---
title: "Level book, B and C, RL 50.250 base"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:level-book-15m-b-c-rl-50-800-51-000"
source: https://www.bctengineer.com/q/semester_3/surveying_i/level-book-15m-b-c-rl-50-800-51-000
site: BCT Engineering
---

# Level book, B and C, RL 50.250 base

Surveying I · Chapter 5: Leveling

- Asked in: 2059 Chaitra, 10 marks
- Video: [Rise and Fall Method | Calculation of Reduced Level by Rise and Fall Method in Levelling | Surveying](https://www.youtube.com/watch?v=P_nn1ehp_LI) — CivilConstruction
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/level-book-15m-b-c-rl-50-800-51-000
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

The following consecutive staff readings were taken on pegs at 15 m interval on a continuously sloping ground: 0.895, 1.305, 2.800, 1.960, 2.690, 3.255, 2.120, 2.825, 3.450, 3.895, 1.685, 2.050 (st. A). RL of station A where the reading 2.050 was taken is known to be 50.250 m. From the last position of the instrument two stations B and C with RL 50.800 and 51.000 respectively are to be established without disturbing the instrument. Work out the staff readings at B and C and complete all the work in level book form.

## Short answer

Working backward from A (RL 50.250 m) by rise and fall, RL of first point is 55.590 m. With last instrument HI = 52.300 m, staff reading at B = 1.500 m, at C = 1.300 m.

## Answer

**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).

| Point | Chainage | BS | IS | FS | HI | Rise | Fall | RL (m) |
|---|---|---|---|---|---|---|---|---|
| 1 | 0 | 0.895 |  |  | 56.485 |  |  | 55.590 |
| 2 | 15 |  | 1.305 |  |  |  | 0.410 | 55.180 |
| 3 | 30 | 1.960 |  | 2.800 | 55.645 |  | 1.495 | 53.685 |
| 4 | 45 |  | 2.690 |  |  |  | 0.730 | 52.955 |
| 5 | 60 | 2.120 |  | 3.255 | 54.510 |  | 0.565 | 52.390 |
| 6 | 75 |  | 2.825 |  |  |  | 0.705 | 51.685 |
| 7 | 90 |  | 3.450 |  |  |  | 0.625 | 51.060 |
| 8 | 105 | 1.685 |  | 3.895 | 52.300 |  | 0.445 | 50.615 |
| A | 120 |  |  | 2.050 |  |  | 0.365 | 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is HI of the last set-up found?
   A. 55.590 + 0.895 = 56.485 m
   B. 50.250 + 2.050 = 52.300 m
   C. 50.250 - 2.050 = 48.200 m
   D. 50.250 + 1.685 = 51.935 m
   Answer: B. HI = RL of A + FS on A = 52.300 m, which agrees with the HI of the last set-up. (Key point: HI of last set-up)

2. B must be at RL 50.800 m. What staff reading is needed?
   A. 1.300 m
   B. 1.500 m
   C. 2.050 m
   D. 0.500 m
   Answer: B. 52.300 - 50.800 = 1.500 m. (Key point: Reading at B)

3. C must be at RL 51.000 m. What staff reading is needed?
   A. 0.700 m
   B. 1.000 m
   C. 1.300 m
   D. 1.500 m
   Answer: C. 52.300 - 51.000 = 1.300 m. (Key point: Reading at C)

4. What are the three equal values in the arithmetic check?
   A. +5.340 m
   B. 0.000 m
   C. -12.000 m
   D. -5.340 m
   Answer: D. Sum BS - sum FS = 6.660 - 12.000 = -5.340; last RL - first RL = 50.250 - 55.590 = -5.340. (Key point: Arithmetic check)

## Close variants set in other exams

- [Level book, staff readings at B and C (1251.250)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-15m-b-c-rl-1250-800-1251-250) — Surveying I, chapter 5, asked 3×
- [Level book, staff readings at B and C (1251.000)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-15m-b-c-rl-1250-800-1251-000) — Surveying I, chapter 5, asked 3×

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