Destination Point from Bearing and Distance
Find the coordinates you reach from a start point on a given bearing and distance, in decimal degrees and DMS, with a downloadable KML placemark.
Destination point
41.3456363, -73.1589503
- Degrees, minutes, seconds
- 41°20'44.29"N 73°9'32.22"W
- Degrees and decimal minutes
- 41°20.7382'N 73°9.5370'W
- Bearing on arrival
- 45.56° NE
- Bearing back to the start
- 225.56° SW
- Distance travelled
- 100.000 km · 62.137 mi · 53.996 nmi
- Angular distance
- 0.89932°
Solved on a sphere of mean radius 6,371.0088 km. The bearing on arrival differs from the bearing you set because a great circle crosses each meridian at a different angle.
Bearing and waypoints along the way
| Along the leg | Distance | Position |
|---|---|---|
| 25% | 25.00 km | 40.871588, -73.795760 |
| 50% | 50.00 km | 41.029993, -73.584510 |
| 75% | 75.00 km | 41.188011, -73.372243 |
| 100% | 100.00 km | 41.345636, -73.158950 |
What is the Destination Point from Bearing and Distance?
Give this calculator a starting coordinate, an initial bearing and a distance and it solves the direct geodesic problem on a sphere: φ₂ = asin(sin φ₁ cos δ + cos φ₁ sin δ cos θ), with the matching longitude from atan2.
- Solves the direct geodesic problem on a sphere of mean radius 6,371.0088 km
- Distance in kilometres, miles, nautical miles, metres or feet
- Destination shown in decimal degrees, DMS and degrees-decimal-minutes
- Reports the arrival bearing and the back bearing, not just your input bearing
- Compass rose drawing plus quarter-way, half-way and three-quarter waypoints
- Downloads a KML file with both the start and the destination placemarked
How to use the Destination Point from Bearing and Distance
- 1
Enter the starting latitude and longitude in decimal degrees or DMS.
- 2
Enter the initial bearing in degrees clockwise from true north, and the distance to travel.
- 3
Pick the distance unit — kilometres, miles, nautical miles, metres or feet.
- 4
Read the destination in decimal degrees, DMS and DDM, with the arrival and back bearings.
- 5
Download the KML placemark, or copy the coordinate pair straight out.
About the Destination Point from Bearing and Distance
Give this calculator a starting coordinate, an initial bearing and a distance and it solves the direct geodesic problem on a sphere: φ₂ = asin(sin φ₁ cos δ + cos φ₁ sin δ cos θ), with the matching longitude from atan2. You get the destination in decimal degrees, degrees-minutes-seconds and degrees-decimal-minutes, plus a KML placemark you can open in Google Earth.
It is the calculation behind plotting a leg from a fix, projecting a survey line, setting a search pattern, or working out where a bearing and range from a radar contact actually puts something. Distance can be entered in kilometres, miles, nautical miles, metres or feet.
Because a great circle crosses each meridian at a different angle, the tool also reports the bearing you arrive on and the bearing back to your start — which are not simply your original bearing plus 180°. Everything is computed locally in your browser and nothing is uploaded.
Frequently asked questions
How do you find a destination from a bearing and distance?
Convert the distance to an angular distance by dividing by the Earth's radius, then apply the spherical law of cosines for the new latitude and atan2 for the new longitude. This tool does that and normalises the longitude back into ±180° so it never returns something like 190°.
Is the bearing true or magnetic?
True — measured from geographic north. To steer it on a magnetic compass you must apply the local magnetic declination, which varies by location and drifts a fraction of a degree each year. Check a current declination model before navigating on it.
Why is the arrival bearing different from the bearing I entered?
On a sphere a great circle is not a straight line relative to the meridians, so your heading drifts as you go. Over a short leg the difference is negligible; over a few thousand kilometres east or west it can be tens of degrees, which is why the tool shows it.
How accurate is a spherical calculation compared with an ellipsoid?
Spherical great-circle maths is typically within about 0.3% of the ellipsoidal Vincenty result — a few hundred metres over 100 km. That is fine for planning and plotting, but for survey-grade work use a geodesic solver on the WGS-84 ellipsoid instead.
Can I use this to draw a circle of a fixed radius?
Yes, indirectly — run the same distance at bearings 0°, 10°, 20° and so on and you get points on a circle of that radius. If you want the whole ring in one step, the KML circle generator builds the polygon for you.
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