I'm trying to generate some points at random distances away from a fixed point using GPS.
How can I add distance in meters to a GPS coordinate?
I've looked at UTM to GPS conversion but is there a simpler method to achieve this?
I'm working on Android platform just in case.
Cheers,
fgs
P0(lat0,lon0) : initial position (unit : degrees)
dx,dy : random offsets from your initial position in meters
You can use an approximation to compute the position of the randomized position:
lat = lat0 + (180/pi)*(dy/6378137)
lon = lon0 + (180/pi)*(dx/6378137)/cos(lat0)
This is quite precise as long as the random distance offset is below 10-100 km
Edit: of course in Java Math.cos() expects radians so do use Math.cos(Math.PI/180.0*lat0) if lat0 is in degrees as assumed above.
To take a square I'm using this:
private double[] getBoundingBox(final double pLatitude, final double pLongitude, final int pDistanceInMeters) {
final double[] boundingBox = new double[4];
final double latRadian = Math.toRadians(pLatitude);
final double degLatKm = 110.574235;
final double degLongKm = 110.572833 * Math.cos(latRadian);
final double deltaLat = pDistanceInMeters / 1000.0 / degLatKm;
final double deltaLong = pDistanceInMeters / 1000.0 / degLongKm;
final double minLat = pLatitude - deltaLat;
final double minLong = pLongitude - deltaLong;
final double maxLat = pLatitude + deltaLat;
final double maxLong = pLongitude + deltaLong;
boundingBox[0] = minLat;
boundingBox[1] = minLong;
boundingBox[2] = maxLat;
boundingBox[3] = maxLong;
return boundingBox;
}
This returns an array with 4 coordinates, with them you can make a square with your original point in center.
A detailed outline is given at http://www.movable-type.co.uk/scripts/latlong.html.
If you, somewhere, need to interconvert longitude/latitude to UTM coordinates (the ones used in GPS) you may want to have a look at http://www.uwgb.edu/dutchs/UsefulData/UTMFormulas.htm
If you want to go east or north or west or south you can use this:
#SuppressLint("DefaultLocale")
public static double go_mock_loc(double xx_lat,double xx_long,double xx_dinstance,String Direction)
{
// double xx_lat= 45.815005;
// double xx_long= 15.978501;
// int xx_dinstance=500;
int equator_circumference=6371000;
int polar_circumference=6356800;
double m_per_deg_long = 360 / polar_circumference;
double rad_lat=(xx_lat* (Math.PI) / 180);
double m_per_deg_lat = 360 / ( Math.cos(rad_lat) * equator_circumference);
double deg_diff_long = xx_dinstance * m_per_deg_long;
double deg_diff_lat = xx_dinstance * m_per_deg_lat;
double xx_north_lat = xx_lat + deg_diff_long;
//double xx_north_long= xx_long;
double xx_south_lat = xx_lat - deg_diff_long;
//double xx_south_long= xx_long;
//double xx_east_lat = xx_lat;
double xx_east_long= xx_long + deg_diff_lat;
//double xx_west_lat = xx_lat;
double xx_west_long= xx_long - deg_diff_lat;
if (Direction.toUpperCase().contains("NORTH")) {
return xx_north_lat;
} else if (Direction.toUpperCase().contains("SOUTH"))
{
return xx_south_lat;
} else if (Direction.toUpperCase().contains("EAST"))
{
return xx_east_long;
} else if (Direction.toUpperCase().contains("WEST"))
{
return xx_west_long;
}
else
return 0;
}
I found that solution of #Bogdan Khrystov is very well.
So here is C# version of his solution.
public enum GeoDirection
{
NORTH = 1, SOUTH = 2, EAST = 3, WEST = 4
}
public static Tuple<double, double> AddDistanceInMeters(double latitude, double longitude, int distanceInMeters, GeoDirection direction)
{
var equatorCircumference = 6371000;
var polarCircumference = 6356800;
var mPerDegLong = 360 / (double)polarCircumference;
var radLat = latitude * Math.PI / 180;
var mPerDegLat = 360 / (Math.Cos(radLat) * equatorCircumference);
var degDiffLong = distanceInMeters * mPerDegLong;
var degDiffLat = distanceInMeters * mPerDegLat;
var xxNorthLat = latitude + degDiffLong;
var xxSouthLat = latitude - degDiffLong;
var xxEastLong = longitude + degDiffLat;
var xxWestLong = longitude - degDiffLat;
switch (direction)
{
case GeoDirection.NORTH:
return new Tuple<double, double>(xxNorthLat, longitude);
case GeoDirection.SOUTH:
return new Tuple<double, double>(xxSouthLat, longitude);
case GeoDirection.EAST:
return new Tuple<double, double>(latitude, xxEastLong);
case GeoDirection.WEST:
return new Tuple<double, double>(latitude, xxWestLong);
default:
return null;
}
}
rewrite #Ersin Gülbahar answer in Kotlin:
object LocationUtil {
enum class Direction {
NORTH, SOUTH, EAST, WEST
}
fun addDistanceInMeters(
latitude: Double,
longitude: Double,
distanceInMeters: Int,
direction: Direction
): Pair<Double, Double> {
val equatorCircumference = 6371000
val polarCircumference = 6356800
val mPerDegLong = (360 / polarCircumference.toDouble())
val radLat = latitude * Math.PI / 180
val mPerDegLat = 360 / (Math.cos(radLat) * equatorCircumference)
val degDiffLong = distanceInMeters * mPerDegLong
val degDiffLat = distanceInMeters * mPerDegLat
val xxNorthLat = latitude + degDiffLong
val xxSouthLat = latitude - degDiffLong
val xxEastLong = longitude + degDiffLat
val xxWestLong = longitude - degDiffLat
return when (direction) {
Direction.NORTH -> Pair(xxNorthLat, longitude)
Direction.SOUTH -> Pair(xxSouthLat, longitude)
Direction.EAST -> Pair(latitude, xxEastLong)
Direction.WEST -> Pair(latitude, xxWestLong)
}
}
}
This code splits the line between two coordinates in n segments. Replace the delta calculation by your fixed distance
#Override
public void split(Coordinates p1, Coordinates p2, int segments) {
double φ1 = Math.toRadians(p1.getLat());
double λ1 = Math.toRadians(p1.getLon());
double φ2 = Math.toRadians(p2.getLat());
double λ2 = Math.toRadians(p2.getLon());
double xDelta = (φ2 - φ1) / segments;
double yDelta = (λ2 - λ1) / segments;
for (int i = 0; i < segments; i++){
double x = φ1 + i * xDelta;
double y = λ1 + i * yDelta;
double xc = Math.toDegrees(x);
double yc = Math.toDegrees(y);
System.out.println(xc+","+yc);
}
}
Combining answers from #Ersin Gülbahar and #Stéphane above, I came up with this solution in Flutter/Dart:
import 'dart:math' as math;
enum Direction { north, south, east, west }
double moveCoordinate(
double latitude, double longitude, double distanceToMoveInMeters, Direction directionToMove) {
const earthEquatorRadius = 6378137;
final latitudeOffset = (180 / math.pi) * (distanceToMoveInMeters / earthEquatorRadius);
final longitudeOffset = (180 / math.pi) *
(distanceToMoveInMeters / earthEquatorRadius) /
math.cos(math.pi / 180 * latitude);
switch (directionToMove) {
case Direction.north:
return latitude + latitudeOffset;
case Direction.south:
return latitude - latitudeOffset;
case Direction.east:
return longitude + longitudeOffset;
case Direction.west:
return longitude - longitudeOffset;
}
return 0;
}
This works, tested. The code is C# but you can easily change it to another language
private PointLatLng NewPositionBasedOnDistanceAngle(PointLatLng org, double distance, double bearing)
{
double rad = bearing * Math.PI / 180; //to radians
double lat1 = org.Lat * Math.PI / 180; //to radians
double lng1 = org.Lng * Math.PI / 180; //to radians
double lat = Math.Asin(Math.Sin(lat1) * Math.Cos(distance / 6378137) + Math.Cos(lat1) * Math.Sin(distance / 6378137) * Math.Cos(rad));
double lng = lng1 + Math.Atan2(Math.Sin(rad) * Math.Sin(distance / 6378137) * Math.Cos(lat1), Math.Cos(distance / 6378137) - Math.Sin(lat1) * Math.Sin(lat));
return new PointLatLng(lat * 180 / Math.PI, lng * 180 / Math.PI); // to degrees
}
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;
As title said, I have current location, and want to know the bearing in degree from my current location to other location. I have read this post, is the value return by location.getBearing() my answer?
Let say it simply: from the picture, I expect the value of 45 degree.
I was having trouble doing this and managed to figure it out converting the C? approach to working out the bearing degree.
I worked this out using 4 coordinate points
Start Latitude
Start Longitude
End Latitude
End Longitude
Here's the code:
protected double bearing(double startLat, double startLng, double endLat, double endLng){
double latitude1 = Math.toRadians(startLat);
double latitude2 = Math.toRadians(endLat);
double longDiff= Math.toRadians(endLng - startLng);
double y= Math.sin(longDiff)*Math.cos(latitude2);
double x=Math.cos(latitude1)*Math.sin(latitude2)-Math.sin(latitude1)*Math.cos(latitude2)*Math.cos(longDiff);
return (Math.toDegrees(Math.atan2(y, x))+360)%360;
}
Just change the variable names were needed.
Output bearing to a TextView
Hope it helped!
Location have method bearingTo:
float bearingTo(Location dest)
Location from = new Location(LocationManager.GPS_PROVIDER);
Location to = new Location(LocationManager.GPS_PROVIDER);
from.setLatitude(0);
from.setLongitude(0);
to.setLongitude(10);
to.setLatitude(10);
float bearingTo = from.bearingTo(to);
Here is the code for calculating bearing angle between two points(startPoint, endPoint):
public float CalculateBearingAngle(double startLatitude,double startLongitude, double endLatitude, double endLongitude){
double Phi1 = Math.toRadians(startLatitude);
double Phi2 = Math.toRadians(endLatitude);
double DeltaLambda = Math.toRadians(endLongitude - startLongitude);
double Theta = atan2((sin(DeltaLambda)*cos(Phi2)) , (cos(Phi1)*sin(Phi2) - sin(Phi1)*cos(Phi2)*cos(DeltaLambda)));
return (float)Math.toDegrees(Theta);
}
Call for function:
float angle = CalculateBearingAngle(startLatitude, startLongitude, endLatitude, endLongitude);
I found best method for calculating "Bearing between two point " from this answer and convert it to java from python. And this method work like charm for me!
here is the code:
public static double getBearing(double startLat, double startLng, double endLat, double endLng) {
double latitude1 = Math.toRadians(startLat);
double longitude1 = Math.toRadians(-startLng);
double latitude2 = Math.toRadians(endLat);
double longitude2 = Math.toRadians(-endLng);
double dLong = longitude2 - longitude1;
double dPhi = Math.log(Math.tan(latitude2 / 2.0 + Math.PI / 4.0) / Math.tan(latitude1 / 2.0 + Math.PI / 4.0));
if (abs(dLong) > Math.PI)
if (dLong > 0.0)
dLong = -(2.0 * Math.PI - dLong);
else
dLong = (2.0 * Math.PI + dLong);
return (Math.toDegrees(Math.atan2(dLong, dPhi)) + 360.0) % 360.0;
}