Destination Point from Bearing Calculator
Given a starting position, a bearing to set off on, and a distance to travel, you can compute exactly where you will end up. This is the direct, or forward, geodesic problem, and it is the basis of dead reckoning in navigation. This tool follows the great circle from the start point along the given initial bearing for the given distance and returns the destination latitude and longitude. Enter the start coordinates in decimal degrees (negative for south and west), the initial bearing in degrees true (measured clockwise from north), and the distance in kilometres. It uses a spherical Earth of radius 6,371 kilometres.
Destination point formula
ad = distance / R (angular distance, R = 6,371 km)
lat2 = asin(sin(lat1) * cos(ad) + cos(lat1) * sin(ad) * cos(bearing))
lon2 = lon1 + atan2(sin(bearing) * sin(ad) * cos(lat1), cos(ad) - sin(lat1) * sin(lat2))
The distance is first turned into an angular distance by dividing by the Earth's radius. Spherical trigonometry then gives the destination latitude from the start latitude, the angular distance, and the bearing, and the longitude from the change in longitude that the path produces. The longitude is normalised to the range minus 180 to 180 degrees. The bearing used is the initial heading; the great-circle path's bearing changes along the way.
Worked example
Starting at the equator point (0, 0) and heading due east (bearing 090) for 100 km: the angular distance is 100 / 6,371 = 0.015697 radians. Because the path starts on the equator heading east, the destination stays on the equator at latitude 0.00 and moves to longitude 0.90 degrees east. From New York (40.7128, -74.0060) on the same 090 bearing for 100 km, the destination is about latitude 40.71 and longitude -72.82, roughly 1.18 degrees of longitude east.
Frequently asked questions
What is the direct geodesic problem?
The direct, or forward, geodesic problem asks: starting from a known point and travelling a given distance on a given bearing, where do you arrive? It is the counterpart to the inverse problem, which asks for the distance and bearing between two known points. This calculator solves the direct problem on a spherical Earth.
Is the bearing constant along the path?
No. The calculator follows a great circle, the shortest path, whose bearing changes continuously as it crosses meridians. The value you enter is the initial bearing at the start point. If you instead held a constant compass bearing you would follow a rhumb line, a different and slightly longer path, ending at a different point.
What Earth radius does it use?
It uses the mean spherical Earth radius of 6,371 kilometres, the standard value for great-circle navigation calculations. This treats the Earth as a perfect sphere, which is accurate to within about 0.3 percent. For survey-grade results an ellipsoidal model such as WGS84 with Vincenty's or Karney's method would be used instead.
How do I enter a southerly or westerly start?
Use negative numbers: a southern latitude and a western longitude are both negative. The bearing is always 0 to 360 degrees clockwise from true north regardless of hemisphere. The destination longitude is returned in the range minus 180 to 180 degrees, so westerly results come out negative.
Official sources
- NOAA National Geodetic Survey: geodetic computations and the direct problem.
- U.S. Geological Survey: coordinate systems and mapping.
Reviewed by the CalculatorHub team, edited by James Graham, 19 June 2026. See our methodology.