- 2074 Chaitra · 6 marks
Level book with missing data, RL 1249.260
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.
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
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 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 .
Finding the missing values
- B (FS): fall of 0.750 from A, so FS . RL of B .
- C (fall): BS at B is 1.755 (new set-up), IS at C is 1.950: fall ; RL of C .
- D (rise): FS at D 1.920 against IS at C 1.950: rise ; RL of D .
- D (BS): E is in the next set-up with D as BS: rise at E , so BS at D ; RL of E .
- F (IS): rise at F is 1.000 from E: IS at F ; RL of F .
- G (fall): FS at G 2.185 against IS at F 1.340: fall ; RL of G .
- H (rise): BS at G 1.850 and IS at H : rise ; RL of H .
- I (IS): fall of 2.820 from H: IS at I ; RL of I .
- J (fall): FS at J 1.895 against IS at I 1.245: fall ; RL of J .
- J (BS): , so BS at J .
- K (rise): BS at J 2.100 and FS at K : rise ; RL of K .
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
, so .
.
Last RL first RL . 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.
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- Closed loop level book, F BS 1.372 · Surveying I, chapter 5
- Closed loop level book, F BS blank · Surveying I, chapter 5
- Level book, RL of B 1180.750 m · Surveying I, chapter 5 · asked 2×
- Level book, B and C, RL 50.250 base · Surveying I, chapter 5 · asked 3×
- Level book, 5 m staff, B at 1200.800 · Surveying I, chapter 5 · asked 2×
- Level book, staff readings at B and C (1251.250) · Surveying I, chapter 5 · asked 3×
- RL of hill station P from stations R and Q · Surveying II, chapter 3
- RL of building top from stations A and B · Surveying II, chapter 3
- RL of temple top from stations A and B · Surveying II, chapter 3
- Crest level of fall, Q 30 m3/s · Irrigation and Drainage Engineering, chapter 7