Level book, staff readings at B and C (1251.000)

Assumptions: 12 readings at 15 m interval. The instrument was moved whenever the reading nearly reached the limit of the 4 m staff and the next reading drops sharply. The shifts are therefore taken after the readings 2.805, 3.255 and 3.895 (the change points), and 1.685 is the last back sight. Set-ups: (0.895, 1.305, 2.805), (0.965, 2.695, 3.255), (1.120, 2.825, 3.450, 3.895), (1.685, 2.050). The last reading 2.050 is on A, whose RL is 1250.250 m, so levels are worked from A backwards (rise/fall is first found with an arbitrary RL and then shifted so that A = 1250.250).

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

Arithmetic checks

$$\begin{aligned} \Sigma BS - \Sigma FS &= 4.665 - 12.005 = -7.340 \text{ m}\\ \Sigma \text{Rise} - \Sigma \text{Fall} &= 0.000 - 7.340 = -7.340 \text{ m}\\ \text{Last RL} - \text{First RL} &= 1250.250 - 1257.590 = -7.340 \text{ m} \end{aligned}$$

All three are equal, so the arithmetic is correct.

Setting out B and C from the last instrument position

HI of the last set-up = RL of point 8 + BS = 1250.615 + 1.685 = 1252.300 m (check: RL of A + FS on A = 1250.250 + 2.050 = 1252.300 m).

Staff reading required = HI - required RL.

$$\begin{aligned} \text{Reading at B} &= 1252.300 - 1250.800 = 1.500 \text{ m}\\ \text{Reading at C} &= 1252.300 - 1251.000 = 1.300 \text{ m} \end{aligned}$$

Both readings are within the 4 m staff length, so B and C can be set out without shifting the instrument. A peg is driven at B and C until the staff reads these values. In the level book, B and C are entered as intermediate sights of the last set-up: B (IS 1.500, RL 1250.800 m) and C (IS 1.300, RL 1251.000 m).

Answer: staff reading at B = 1.500 m, at C = 1.300 m. RL of the first point = 1257.590 m.