--- title: "Closed loop level book, F BS 1.372" subject: "Surveying I" chapter: 5 key: "semester_3/surveying_i:closed-loop-level-book-missing-f-bs-1-372" source: https://www.bctengineer.com/q/semester_3/surveying_i/closed-loop-level-book-missing-f-bs-1-372 site: BCT Engineering --- # Closed loop level book, F BS 1.372 Surveying I · Chapter 5: Leveling - Asked in: 2069 Chaitra, 10 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/closed-loop-level-book-missing-f-bs-1-372 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5 ## Question A levelling operation is carried out in a closed loop. Fill all the missing data of a levelling field book given below: | Station | BS | IS | FS | Rise | Fall | RL | |---|---|---|---|---|---|---| | A | ? | | | | | ? | | B | | 2.572 | | | 0.319 | 295.909 | | C | ? | | 1.987 | ? | | ? | | D | | 0.918 | | | 0.236 | ? | | E | ? | | ? | 1.433 | | ? | | F | 1.372 | | 2.115 | ? | | 298.848 | | G | | 1.750 | | | ? | ? | | H | ? | | 2.057 | | ? | ? | | A | | | 1.456 | | 1.847 | ? | ## Short answer Using rise and fall method, BS: A=2.253 m, C=0.682 m, E=3.272 m, H=-0.391 m; FS at E=-0.515 m (inverted staff). RLs: A 296.228 m, C 296.494 m, D 296.258 m, E 297.691 m, G 298.470 m, H 298.163 m, closing A 296.316 m. Arithmetic checks verified. ## Answer **Method (rise and fall).** A fall at a station means its staff reading is larger than the previous reading on the same set-up; a rise means it is smaller. The same relations are used backwards to find the missing readings. 1. **A:** B has IS 2.572 and a fall of 0.319, so the reading at A (BS) = 2.572 - 0.319 = 2.253 m. RL of A = RL of B + fall = 295.909 + 0.319 = 296.228 m. 2. **C:** FS 1.987 is smaller than 2.572, so rise = 2.572 - 1.987 = 0.585 m, RL of C = 295.909 + 0.585 = 296.494 m. 3. **C (BS) and D:** D has IS 0.918 and fall 0.236 on the same set-up as the BS at C, so BS at C = 0.918 - 0.236 = 0.682 m. RL of D = 296.494 - 0.236 = 296.258 m. 4. **E:** rise 1.433 from D, so RL of E = 296.258 + 1.433 = 297.691 m and FS at E = 0.918 - 1.433 = -0.515 m. The negative reading means the staff was held *inverted* (touching a roof or beam), and the reading is booked with a minus sign. 5. **E (BS):** RL of F = 298.848, so the rise E to F = 298.848 - 297.691 = 1.157 m, and BS at E = 1.157 + 2.115 = 3.272 m. 6. **G and H:** BS at F is given as 1.372 m. G has IS 1.750, which is larger, so fall F to G = 1.750 - 1.372 = 0.378 m and RL of G = 298.848 - 0.378 = 298.470 m. At H, FS 2.057 gives a fall of 2.057 - 1.750 = 0.307 m, so RL of H = 298.470 - 0.307 = 298.163 m. 7. **H (BS) and A:** the final FS at A is 1.456 with a fall of 1.847, so BS at H = 1.456 - 1.847 = -0.391 m (inverted staff) and the closing RL of A = 298.163 - 1.847 = 296.316 m. #### Completed field book | Station | BS | IS | FS | Rise | Fall | RL (m) | |---|---|---|---|---|---|---| | A | **2.253** | | | | | **296.228** | | B | | 2.572 | | | 0.319 | 295.909 | | C | **0.682** | | 1.987 | **0.585** | | **296.494** | | D | | 0.918 | | | 0.236 | **296.258** | | E | **3.272** | | **-0.515** | 1.433 | | **297.691** | | F | 1.372 | | 2.115 | **1.157** | | 298.848 | | G | | 1.750 | | | **0.378** | **298.470** | | H | **-0.391** | | 2.057 | | **0.307** | **298.163** | | A | | | 1.456 | | 1.847 | **296.316** | #### Arithmetic check $$ \begin{aligned} \Sigma BS &= 2.253+0.682+3.272+1.372-0.391 = 7.188\\ \Sigma FS &= 1.987-0.515+2.115+2.057+1.456 = 7.100\\ \Sigma \text{Rise} &= 0.585+1.433+1.157 = 3.175\\ \Sigma \text{Fall} &= 0.319+0.236+0.378+0.307+1.847 = 3.087 \end{aligned} $$ $\Sigma BS - \Sigma FS = 0.088$ m, $\Sigma \text{Rise} - \Sigma \text{Fall} = 0.088$ m and last RL - first RL = 296.316 - 296.228 = 0.088 m. The three values agree, so the arithmetic is correct. **Answer:** missing values are as shown in bold. BS: A 2.253, C 0.682, E 3.272, H -0.391 m; FS at E = -0.515 m. RLs: A 296.228, C 296.494, D 296.258, E 297.691, G 298.470, H 298.163, A (closing) 296.316 m. *Note on reading:* with the printed F BS of 1.372 m the loop returns to A at 296.316 m, a closing error of 0.088 m. The loop would close exactly if F BS were 1.284 m (the value needed to return to 296.228 m), so 1.372 may be a misprint of the scanned value. The table above uses the value as printed. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. BS at F is 1.372 and IS at G is 1.750. What is the fall F to G and RL of G? A. Fall 0.378 m, RL 299.226 m B. Fall 0.378 m, RL 298.470 m C. Fall 0.466 m, RL 298.382 m D. Rise 0.378 m, RL 299.226 m Answer: B. Larger reading means fall: 1.750 - 1.372 = 0.378 m, so RL G = 298.848 - 0.378 = 298.470 m. (Key point: Fall F to G) 2. The IS at G is 1.750 and the FS at H is 2.057 on the same set-up. What is RL of H? A. 296.316 m B. 298.163 m C. 298.777 m D. 298.470 m Answer: B. Fall G to H = 2.057 - 1.750 = 0.307 m, so RL H = 298.470 - 0.307 = 298.163 m. (Key point: RL of H) 3. With the printed F BS of 1.372, what is the closing RL of A and what does it mean? A. 296.140 m, a closing error of 0.088 m B. 296.228 m, perfect closure C. 296.316 m, a closing error of 0.088 m D. 296.316 m, no error because three checks agree Answer: C. Closing RL = 298.163 - 1.847 = 296.316 m, which differs from 296.228 m by 0.088 m. (Key point: Closing RL of A) 4. Which statement is right about the arithmetic check in this book? A. The check fails because the sums are not zero B. Only the BS and FS sums need to agree C. Sum BS - sum FS = 0.088 m, sum rise - sum fall = 0.088 m, last RL - first RL = 0.088 m; all agree so the arithmetic is correct D. The check is only valid when the loop closes exactly Answer: C. The check requires the three differences to be equal to each other, not zero; a misclosure still shows as equal values. (Key point: Arithmetic check) ## Related in Surveying I - [Closed loop level book, F BS blank](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aclosed-loop-level-book-missing-a-295-909) — Surveying I, chapter 5 - [Level book with missing data, RL 1249.260](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-missing-data-ba-1249-260) — Surveying I, chapter 5 - [Level book, RL of B 1180.750 m](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-dumpy-4m-staff-b-1180-750) — Surveying I, chapter 5, asked 2× - [Reciprocal levelling; RL of closed traverse stations](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Areciprocal-leveling-closed-traverse-bcdefgb) — Surveying I, chapter 5 - [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