--- title: "Level book, elevation of B 1120.450" subject: "Surveying I" chapter: 5 key: "semester_3/surveying_i:level-book-b-1120-450-4m-staff-30m" source: https://www.bctengineer.com/q/semester_3/surveying_i/level-book-b-1120-450-4m-staff-30m site: BCT Engineering --- # Level book, elevation of B 1120.450 Surveying I · Chapter 5: Leveling - Asked in: 2074 Asoj, 8 marks; 2063 Baisakh, 10 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/level-book-b-1120-450-4m-staff-30m - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5 ## Question The following consecutive readings were taken with a level and a 4 m levelling staff on continuously sloping ground at a common interval of 30 m. 0.585 on A, 0.936, 1.953, 2.846, 3.644, 3.938, 0.962, 1.035, 1.689, 2.534, 3.844, 0.956, 1.579, 3.016 on B. The elevation of B was 1120.450. Make up the level field book and apply the usual checks. Find the gradient between the first and last point. ## Short answer RL of A = 1128.745 m (obtained by adding ΣFS–ΣBS to RL of B). Distance between A and B is 330 m (11 intervals of 30 m). Fall = 8.295 m, giving a falling gradient of 1 in 39.8. ## Answer **Reading the data.** The staff is 4 m long, so the readings 3.938 and 3.844 (nearly 4 m) show where the instrument was shifted. The sequence is therefore: - Setup 1: BS 0.585 on A; IS 0.936, 1.953, 2.846, 3.644; FS 3.938 on change point CP1. - Setup 2: BS 0.962 on CP1; IS 1.035, 1.689, 2.534; FS 3.844 on change point CP2. - Setup 3: BS 0.956 on CP2; IS 1.579; FS 3.016 on B. The ground slopes continuously down (the readings increase). The RL of B is known, so the RL of A is found first: $$ \text{RL of A} = \text{RL of B} + (\Sigma FS - \Sigma BS) = 1120.450 + (10.798 - 2.503) = 1128.745\ \text{m} $$ #### Level field book (rise and fall method) | Station | BS | IS | FS | Rise | Fall | RL (m) | |---|---|---|---|---|---|---| | A | 0.585 | | | | | 1128.745 | | 2 | | 0.936 | | | 0.351 | 1128.394 | | 3 | | 1.953 | | | 1.017 | 1127.377 | | 4 | | 2.846 | | | 0.893 | 1126.484 | | 5 | | 3.644 | | | 0.798 | 1125.686 | | 6 (CP1) | 0.962 | | 3.938 | | 0.294 | 1125.392 | | 7 | | 1.035 | | | 0.073 | 1125.319 | | 8 | | 1.689 | | | 0.654 | 1124.665 | | 9 | | 2.534 | | | 0.845 | 1123.820 | | 10 (CP2) | 0.956 | | 3.844 | | 1.310 | 1122.510 | | 11 | | 1.579 | | | 0.623 | 1121.887 | | B | | | 3.016 | | 1.437 | 1120.450 | #### Arithmetic check $\Sigma BS - \Sigma FS = 2.503 - 10.798 = -8.295$; $\Sigma \text{Rise} - \Sigma \text{Fall} = 0.000 - 8.295 = -8.295$; last RL $-$ first RL $= 1120.450 - 1128.745 = -8.295$. All three are equal, so the reduction is correct. Working down the rises and falls from A brings the RL back to the given value of B (1120.450 m), which confirms the arithmetic of the table. #### Gradient between the first and last point The FS and BS at a change point are taken on the same point, so the 14 readings fall on 12 points at 30 m spacing: 11 intervals, a distance of $11 \times 30 = 330$ m. $$ \text{Fall} = 1128.745 - 1120.450 = 8.295\ \text{m}, \qquad \text{Gradient} = \frac{8.295}{330} = \frac{1}{39.8} $$ **Answer: RL of A = 1128.745 m; falling gradient from A to B of about 1 in 40 (1 in 39.8).** *Note:* if each of the 14 readings were counted as a separate 30 m station (13 intervals, 390 m), the gradient would be 1 in 47.0. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the level book, how are the change points identified from the readings on a 4 m staff? A. Readings close to 4 m (3.938, 3.844) show the instrument was shifted, so they are FS readings on change points B. The smallest readings are always the change points C. Every fifth reading is a change point D. The readings equal to 1.000 m mark the change points Answer: A. A reading of nearly 4 m is at the top limit of the staff, so the next reading must come from a new setup; these are FS on change points. (Key point: Identifying change points) 2. How is the RL of the first point A found when only the RL of B is given? A. RL of A = RL of B - sum BS B. RL of A = RL of B + (sum FS - sum BS) C. RL of A = RL of B - (sum FS - sum BS) D. RL of A = RL of B + sum of all readings Answer: B. Last RL - first RL = sum BS - sum FS, so first RL = last RL + (sum FS - sum BS) = 1120.450 + 8.295 = 1128.745 m. (Key point: Finding RL of A) 3. Which three quantities must be equal in the arithmetic check of the rise and fall method? A. Sum IS - sum FS, sum Fall, first RL B. Sum BS, sum FS, sum IS C. Sum BS - sum FS, sum Rise - sum Fall, last RL - first RL D. Sum BS + sum FS, sum IS, sum Rise Answer: C. All three equal -8.295 m here, which proves the reductions are arithmetically correct. (Key point: Arithmetic check) 4. Why is the distance between A and B taken as 11 x 30 = 330 m although there are 14 readings? A. Three readings are rejected as mistakes B. Each setup adds no distance C. The staff length reduces the distance D. The FS and BS at a change point are on the same point, so the 14 readings fall on 12 points (11 intervals) Answer: D. Two change points each use two readings on one point, so 14 readings give 12 points and 11 intervals of 30 m. 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