--- title: "RL of B for bridge axis AB (58.60 m)" subject: "Surveying I" chapter: 5 key: "semester_3/surveying_i:bridge-axis-ab-58-60m-rl-b" source: https://www.bctengineer.com/q/semester_3/surveying_i/bridge-axis-ab-58-60m-rl-b site: BCT Engineering --- # RL of B for bridge axis AB (58.60 m) Surveying I · Chapter 5: Leveling - Asked in: 2076 Asoj, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/bridge-axis-ab-58-60m-rl-b - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5 ## Question Determine the RL of station B of a bridge axis AB of axis length 58.60 m from the following information. If RL of station A was 1295 m. Apply necessary check. | Instrument near to | Sighted to | Staff reading T | M | B | |---|---|---|---|---| | A | A | 1.57 | 1.559 | 1.548 | | A | B | 1.9 | 1.585 | 1.271 | | B | A | 1.96 | 1.647 | 1.335 | | B | B | 1.685 | 1.671 | 1.659 | ## Short answer True difference of level is the mean of two apparent differences: −0.025 m, so B is lower than A. RL of B = 1294.975 m. ## Answer The axis AB is crossed by reciprocal levelling: two sets of readings are taken, one with the instrument near A and one near B. The mean of the two apparent differences gives the true difference of level, free from collimation, curvature and refraction errors. #### Check on the hair readings For each pair of readings, the middle hair should equal the mean of the top and bottom hairs: | Instrument near | Staff on | Top | Middle | Bottom | (T + B)/2 | Remark | |---|---|---|---|---|---|---| | A | A | 1.570 | 1.559 | 1.548 | 1.559 | agrees | | A | B | 1.900 | 1.585 | 1.271 | 1.5855 | agrees (0.5 mm) | | B | A | 1.960 | 1.647 | 1.335 | 1.6475 | agrees (0.5 mm) | | B | B | 1.685 | 1.671 | 1.659 | 1.672 | agrees (1 mm) | The hair readings are consistent, so the middle-hair readings are used. #### Apparent differences (rise of B above A $=$ reading on A $-$ reading on B) $$ H_1 = 1.559 - 1.585 = -0.026\ \text{m} \quad (\text{instrument near A}) $$ $$ H_2 = 1.647 - 1.671 = -0.024\ \text{m} \quad (\text{instrument near B}) $$ The two values agree within 2 mm, so the work is acceptable. #### True difference and RL of B $$ H = \frac{H_1 + H_2}{2} = \frac{-0.026 + (-0.024)}{2} = -0.025\ \text{m}\ (\text{B is lower than A}) $$ $$ \text{RL of B} = 1295.000 - 0.025 = 1294.975\ \text{m} $$ **Check:** the error in the far reading is $e = (H_2 - H_1)/2 = +0.001$ m, which is only 1 mm in about 60 m, so the instrument is well adjusted. The mean of the top and bottom hairs gives $-0.0265$ and $-0.0245$, mean $-0.0255$, i.e. RL of B = 1294.9745 m, which agrees with 1294.975 m to 0.5 mm. **Answer: RL of B = 1294.975 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What check is applied to the top, middle and bottom hair readings? A. Middle hair should equal the mean of top and bottom hairs B. Top hair should equal bottom hair C. Sum of the three hairs should be zero D. Middle hair should equal top hair minus bottom hair Answer: A. For a correctly read staff, M = (T + B)/2, within about 1 mm; this detects reading and booking mistakes. (Key point: Hair check) 2. What are H1 and H2 from the middle hair readings? A. H1 = +0.026 m and H2 = +0.024 m B. H1 = 1.559 - 1.585 = -0.026 m and H2 = 1.647 - 1.671 = -0.024 m C. H1 = 1.585 - 1.559 = +0.026 m and H2 = -0.024 m D. H1 = -0.026 m and H2 = -0.251 m Answer: B. Rise of B above A = reading on A - reading on B for each set; both are small falls. (Key point: Apparent differences) 3. What is RL of B (RL of A = 1295 m)? A. 1295.025 m B. 1294.950 m C. 1294.975 m (mean H = -0.025 m, B lower) D. 1295.000 m Answer: C. H = (-0.026 - 0.024)/2 = -0.025 m, so RL of B = 1295.000 - 0.025 = 1294.975 m. (Key point: RL of B) 4. The far-reading error e = (H2 - H1)/2 = +0.001 m over about 60 m. What does this show? A. The work must be rejected B. Curvature has not been corrected C. The RL of A is wrong D. The instrument is well adjusted and the two sets agree Answer: D. A 1 mm discrepancy over about 60 m is tiny, so the two sets agree and the result is acceptable. 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