Get lat/long given current point, distance and bearing

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)

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

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:

• 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

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;
 }