Training Navigation Coordinates, Bearings & Distance
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Coordinates, Bearings & Distance

30 min Navigation

Coordinates, Bearings, and Distance

Navigation starts with a position. On a flat map we use $(x, y)$ cartesian coordinates; on the Earth we use latitude $\phi$ (north–south, 0° at equator, ±90° at poles) and longitude $\lambda$ (east–west, 0° at Greenwich, ±180°). A bearing (or heading) is the clockwise angle from north, measured 0°–360°, with north = 0°, east = 90°, south = 180°, west = 270°.

Over short distances, a flat-earth approximation works fine: meters-east and meters-north are just $\Delta\lambda R\cos\phi$ and $\Delta\phi R$, where $R \approx 6371$ km is Earth's mean radius. For long-range work (trans-oceanic flights, ships), we must use the spherical haversine formula to get true great-circle distance.

We introduce haversine, bearing between two points, and practice converting between latitude/longitude pairs and local east-north-up (ENU) offsets.

Haversine Distance

$$a = \sin^2\!\left(\frac{\Delta\phi}{2}\right) + \cos\phi_1\cos\phi_2\sin^2\!\left(\frac{\Delta\lambda}{2}\right)$$

$$d = 2R\,\arcsin(\sqrt{a})$$

with $\phi$ in radians, $R = 6371\,\text{km}$.

Initial Bearing (forward azimuth)

$$\theta = \operatorname{atan2}\bigl(\sin\Delta\lambda\cos\phi_2,\; \cos\phi_1\sin\phi_2 - \sin\phi_1\cos\phi_2\cos\Delta\lambda\bigr)$$

Result is in radians; convert to compass bearing by $(\theta \bmod 2\pi) \cdot 180/\pi$.

Example 1 — Short-range flat approximation

From $A=(40.000°\,\text{N}, 105.000°\,\text{W})$ to $B=(40.010°\,\text{N}, 104.990°\,\text{W})$, estimate the distance in meters.

$\Delta\phi = 0.010° = 1.745\times 10^{-4}\,\text{rad}$; north offset = $R\Delta\phi$ = $6.371\times 10^6 \cdot 1.745\times 10^{-4} \approx 1112\,\text{m}$.

$\Delta\lambda = 0.010°$; east offset = $R\cos\phi\,\Delta\lambda = 6.371\times 10^6 \cdot \cos(40°) \cdot 1.745\times 10^{-4} \approx 852\,\text{m}$.

Distance $\approx \sqrt{1112^2 + 852^2} \approx 1401\,\text{m}$.

Example 2 — Haversine between NYC and London

NYC = $(40.713°\,\text{N}, 74.006°\,\text{W})$, London = $(51.507°\,\text{N}, 0.128°\,\text{W})$. Find the great-circle distance.

Convert to radians: $\phi_1=0.7106$, $\phi_2=0.8990$, $\Delta\phi=0.1884$, $\Delta\lambda=1.2891$.

$a = \sin^2(0.0942) + \cos(0.7106)\cos(0.8990)\sin^2(0.6445)$

$= (0.0940)^2 + (0.7559)(0.6227)(0.6008)^2 = 0.00884 + 0.16991 = 0.17875$.

$d = 2(6371)\arcsin(\sqrt{0.17875}) = 12742 \cdot \arcsin(0.4228) = 12742 \cdot 0.4363 \approx 5559\,\text{km}$.

Example 3 — Initial bearing NYC → London

Using the points from Example 2, find the initial bearing from NYC to London.

$\sin\Delta\lambda\cos\phi_2 = \sin(1.2891)(0.6227) = 0.9607 \cdot 0.6227 = 0.5983$.

$\cos\phi_1\sin\phi_2 - \sin\phi_1\cos\phi_2\cos\Delta\lambda = 0.7559(0.7823) - 0.6547(0.6227)(0.2773) = 0.5914 - 0.1130 = 0.4784$.

$\theta = \operatorname{atan2}(0.5983, 0.4784) = 0.896$ rad $\approx 51.3°$. Bearing NE — quarter-turn toward the north-east.

Interactive Demo: Haversine Distance & Bearing
distance =5570km
= 3008nmi
initial bearing =51.2°
compass rose ≈NE

Practice Problems

1. What is the compass bearing due south?
2. Compute the meters-north for a 0.005° change in latitude.
3. If $\phi = 60°$, how many meters is $0.001°$ of longitude?
4. State the haversine formula.
5. Why must we use cosine of latitude when converting longitude to distance?
6. Name one situation where the flat-earth approximation breaks down.
Show Answer Key

1. 180°.

2. $R\cdot 0.005\cdot\pi/180 = 6371000\cdot 8.727\times 10^{-5} \approx 556\,\text{m}$.

3. $R\cos 60°\cdot 1.745\times 10^{-5} \approx 55.6\,\text{m}$.

4. $d = 2R\arcsin\sqrt{\sin^2(\Delta\phi/2) + \cos\phi_1\cos\phi_2\sin^2(\Delta\lambda/2)}$.

5. Meridians converge toward the poles, so an arc of longitude covers less distance at higher latitude.

6. Transoceanic or polar routes — great-circle curvature introduces multi-percent errors over thousands of km.

Dead Reckoning & Position Integration