Level book TBM1-TBM2 with (-)1.685 reading

Entering the readings. The level was changed after the 4th and 8th readings, so there are three set-ups:

• Set-up 1: BS 2.191 on TBM1; IS 2.505, 2.325; FS 1.496 on CP1 (4th reading).
• Set-up 2: BS 3.019 on CP1; IS 2.513 and $-1.685$; FS 2.811 on CP2 (8th reading).
• Set-up 3: BS 1.752 on CP2; FS 3.824 on TBM2.

The reading $-1.685$ m is taken with the staff held inverted (for example against a ceiling or beam): the point is 1.685 m above the line of sight, so it is entered with a negative sign and the usual rule RL $=$ HI $-$ reading still applies. RL of TBM1 $= 1449.870$ m; RL of TBM2 (known) $= 1448.710$ m.

Level field book (rise and fall method)

Arithmetic check

$\Sigma BS - \Sigma FS = 6.962 - (8.131) = -1.169$ ; $\Sigma \text{Rise} - \Sigma \text{Fall} = 5.713 - 6.882 = -1.169$ ; last RL $-$ first RL $= 1448.701 - 1449.870 = -1.169$

All three are equal, so the reduction is correct.

Adjustment of misclosure

Computed RL of the last point = 1448.701 m; known RL = 1448.710 m.

$$\text{Closing error} = 1448.701 - 1448.710 = -0.009\ \text{m}$$

The correction at a point is $-(-0.009) \times n / 7$, where $n$ is the number of intervals from TBM1. The distances between the points are not given, so the 8 staff points are taken at equal spacing (the 'Interval no.' column counts the intervals from TBM1, 0 to 7).

Answer: the computed RL of TBM2 is 1448.701 m against the known 1448.710 m, so the closing error is -0.009 m. Adjusted RLs: TBM1 1449.870, point 2 1449.557, point 3 1449.739, CP1 1450.569, point 6 1451.076, point 7 1455.275, CP2 1450.781, TBM2 1448.710 m.