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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.

StationsBSISFSRise (+)Fall (-)RL (m)Remarks
A3.2501249.260
B1.755?0.750CP1
C1.950?
D?1.920?CP2
E2.3401.500
F?1.000
G1.8502.185?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 ±\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 ΣBS=12.795\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= 3.250 + 0.750 = 4.000. RL of B =1249.260−0.750=1248.510= 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= 1.950 - 1.755 = 0.195; RL of C =1248.510−0.195=1248.315= 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= 1.950 - 1.920 = 0.030; RL of D =1248.345= 1248.345.
  4. D (BS): E is in the next set-up with D as BS: rise at E =BSD−2.340=1.500= BS_D - 2.340 = 1.500, so BS at D =2.340+1.500=3.840= 2.340 + 1.500 = 3.840; RL of E =1249.845= 1249.845.
  5. F (IS): rise at F is 1.000 from E: IS at F =2.340−1.000=1.340= 2.340 - 1.000 = 1.340; RL of F =1250.845= 1250.845.
  6. G (fall): FS at G 2.185 against IS at F 1.340: fall =2.185−1.340=0.845= 2.185 - 1.340 = 0.845; RL of G =1250.000= 1250.000.
  7. H (rise): BS at G 1.850 and IS at H −1.575-1.575: rise =1.850−(−1.575)=3.425= 1.850 - (-1.575) = 3.425; RL of H =1253.425= 1253.425.
  8. I (IS): fall of 2.820 from H: IS at I =−1.575+2.820=1.245= -1.575 + 2.820 = 1.245; RL of I =1250.605= 1250.605.
  9. J (fall): FS at J 1.895 against IS at I 1.245: fall =1.895−1.245=0.650= 1.895 - 1.245 = 0.650; RL of J =1249.955= 1249.955.
  10. J (BS): ΣBS=12.795\Sigma BS = 12.795, so BS at J =12.795−(3.250+1.755+3.840+1.850)=2.100= 12.795 - (3.250 + 1.755 + 3.840 + 1.850) = 2.100.
  11. K (rise): BS at J 2.100 and FS at K −1.350-1.350: rise =2.100−(−1.350)=3.450= 2.100 - (-1.350) = 3.450; RL of K =1253.405= 1253.405.

Completed level book

StationBSISFSRiseFallRL (m)Remarks
A3.2501249.260
B1.7554.0000.7501248.510CP1
C1.9500.1951248.315
D3.8401.9200.0301248.345CP2
E2.3401.5001249.845
F1.3401.0001250.845
G1.8502.1850.8451250.000CP3
H(−) 1.5753.4251253.425
I1.2452.8201250.605
J2.1001.8950.6501249.955CP4
K(−) 1.3503.4501253.405

Check

ΣFS=4.000+1.920+2.185+1.895−1.350=8.650\Sigma FS = 4.000 + 1.920 + 2.185 + 1.895 - 1.350 = 8.650, so ΣBS−ΣFS=12.795−8.650=4.145\Sigma BS - \Sigma FS = 12.795 - 8.650 = 4.145.

ΣRise−ΣFall=9.405−5.260=4.145\Sigma \text{Rise} - \Sigma \text{Fall} = 9.405 - 5.260 = 4.145.

Last RL −- first RL =1253.405−1249.260=4.145= 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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