Given an existing point in lat/long, distance in (in KM) and bearing (in degrees converted to radians), I would like to calculate the new lat/long. This site crops up over and over again, but I just can’t get the formula to work for me.

The formulas as taken the above link are:

lat2 = asin(sin(lat1)*cos(d/R) + cos(lat1)*sin(d/R)*cos(θ)) lon2 = lon1 + atan2(sin(θ)*sin(d/R)*cos(lat1), cos(d/R)−sin(lat1)*sin(lat2))

The above formula is for MSExcel where-

asin = arc sin() d = distance (in any unit) R = Radius of the earth (in the same unit as above) and hence d/r = is the angular distance (in radians) atan2(a,b) = arc tan(b/a) θ is the bearing (in radians, clockwise from north);

Here’s the code I’ve got in Python.

import math R = 6378.1 #Radius of the Earth brng = 1.57 #Bearing is 90 degrees converted to radians. d = 15 #Distance in km #lat2 52.20444 - the lat result I'm hoping for #lon2 0.36056 - the long result I'm hoping for. lat1 = 52.20472 * (math.pi * 180) #Current lat point converted to radians lon1 = 0.14056 * (math.pi * 180) #Current long point converted to radians lat2 = math.asin( math.sin(lat1)*math.cos(d/R) + math.cos(lat1)*math.sin(d/R)*math.cos(brng)) lon2 = lon1 + math.atan2(math.sin(brng)*math.sin(d/R)*math.cos(lat1), math.cos(d/R)-math.sin(lat1)*math.sin(lat2)) print(lat2) print(lon2)

I get

lat2 = 0.472492248844 lon2 = 79.4821662373

## Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

### Method 1

Needed to convert answers from radians back to degrees. Working code below:

import math R = 6378.1 #Radius of the Earth brng = 1.57 #Bearing is 90 degrees converted to radians. d = 15 #Distance in km #lat2 52.20444 - the lat result I'm hoping for #lon2 0.36056 - the long result I'm hoping for. lat1 = math.radians(52.20472) #Current lat point converted to radians lon1 = math.radians(0.14056) #Current long point converted to radians lat2 = math.asin( math.sin(lat1)*math.cos(d/R) + math.cos(lat1)*math.sin(d/R)*math.cos(brng)) lon2 = lon1 + math.atan2(math.sin(brng)*math.sin(d/R)*math.cos(lat1), math.cos(d/R)-math.sin(lat1)*math.sin(lat2)) lat2 = math.degrees(lat2) lon2 = math.degrees(lon2) print(lat2) print(lon2)

### Method 2

The geopy library supports this:

import geopy from geopy.distance import VincentyDistance # given: lat1, lon1, b = bearing in degrees, d = distance in kilometers origin = geopy.Point(lat1, lon1) destination = VincentyDistance(kilometers=d).destination(origin, b) lat2, lon2 = destination.latitude, destination.longitude

Found via https://stackoverflow.com/a/4531227/37610

### Method 3

This question is known as the **direct problem** in the study of geodesy.

This is indeed a very popular question and one that is a constant cause of confusion. The reason is that most people are looking for a simple and straight-forward answer. But there is none, because most people asking this question are not supplying enough information, simply because they are not aware that:

- Earth is not a perfect sphere, since it is flattened/compressed by it poles
- Because of (1) earth does not have a constant Radius,
`R`

. See here. - Earth is not perfectly smooth (variations in altitude) etc.
- Due to tectonic plate movement, a geographic point’s lat/lon position may change by several millimeters (at least), every year.

Therefore there are many different assumptions used in the various geometric models that apply differently, depending on your needed accuracy. So to answer the question you need to consider to what **accuracy** you would like to have your result.

**Some examples:**

- I’m just looking for an approximate location to the nearest few kilometers for small (
**<**100 km) distances of in`latitudes`

between`0-70 deg`

**N|S**. (Earth is ~flat model.) - I want an answer that is good anywhere on the globe, but only accurate to about a few meters
- I want a super accurate positioning that is valid down to atomic scales of
`nanometers`

[nm]. - I want answers that is very fast and easy to calculate and not computationally intensive.

So you can have many choices in which algorithm to use. In addition each programming language has it’s own implementation or “package” multiplied by number of models and the model developers specific needs. For all practical purposes here, it pays off to ignore any other language apart `javascript`

, since it very closely resemble pseudo-code by its nature. Thus it can be easily converted to any other language, with minimal changes.

Then the main models are:

`Euclidian/Flat earth model`

: good for very short distances under ~10 km`Spherical model`

: good for large longitudinal distances, but with small latitudinal difference. Popular model:- Haversine:
**meter**accuracy on [km] scales, very simple code.

- Haversine:
`Ellipsoidal models`

: Most accurate at any lat/lon and distance, but is still a numerical approximation that depend on what accuracy you need. Some popular models are:- Lambert:
**~10 meter**precision over 1000’s of*km*. - Paul D.Thomas: Andoyer-Lambert approximation
- Vincenty:
**millimeter**precision and computational efficiency - Kerney:
**nanometer**precision

- Lambert:

**References:**

- https://en.wikipedia.org/wiki/Reference_ellipsoid
- https://en.wikipedia.org/wiki/Haversine_formula
- https://en.wikipedia.org/wiki/Earth_ellipsoid
- https://en.wikipedia.org/wiki/Geodesics_on_an_ellipsoid
- https://en.wikipedia.org/wiki/Vincenty%27s_formulae
- https://geographiclib.sourceforge.io/scripts/geod-calc.html

### Method 4

May be a bit late for answering, but after testing the other answers, it appears they don’t work correctly. Here is a PHP code we use for our system. Working in all directions.

PHP code:

lat1 = latitude of start point in degrees

long1 = longitude of start point in degrees

d = distance in KM

angle = bearing in degrees

function get_gps_distance($lat1,$long1,$d,$angle) { # Earth Radious in KM $R = 6378.14; # Degree to Radian $latitude1 = $lat1 * (M_PI/180); $longitude1 = $long1 * (M_PI/180); $brng = $angle * (M_PI/180); $latitude2 = asin(sin($latitude1)*cos($d/$R) + cos($latitude1)*sin($d/$R)*cos($brng)); $longitude2 = $longitude1 + atan2(sin($brng)*sin($d/$R)*cos($latitude1),cos($d/$R)-sin($latitude1)*sin($latitude2)); # back to degrees $latitude2 = $latitude2 * (180/M_PI); $longitude2 = $longitude2 * (180/M_PI); # 6 decimal for Leaflet and other system compatibility $lat2 = round ($latitude2,6); $long2 = round ($longitude2,6); // Push in array and get back $tab[0] = $lat2; $tab[1] = $long2; return $tab; }

### Method 5

I ported answer by Brad to vanilla JS answer, with no Bing maps dependency

https://jsfiddle.net/kodisha/8a3hcjtd/

```
// ----------------------------------------
// Calculate new Lat/Lng from original points
// on a distance and bearing (angle)
// ----------------------------------------
let llFromDistance = function(latitude, longitude, distance, bearing) {
// taken from: https://stackoverflow.com/a/46410871/13549
// distance in KM, bearing in degrees
const R = 6378.1; // Radius of the Earth
const brng = bearing * Math.PI / 180; // Convert bearing to radian
let lat = latitude * Math.PI / 180; // Current coords to radians
let lon = longitude * Math.PI / 180;
// Do the math magic
lat = Math.asin(Math.sin(lat) * Math.cos(distance / R) + Math.cos(lat) * Math.sin(distance / R) * Math.cos(brng));
lon += Math.atan2(Math.sin(brng) * Math.sin(distance / R) * Math.cos(lat), Math.cos(distance / R) - Math.sin(lat) * Math.sin(lat));
// Coords back to degrees and return
return [(lat * 180 / Math.PI), (lon * 180 / Math.PI)];
}
let pointsOnMapCircle = function(latitude, longitude, distance, numPoints) {
const points = [];
for (let i = 0; i <= numPoints - 1; i++) {
const bearing = Math.round((360 / numPoints) * i);
console.log(bearing, i);
const newPoints = llFromDistance(latitude, longitude, distance, bearing);
points.push(newPoints);
}
return points;
}
const points = pointsOnMapCircle(41.890242042122836, 12.492358982563019, 0.2, 8);
let geoJSON = {
"type": "FeatureCollection",
"features": []
};
points.forEach((p) => {
geoJSON.features.push({
"type": "Feature",
"properties": {},
"geometry": {
"type": "Point",
"coordinates": [
p[1],
p[0]
]
}
});
});
document.getElementById('res').innerHTML = JSON.stringify(geoJSON, true, 2);
```

In addition, I added `geoJSON`

export, so you can simply paste resulting geoJSON to: `http://geojson.io/#map=17/41.89017/12.49171`

to see the results instantly.

### Method 6

Quick way using geopy

from geopy import distance #distance.distance(unit=15).destination((lat,lon),bering) #Exemples distance.distance(nautical=15).destination((-24,-42),90) distance.distance(miles=15).destination((-24,-42),90) distance.distance(kilometers=15).destination((-24,-42),90)

### Method 7

lon1 and lat1 in degrees

brng = bearing in radians

d = distance in km

R = radius of the Earth in km

lat2 = math.degrees((d/R) * math.cos(brng)) + lat1 long2 = math.degrees((d/(R*math.sin(math.radians(lat2)))) * math.sin(brng)) + long1

I implemented your algorithm and mine in PHP and benchmarked it. This version ran in about 50% of the time. The results generated were identical, so it seems to be mathematically equivalent.

I didn’t test the python code above so there might be syntax errors.

### Method 8

I ported the Python to Javascript. This returns a Bing Maps `Location`

object, you can change to whatever you like.

getLocationXDistanceFromLocation: function(latitude, longitude, distance, bearing) { // distance in KM, bearing in degrees var R = 6378.1, // Radius of the Earth brng = Math.radians(bearing) // Convert bearing to radian lat = Math.radians(latitude), // Current coords to radians lon = Math.radians(longitude); // Do the math magic lat = Math.asin(Math.sin(lat) * Math.cos(distance / R) + Math.cos(lat) * Math.sin(distance / R) * Math.cos(brng)); lon += Math.atan2(Math.sin(brng) * Math.sin(distance / R) * Math.cos(lat), Math.cos(distance/R)-Math.sin(lat)*Math.sin(lat)); // Coords back to degrees and return return new Microsoft.Maps.Location(Math.degrees(lat), Math.degrees(lon)); },

### Method 9

Also late but for those who might find this, you will get more accurate results using the geographiclib library. Check out the geodesic problem descriptions and the JavaScript examples for an easy introduction to how to use to answer the subject question as well as many others. Implementations in a variety of languages including Python. Far better than coding your own if you care about accuracy; better than VincentyDistance in the earlier “use a library” recommendation. As the documentation says: “The emphasis is on returning accurate results with errors close to round-off (about 5–15 nanometers).”

### Method 10

Just interchange the values in the atan2(y,x) function. Not atan2(x,y)!

### Method 11

I ported the answer from @David M to java if anyone wanted this… I do get a slight different result of 52.20462299620793, 0.360433887489931

double R = 6378.1; //Radius of the Earth double brng = 1.57; //Bearing is 90 degrees converted to radians. double d = 15; //Distance in km double lat2 = 52.20444; // - the lat result I'm hoping for double lon2 = 0.36056; // - the long result I'm hoping for. double lat1 = Math.toRadians(52.20472); //Current lat point converted to radians double lon1 = Math.toRadians(0.14056); //Current long point converted to radians lat2 = Math.asin( Math.sin(lat1)*Math.cos(d/R) + Math.cos(lat1)*Math.sin(d/R)*Math.cos(brng)); lon2 = lon1 + Math.atan2(Math.sin(brng)*Math.sin(d/R)*Math.cos(lat1), Math.cos(d/R)-Math.sin(lat1)*Math.sin(lat2)); lat2 = Math.toDegrees(lat2); lon2 = Math.toDegrees(lon2); System.out.println(lat2 + ", " + lon2);

### Method 12

Thanks to @kodisha, here is a Swift version, but with improved and more precise calculation for Earth radius:

extension CLLocationCoordinate2D { func earthRadius() -> CLLocationDistance { let earthRadiusInMetersAtSeaLevel = 6378137.0 let earthRadiusInMetersAtPole = 6356752.314 let r1 = earthRadiusInMetersAtSeaLevel let r2 = earthRadiusInMetersAtPole let beta = latitude let earthRadiuseAtGivenLatitude = ( ( pow(pow(r1, 2) * cos(beta), 2) + pow(pow(r2, 2) * sin(beta), 2) ) / ( pow(r1 * cos(beta), 2) + pow(r2 * sin(beta), 2) ) ) .squareRoot() return earthRadiuseAtGivenLatitude } func locationByAdding( distance: CLLocationDistance, bearing: CLLocationDegrees ) -> CLLocationCoordinate2D { let latitude = self.latitude let longitude = self.longitude let earthRadiusInMeters = self.earthRadius() let brng = bearing.degreesToRadians var lat = latitude.degreesToRadians var lon = longitude.degreesToRadians lat = asin( sin(lat) * cos(distance / earthRadiusInMeters) + cos(lat) * sin(distance / earthRadiusInMeters) * cos(brng) ) lon += atan2( sin(brng) * sin(distance / earthRadiusInMeters) * cos(lat), cos(distance / earthRadiusInMeters) - sin(lat) * sin(lat) ) let newCoordinate = CLLocationCoordinate2D( latitude: lat.radiansToDegrees, longitude: lon.radiansToDegrees ) return newCoordinate } } extension FloatingPoint { var degreesToRadians: Self { self * .pi / 180 } var radiansToDegrees: Self { self * 180 / .pi } }

### Method 13

Here is a PHP version based on Ed Williams Aviation Formulary. Modulus is handled a little different in PHP. This works for me.

function get_new_waypoint ( $lat, $lon, $radial, $magvar, $range ) { // $range in nm. // $radial is heading to or bearing from // $magvar for local area. $range = $range * pi() /(180*60); $radial = $radial - $magvar ; if ( $radial < 1 ) { $radial = 360 + $radial - $magvar; } $radial = deg2rad($radial); $tmp_lat = deg2rad($lat); $tmp_lon = deg2rad($lon); $new_lat = asin(sin($tmp_lat)* cos($range) + cos($tmp_lat) * sin($range) * cos($radial)); $new_lat = rad2deg($new_lat); $new_lon = $tmp_lon - asin(sin($radial) * sin($range)/cos($new_lat))+ pi() % 2 * pi() - pi(); $new_lon = rad2deg($new_lon); return $new_lat." ".$new_lon; }

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0