Level book and longitudinal section, RL 525.50

Assumptions: the 4th and 8th readings (3.455 and 3.650) are foresights on change points, and the next readings (1.30 and 0.855) are backsights on the same points after the instrument is shifted. So 11 readings give 9 points at 20 m spacing, chainage 200 m to 360 m.

HI is shown on the row where a new set-up starts. RL = previous RL + rise - fall.

Arithmetic checks

$$\begin{aligned} \Sigma BS - \Sigma FS &= 2.670 - 10.210 = -7.540 \text{ m}\\ \Sigma \text{Rise} - \Sigma \text{Fall} &= 0.000 - 7.540 = -7.540 \text{ m}\\ \text{Last RL} - \text{First RL} &= 517.960 - 525.500 = -7.540 \text{ m} \end{aligned}$$

All three are equal, so the arithmetic is correct.

Longitudinal section

Plot chainage on the horizontal scale (say 1 cm = 20 m) and RL on the vertical scale (say 1 cm = 1 m, exaggerated for clarity); join the points by a straight line (ground profile). A text sketch of the profile is shown below (each column is one point, * marks the RL).

 525.50|*
 524.81|
 524.13|     *
 523.44|          *
 522.76|               *
 522.07|
 521.39|                    *
 520.70|                         *
 520.02|                              *
 519.33|
 518.65|                                   *
 517.96|                                        *
       +---------------------------------------------
        200  220  240  260  280  300  320  340  360  

The ground falls continuously from 525.500 m at chainage 200 m to 517.960 m at chainage 360 m, an average fall of 7.540 m in 160 m (about 1 in 21.2).

Answer: RLs at chainage 200, 220, ..., 360 m = 525.500, 524.295, 523.515, 522.560, 521.400, 520.910, 520.210, 518.635, 517.960 m.