# Get lat/long given current point, distance and bearing

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```
Contents

### 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.

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```

### 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:

1. Earth is not a perfect sphere, since it is flattened/compressed by it poles
2. Because of (1) earth does not have a constant Radius, `R`. See here.
3. Earth is not perfectly smooth (variations in altitude) etc.
4. 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.
• `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

References:

### 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)
{
\$R = 6378.14;

\$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

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.

Result:

### 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

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

// 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.

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 {

let beta = latitude

( 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()

}

distance: CLLocationDistance,
bearing: CLLocationDegrees
) -> CLLocationCoordinate2D {
let latitude = self.latitude
let longitude = self.longitude

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(
)

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.
// \$magvar for local area.

\$range = \$range * pi() /(180*60);

if ( \$radial < 1 )
{
}
\$new_lat = asin(sin(\$tmp_lat)* cos(\$range) + cos(\$tmp_lat) * sin(\$range) * cos(\$radial));
\$new_lon = \$tmp_lon - asin(sin(\$radial) * sin(\$range)/cos(\$new_lat))+ pi() % 2 * pi() -  pi();

return \$new_lat." ".\$new_lon;

}```