Level book, elevation of B 1120.450

Reading the data. The staff is 4 m long, so the readings 3.938 and 3.844 (nearly 4 m) show where the instrument was shifted. The sequence is therefore:

• Setup 1: BS 0.585 on A; IS 0.936, 1.953, 2.846, 3.644; FS 3.938 on change point CP1.
• Setup 2: BS 0.962 on CP1; IS 1.035, 1.689, 2.534; FS 3.844 on change point CP2.
• Setup 3: BS 0.956 on CP2; IS 1.579; FS 3.016 on B.

The ground slopes continuously down (the readings increase). The RL of B is known, so the RL of A is found first:

$$\text{RL of A} = \text{RL of B} + (\Sigma FS - \Sigma BS) = 1120.450 + (10.798 - 2.503) = 1128.745\ \text{m}$$

Level field book (rise and fall method)

Arithmetic check

$\Sigma BS - \Sigma FS = 2.503 - 10.798 = -8.295$; $\Sigma \text{Rise} - \Sigma \text{Fall} = 0.000 - 8.295 = -8.295$; last RL $-$ first RL $= 1120.450 - 1128.745 = -8.295$.

All three are equal, so the reduction is correct. Working down the rises and falls from A brings the RL back to the given value of B (1120.450 m), which confirms the arithmetic of the table.

Gradient between the first and last point

The FS and BS at a change point are taken on the same point, so the 14 readings fall on 12 points at 30 m spacing: 11 intervals, a distance of $11 \times 30 = 330$ m.

$$\text{Fall} = 1128.745 - 1120.450 = 8.295\ \text{m}, \qquad \text{Gradient} = \frac{8.295}{330} = \frac{1}{39.8}$$

Answer: RL of A = 1128.745 m; falling gradient from A to B of about 1 in 40 (1 in 39.8).

Note: if each of the 14 readings were counted as a separate 30 m station (13 intervals, 390 m), the gradient would be 1 in 47.0.