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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:

StationBSISFSRiseFallRL
A??
B2.5720.319295.909
C?1.987??
D0.9180.236?
E??1.433?
F1.3722.115?298.848
G1.750??
H?2.057??
A1.4561.847?

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.

  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

StationBSISFSRiseFallRL (m)
A2.253296.228
B2.5720.319295.909
C0.6821.9870.585296.494
D0.9180.236296.258
E3.272-0.5151.433297.691
F1.3722.1151.157298.848
G1.7500.378298.470
H-0.3912.0570.307298.163
A1.4561.847296.316

Arithmetic check

ΣBS=2.253+0.682+3.272+1.372−0.391=7.188ΣFS=1.987−0.515+2.115+2.057+1.456=7.100ΣRise=0.585+1.433+1.157=3.175ΣFall=0.319+0.236+0.378+0.307+1.847=3.087\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}

ΣBS−ΣFS=0.088\Sigma BS - \Sigma FS = 0.088 m, ΣRise−ΣFall=0.088\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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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