Differential levelling loop on BM Gate

Method: differential levelling by the height of instrument method. The BS and FS distances are equal, so the closing error is shared equally among the four set-ups (the correction grows by the same amount at each set-up).

Field notes (BM Gate = 1237.280 m; each FS and the next BS are on the same turning point)

Calculation: HI = RL + BS, and next RL = HI - FS. For example, HI = 1237.280 + 2.863 = 1240.143 m and TP1 = 1240.143 - 2.147 = 1237.996 m.

Check

$$\begin{aligned} \Sigma BS &= 2.863+1.623+2.027+0.933 = 7.446\\ \Sigma FS &= 2.147+2.500+2.410+0.370 = 7.427\\ \Sigma BS - \Sigma FS &= 0.019 \text{ m}\\ \text{Last RL} - \text{first RL} &= 1237.299 - 1237.280 = 0.019 \text{ m} \end{aligned}$$

The arithmetic check is satisfied, i.e. $\Sigma BS - \Sigma FS$ equals the difference of the end RLs.

Misclosure and adjustment

Since the loop starts and ends on the same BM, the RL of BM Gate on closing should be 1237.280 m, but it comes to 1237.299 m.

$$\text{Closing error} = 0.019 \text{ m (too high)}$$

The distances are not given, so the error (19 mm in four set-ups) is taken to be within the permissible limit for ordinary levelling (about $\pm 24\sqrt{K}$ mm, K in km) and is adjusted. Correction per set-up = $-0.019/4 = -0.00475$ m, applied cumulatively: -0.00475, -0.00950, -0.01425 and -0.01900 m to TP1, BMX, TP2 and BM Gate respectively.

Answer: adjusted RLs: TP1 = 1237.991 m, BMX = 1237.110 m, TP2 = 1236.722 m, BM Gate = 1237.280 m. Closing error = 0.019 m.