Destination Longitude from Course and Distance

Also known as destination longitude · longitude reached from bearing and distance · great circle direct problem longitude · point from radial and distance longitude · lat lon given radial and distance · dead reckoning longitude great circle · position after a distance on a bearing · waypoint longitude from course and distance · longitude of a waypoint along a great circle · Aviation Formulary radial and distance

λ2=λ1+atan2(sin⁡θ sin⁡δ cos⁡φ1, cos⁡δ−sin⁡φ1sin⁡φ2),δ=dR\lambda_2 = \lambda_1 + \mathrm{atan2}\left(\sin\theta\,\sin\delta\,\cos\varphi_1,\ \cos\delta - \sin\varphi_1\sin\varphi_2\right),\quad \delta = \frac{d}{R}
sig figs

Worked example: due north along 20 degrees E for 1 000 km → still 20 degrees. Press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Change any unit and the answer follows.

Where Destination Longitude from Course and Distance fits

Missing an input?
Your answer feeds
Learning zone

Destination Longitude from Course and Distance explained

northφ1θdλ2 − λ1λ1λ2

This gives the longitude reached after running a distance dd along the great circle that leaves a start at latitude φ1\varphi_1 and longitude λ1\lambda_1 on an initial true bearing θ\theta. It is the second half of a destination point, the first half being the latitude φ2\varphi_2 from the library's destination-latitude page; here φ2\varphi_2 is found inside the page from the same start, bearing and distance. The distance enters as the angle δ=d/R\delta = d/R at the centre of the Earth. The change in longitude is the angle of a point whose two coordinates are sin⁡θ sin⁡δ cos⁡φ1\sin\theta\,\sin\delta\,\cos\varphi_1 and cos⁡δ−sin⁡φ1sin⁡φ2\cos\delta - \sin\varphi_1\sin\varphi_2; atan2\mathrm{atan2} uses the signs of both to place that angle in the right quadrant, so the result has the right sign as well as the right size.

Watch the sign convention. Ed Williams' Aviation Formulary counts longitude positive to the west, so it subtracts the change from the starting longitude. This page counts longitude positive to the east, as the rest of the library does, so it adds the change instead: λ2=λ1+dlon\lambda_2 = \lambda_1 + \mathrm{dlon}, where dlon is the atan2\mathrm{atan2} term, wrapped into the range from -180 to 180 degrees. Latitude is north positive and the bearing is clockwise from true north.

The bearing is the initial one: along a great circle the course changes as you go, except along the equator and the meridians. Those two exceptions make good checks. Due north along a meridian does not change the longitude, which the formula shows because sin⁡θ\sin\theta is zero for a bearing of 0; the page's example, 1 000 km due north from 40 degrees north on 20 degrees east, returns 20 degrees. Due east along the equator from the Greenwich meridian for a quarter of the Earth's circumference, πR/2\pi R / 2 or 10 007 557.22 m, returns 90 degrees.

The Formulary's own worked example is a waypoint 100 nautical miles out of Los Angeles (33 degrees 57 minutes north, 118 degrees 24 minutes west) on the 66 degree radial (1.150035 radians). It lands at 34 degrees 37 minutes north and 116 degrees 33 minutes west. The page gives -116.5514 degrees, and run backwards from that waypoint it returns the start longitude of -118.4000 degrees. It uses the mean Earth radius of 6 371 008.8 m.

Destination Longitude from Course and Distance formula

λ2=λ1+atan2(sin⁡θ sin⁡δ cos⁡φ1, cos⁡δ−sin⁡φ1sin⁡φ2),δ=dR\lambda_2 = \lambda_1 + \mathrm{atan2}\left(\sin\theta\,\sin\delta\,\cos\varphi_1,\ \cos\delta - \sin\varphi_1\sin\varphi_2\right),\quad \delta = \frac{d}{R}
Where
  • λ2\lambda_2= Longitude reached (°)
  • λ1\lambda_1= Longitude of the start (°)
  • φ1\varphi_1= Latitude of the start (°)
  • θ\theta= Initial true bearing (°)
  • dd= Distance run (nmi)
Where this comes from
  • Williams, E. (2011). Aviation Formulary V1.47, 'Lat/lon given radial and distance', read 23 Oct 2026 at the author's site: 'lat = asin(sin(lat1) cos(d) + cos(lat1) sin(d) cos(tc))', then 'dlon = atan2(sin(tc) sin(d) cos(lat1), cos(d) - sin(lat1) sin(lat))' and 'lon = mod(lon1 - dlon + pi, 2 pi) - pi' (West longitudes positive in that document); and the worked example: a waypoint 100 nm out of LAX (lat1 = 0.592539, lon1 = 2.066470 W) on the radial tc = 1.150035 rad, d = 0.0290888 rad, lands at lat = 0.604180 rad (34 degrees 37 minutes N) and lon = 2.034206 rad (116 degrees 33 minutes W). Read it
Revised 30 October 2026
Cite this page

formula.expert. (2026, October 30). Destination Longitude from Course and Distance Calculator. https://formula.expert/solvers/destination-longitude-from-bearing-distance