--- title: "Differential levelling loop on BM Gate" subject: "Surveying I" chapter: 5 key: "semester_3/surveying_i:differential-leveling-loop-bm-gate" source: https://www.bctengineer.com/q/semester_3/surveying_i/differential-leveling-loop-bm-gate site: BCT Engineering --- # Differential levelling loop on BM Gate Surveying I · Chapter 5: Leveling - Asked in: 2058 Chaitra, 10 marks - Video: [Differential Leveling Sample Problem](https://www.youtube.com/watch?v=R5hovS8G3w8) — RMP - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/differential-leveling-loop-bm-gate - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5 ## Question A differential levelling loop began and closed on BM Gate (elevation 1237.280 m). The BS and FS distances were kept approximately equal. Readings taken in order are 2.863 on BM Gate, 2.147 and 1.623 on TP1, 2.500 and 2.027 on BMX, 2.410 and 0.933 on TP2, and 0.370 on BM Gate. Prepare, check and adjust the notes. ## Short answer Unadjusted RLs are TP1 1237.996, BMX 1237.119, TP2 1236.736, and closing BM Gate 1237.299 m. The closing error of 0.019 m is distributed equally over four setups (−0.00475 m each), giving adjusted RLs: TP1 1237.991, BMX 1237.110, TP2 1236.722, and BM Gate 1237.280 m. ## Answer **Method:** differential levelling by the height of instrument method. The BS and FS distances are equal, so the closing error is shared equally among the four set-ups (the correction grows by the same amount at each set-up). **Field notes** (BM Gate = 1237.280 m; each FS and the next BS are on the same turning point) | Station | BS | FS | HI (m) | RL (m) | Correction (m) | Adjusted RL (m) | |---|---|---|---|---|---|---| | BM Gate | 2.863 | | 1240.143 | 1237.280 | 0 | 1237.280 | | TP1 | 1.623 | 2.147 | 1239.619 | 1237.996 | -0.00475 | 1237.991 | | BMX | 2.027 | 2.500 | 1239.146 | 1237.119 | -0.00950 | 1237.110 | | TP2 | 0.933 | 2.410 | 1237.669 | 1236.736 | -0.01425 | 1236.722 | | BM Gate (closing) | | 0.370 | | 1237.299 | -0.01900 | 1237.280 | Calculation: HI = RL + BS, and next RL = HI - FS. For example, HI = 1237.280 + 2.863 = 1240.143 m and TP1 = 1240.143 - 2.147 = 1237.996 m. #### Check $$ \begin{aligned} \Sigma BS &= 2.863+1.623+2.027+0.933 = 7.446\\ \Sigma FS &= 2.147+2.500+2.410+0.370 = 7.427\\ \Sigma BS - \Sigma FS &= 0.019 \text{ m}\\ \text{Last RL} - \text{first RL} &= 1237.299 - 1237.280 = 0.019 \text{ m} \end{aligned} $$ The arithmetic check is satisfied, i.e. $\Sigma BS - \Sigma FS$ equals the difference of the end RLs. #### Misclosure and adjustment Since the loop starts and ends on the same BM, the RL of BM Gate on closing should be 1237.280 m, but it comes to 1237.299 m. $$ \text{Closing error} = 0.019 \text{ m (too high)} $$ The distances are not given, so the error (19 mm in four set-ups) is taken to be within the permissible limit for ordinary levelling (about $\pm 24\sqrt{K}$ mm, K in km) and is adjusted. Correction per set-up = $-0.019/4 = -0.00475$ m, applied cumulatively: -0.00475, -0.00950, -0.01425 and -0.01900 m to TP1, BMX, TP2 and BM Gate respectively. **Answer:** adjusted RLs: TP1 = 1237.991 m, BMX = 1237.110 m, TP2 = 1236.722 m, BM Gate = 1237.280 m. Closing error = 0.019 m. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The loop closes on BM Gate (1237.280) but the computed RL is 1237.299. The closing error is: A. 0.019 m too low B. 0.299 m C. 0.019 m too high D. 0.000 m Answer: C. 1237.299 - 1237.280 = +0.019 m, so the computed value is too high. (Key point: Closing error) 2. Sum BS = 7.446, sum FS = 7.427. What does the check show? A. It should be zero B. Sum BS - sum FS = 0.019 m = last RL - first RL, so the arithmetic is correct C. The arithmetic is wrong D. Only FS matter Answer: B. The difference equals the difference of end RLs, confirming the arithmetic; the misclosure is a field error. (Key point: Check) 3. How is the 0.019 m error distributed? A. Only to BMX B. Added to every RL C. Equally per set-up, cumulatively: -0.00475, -0.00950, -0.01425, -0.01900 D. All at the last station Answer: C. Since distances are equal, correction per set-up = 0.019/4, applied cumulatively. (Key point: Adjustment) 4. How is TP1 found? A. 1237.280 + 2.147 B. HI = 1237.280 + 2.863 = 1240.143; TP1 = 1240.143 - 2.147 = 1237.996 m C. 1240.143 + 2.147 D. 1237.280 - 2.863 Answer: B. HI = RL + BS and RL = HI - FS. (Key point: HI method) ## 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 - [Closed loop level book, F BS 1.372](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aclosed-loop-level-book-missing-f-bs-1-372) — Surveying I, chapter 5 - [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 - [Fly levelling with inverted staff, BM 1290.500](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Afly-level-inverted-staff-1290-500) — 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 - [Fly levelling, BM 1470 m; set 4 pegs](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Afly-level-set-4-pegs-1470-rising-1-in-200) — Surveying I, chapter 5 ## Related in other subjects - [Zero reflection coefficient; 3rd order LPF equal termination](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Areflection-zero-third-order-equal) — Filter Design, chapter 5, asked 4× - [Open loop versus closed loop control](https://www.bctengineer.com/question.md?key=semester_6%2Fembedded_system%3Aopen-vs-closed-loop) — Embedded System, chapter 7, asked 6× - [Define control system and closed loop block roles](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Adefine-cs-bd-role) — Control Systems, chapter 1, asked 5× - [Why closed loop response is faster than open loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Acl-faster-than-ol) — Control Systems, chapter 1, asked 4×