--- title: "Level book, 20 m interval, RL 1280 to 1279.924" subject: "Surveying I" chapter: 5 key: "semester_3/surveying_i:level-book-20m-interval-1280-1279-924" source: https://www.bctengineer.com/q/semester_3/surveying_i/level-book-20m-interval-1280-1279-924 site: BCT Engineering --- # Level book, 20 m interval, RL 1280 to 1279.924 Surveying I · Chapter 5: Leveling - Asked in: 2078 Kartik, 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/level-book-20m-interval-1280-1279-924 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5 ## Question The following staff readings were taken during a levelling operation at a common interval of 20 m: 1.253, 1.752, 1.005, 0.675, 1.998, 0.825, 1.737, 1.444, 1.619, 0.750 and 2.619 m. The instrument is shifted after the 4th, 6th and 9th readings. The RL of the starting station is 1280 and that of the end station is 1279.924 m respectively. Compute RL, apply necessary check and adjust the RL of each station by any method. ## Short answer Rise and fall computation gives closing error +0.076 m (computed end RL 1280.000 m vs. known 1279.924 m). Adjustment proportional to distance yields adjusted RLs: 1280.000, 1279.490, 1280.226, 1280.545, 1281.708, 1281.990, 1281.804, 1279.924 m. ## Answer **Arranging the readings.** The instrument was shifted after the 4th, 6th and 9th readings, so there are four set-ups: - Set-up 1: BS 1.253 on station 1; IS 1.752, 1.005; FS 0.675 on CP1 (4th reading). - Set-up 2: BS 1.998 on CP1; FS 0.825 on CP2 (6th reading). - Set-up 3: BS 1.737 on CP2; IS 1.444; FS 1.619 on CP3 (9th reading). - Set-up 4: BS 0.750 on CP3; FS 2.619 on the end station. The 11 readings are on 8 stations (FS and next BS at a change point are on one station) at 20 m, so the total length is $7 \times 20 = 140$ m. RL of the start $= 1280.000$ m, RL of the end $= 1279.924$ m (known). #### Level field book (rise and fall method) | Station | BS | IS | FS | Rise | Fall | RL (m) | |---|---|---|---|---|---|---| | 1 | 1.253 | | | | | 1280.000 | | 2 | | 1.752 | | | 0.499 | 1279.501 | | 3 | | 1.005 | | 0.747 | | 1280.248 | | 4 (CP1) | 1.998 | | 0.675 | 0.330 | | 1280.578 | | 5 (CP2) | 1.737 | | 0.825 | 1.173 | | 1281.751 | | 6 | | 1.444 | | 0.293 | | 1282.044 | | 7 (CP3) | 0.750 | | 1.619 | | 0.175 | 1281.869 | | 8 | | | 2.619 | | 1.869 | 1280.000 | #### Arithmetic check $\Sigma BS - \Sigma FS = 5.738 - (5.738) = 0.000$ ; $\Sigma \text{Rise} - \Sigma \text{Fall} = 2.543 - 2.543 = 0.000$ ; last RL $-$ first RL $= 1280.000 - 1280.000 = 0.000$ All three are equal, so the reduction is correct. #### Adjustment of misclosure Computed RL of the last point = 1280.000 m; known RL = 1279.924 m. $$ \text{Closing error} = 1280.000 - 1279.924 = +0.076\ \text{m} $$ The correction at a point is $-(+0.076) \times d / 140$, where $d$ is the distance of the point from the start (proportional to distance). | Station | Distance (m) | Correction (m) | Adjusted RL (m) | |---|---|---|---| | 1 | 0 | 0.0000 | 1280.000 | | 2 | 20 | -0.0109 | 1279.490 | | 3 | 40 | -0.0217 | 1280.226 | | 4 (CP1) | 60 | -0.0326 | 1280.545 | | 5 (CP2) | 80 | -0.0434 | 1281.708 | | 6 | 100 | -0.0543 | 1281.990 | | 7 (CP3) | 120 | -0.0651 | 1281.804 | | 8 | 140 | -0.0760 | 1279.924 | **Answer:** the computed RL of the end station is 1280.000 m, against the known 1279.924 m, so the closing error is +0.076 m (computed is too high). Adjusted RLs of stations 1 to 8 = 1280.000, 1279.490, 1280.226, 1280.545, 1281.708, 1281.990, 1281.804, 1279.924 m. *Note:* the adjustment is proportional to the distance from the start (20 m between stations). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The instrument is shifted after the 4th, 6th and 9th readings. How are the readings assigned? A. Four set-ups; the 4th, 6th and 9th are FS on change points, followed by BS on the same points B. Three set-ups with no change points C. The 4th, 6th and 9th are all IS D. Each shifted reading is both BS on a new point Answer: A. A shift after a reading means that reading is the FS; the next reading is the BS from the new set-up on the same point. (Key point: Set-ups) 2. The computed RL of the end is 1280.000 m and the known RL is 1279.924 m. What is the closing error? A. -0.076 m; computed is too low B. +0.076 m; computed is too high C. 0.000 m, as the checks agree D. +1.000 m Answer: B. Closing error = computed - known = +0.076 m. The arithmetic check agrees, but the field readings do not close on the known RL. (Key point: Closing error) 3. How are the RLs adjusted? A. Equal correction of +0.076 m at every point B. Correction proportional to distance, added C. Correction proportional to distance from the start, subtracted: -0.076 x d/140 D. Only the last point is changed Answer: C. A positive closing error is removed by subtracting a share proportional to distance; the end then equals 1279.924 m. (Key point: Adjustment) 4. What is the correction at station 5 (80 m from the start)? A. -0.0326 m B. -0.0760 m C. -0.0217 m D. -0.0434 m Answer: D. -0.076 x 80/140 = -0.0434 m. (Key point: Correction value) ## Related in Surveying I - [Level book: profile levelling readings](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aprofile-leveling-readings-3rd-7th-9th) — Surveying I, chapter 5 - [Level book, staff readings at B and C (1251.250)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-15m-b-c-rl-1250-800-1251-250) — Surveying I, chapter 5, asked 3× - [Level book, staff readings at B and C (1251.000)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-15m-b-c-rl-1250-800-1251-000) — Surveying I, chapter 5, asked 3× - [Level book, 5 m staff, B at 1200.800](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-5m-staff-20m-points-a-b-c-d) — Surveying I, chapter 5, asked 2× - [Fly levelling, BM 1280 m; falling gradient](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Afly-level-set-4-pegs-1280-falling-1-in-200) — Surveying I, chapter 5 - [Level book, RL of A 380.500, 3.945 on B](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-4m-staff-30m-380-500-minus) — Surveying I, chapter 5, asked 3× ## Related in other subjects - [RL of hill top P (A to B 112 m, RL 1280)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2073m-ab-112) — Surveying II, chapter 3, asked 4× - [RL of hill station P (A-B 120 m, RL 1280)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2074b-ab-120) — Surveying II, chapter 3, asked 2× - [RL of hill station P (A-B 125 m, RL B 1280.50)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-p-rod-2m-at-b) — Surveying II, chapter 3, asked 2×