---
title: "Fly levelling with inverted staff, BM 1290.500"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:fly-level-inverted-staff-1290-500"
source: https://www.bctengineer.com/q/semester_3/surveying_i/fly-level-inverted-staff-1290-500
site: BCT Engineering
---

# Fly levelling with inverted staff, BM 1290.500

Surveying I · Chapter 5: Leveling

- Asked in: 2065 Shrawan, 9 marks
- 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-inverted-staff-1290-500
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

During a fly levelling operation, the following observations were made: BS: (-)1.650, 2.155, 1.405, 2.655, 2.435 m; FS: 2.455, 1.305, 0.555, 2.405 m. The first backsight was taken on a BM of RL 1290.500 m with inverted staff. From the last backsight, it is required to set four pegs each at a distance of 20 m on a falling gradient of 1 in 100. Calculate the RL (reduced levels) of these four pegs.

## Short answer

The inverted first BS (−1.650) is added to BM RL 1290.500 to get HI. Reducing levels yields CP4 RL 1288.345 m. With last HI 1290.780 m and a fall of 0.2 m per 20 m, the RLs of the four pegs are 1288.145, 1287.945, 1287.745, and 1287.545 m.

## Answer

**Method:** HI = RL of the point on which the back sight is taken + back sight reading. An inverted staff reading (on a ceiling or beam above the point) is booked with a minus sign, so the instrument is *lower* than the point: HI = RL + (-1.650).

| Station | BS | FS | HI (m) | RL (m) |
|---|---|---|---|---|
| BM | -1.650 (inverted) | | 1288.850 | 1290.500 |
| CP1 | 2.155 | 2.455 | 1288.550 | 1286.395 |
| CP2 | 1.405 | 1.305 | 1288.650 | 1287.245 |
| CP3 | 2.655 | 0.555 | 1290.750 | 1288.095 |
| CP4 | 2.435 | 2.405 | 1290.780 | 1288.345 |

Station RLs by HI method: RL = HI - FS. For example, CP1 = 1288.850 - 2.455 = 1286.395 m.

#### Check
The last back sight (2.435) belongs to the next set-up, so it is left out of the check:

$$
\Sigma BS - \Sigma FS = 4.565 - 6.720 = -2.155 \text{ m}
$$

Last RL - first RL = 1288.345 - 1290.500 = -2.155 m. The two agree.

#### Setting out four pegs on a falling gradient 1 in 100
The last back sight (2.435 m) is on the last change point, CP4, whose RL is 1288.345 m. The HI of the last set-up is 1288.345 + 2.435 = 1290.780 m. A gradient of 1 in 100 falling means a fall of 1 m per 100 m, i.e. 0.2 m per 20 m peg.

| Peg | Distance from CP4 | RL = 1288.345 - fall (m) | Staff reading = HI - RL (m) |
|---|---|---|---|
| P1 | 20 m | 1288.145 | 2.635 |
| P2 | 40 m | 1287.945 | 2.835 |
| P3 | 60 m | 1287.745 | 3.035 |
| P4 | 80 m | 1287.545 | 3.235 |

**Answer:** RLs of the four pegs = 1288.145, 1287.945, 1287.745, 1287.545 m (staff readings 2.635, 2.835, 3.035, 3.235 m).

## Check yourself

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

1. The first BS is -1.650 m, with inverted staff, on BM 1290.500. What is HI?
   A. 1292.150 m
   B. 1288.850 m
   C. 1289.500 m
   D. 1290.500 m
   Answer: B. HI = RL + BS = 1290.500 + (-1.650) = 1288.850 m; the instrument is below the point. (Key point: HI with inverted staff)

2. A falling gradient of 1 in 100 with pegs 20 m apart means each peg is lower than the previous by:
   A. 0.020 m
   B. 0.100 m
   C. 0.200 m
   D. 2.000 m
   Answer: C. 20/100 = 0.2 m fall per peg. (Key point: Gradient fall)

3. The HI of the last set-up is 1290.780 m. What reading is needed at P1 (RL 1288.145)?
   A. 2.835 m
   B. 2.635 m
   C. 2.435 m
   D. 1.288 m
   Answer: B. Reading = HI - RL = 1290.780 - 1288.145 = 2.635 m. Readings grow by 0.2 m for each further peg. (Key point: Staff reading for pegs)

4. Why is the last BS (2.435) left out of the arithmetic check?
   A. It is an error
   B. It is an IS
   C. It belongs to the next set-up used for setting out pegs and has no matching FS
   D. It is always ignored in any level book
   Answer: C. The last BS starts a new set-up with no FS, so only the 4 BS with 4 FS enter the check: -2.155 m. (Key point: Check exclusion)

## 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, 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
- [Differential levelling loop on BM Gate](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Adifferential-leveling-loop-bm-gate) — Surveying I, chapter 5
- [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

- [Errors encountered during file operation](https://www.bctengineer.com/question.md?key=semester_3%2Fobject_oriented_programming%3Afile-errors) — Object Oriented Programming, chapter 8
- [RL of hill top P (A to B 112 m, RL 1290)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2077c-ab-112) — Surveying II, chapter 3, asked 4×