Level book, B and C, RL 50.250 base

Assumptions: 12 readings at 15 m interval. On continuously sloping ground the readings keep increasing within one set-up, so the instrument was moved wherever the next reading drops. The shifts are therefore taken after the readings 2.800, 3.255 and 3.895 (the change points), and 1.685 is the last back sight. Set-ups: (0.895, 1.305, 2.800), (1.960, 2.690, 3.255), (2.120, 2.825, 3.450, 3.895), (1.685, 2.050). The last reading 2.050 is on A, whose RL is 50.250 m, so levels are worked from A backwards (rise/fall is first found with an arbitrary RL and then shifted so that A = 50.250).

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

Arithmetic checks

$$\begin{aligned} \Sigma BS - \Sigma FS &= 6.660 - 12.000 = -5.340 \text{ m}\\ \Sigma \text{Rise} - \Sigma \text{Fall} &= 0.000 - 5.340 = -5.340 \text{ m}\\ \text{Last RL} - \text{First RL} &= 50.250 - 55.590 = -5.340 \text{ m} \end{aligned}$$

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 = 50.615 + 1.685 = 52.300 m (check: RL of A + FS on A = 50.250 + 2.050 = 52.300 m).

Staff reading required = HI - required RL.

$$\begin{aligned} \text{Reading at B} &= 52.300 - 50.800 = 1.500 \text{ m}\\ \text{Reading at C} &= 52.300 - 51.000 = 1.300 \text{ m} \end{aligned}$$

Both readings are within the 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 50.800 m) and C (IS 1.300, RL 51.000 m).

Answer: staff reading at B = 1.500 m, at C = 1.300 m. RL of the first point = 55.590 m.