Level book, 20 m interval, RL 1280 to 1279.924

Arranging the readings. The instrument was shifted after the 4th, 6th and 9th readings, so there are four set-ups:

• Set-up 1: BS 1.253 on station 1; IS 1.752, 1.005; FS 0.675 on CP1 (4th reading).
• Set-up 2: BS 1.998 on CP1; FS 0.825 on CP2 (6th reading).
• Set-up 3: BS 1.737 on CP2; IS 1.444; FS 1.619 on CP3 (9th reading).
• Set-up 4: BS 0.750 on CP3; FS 2.619 on the end station.

The 11 readings are on 8 stations (FS and next BS at a change point are on one station) at 20 m, so the total length is $7 \times 20 = 140$ m. RL of the start $= 1280.000$ m, RL of the end $= 1279.924$ m (known).

Level field book (rise and fall method)

Arithmetic check

$\Sigma BS - \Sigma FS = 5.738 - (5.738) = 0.000$ ; $\Sigma \text{Rise} - \Sigma \text{Fall} = 2.543 - 2.543 = 0.000$ ; last RL $-$ first RL $= 1280.000 - 1280.000 = 0.000$

All three are equal, so the reduction is correct.

Adjustment of misclosure

Computed RL of the last point = 1280.000 m; known RL = 1279.924 m.

$$\text{Closing error} = 1280.000 - 1279.924 = +0.076\ \text{m}$$

The correction at a point is $-(+0.076) \times d / 140$, where $d$ is the distance of the point from the start (proportional to distance).

Answer: the computed RL of the end station is 1280.000 m, against the known 1279.924 m, so the closing error is +0.076 m (computed is too high). Adjusted RLs of stations 1 to 8 = 1280.000, 1279.490, 1280.226, 1280.545, 1281.708, 1281.990, 1281.804, 1279.924 m.

Note: the adjustment is proportional to the distance from the start (20 m between stations).