---
title: "Fly levelling, BM 1470 m; set 4 pegs"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:fly-level-set-4-pegs-1470-rising-1-in-200"
source: https://www.bctengineer.com/q/semester_3/surveying_i/fly-level-set-4-pegs-1470-rising-1-in-200
site: BCT Engineering
---

# Fly levelling, BM 1470 m; set 4 pegs

Surveying I · Chapter 5: Leveling

- Asked in: 2076 Chaitra, 6 marks
- Video: [Leveling in Civil Engineering | Fly Levelling in Surveying | How to Transfer Level using Auto Level💯](https://www.youtube.com/watch?v=A5kSTw8jA7U) — Civil Survey Training institute 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/fly-level-set-4-pegs-1470-rising-1-in-200
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

During fly levelling the following note is made. B.S: 0.62, 2.05, 1.42, 2.63 and 2.42 m. F.S.: 2.44, 1.35, 0.53 and 2.41 m. The first B.S was taken on a BM of RL 1470 m. From the last B.S. it is required to set 4 pegs each at a distance of 30 m on a rising gradient of 1 in 200. Enter these notes in the form of a standard level book and calculate the R.L. of the top of each peg by the rise and fall method. Also calculate the staff readings on each peg.

## Short answer

From BM 1470.000 m, RL of CP4 is 1469.99 m (HI 1472.41 m). For a rising 1 in 200 gradient, peg RLs at 30 m intervals: 1470.14, 1470.29, 1470.44, 1470.59 m; corresponding staff readings: 2.27, 2.12, 1.97, 1.82 m.

## Answer

**Entering the notes.** There are 5 BS and 4 FS readings, so the instrument was set up five times. The first BS (0.62 m) is on the BM (RL 1470 m); the readings in order are BS 0.62, FS 2.44; BS 2.05, FS 1.35; BS 1.42, FS 0.53; BS 2.63, FS 2.41; and the last BS 2.42 m is on the last change point CP4, from which the pegs are set out.

#### Level field book (rise and fall method)

| Station | BS | FS | Rise | Fall | RL (m) | HI (m) |
|---|---|---|---|---|---|---|
| BM | 0.620 |  |  |  | 1470.000 | 1470.620 |
| CP1 | 2.050 | 2.440 |  | 1.820 | 1468.180 | 1470.230 |
| CP2 | 1.420 | 1.350 | 0.700 |  | 1468.880 | 1470.300 |
| CP3 | 2.630 | 0.530 | 0.890 |  | 1469.770 | 1472.400 |
| CP4 | 2.420 | 2.410 | 0.220 |  | 1469.990 | 1472.410 |

#### Arithmetic check

$\Sigma BS - \Sigma FS$ (set-ups 1 to 4) $= (0.62+2.05+1.42+2.63) - (2.44+1.35+0.53+2.41) = 6.72 - 6.73 = -0.01$ m.

$\Sigma \text{Rise} - \Sigma \text{Fall} = 1.81 - 1.82 = -0.01$ m, and RL of CP4 $-$ RL of BM $= 1469.99 - 1470.00 = -0.01$ m. All agree.

#### Setting out 4 pegs at 30 m on a rising gradient of 1 in 200

HI of the last set-up $= 1469.99 + 2.42 = 1472.41$ m. Rise per 30 m $= 30/200 = 0.15$ m. The pegs are at 30, 60, 90 and 120 m from CP4.

RL of a peg $=$ RL of CP4 $+ 0.15 \times n$, and staff reading $=$ HI $-$ RL.

| Peg | Distance from CP4 (m) | RL (m) | Staff reading (m) |
|---|---|---|---|
| 1 | 30 | 1470.14 | 2.27 |
| 2 | 60 | 1470.29 | 2.12 |
| 3 | 90 | 1470.44 | 1.97 |
| 4 | 120 | 1470.59 | 1.82 |

**Answer:** RL of CP4 = 1469.99 m (HI = 1472.41 m). RLs of the tops of the 4 pegs = 1470.14, 1470.29, 1470.44, 1470.59 m; staff readings on the pegs = 2.27, 2.12, 1.97, 1.82 m.

## Check yourself

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

1. In the rise and fall method, how is the rise or fall between consecutive points found?
   A. Previous reading minus present reading from the same set-up; positive is a rise
   B. Present minus previous; positive is a rise
   C. BS minus IS always
   D. Sum of BS and FS
   Answer: A. A smaller reading means the point is higher, so previous - present > 0 is a rise. (Key point: Rise/fall rule)

2. The RL of CP4 is 1469.99 m and the last BS is 2.42 m. What is the HI?
   A. 1467.57 m
   B. 1472.41 m
   C. 1469.99 m
   D. 1470.14 m
   Answer: B. HI = 1469.99 + 2.42 = 1472.41 m. (Key point: HI)

3. On a rising gradient of 1 in 200 with 30 m spacing, what are the first peg RL and staff reading?
   A. RL 1470.14 m, staff reading 2.57 m
   B. RL 1469.84 m, staff reading 2.57 m
   C. RL 1470.14 m, staff reading 2.27 m
   D. RL 1470.14 m, staff reading 1.82 m
   Answer: C. Rise per peg = 30/200 = 0.15 m; RL = 1469.99 + 0.15 = 1470.14 m; reading = 1472.41 - 1470.14 = 2.27 m. (Key point: Peg RL)

4. SumBS - SumFS for the first four set-ups is -0.01 m. What must it equal?
   A. Zero, if the work is correct
   B. SumIS
   C. RL of CP4 + RL of BM
   D. SumRise - SumFall and RL of CP4 - RL of BM
   Answer: D. All three equal -0.01 m (1469.99 - 1470.00), verifying the arithmetic. (Key point: Check)

## Related in Surveying I

- [Fly levelling, BM 1280 m; falling gradient](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Afly-level-set-4-pegs-1280-falling-1-in-200) — Surveying I, chapter 5
- [Fly levelling for road pegs, BM 3010 m](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Afly-level-road-pegs-3010m) — Surveying I, chapter 5
- [Fly levelling with inverted staff, BM 1290.500](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Afly-level-inverted-staff-1290-500) — Surveying I, chapter 5
- [Fly levelling from BM RL 1291.610](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Afly-level-bm-1291-610) — Surveying I, chapter 5
- [Link traverse ABCDEF, last line 108°05′40″](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alink-traverse-abcdef-108-05-40) — Surveying I, chapter 4, asked 3×
- [Reciprocal levelling AB 250 m, RL 1450.500](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Areciprocal-leveling-ab-250m-1-252) — Surveying I, chapter 5

## Related in other subjects

- [RL of hill top P (A to B = 53 m)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2078p-ab-53) — Surveying II, chapter 3