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Aerospace / Navigation · Initial great-circle heading between two points

Parameters

InputSymbolUnitDefaultDescription
lat1lat1deg0.0Start latitude
lon1lon1deg0.0Start longitude
lat2lat2deg1.0End latitude
lon2lon2deg1.0End longitude
OutputSymbolUnitDescription
hdgψdegHeading

The science & history

Understanding the Parameters

Derivation (Approaching a Proof)

The initial bearing is the angle at the origin vertex of the spherical triangle formed by the origin, the destination, and the North Pole. Work it out with vectors, which cleanly yields the $\operatorname{atan2}$ arguments.

At the origin, set up the local tangent-plane axes: north (toward the pole along the meridian) and east (perpendicular, toward increasing longitude). The bearing $\psi$ is the angle from north to the initial direction of the great circle, measured toward east:

$$\tan\psi = \frac{(\text{east component of the initial direction})}{(\text{north component})}.$$

The initial direction of the great circle toward the destination, projected onto these axes, gives (from spherical trigonometry / the derivative of the position along the geodesic):

$$\text{east} = \sin\Delta\lambda\,\cos\phi_2, \qquad \text{north} = \cos\phi_1\sin\phi_2 - \sin\phi_1\cos\phi_2\cos\Delta\lambda,$$

where $\Delta\lambda = \lambda_2 - \lambda_1$. The east component is largest when the destination is due east and far in longitude; the north component is the "how much more poleward" term. Taking the four-quadrant arctangent:

$$\psi = \operatorname{atan2}(\text{east}, \text{north}) = \operatorname{atan2}\!\big(\sin\Delta\lambda\cos\phi_2,\; \cos\phi_1\sin\phi_2 - \sin\phi_1\cos\phi_2\cos\Delta\lambda\big),$$

and finally normalise into $[0^\circ,360^\circ)$ by adding $360^\circ$ and taking the modulus. $\qquad\blacksquare$

Sanity checks. Two points on the equator ($\phi_1=\phi_2=0$) give north $= 0$ and east $=\sin\Delta\lambda$, so $\psi = 90^\circ$ (due east) for an eastward destination — correct. A destination due north on the same meridian ($\Delta\lambda=0$, $\phi_2>\phi_1$) gives east $=0$ and north $>0$, so $\psi=0^\circ$ — correct.

Dimensional check. Both $\operatorname{atan2}$ arguments are products of sines/cosines (dimensionless), so $\psi$ is a pure angle ✓.

History and Development

Related Concepts: Great-Circle Distance, Cross Track Error, Magnetic Heading, Time To Go, Ground Track Velocity

Notes: Registry latex is the placeholder $\operatorname{atan2}(\ldots)$; real form is the four-quadrant arctangent above. True bearing, clockwise from north, $[0^\circ,360^\circ)$; apply declination (Magnetic Heading) for magnetic. Initial only — a great circle's heading changes continuously along the route (unlike a constant-heading rhumb line, which is longer). $\operatorname{atan2}$ (not $\arctan$) for correct quadrant. Inputs in degrees.

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