Reciprocal levelling; RL of closed traverse stations

Reciprocal levelling is used to carry the RL across the river from A to B, because a level set midway would be impossible. The error in the far reading (collimation, curvature and refraction) is the same in both sets, so it cancels in the mean.

Notation: level near A: staff on A (near) $a_1 = 1.379$, staff on B (far) $b_1 = 2.918$. Level near B: staff on A (far) $a_2 = 1.210$, staff on B (near) $b_2 = 2.739$. Rise of B above A is BS $-$ FS.

Step 1: Apparent differences of level

$$H_1 = a_1 - b_1 = 1.379 - 2.918 = -1.539\ \text{m}$$ $$H_2 = a_2 - b_2 = 1.210 - 2.739 = -1.529\ \text{m}$$

Step 2: True difference of level and error

$$H = \frac{H_1 + H_2}{2} = \frac{-1.539 + (-1.529)}{2} = -1.534\ \text{m}$$ $$e = \frac{H_2 - H_1}{2} = \frac{-1.529 - (-1.539)}{2} = +0.005\ \text{m} \quad (\text{error in the far reading over 1200 m})$$

B is lower than A by 1.534 m.

Step 3: True RL of B

$$\text{RL of B} = 1200.000 - 1.534 = 1198.466\ \text{m}$$

Step 4: Correct RLs of the stations of the traverse BCDEFGB

The traverse RLs were calculated from an assumed RL of B = 1198.406 m. The true RL of B is 1198.466 m, so every station is corrected by

$$1198.466 - 1198.406 = +0.060\ \text{m}$$

The traverse is closed on B, so the lengths of the lines (50, 54, 64, 60, 58, 40 m) are not needed for this constant shift. If the loop levelling returns to B with a misclosure, it is distributed in proportion to the length of the lines and added to these values.

Answer: true RL of B = 1198.466 m; each calculated RL is increased by 0.060 m, giving C = 1201.560, D = 1201.660, E = 1201.960, F = 1201.760, G = 1200.860 m.