--- title: "RL of temple top with refraction correction" subject: "Surveying II" chapter: 3 key: "semester_4/surveying_ii:temple-top-rl-refraction-curvature" source: https://www.bctengineer.com/q/semester_4/surveying_ii/temple-top-rl-refraction-curvature site: BCT Engineering --- # RL of temple top with refraction correction Surveying II · Chapter 3: Trigonometric Leveling - Asked in: 2071 Magh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/temple-top-rl-refraction-curvature - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-3 ## Question The top of temple was sighted from two stations A and B at very different level. The observed vertical angle from A and B to top of temple 'P' were 30°36' and 20°12' respectively. The vertical angle B to A to a vane at 1.5 m above the foot of the vane was 4°15'. The height of instrument at A and B were 1.47 m and 1.42 m. The distance between two instrument station was 112 m. RL of B was 1280.00 m. Find the RL of the top of the temple. Also apply the correction for refraction and curvature. ## Short answer Without correction: axis B = 1281.420 m, axis A = 1289.713 m, x = 259.290 m, RL P = 1376.820 m. Applying C = 0.0673 d^2 gives x = 259.300 m and RL P = 1376.828 m. ## Answer Let B (RL 1280.00 m, lower station) be the farther station and A the nearer one: the sight from B to A is an angle of elevation, so A is higher. D = 112 m is taken as the horizontal distance AB. Curvature and refraction correction: $C = 0.0673\,d^2$ m with $d$ in km (added to the height of a sighted point). #### (a) Without correction - Axis of B $= 1280.000 + 1.42 = 1281.420$ m - Sight B to the vane at A (1.5 m above foot): $\alpha = 4^\circ15'$. Vane level $= 1281.420 + 112\tan 4^\circ15' = 1289.743$ m, ground A $= 1288.243$ m - Axis of A $= 1288.243 + 1.47 = 1289.713$ m Let $x$ = horizontal distance from B to the temple. $$ x = \frac{(1289.713 - 1281.420) - 112\tan 30^\circ36'}{\tan 20^\circ12' - \tan 30^\circ36'} = \frac{8.293 - 66.237}{0.36793 - 0.59140} = 259.290\ \text{m} $$ $$ RL_P = 1281.420 + 259.290\tan 20^\circ12' = 1376.820\ \text{m} $$ #### (b) With curvature and refraction Each calculated height is increased by $0.0673\,(d/1000)^2$. For the sight B to A, $d = 112$ m: $C = 0.0008$ m, so the axis of A becomes 1289.714 m. The height of P must be the same from both stations: $$ 1281.420 + x\tan 20^\circ12' + C(x) = 1289.714 + (x - 112)\tan 30^\circ36' + C(x-112) $$ Solving (the correction terms are small, so a few iterations are enough): $$ x = 259.300\ \text{m}, \qquad C(x) = 0.0045\ \text{m} $$ $$ RL_P = 1281.420 + 259.300\tan 20^\circ12' + 0.005 = 1376.828\ \text{m} $$ **Answer: RL of the top of the temple = 1376.820 m without correction; 1376.828 m after applying curvature and refraction.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The vertical angle from B to the vane at A is an elevation of 4°15'. What does the answer conclude? A. B is higher than A B. A and B are at the same level C. A is higher than B, so B (RL 1280) is the farther station and A the nearer D. The temple is lower than B Answer: C. An upward sight from B means A is higher. B is taken as the farther station. (Key point: Which station is higher) 2. What is the curvature and refraction correction C = 0.0673 d^2 (d in km) for d = 112 m? A. 0.0075 m B. 0.0008 m C. 0.84 m D. 8.44 m Answer: B. d = 0.112 km, so C = 0.0673 x 0.012544 = 0.0008 m. (Key point: Size of the correction) 3. How do the corrections change the RL of the temple top? A. They are subtracted, so the RL is lower B. They make no difference C. They raise it by about 1 m D. They are added to the heights, so the RL rises slightly: 1376.828 m against 1376.820 m Answer: D. The correction is added to each computed height, increasing the RL by about 0.008 m here. (Key point: Effect on the RL) 4. Why is x found by a few iterations when corrections are applied? A. Because the correction C depends on the distance x, which is itself unknown B. Because the angles are unknown C. Because the HI is unknown D. Because the distance D is unknown Answer: A. C(x) and C(x - 112) contain x. Since they are small, a first value of x is refined a few times. 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