Celestial Navigation
Celestial Navigation — Altitude, Azimuth, Position
For centuries, mariners found their position by measuring the angle of celestial bodies above the horizon. A sextant gives the observed altitude $H_o$; the almanac gives the declination $\delta$ and Greenwich hour angle (GHA) of the body at the observation time. From these three numbers and the navigator's dead-reckoning estimate, one can compute a line of position (LOP).
The key equation is the altitude formula: $\sin H_c = \sin\phi\sin\delta + \cos\phi\cos\delta\cos(\text{LHA})$, where LHA = GHA $-$ longitude (east positive). The difference $H_o - H_c$ (in arc-minutes) is the intercept: one minute equals one nautical mile toward or away from the body's subpoint.
Two or more LOPs — from different bodies, or the same body observed an hour apart — cross at the fix. We will work one full reduction: compute $H_c$, the azimuth $Z$, and plot the intercept.
$$\sin H_c = \sin\phi\sin\delta + \cos\phi\cos\delta\cos(\text{LHA})$$
$$\cos Z = \frac{\sin\delta - \sin\phi\sin H_c}{\cos\phi\cos H_c}$$
Azimuth $Z$ is measured from north; ambiguity resolved by sign of sin(LHA).
1. Assume DR position $(\phi_D, \lambda_D)$. 2. Compute $H_c$ and $Z$ at this position. 3. Observe $H_o$. 4. Intercept = $H_o - H_c$ in arc-min = distance in nmi: positive = toward the body's subpoint, negative = away.
At local apparent noon the sun's observed altitude is $H_o = 45° 30'$ and the sun's declination is $\delta = +12° 00'$. Find latitude.
At local noon, the sun is due south (or north); LHA = 0 and the altitude equation collapses to $H_c = 90° - |\phi - \delta|$, so $|\phi - \delta| = 90° - H_o = 44° 30'$.
Assuming observer is north of the sun's subpoint: $\phi = \delta + 44° 30' = 12° + 44° 30' = 56° 30'\,\text{N}$.
DR position $\phi = 40°\,\text{N}$, $\lambda = 70°\,\text{W}$. Star has $\delta = 30°\,\text{N}$, GHA = $100°$. Find $H_c$.
LHA = GHA − λ = 100° − (−70°) = 170°. Wait: λ west is negative in convention; LHA = GHA + λ (east positive) = 100 + (−70) = 30°.
$\sin H_c = \sin 40°\sin 30° + \cos 40°\cos 30°\cos 30°$
$= (0.6428)(0.5) + (0.7660)(0.8660)(0.8660) = 0.3214 + 0.5744 = 0.8958$.
$H_c = \arcsin 0.8958 \approx 63.6° = 63° 36'$.
If $H_o$ in Example 2 is $63° 48'$, compute intercept and direction.
$H_o - H_c = 63°48' - 63°36' = +12'$. Intercept = $+12\,\text{nmi}$ toward the star's subpoint.
Practice Problems
Show Answer Key
1. $\sin H_c = \sin\phi\sin\delta + \cos\phi\cos\delta\cos(\text{LHA})$.
2. The body is on the observer's meridian, so $\cos(\text{LHA}) = 1$ and the equation reduces to $H_c = 90° - |\phi - \delta|$.
3. Intercept is 15 nmi away from the body's subpoint.
4. From true north (or sometimes south in the Southern Hemisphere), measured clockwise.
5. It tabulates the declination and GHA of the sun, moon, planets, and 57 stars at hourly intervals.
6. Independence from GPS jamming/spoofing; backup for submarines and military aviation.