Level book, RL of B 1180.750 m

Reading of the data. The first reading is printed as 0.680 [?]; it is read as 0.680 m. The printed "(B)" stands against the change point where the 10th reading (FS 3.540) and the 11th reading (BS 0.835) are taken, so the point B (RL 1180.750 m) is this change point (CP2); the RLs of all the other points are found from it. If the 1180.750 m belonged to the first point instead, every RL below would change by the same constant and the gradient would be unchanged.

The staff is 4 m, so the instrument was shifted after the readings 3.380 and 3.540 (6th and 10th readings). The set-ups are:

• Set-up 1: BS 0.680; IS 1.455, 1.855, 2.880, 2.800; FS 3.380 on CP1.
• Set-up 2: BS 1.055 on CP1; IS 1.860, 2.265; FS 3.540 on CP2 (point B).
• Set-up 3: BS 0.835 on B; IS 0.945, 1.530; FS 2.445 on the last point.

The 14 readings are on 12 points, 30 m apart. Taking B (9th point) as RL 1180.750 m, the HI of set-up 3 is $1180.750 + 0.835 = 1181.585$ m, and the other RLs follow by working back through the rises and falls.

Level field book (rise and fall method)

Arithmetic check

$\Sigma BS - \Sigma FS = 2.570 - (9.365) = -6.795$ ; $\Sigma \text{Rise} - \Sigma \text{Fall} = 0.080 - 6.875 = -6.795$ ; last RL $-$ first RL $= 1179.140 - 1185.935 = -6.795$

All three are equal, so the reduction is correct.

Gradient of the line joining the first and last points

Distance $= 11 \times 30 = 330$ m.

$$\text{Fall} = 1185.935 - 1179.140 = 6.795\ \text{m}, \qquad \text{Gradient} = \frac{6.795}{330} = \frac{1}{48.6}\ (\text{falling})$$

Answer: RLs of the points 1 to 12 = 1185.935, 1185.160, 1184.760, 1183.735, 1183.815, 1183.235, 1182.430, 1182.025, 1180.750, 1180.640, 1180.055, 1179.140 m. The line from the first to the last point falls at a gradient of 1 in 48.6 (about 1 in 49).