--- title: "Level book with missing data, RL 1249.260" subject: "Surveying I" chapter: 5 key: "semester_3/surveying_i:level-book-missing-data-ba-1249-260" source: https://www.bctengineer.com/q/semester_3/surveying_i/level-book-missing-data-ba-1249-260 site: BCT Engineering --- # Level book with missing data, RL 1249.260 Surveying I · Chapter 5: Leveling - Asked in: 2074 Chaitra, 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-missing-data-ba-1249-260 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5 ## Question A page of a level field book with some missing data is given below. Find those missing data and calculate the reduced levels of all the points. | Stations | BS | IS | FS | Rise (+) | Fall (-) | RL (m) | Remarks | |---|---|---|---|---|---|---|---| | A | 3.250 | | | | | 1249.260 | | | B | 1.755 | | ? | | 0.750 | | CP1 | | C | | 1.950 | | | ? | | | | D | ? | | 1.920 | ? | | | CP2 | | E | | 2.340 | | 1.500 | | | | | F | | ? | | 1.000 | | | | | G | 1.850 | | 2.185 | | ? | | CP3 | | H | | (-) 1.575 | | ? | | | | | I | | ? | | | 2.820 | | | | J | ? | | 1.895 | | ? | | CP4 | | K | | | (-) 1.350 | ? | | | | ΣBS = 12.795 ## Short answer Missing data: FS(B)=4.000, Fall(C)=0.195, Rise(D)=0.030, BS(D)=3.840, IS(F)=1.340, Fall(G)=0.845, Rise(H)=3.425, IS(I)=1.245, Fall(J)=0.650, BS(J)=2.100, Rise(K)=3.450. RLs: A 1249.260, B 1248.510, C 1248.315, D 1248.345, E 1249.845, F 1250.845, G 1250.000, H 1253.425, I 1250.605, J 1249.955, K 1253.405 m. ## Answer **Method.** In the rise and fall method each rise or fall is the difference between **two consecutive staff readings taken from the same set-up** (previous reading $-$ present reading: positive is a rise, negative a fall), and RL $=$ previous RL $\pm$ rise or fall. A change point has an FS (from the old set-up) and a BS (from the new set-up). The missing values are found in order, then checked with $\Sigma BS = 12.795$. #### Finding the missing values 1. **B (FS):** fall of 0.750 from A, so FS $= 3.250 + 0.750 = 4.000$. RL of B $= 1249.260 - 0.750 = 1248.510$. 2. **C (fall):** BS at B is 1.755 (new set-up), IS at C is 1.950: fall $= 1.950 - 1.755 = 0.195$; RL of C $= 1248.510 - 0.195 = 1248.315$. 3. **D (rise):** FS at D 1.920 against IS at C 1.950: rise $= 1.950 - 1.920 = 0.030$; RL of D $= 1248.345$. 4. **D (BS):** E is in the next set-up with D as BS: rise at E $= BS_D - 2.340 = 1.500$, so BS at D $= 2.340 + 1.500 = 3.840$; RL of E $= 1249.845$. 5. **F (IS):** rise at F is 1.000 from E: IS at F $= 2.340 - 1.000 = 1.340$; RL of F $= 1250.845$. 6. **G (fall):** FS at G 2.185 against IS at F 1.340: fall $= 2.185 - 1.340 = 0.845$; RL of G $= 1250.000$. 7. **H (rise):** BS at G 1.850 and IS at H $-1.575$: rise $= 1.850 - (-1.575) = 3.425$; RL of H $= 1253.425$. 8. **I (IS):** fall of 2.820 from H: IS at I $= -1.575 + 2.820 = 1.245$; RL of I $= 1250.605$. 9. **J (fall):** FS at J 1.895 against IS at I 1.245: fall $= 1.895 - 1.245 = 0.650$; RL of J $= 1249.955$. 10. **J (BS):** $\Sigma BS = 12.795$, so BS at J $= 12.795 - (3.250 + 1.755 + 3.840 + 1.850) = 2.100$. 11. **K (rise):** BS at J 2.100 and FS at K $-1.350$: rise $= 2.100 - (-1.350) = 3.450$; RL of K $= 1253.405$. #### Completed level book | Station | BS | IS | FS | Rise | Fall | RL (m) | Remarks | |---|---|---|---|---|---|---|---| | A | 3.250 | | | | | 1249.260 | | | B | 1.755 | | **4.000** | | 0.750 | 1248.510 | CP1 | | C | | 1.950 | | | **0.195** | 1248.315 | | | D | **3.840** | | 1.920 | **0.030** | | 1248.345 | CP2 | | E | | 2.340 | | 1.500 | | 1249.845 | | | F | | **1.340** | | 1.000 | | 1250.845 | | | G | 1.850 | | 2.185 | | **0.845** | 1250.000 | CP3 | | H | | (−) 1.575 | | **3.425** | | 1253.425 | | | I | | **1.245** | | | 2.820 | 1250.605 | | | J | **2.100** | | 1.895 | | **0.650** | 1249.955 | CP4 | | K | | | (−) 1.350 | **3.450** | | 1253.405 | | #### Check $\Sigma FS = 4.000 + 1.920 + 2.185 + 1.895 - 1.350 = 8.650$, so $\Sigma BS - \Sigma FS = 12.795 - 8.650 = 4.145$. $\Sigma \text{Rise} - \Sigma \text{Fall} = 9.405 - 5.260 = 4.145$. Last RL $-$ first RL $= 1253.405 - 1249.260 = 4.145$. All three are equal, so the book is correct. **Answer:** RLs: A 1249.260, B 1248.510, C 1248.315, D 1248.345, E 1249.845, F 1250.845, G 1250.000, H 1253.425, I 1250.605, J 1249.955, K 1253.405 m. Missing data: FS(B) = 4.000, Fall(C) = 0.195, Rise(D) = 0.030, BS(D) = 3.840, IS(F) = 1.340, Fall(G) = 0.845, Rise(H) = 3.425, IS(I) = 1.245, Fall(J) = 0.650, BS(J) = 2.100, Rise(K) = 3.450. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are missing rises and falls found? A. From the difference of two consecutive readings of the same set-up (previous minus present) B. From any two readings in the book C. From the difference of RLs of consecutive stations only D. From the sum of BS and FS Answer: A. Rise or fall is the difference of consecutive readings from one set-up, so the BS at a change point pairs with the following IS. (Key point: Rise and fall) 2. The fall from A to B is 0.750 m and BS at A is 3.250 m. What is FS at B? A. 2.500 m B. 4.000 m C. 3.250 m D. 0.750 m Answer: B. FS at B = 3.250 + 0.750 = 4.000 m, as a fall means a larger reading. (Key point: FS at B) 3. How is BS at J found? A. By the fall at I B. By FS at J = 1.895 C. ΣBS - the other BS = 12.795 - (3.250 + 1.755 + 3.840 + 1.850) = 2.100 m D. By adding all IS Answer: C. The total of backsights is given, so the remaining BS follows by subtraction. (Key point: BS at J) 4. BS at G is 1.850 m and IS at H is -1.575 m. What is the rise at H? A. 0.275 m B. -3.425 m C. 1.575 m D. 3.425 m Answer: D. Rise = 1.850 - (-1.575) = 3.425 m. 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