---
title: "Level book, staff readings at B and C (1251.250)"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:level-book-15m-b-c-rl-1250-800-1251-250"
source: https://www.bctengineer.com/q/semester_3/surveying_i/level-book-15m-b-c-rl-1250-800-1251-250
site: BCT Engineering
---

# Level book, staff readings at B and C (1251.250)

Surveying I · Chapter 5: Leveling

- Asked in: 2067 Asar, 10 marks
- 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-1250-800-1251-250
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

The following consecutive staff readings were taken by a level and 4 m levelling staff on pegs at 15 m interval on a continuously sloping ground: 0.895, 1.305, 2.805, 0.965, 2.695, 3.255, 1.120, 2.825, 3.450, 3.895, 1.685, 2.050 (Station A). The RL of station A where the reading 2.050 was taken is known to be 1250.250 m. From the last position of the instrument, two stations B and C with RLs 1250.800 and 1251.250 m respectively are to be established without disturbing the instrument. Work out the required staff readings at stations B and C and complete all the works in level book form.

## Short answer

Working backward from A (RL 1250.250 m) using rise and fall, RL of first point is 1257.590 m. From last instrument set-up (HI 1252.300 m), required staff reading at B = 1.500 m, at C = 1.050 m.

## Answer

**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).

| Point | Chainage | BS | IS | FS | HI | Rise | Fall | RL (m) |
|---|---|---|---|---|---|---|---|---|
| 1 | 0 | 0.895 |  |  | 1258.485 |  |  | 1257.590 |
| 2 | 15 |  | 1.305 |  |  |  | 0.410 | 1257.180 |
| 3 | 30 | 0.965 |  | 2.805 | 1256.645 |  | 1.500 | 1255.680 |
| 4 | 45 |  | 2.695 |  |  |  | 1.730 | 1253.950 |
| 5 | 60 | 1.120 |  | 3.255 | 1254.510 |  | 0.560 | 1253.390 |
| 6 | 75 |  | 2.825 |  |  |  | 1.705 | 1251.685 |
| 7 | 90 |  | 3.450 |  |  |  | 0.625 | 1251.060 |
| 8 | 105 | 1.685 |  | 3.895 | 1252.300 |  | 0.445 | 1250.615 |
| A | 120 |  |  | 2.050 |  |  | 0.365 | 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.

$$
\begin{aligned}
\text{Reading at B} &= 1252.300 - 1250.800 = 1.500 \text{ m}\\
\text{Reading at C} &= 1252.300 - 1251.250 = 1.050 \text{ m}
\end{aligned}
$$

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 1.050, RL 1251.250 m).

**Answer:** staff reading at B = **1.500 m**, at C = **1.050 m**. RL of the first point = 1257.590 m.

## Check yourself

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

1. B must be at RL 1250.800 m. What staff reading is required?
   A. 0.550 m
   B. 1.500 m
   C. 1.050 m
   D. 2.050 m
   Answer: B. 1252.300 - 1250.800 = 1.500 m. (Key point: Reading at B)

2. C must be at RL 1251.250 m. What staff reading is required?
   A. 2.550 m
   B. 1.050 m
   C. 1.300 m
   D. 0.450 m
   Answer: B. 1252.300 - 1251.250 = 1.050 m. A higher RL needs a smaller reading. (Key point: Reading at C)

3. What are the three equal values in the arithmetic check?
   A. -12.005 m
   B. -7.340 m
   C. 0.000 m
   D. +7.340 m
   Answer: B. Sum BS - sum FS = 4.665 - 12.005 = -7.340, equal to sum rise - sum fall and last RL - first RL. (Key point: Arithmetic check)

## Close variants set in other exams

- [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×
- [Level book, B and C, RL 50.250 base](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-15m-b-c-rl-50-800-51-000) — Surveying I, chapter 5, asked 3×

## Related in Surveying I

- [Level book, 5 m staff, B at 1200.800](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-5m-staff-20m-points-a-b-c-d) — Surveying I, chapter 5, asked 2×
- [Level book: profile levelling readings](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Aprofile-leveling-readings-3rd-7th-9th) — Surveying I, chapter 5
- [Level book, RL of A 380.500, 3.945 on B](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-4m-staff-30m-380-500-minus) — Surveying I, chapter 5, asked 3×
- [Level book, RL of B 1380.500](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-4m-staff-b-1380-500) — Surveying I, chapter 5, asked 3×
- [Level book, RL of A 380.500, 3.845 on B](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-4m-staff-a-380-500-3-845) — Surveying I, chapter 5, asked 3×