--- title: "Reciprocal levelling AB 250 m, RL 1450.500" subject: "Surveying I" chapter: 5 key: "semester_3/surveying_i:reciprocal-leveling-ab-250m-1-252" source: https://www.bctengineer.com/q/semester_3/surveying_i/reciprocal-leveling-ab-250m-1-252 site: BCT Engineering --- # Reciprocal levelling AB 250 m, RL 1450.500 Surveying I · Chapter 5: Leveling - Asked in: 2075 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/reciprocal-leveling-ab-250m-1-252 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5 ## Question The following staff readings were taken during a reciprocal levelling: | Instrument at near | Staff reading on A | Staff reading on B | |---|---|---| | A | 1.252 | 1.052 | | B | 1.419 | 1.253 | If the distance AB is 250 m, compute the RL of B. If RL of A is 1450.500 m, find the combined correction, collimation error and the correct reading for A during the second setup. ## Short answer RL of B = 1450.683 m. Combined correction = 0.0042 m. Collimation error = −0.0212 m in 250 m (downward). Correct reading on A in second setup = 1.436 m. ## Answer **Notation:** instrument near A: staff on A (near) $a_1 = 1.252$, staff on B (far) $b_1 = 1.052$. Instrument near B: staff on A (far) $a_2 = 1.419$, staff on B (near) $b_2 = 1.253$. Rise of B above A $=$ BS $-$ FS. AB = 250 m, R = 6370 km (assumed). #### RL of B $$ H_1 = a_1 - b_1 = 1.252 - 1.052 = +0.200\ \text{m}, \qquad H_2 = a_2 - b_2 = 1.419 - 1.253 = +0.166\ \text{m} $$ $$ H = \frac{H_1 + H_2}{2} = \frac{0.200 + 0.166}{2} = +0.183\ \text{m}\ (\text{B higher than A}) $$ $$ \text{RL of B} = 1450.500 + 0.183 = 1450.683\ \text{m} $$ #### Total error in the far reading $$ e = \frac{H_2 - H_1}{2} = \frac{0.166 - 0.200}{2} = -0.017\ \text{m} $$ The negative sign means the far readings are too small by 0.017 m. #### Combined curvature and refraction correction (250 m) $$ \frac{d^2}{2R} - \frac{1}{7}\frac{d^2}{2R} = \frac{6}{7} \times \frac{0.25^2}{2 \times 6370} \times 1000 = 0.00420\ \text{m} \approx 0.0042\ \text{m} $$ (Curvature and refraction together make the far readings too large, by 0.0042 m.) #### Collimation error $$ \text{Collimation error} = e - (\text{C \& R}) = -0.0170 - 0.0042 = -0.0212\ \text{m in 250 m} $$ That is 8.5 mm per 100 m (about 1 in 11,800); the sign is negative, so the line of collimation is inclined **downward**. #### Correct reading on A in the second set-up The far reading on A (1.419 m) is too small by the total error of 0.017 m. The reading free of collimation, curvature and refraction errors is $$ a_2 - e = 1.419 - (-0.017) = 1.436\ \text{m} $$ (If only the collimation error is removed, with curvature and refraction left in, the reading is $1.419 + 0.0212 = 1.4402$ m.) **Answer:** RL of B = 1450.683 m; combined correction = 0.0042 m; collimation error = −0.0212 m in 250 m (downward); correct reading on A in the second set-up = 1.436 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are H1, H2 and RL of B (RL of A = 1450.500 m)? A. H1 = +0.200, H2 = +0.166, H = +0.183, RL of B = 1450.683 m B. H = +0.366, RL of B = 1450.866 m C. H = -0.183, RL of B = 1450.317 m D. H = +0.017, RL of B = 1450.517 m Answer: A. H = (0.200 + 0.166)/2 = +0.183 m, B higher than A, so RL of B = 1450.500 + 0.183 = 1450.683 m. (Key point: RL of B) 2. What is the total error in the far reading and what does its sign show? A. e = +0.017 m; far readings are too large B. e = (H2 - H1)/2 = -0.017 m; far readings are too small C. e = -0.034 m; far readings are too small D. e = +0.183 m; far readings are too large Answer: B. e = (0.166 - 0.200)/2 = -0.017 m. The negative sign means far readings are too small. (Key point: Total error) 3. Curvature and refraction (0.0042 m) make far readings too large. How is the collimation error then found? A. Collimation = e + (C & R) = -0.0128 m B. Collimation = e = -0.017 m C. Collimation = e - (C & R) = -0.017 - 0.0042 = -0.0212 m in 250 m D. Collimation = C & R - e = +0.0212 m Answer: C. e = collimation + C&R, so collimation = -0.017 - 0.0042 = -0.0212 m. Negative: line of collimation inclined downward. (Key point: Collimation error) 4. What is the correct reading on A in the second set-up, free from all errors? A. 1.402 m B. 1.440 m C. 1.419 m D. 1.436 m Answer: D. Correct reading = a2 - e = 1.419 - (-0.017) = 1.436 m. 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