In a triangulation exercise, the horizontal distance between two points P and Q was found to be 12380.56 m. The average elevation along the line PQ was 748.82 m above the reference ellipsoid. The reduced horizontal distance between P and Q over the reference ellipsoid is ______ m (Rounded off to two decimal places). Consider the radius of Earth to be 6378 km.
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A distance measured at an elevation h above the ellipsoid must be scaled down by the factor R divided by (R plus h) to get the equivalent distance on the ellipsoid.
Step 1: Understand why elevation shrinks a measured distance. Picture the Earth as a sphere of radius $R$ centred at $O$. A line measured at the ground surface, at height $h$ above the reference ellipsoid, subtends the same central angle $\alpha$ at $O$ as the corresponding segment on the ellipsoid itself, because both are, to first order, radiating segments over the same small angle. Using length $\approx$ radius $\times$ angle: \[ D = (R+h)\,\alpha , \qquad D_0 = R\,\alpha \] Dividing the two relations eliminates the unknown angle $\alpha$: \[ \frac{D_0}{D} = \frac{R}{R+h} \implies D_0 = D\cdot\frac{R}{R+h} \] Step 2: Plug in the numbers. $D = 12380.56$ m, $R = 6378\times10^3$ m, $h = 748.82$ m. \[ D_0 = 12380.56 \times \frac{6378000}{6378748.82} \] Step 3: Compute the correction as a small fractional loss. Since $h \ll R$, the fractional reduction is approximately $h/R$: \[ \frac{h}{R} = \frac{748.82}{6378000} = 1.1740\times10^{-4} \] The distance lost due to elevation is \[ D \times \frac{h}{R} = 12380.56 \times 1.1740\times10^{-4} = 1.45 \text{ m} \] Step 4: Subtract this loss from the measured distance. \[ D_0 = D - 1.45 = 12380.56 - 1.45 = 12379.11 \text{ m} \] Step 5: Final answer. \[ \boxed{D_0 \approx 12379.11 \text{ m}} \]
