---
title: "Fly levelling, BM 1280 m; falling gradient"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:fly-level-set-4-pegs-1280-falling-1-in-200"
source: https://www.bctengineer.com/q/semester_3/surveying_i/fly-level-set-4-pegs-1280-falling-1-in-200
site: BCT Engineering
---

# Fly levelling, BM 1280 m; falling gradient

Surveying I · Chapter 5: Leveling

- Asked in: 2076 Asoj, 6 marks
- Video: [Levelling Problem No 8 - Fly Levelling Missing Values Calculation ( Falling Ground )](https://www.youtube.com/watch?v=GaBbF3-1bLk) — Stan Academy
- 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-1280-falling-1-in-200
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

During fly levelling, the following staff readings were noted: BS = 0.63, 2.05, -2.424, (B) and 2.56 m; FS = 2.444, 1.35 and -2.42 m. The (B) was taken on a BM of RL 1280.00 m. From the last point, it is required to set up 4 pegs each at 30 m interval on a falling gradient of 1 in 200.
a) Prepare the level book and calculate the RL of the top of each peg by rise and fall method.
b) Also calculate the staff readings on each peg and apply usual checks.

## Short answer

Fly level from BM 1280.00 m gives CP3 RL 1278.882 m and HI 1281.442 m. On a falling 1 in 200 gradient, peg RLs (30 m spacing): 1278.732, 1278.582, 1278.432, 1278.282 m; staff readings increase: 2.71, 2.86, 3.01, 3.16 m.

## Answer

**Reading of the notes.** The printed "(B)" is taken as the mark of the first BS, which is on the BM of RL 1280.00 m. The four BS readings are 0.63, 2.05, $-2.424$ and 2.56 m, and the three FS readings are 2.444, 1.35 and $-2.42$ m (negative readings are taken with the staff held inverted). The set-ups are BS 0.63, FS 2.444; BS 2.05, FS 1.35; BS $-2.424$, FS $-2.42$; and the last BS 2.56 m on the last change point CP3, from which the pegs are set out.

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

| Station | BS | FS | Rise | Fall | RL (m) | HI (m) |
|---|---|---|---|---|---|---|
| BM | 0.630 |  |  |  | 1280.000 | 1280.630 |
| CP1 | 2.050 | 2.444 |  | 1.814 | 1278.186 | 1280.236 |
| CP2 | -2.424 | 1.350 | 0.700 |  | 1278.886 | 1276.462 |
| CP3 | 2.560 | -2.420 |  | 0.004 | 1278.882 | 1281.442 |

#### Arithmetic check

$\Sigma BS - \Sigma FS$ (set-ups 1 to 3) $= (0.63+2.05-2.424) - (2.444+1.35-2.42) = 0.256 - 1.374 = -1.118$ m.

$\Sigma \text{Rise} - \Sigma \text{Fall} = 0.700 - 1.818 = -1.118$ m, and RL of CP3 $-$ RL of BM $= 1278.882 - 1280.000 = -1.118$ m. All agree.

#### b) Pegs on a falling gradient of 1 in 200

HI of the last set-up $= 1278.882 + 2.56 = 1281.442$ m. Fall per 30 m $= 30/200 = 0.15$ m. The pegs are at 30, 60, 90 and 120 m from CP3.

RL of a peg $=$ RL of CP3 $- 0.15 \times n$, and staff reading $=$ HI $-$ RL (the readings increase by 0.15 m per peg).

| Peg | Distance from CP3 (m) | RL (m) | Staff reading (m) |
|---|---|---|---|
| 1 | 30 | 1278.732 | 2.710 |
| 2 | 60 | 1278.582 | 2.860 |
| 3 | 90 | 1278.432 | 3.010 |
| 4 | 120 | 1278.282 | 3.160 |

Check: the staff reading on the first peg $= 2.56 + 0.15 = 2.71$ m, and each following peg adds 0.15 m, so the readings are 2.71, 2.86, 3.01, 3.16 m.

**Answer:** RL of CP3 = 1278.882 m (HI = 1281.442 m). RLs of the 4 pegs = 1278.732, 1278.582, 1278.432, 1278.282 m; staff readings on the pegs = 2.71, 2.86, 3.01, 3.16 m.

## Check yourself

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

1. BS -2.424 and FS -2.42 are given. What do negative readings mean?
   A. Staff held inverted; the point is above the line of sight
   B. Staff held upright but the reading is an error
   C. Points below the instrument
   D. Readings to be ignored in the check
   Answer: A. Negative readings come from an inverted staff, and the usual calculations apply with their sign. (Key point: Negative readings)

2. RL of CP3 = 1278.882 m and the last BS = 2.56 m. What is the HI?
   A. 1276.322 m
   B. 1281.442 m
   C. 1278.882 m
   D. 1280.236 m
   Answer: B. HI = 1278.882 + 2.56 = 1281.442 m. (Key point: HI)

3. On a falling gradient of 1 in 200 and 30 m spacing, how do the staff readings on pegs change?
   A. They decrease by 0.15 m per peg
   B. They increase by 6.67 m per peg
   C. They increase by 0.15 m per peg: 2.71, 2.86, 3.01, 3.16 m
   D. They remain 2.56 m
   Answer: C. RL falls 0.15 m per peg, so HI - RL grows by 0.15 m. First reading = 2.56 + 0.15 = 2.71 m. (Key point: Falling gradient)

4. SumBS - SumFS for three set-ups is -1.118 m. What else must equal it?
   A. SumRise + SumFall
   B. The gradient
   C. RL of CP3 only
   D. SumRise - SumFall and RL of CP3 - RL of BM
   Answer: D. 0.700 - 1.818 = -1.118 m and 1278.882 - 1280.000 = -1.118 m. (Key point: Check)

## Related in Surveying I

- [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, BM 1470 m; set 4 pegs](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Afly-level-set-4-pegs-1470-rising-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 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
- [Level book, 20 m interval, RL 1280 to 1279.924](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-20m-interval-1280-1279-924) — Surveying I, chapter 5
- [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

## Related in other subjects

- [RL of hill top P (A to B 112 m, RL 1280)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2073m-ab-112) — Surveying II, chapter 3, asked 4×
- [Gradient A to B, RL of P = 1234.560](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Agradient-ab-tacheometer-100-0-p) — Surveying II, chapter 2, asked 2×