Slope Distance to Horizontal Distance: 18-Geom-A1 Surveying, May 2019, Q3

Question 3: Slope Distance to Horizontal Distance (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Vertical (zenith-complement) angle $\alpha = +3^\circ27'30''$; slope distance $S = 17728.947$ m; ellipsoidal height of A, $h_A = 1000.55$ m.

Find. The horizontal distance $H$ from A to B.

Vertical section A–B: the measured slope distance S and vertical angle α resolve into the horizontal distance H = S cos α.

Approach. Resolve the slope distance onto the horizontal with the vertical angle; then note (Check) the second-order reduction to the ellipsoid that the given height enables.

Convert the vertical angle to decimal degrees. $$ \alpha = 3 + \tfrac{27}{60} + \tfrac{30}{3600} = 3.45833^\circ. $$
Project the slope distance onto the horizontal. With the vertical angle measured at the instrument, $$ H = S\cos\alpha = 17728.947\,\cos(3.45833^\circ) = 17728.947 \times 0.998179 $$ $$ \boxed{H \approx 17696.66\text{ m}} $$

Where the ellipsoidal height enters. $H = S\cos\alpha$ is the horizontal distance in the plane through A, and it does not depend on $h_A$. On a 17.7 km line the Earth’s curvature makes that plane distance differ from the geodetic distance on the ellipsoid, and the given height is what allows the reduction. B lies about $S\sin\alpha = 1069.46$ m above A, plus about 24.6 m of curvature, so $h_B \approx 2094.6$ m. The standard chord reduction ($R \approx 6\,371\,000$ m) then gives $$ c_0 = \sqrt{\frac{S^2-\Delta h^2}{(1+h_A/R)(1+h_B/R)}} = 17690.86\text{ m}, \qquad S_{ell} = 2R\sin^{-1}\!\frac{c_0}{2R} \approx 17690.87\text{ m}. $$ The question asks for the horizontal distance, so $H$ is boxed. The ellipsoid distance is what a candidate should quote if the examiner intends a geodetic (mapping) distance.

Quantity	Value
Vertical angle $\alpha$	$3.45833^\circ$
Horizontal distance $H = S\cos\alpha$	$17696.66$ m
Height of B, $h_B$ (incl. curvature)	$\approx 2094.6$ m
Distance reduced to the ellipsoid, $S_{ell}$	$\approx 17690.87$ m

Concept & Theory

Topic. Reduction of an EDM slope distance to horizontal (and onward to the ellipsoid).

Key relations. $H = S\cos\alpha$ (vertical angle) or $H = S\sin z$ (zenith angle $z$); ellipsoidal reduction $S_{ell}=H\,R/(R+h_m)$.

Why this works. The total station measures the straight-line (slope) range and the vertical angle in the same vertical plane, so the horizontal leg is simply the slope times the cosine of the vertical angle. The height correction accounts for the fact that a horizontal arc at elevation is longer than its projection onto the reference ellipsoid.

Common pitfall. Mixing a vertical angle (from horizontal) with a zenith angle (from the plumb line): $H=S\cos\alpha=S\sin z$. Neglecting Earth curvature and refraction on very long trigonometric-levelling lines is acceptable for the horizontal distance but not for the height difference.

Source. Wolf & Ghilani, Elementary Surveying, 15th ed., §6.13–6.14 (reduction of EDM distances).

Question 3 of 9: Slope Distance to Horizontal Distance

Question 3: Slope Distance to Horizontal Distance (10 marks)