- 2069 Chaitra · 10 marks
Closed loop level book, F BS 1.372
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.
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 | ? |
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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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.
- 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
m, 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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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