Question

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.

Hint:

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.

Answer 1

Correct Answer: 12250

Step 1: Recall the reduction-to-ellipsoid formula.
A horizontal distance measured at an average elevation \(h\) above the reference ellipsoid needs to be reduced to the equivalent distance on the ellipsoid surface, because the measured line is effectively a chord at radius \((R+h)\) while the required distance corresponds to radius \(R\). For a small elevation compared to the Earth's radius, the reduced distance is \[ D_0 = D \times \frac{R}{R+h} \] where \(D\) is the measured horizontal distance, \(R\) is the mean radius of the Earth, and \(h\) is the average elevation of the line above the ellipsoid.
Step 2: Substitute the given values.
Here \(D = 12380.56\) m, \(h = 748.82\) m, and \(R = 6378 \text{ km} = 6{,}378{,}000\) m. So \[ D_0 = 12380.56 \times \frac{6378000}{6378000+748.82} = 12380.56 \times \frac{6378000}{6378748.82} \]
Step 3: Evaluate the reduction factor.
\[ \frac{6378000}{6378748.82} = 1 - \frac{748.82}{6378748.82} = 1 - 0.00011740 = 0.99988260 \]
Step 4: Compute the reduced distance.
\[ D_0 = 12380.56 \times 0.99988260 = 12380.56 - 12380.56\times0.00011740 \] \[ 12380.56 \times 0.00011740 \approx 1.45 \text{ m} \] \[ D_0 \approx 12380.56 - 1.45 = 12379.11 \text{ m} \]
Step 5: State the result.
\[ \boxed{D_0 \approx 12379.11 \text{ m}} \]