff.long/2)^2
b <- 2 * asin(pmin(1, sqrt(a)))
d = R * b
return(d)
}

The haversine is definitely a good formula for probably most cases, other answers already include it so I am not going to take the space. But it is important to note that no matter what formula is used (yes not just one). Because of the huge range of accuracy possible as well as the computation time required. The choice of formula requires a bit more thought than a simple no brainer answer.
This posting from a person at nasa, is the best one I found at discussing the options
http://www.cs.nyu.edu/visual/home/proj/tiger/gisfaq.html
For example, if you are just sorting rows by distance in a 100 miles radius. The flat earth formula will be much faster than the haversine.
HalfPi = 1.5707963;
R = 3956; /* the radius gives you the measurement unit*/
a = HalfPi - latoriginrad;
b = HalfPi - latdestrad;
u = a * a + b * b;
v = - 2 * a * b * cos(longdestrad - longoriginrad);
c = sqrt(abs(u + v));
return R * c;
Notice there is just one cosine and one square root. Vs 9 of them on the Haversine formula.

There could be a simpler solution, and more correct: The perimeter of earth is 40,000Km at the equator, about 37,000 on Greenwich (or any longitude) cycle. Thus:
pythagoras = function (lat1, lon1, lat2, lon2) {
function sqr(x) {return x * x;}
function cosDeg(x) {return Math.cos(x * Math.PI / 180.0);}
var earthCyclePerimeter = 40000000.0 * cosDeg((lat1 + lat2) / 2.0);
var dx = (lon1 - lon2) * earthCyclePerimeter / 360.0;
var dy = 37000000.0 * (lat1 - lat2) / 360.0;
return Math.sqrt(sqr(dx) + sqr(dy));
};
I agree that it should be fine-tuned as, I myself said that it's an ellipsoid, so the radius to be multiplied by the cosine varies. But it's a bit more accurate. Compared with Google Maps and it did reduce the error significantly.

pip install haversine
Python implementation
Origin is the center of the contiguous United States.
from haversine import haversine, Unit
origin = (39.50, 98.35)
paris = (48.8567, 2.3508)
haversine(origin, paris, unit=Unit.MILES)
To get the answer in kilometers simply set unit=Unit.KILOMETERS (that's the default).

There is some errors in the code provided, I've fixed it below.
All the above answers assumes the earth is a sphere. However, a more accurate approximation would be that of an oblate spheroid.
a= 6378.137#equitorial radius in km
b= 6356.752#polar radius in km
def Distance(lat1, lons1, lat2, lons2):
lat1=math.radians(lat1)
lons1=math.radians(lons1)
R1=(((((a**2)*math.cos(lat1))**2)+(((b**2)*math.sin(lat1))**2))/((a*math.cos(lat1))**2+(b*math.sin(lat1))**2))**0.5 #radius of earth at lat1
x1=R1*math.cos(lat1)*math.cos(lons1)
y1=R1*math.cos(lat1)*math.sin(lons1)
z1=R1*math.sin(lat1)
lat2=math.radians(lat2)
lons2=math.radians(lons2)
R2=(((((a**2)*math.cos(lat2))**2)+(((b**2)*math.sin(lat2))**2))/((a*math.cos(lat2))**2+(b*math.sin(lat2))**2))**0.5 #radius of earth at lat2
x2=R2*math.cos(lat2)*math.cos(lons2)
y2=R2*math.cos(lat2)*math.sin(lons2)
z2=R2*math.sin(lat2)
return ((x1-x2)**2+(y1-y2)**2+(z1-z2)**2)**0.5

I don't like adding yet another answer, but the Google maps API v.3 has spherical geometry (and more). After converting your WGS84 to decimal degrees you can do this:
<script src="http://maps.google.com/maps/api/js?sensor=false&libraries=geometry" type="text/javascript"></script>
distance = google.maps.geometry.spherical.computeDistanceBetween(
new google.maps.LatLng(fromLat, fromLng),
new google.maps.LatLng(toLat, toLng));
No word about how accurate Google's calculations are or even what model is used (though it does say "spherical" rather than "geoid". By the way, the "straight line" distance will obviously be different from the distance if one travels on the surface of the earth which is what everyone seems to be presuming.

You can use the build in CLLocationDistance to calculate this:
CLLocation *location1 = [[CLLocation alloc] initWithLatitude:latitude1 longitude:longitude1];
CLLocation *location2 = [[CLLocation alloc] initWithLatitude:latitude2 longitude:longitude2];
[self distanceInMetersFromLocation:location1 toLocation:location2]
- (int)distanceInMetersFromLocation:(CLLocation*)location1 toLocation:(CLLocation*)location2 {
CLLocationDistance distanceInMeters = [location1 distanceFromLocation:location2];
return distanceInMeters;
}
In your case if you want kilometers just divide by 1000.

As pointed out, an accurate calculation should take into account that the earth is not a perfect sphere. Here are some comparisons of the various algorithms offered here:
geoDistance(50,5,58,3)
Haversine: 899 km
Maymenn: 833 km
Keerthana: 897 km
google.maps.geometry.spherical.computeDistanceBetween(): 900 km
geoDistance(50,5,-58,-3)
Haversine: 12030 km
Maymenn: 11135 km
Keerthana: 10310 km
google.maps.geometry.spherical.computeDistanceBetween(): 12044 km
geoDistance(.05,.005,.058,.003)
Haversine: 0.9169 km
Maymenn: 0.851723 km
Keerthana: 0.917964 km
google.maps.geometry.spherical.computeDistanceBetween(): 0.917964 km
geoDistance(.05,80,.058,80.3)
Haversine: 33.37 km
Maymenn: 33.34 km
Keerthana: 33.40767 km
google.maps.geometry.spherical.computeDistanceBetween(): 33.40770 km
Over small distances, Keerthana's algorithm does seem to coincide with that of Google Maps. Google Maps does not seem to follow any simple algorithm, suggesting that it may be the most accurate method here.
Anyway, here is a Javascript implementation of Keerthana's algorithm:
function geoDistance(lat1, lng1, lat2, lng2){
const a = 6378.137; // equitorial radius in km
const b = 6356.752; // polar radius in km
var sq = x => (x*x);
var sqr = x => Math.sqrt(x);
var cos = x => Math.cos(x);
var sin = x => Math.sin(x);
var radius = lat => sqr((sq(a*a*cos(lat))+sq(b*b*sin(lat)))/(sq(a*cos(lat))+sq(b*sin(lat))));
lat1 = lat1 * Math.PI / 180;
lng1 = lng1 * Math.PI / 180;
lat2 = lat2 * Math.PI / 180;
lng2 = lng2 * Math.PI / 180;
var R1 = radius(lat1);
var x1 = R1*cos(lat1)*cos(lng1);
var y1 = R1*cos(lat1)*sin(lng1);
var z1 = R1*sin(lat1);
var R2 = radius(lat2);
var x2 = R2*cos(lat2)*cos(lng2);
var y2 = R2*cos(lat2)*sin(lng2);
var z2 = R2*sin(lat2);
return sqr(sq(x1-x2)+sq(y1-y2)+sq(z1-z2));
}

Here is a typescript implementation of the Haversine formula
static getDistanceFromLatLonInKm(lat1: number, lon1: number, lat2: number, lon2: number): number {
var deg2Rad = deg => {
return deg * Math.PI / 180;
}
var r = 6371; // Radius of the earth in km
var dLat = deg2Rad(lat2 - lat1);
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;
}

Here is the SQL Implementation to calculate the distance in km,
SELECT UserId, ( 3959 * acos( cos( radians( your latitude here ) ) * cos( radians(latitude) ) *
cos( radians(longitude) - radians( your longitude here ) ) + sin( radians( your latitude here ) ) *
sin( radians(latitude) ) ) ) AS distance FROM user HAVING
distance < 5 ORDER BY distance LIMIT 0 , 5;
For further details in the implementation by programming langugage, you can just go through the php script given here

This script [in PHP] calculates distances between the two points.
public static function getDistanceOfTwoPoints($source, $dest, $unit='K') {
$lat1 = $source[0];
$lon1 = $source[1];
$lat2 = $dest[0];
$lon2 = $dest[1];
$theta = $lon1 - $lon2;
$dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);
$miles = $dist * 60 * 1.1515;
$unit = strtoupper($unit);
if ($unit == "K") {
return ($miles * 1.609344);
}
else if ($unit == "M")
{
return ($miles * 1.609344 * 1000);
}
else if ($unit == "N") {
return ($miles * 0.8684);
}
else {
return $miles;
}
}

here is an example in postgres sql (in km, for miles version, replace 1.609344 by 0.8684 version)
CREATE OR REPLACE FUNCTION public.geodistance(alat float, alng float, blat
float, blng float)
RETURNS float AS
$BODY$
DECLARE
v_distance float;
BEGIN
v_distance = asin( sqrt(
sin(radians(blat-alat)/2)^2
+ (
(sin(radians(blng-alng)/2)^2) *
cos(radians(alat)) *
cos(radians(blat))
)
)
) * cast('7926.3352' as float) * cast('1.609344' as float) ;
RETURN v_distance;
END
$BODY$
language plpgsql VOLATILE SECURITY DEFINER;
alter function geodistance(alat float, alng float, blat float, blng float)
owner to postgres;

Java implementation in according Haversine formula
double calculateDistance(double latPoint1, double lngPoint1,
double latPoint2, double lngPoint2) {
if(latPoint1 == latPoint2 && lngPoint1 == lngPoint2) {
return 0d;
}
final double EARTH_RADIUS = 6371.0; //km value;
//converting to radians
latPoint1 = Math.toRadians(latPoint1);
lngPoint1 = Math.toRadians(lngPoint1);
latPoint2 = Math.toRadians(latPoint2);
lngPoint2 = Math.toRadians(lngPoint2);
double distance = Math.pow(Math.sin((latPoint2 - latPoint1) / 2.0), 2)
+ Math.cos(latPoint1) * Math.cos(latPoint2)
* Math.pow(Math.sin((lngPoint2 - lngPoint1) / 2.0), 2);
distance = 2.0 * EARTH_RADIUS * Math.asin(Math.sqrt(distance));
return distance; //km value
}

I made a custom function in R to calculate haversine distance(km) between two spatial points using functions available in R base package.
custom_hav_dist <- function(lat1, lon1, lat2, lon2) {
R <- 6371
Radian_factor <- 0.0174533
lat_1 <- (90-lat1)*Radian_factor
lat_2 <- (90-lat2)*Radian_factor
diff_long <-(lon1-lon2)*Radian_factor
distance_in_km <- 6371*acos((cos(lat_1)*cos(lat_2))+
(sin(lat_1)*sin(lat_2)*cos(diff_long)))
rm(lat1, lon1, lat2, lon2)
return(distance_in_km)
}
Sample output
custom_hav_dist(50.31,19.08,54.14,19.39)
[1] 426.3987
PS: To calculate distances in miles, substitute R in function (6371) with 3958.756 (and for nautical miles, use 3440.065).

To calculate the distance between two points on a sphere you need to do the Great Circle calculation.
There are a number of C/C++ libraries to help with map projection at MapTools if you need to reproject your distances to a flat surface. To do this you will need the projection string of the various coordinate systems.
You may also find MapWindow a useful tool to visualise the points. Also as its open source its a useful guide to how to use the proj.dll library, which appears to be the core open source projection library.

Here is my java implementation for calculation distance via decimal degrees after some search. I used mean radius of world (from wikipedia) in km. İf you want result miles then use world radius in miles.
public static double distanceLatLong2(double lat1, double lng1, double lat2, double lng2)
{
double earthRadius = 6371.0d; // KM: use mile here if you want mile result
double dLat = toRadian(lat2 - lat1);
double dLng = toRadian(lng2 - lng1);
double a = Math.pow(Math.sin(dLat/2), 2) +
Math.cos(toRadian(lat1)) * Math.cos(toRadian(lat2)) *
Math.pow(Math.sin(dLng/2), 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
return earthRadius * c; // returns result kilometers
}
public static double toRadian(double degrees)
{
return (degrees * Math.PI) / 180.0d;
}

Here's the accepted answer implementation ported to Java in case anyone needs it.
package com.project529.garage.util;
/**
* Mean radius.
*/
private static double EARTH_RADIUS = 6371;
/**
* Returns the distance between two sets of latitudes and longitudes in meters.
* <p/>
* Based from the following JavaScript SO answer:
* http://stackoverflow.com/questions/27928/calculate-distance-between-two-latitude-longitude-points-haversine-formula,
* which is based on https://en.wikipedia.org/wiki/Haversine_formula (error rate: ~0.55%).
*/
public double getDistanceBetween(double lat1, double lon1, double lat2, double lon2) {
double dLat = toRadians(lat2 - lat1);
double dLon = toRadians(lon2 - lon1);
double a = Math.sin(dLat / 2) * Math.sin(dLat / 2) +
Math.cos(toRadians(lat1)) * Math.cos(toRadians(lat2)) *
Math.sin(dLon / 2) * Math.sin(dLon / 2);
double c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
double d = EARTH_RADIUS * c;
return d;
}
public double toRadians(double degrees) {
return degrees * (Math.PI / 180);
}

For those looking for an Excel formula based on WGS-84 & GRS-80 standards:
=ACOS(COS(RADIANS(90-Lat1))*COS(RADIANS(90-Lat2))+SIN(RADIANS(90-Lat1))*SIN(RADIANS(90-Lat2))*COS(RADIANS(Long1-Long2)))*6371
Source

there is a good example in here to calculate distance with PHP http://www.geodatasource.com/developers/php :
function distance($lat1, $lon1, $lat2, $lon2, $unit) {
$theta = $lon1 - $lon2;
$dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) + cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);
$miles = $dist * 60 * 1.1515;
$unit = strtoupper($unit);
if ($unit == "K") {
return ($miles * 1.609344);
} else if ($unit == "N") {
return ($miles * 0.8684);
} else {
return $miles;
}
}

Here is the implementation VB.NET, this implementation will give you the result in KM or Miles based on an Enum value you pass.
Public Enum DistanceType
Miles
KiloMeters
End Enum
Public Structure Position
Public Latitude As Double
Public Longitude As Double
End Structure
Public Class Haversine
Public Function Distance(Pos1 As Position,
Pos2 As Position,
DistType As DistanceType) As Double
Dim R As Double = If((DistType = DistanceType.Miles), 3960, 6371)
Dim dLat As Double = Me.toRadian(Pos2.Latitude - Pos1.Latitude)
Dim dLon As Double = Me.toRadian(Pos2.Longitude - Pos1.Longitude)
Dim a As Double = Math.Sin(dLat / 2) * Math.Sin(dLat / 2) + Math.Cos(Me.toRadian(Pos1.Latitude)) * Math.Cos(Me.toRadian(Pos2.Latitude)) * Math.Sin(dLon / 2) * Math.Sin(dLon / 2)
Dim c As Double = 2 * Math.Asin(Math.Min(1, Math.Sqrt(a)))
Dim result As Double = R * c
Return result
End Function
Private Function toRadian(val As Double) As Double
Return (Math.PI / 180) * val
End Function
End Class

I condensed the computation down by simplifying the formula.
Here it is in Ruby:
include Math
earth_radius_mi = 3959
radians = lambda { |deg| deg * PI / 180 }
coord_radians = lambda { |c| { :lat => radians[c[:lat]], :lng => radians[c[:lng]] } }
# from/to = { :lat => (latitude_in_degrees), :lng => (longitude_in_degrees) }
def haversine_distance(from, to)
from, to = coord_radians[from], coord_radians[to]
cosines_product = cos(to[:lat]) * cos(from[:lat]) * cos(from[:lng] - to[:lng])
sines_product = sin(to[:lat]) * sin(from[:lat])
return earth_radius_mi * acos(cosines_product + sines_product)
end

function getDistanceFromLatLonInKm(lat1,lon1,lat2,lon2,units) {
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;
var miles = d / 1.609344;
if ( units == 'km' ) {
return d;
} else {
return miles;
}}
Chuck's solution, valid for miles also.

In Mysql use the following function pass the parameters as using POINT(LONG,LAT)
CREATE FUNCTION `distance`(a POINT, b POINT)
RETURNS double
DETERMINISTIC
BEGIN
RETURN
GLength( LineString(( PointFromWKB(a)), (PointFromWKB(b)))) * 100000; -- To Make the distance in meters
END;

Trying to write some maths from Wikipedia in Javascript (sun position computation)

I am trying to approximate the position of the sun in XYZ for a threejs project.
I am following the maths found here: http://en.wikipedia.org/wiki/Position_of_the_Sun
Following the above, I have written the following Javascript code:
var n = ((2440587.5 + (this.datemillis / 8.64E7)) - 2451545);
var L = 280.460 + 0.9856474 * n;
var g = 357.528 + 0.9856003 * n;
L = (L + 360) % 360;
g = (g + 360) % 60;
var lambda = L + 1.915 * Math.sin(g) + 0.0020 * Math.sin(2 * g);
var r = 1.00014 - 0.01671 * Math.cos(g) - 0.00014 * Math.cos(2 * g);
var e = 23.439 - 0.0000004 * n;
var x = (r * this.constants.EARTH_RADIUS * 2) * Math.cos(lambda);
var y = (r * this.constants.EARTH_RADIUS * 2) * Math.cos(e) * Math.sin(lambda);
var z = (r * this.constants.EARTH_RADIUS * 2) * Math.sin(e) * Math.sin(lambda);
this.datemillis is returned by the getMillisecond function of the Javascript date object. It is updated each frame so that time advances at about 1 hour every 2 seconds.
However something must not be correct as this does not produce the expected result. When I apply the computed x y z coordinates to my sun in my threejs project, I can see the sun rotate around the earth (sitting in 0,0,0) but at a very slow rate (rotating the earth in a few days instead of 24 hours).
I'm thinking it might have something to do with the angle calculations that I'm not doing correctly (degrees/radians?) but I'm not very good at maths so I don't really know what I'm doing so maybe I just misinterpreted the Wiki calculations.
If somebody could spot something obvious I'm doing wrong and help me fix this, would be greatly appreciated!
Thanks
EDIT: so my sun currently is not rotating around the earth in a continous way - it rotates clockwise/counterclockwise alternatively and sometimes jumps positions...

I suggest this to get the Julian Date, from Calculating Jday(Julian Day) in javascript
var today = Date();
var JD = Math.floor((today / 86400000) - (today.getTimezoneOffset()/1440) + 2440587.5);
Add to JD the desired amount of days and increment that value at the desired speed. Note that if you add 1 day each millisecond you'll get 1000 days per second, not 1 hour every 2 seconds.
JD += offset;
Then go on with the wikipedia recipe:
var n = JD - 2451545;
//...
To put L and g in the range 0-360 (you have an error here) use
L = L % 360 + ( L < 0 ? 360 : 0 );
g = g % 360 + ( g < 0 ? 360 : 0 );
The wikipedia formulas express angles in degrees. However JavaScript trigonometric functions cos and sin expect radians.
Just write a "degrees" version of them:
function cosD( deg ) {
return Math.cos( deg * Math.PI / 180.0 );
}
function sinD( deg ) {
return Math.sin( deg * Math.PI / 180.0 );
}
Then use sinD() and cosD() in subsequent calculations.
var r = 1.00014 - 0.01671 * cosD(g) - 0.00014 * cosD(2 * g);
var e = 23.439 - 0.0000004 * n;
var x = (r * this.constants.EARTH_RADIUS * 2) * cosD(lambda);
var y = (r * this.constants.EARTH_RADIUS * 2) * cosD(e) * sinD(lambda);
var z = (r * this.constants.EARTH_RADIUS * 2) * sinD(e) * sinD(lambda);

I cannot answer your question but I do know this is a solved problem in threejs. There is an example running in an architecture/engineering workflow on Github on this topic. The sun position code is here https://github.com/radio412/viewer/blob/gh-pages/sun-position.js
You can see it being tapped for a directional light in threejs at line 108 here: https://github.com/radio412/viewer/blob/gh-pages/va3c-viewer.js

Location-based networking website for students

I want to make a location-based academic website(bootstrap project ) where a said question will be visible only to those present in the pre-specified radius. I can code in HTML,CSS and currently learning JavaScript and PHP. Anyone willing to give me some pointers? (code resources, relevant templates or useful literature)

You can find the distances beteen locations using the Haversine formula. The following javascript functions are used to calculate the distance between the geolocated coordinates and that of the center location. Only if the distance is less than radius allow access.
function deg2rad(degrees){
radians = degrees * (Math.PI/180);
return radians;
}
function Haversine(lat1,lon1,lat2,lon2) {
deltaLat = lat2 - lat1 ;
deltaLon = lon2 - lon1 ;
earthRadius = 3959; // in miles 6371 in meters.
alpha = deltaLat/2;
beta = deltaLon/2;
a = Math.sin(deg2rad(alpha)) * Math.sin(deg2rad(alpha)) + Math.cos(deg2rad(lat1)) * Math.cos(deg2rad(lat2)) * Math.sin(deg2rad(beta)) * Math.sin(deg2rad(beta)) ;
c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));
distance = earthRadius * c;
return distance.toFixed(2);
}