9.2 Determination of Latitude, Longitude, and Solar/Polaris Azimuth by Astronomical Observations

9.2 Determination of Latitude, Longitude, and Solar/Polaris Azimuth by Astronomical Observations

Field determinations of latitude ($\phi$), longitude ($\lambda$), and true azimuth ($A$) serve as independent checks on terrestrial triangulation and GNSS networks, eliminating cumulative baseline orientation drift.

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1. The Astronomical Triangle ($P-Z-S$)

The spherical triangle formed on the celestial sphere by the North Celestial Pole ($P$), the observer's Zenith ($Z$), and the celestial body ($S$) is known as the Astronomical Triangle ($P-Z-S$).

Vertices, Sides, and Angles

Fundamental Spherical Trigonometry Formulas

Applying the spherical Law of Cosines to the $P-Z-S$ triangle yields the core equations of astronomical positioning:

$$\cos z = \sin \phi \sin \delta + \cos \phi \cos \delta \cos t$$ $$\sin \delta = \sin \phi \sin h + \cos \phi \cos h \cos A$$

Solving for Azimuth Angle ($\cos A$):

$$\cos A = \frac{\sin \delta - \sin \phi \sin h}{\cos \phi \cos h}$$

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2. Determination of Latitude ($\phi$) by Meridian Altitude

Observing a star or the Sun at the instant it crosses the observer's principal meridian ($t = 0^h$ or $t = 12^h$) provides the simplest and most precise direct method for latitude determination. At transit, the astronomical triangle collapses into a single vertical meridian arc.

Latitude Formulas by Transit Type

1. Upper Transit South of Zenith ($\delta < \phi$): $$\phi = \delta + z = \delta + (90^\circ - h)$$
2. Upper Transit North of Zenith ($\delta > \phi$): $$\phi = \delta - z = \delta - (90^\circ - h)$$
3. Lower Transit of Circumpolar Star ($t = 12^h$): $$\phi = h + (90^\circ - \delta) = h + p$$

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3. Solar Azimuth Determination Methods

Determining true azimuth using solar observations requires measuring horizontal angles between a terrestrial survey line (mark) and the Sun, alongside vertical altitude readings or precise time records.

Method A: Altitude Method (Solar Altitude)

The Altitude Method requires measuring the Sun's vertical altitude $h$, horizontal angle to the mark, and extracting solar declination $\delta$ from an ephemeris for the instant of observation.

$$\cos A = \frac{\sin \delta - \sin \phi \sin h}{\cos \phi \cos h}$$

• Advantage: Does not require split-second time sync (time only needed to interpolate solar declination $\delta$).
• Disadvantage: Highly sensitive to vertical angle errors and atmospheric refraction near the horizon. Best performed when solar altitude is between $20^\circ$ and $40^\circ$.

Method B: Hour Angle Method (Solar Hour Angle)

The Hour Angle Method requires precise epoch recording (accurate to $\le 0.1^s$) of the solar observation, calculating the local hour angle $t$.

$$\tan A = \frac{\sin t}{\sin \phi \cos t - \cos \phi \tan \delta}$$

• Advantage: Independent of vertical angle measurement errors and atmospheric refraction.
• Disadvantage: Requires precise epoch timing (a $1.0^s$ clock error produces up to $15''$ azimuth error at equatorial latitudes).

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4. Polaris Observations for True Azimuth

Polaris ($\alpha$ Ursae Minoris) is the preferred celestial body for high-precision azimuth determination in the northern hemisphere because of its proximity to the North Celestial Pole ($\delta \approx +89^\circ 20'$, polar distance $p \approx 0.67^\circ$).

Polaris at Elongation (Eastern / Western)

Elongation occurs when Polaris reaches its maximum angular distance east or west of the north meridian. At elongation, the parallactic angle $\eta = 90^\circ$, making the vertical circle tangent to the hour circle.

$$\sin A_{\text{max}} = \frac{\sin p}{\cos \phi} = \frac{\cos \delta}{\cos \phi}$$ $$\cos t_{\text{elong}} = \frac{\tan \phi}{\tan \delta}$$

• Practical Significance: At elongation, Polaris moves almost purely vertically for several minutes, allowing the engineer ample time to bisect the star with high accuracy without azimuth changing rapidly.

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5. Field Corrections to Astronomical Observations

Observed celestial angles must be corrected for atmospheric and geometric effects prior to solving the $P-Z-S$ triangle.

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6. Worked Computational Example

Problem: A survey team in Clark, Pampanga ($\phi = 15^\circ 10' 00'' \text{ N}$) performs a solar altitude observation to determine the true azimuth of line AB.

• Corrected solar altitude: $h = 32^\circ 15' 00''$
• Solar declination from ephemeris: $\delta = +18^\circ 45' 00''$ Calculate the true azimuth ($A$) of the Sun.

Solution:

Step 1: Identify Input Parameters

• $\phi = 15^\circ 10' 00'' \implies \sin \phi = 0.261628, \cos \phi = 0.965169$
• $h = 32^\circ 15' 00'' \implies \sin h = 0.533615, \cos h = 0.845726$
• $\delta = +18^\circ 45' 00'' \implies \sin \delta = 0.321439$

Step 2: Compute Terms of the Altitude Azimuth Formula

$$\cos A = \frac{\sin \delta - \sin \phi \sin h}{\cos \phi \cos h}$$

Numerator:

$$\text{Num} = 0.321439 - (0.261628 \times 0.533615) = 0.321439 - 0.139609 = 0.181830$$

Denominator:

$$\text{Den} = 0.965169 \times 0.845726 = 0.816268$$

Step 3: Solve for $\cos A$ and $A$

$$\cos A = \frac{0.181830}{0.816268} = 0.222758$$ $$A = \arccos(0.222758) = 77.1264^\circ = 77^\circ 07' 35''$$

Since the observation was taken in the morning (Sun East of Meridian), the True Solar Azimuth is $A = 77^\circ 07' 35''$ (clockwise from North).