---
title: "Level book, benchmark RL 100.00 at 3rd reading"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:level-book-instrument-shifted-after-3rd-6th"
source: https://www.bctengineer.com/q/semester_3/surveying_i/level-book-instrument-shifted-after-3rd-6th
site: BCT Engineering
---

# Level book, benchmark RL 100.00 at 3rd reading

Surveying I · Chapter 5: Leveling

- Asked in: 2073 Shrawan, 6 marks
- Video: [rise and fall method in levelling | determination of reduced levels | surveying | Animated video](https://www.youtube.com/watch?v=uu55PP8M9zU) — Civil Engineering - Shylus Updates
- 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-instrument-shifted-after-3rd-6th
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

The consecutive readings taken during a levelling operation are as follows: 0.685, 1.315, -1.825, -0.635, 1.205, 1.235, 2.631, 1.355, -2.015. The instrument was shifted after the third and sixth readings. The third reading was taken to a benchmark of assumed elevation 100.00. Find the reduced levels of other points.

## Short answer

Using height of instrument method, RLs of stations 1 to 7 are 97.490, 96.860, 100.000 (BM), 98.160, 98.130, 99.406, 102.776 m.

## Answer

**Arranging the readings.** The instrument was shifted after the 3rd and 6th readings, so there are three set-ups. The 3rd reading is on the benchmark (assumed RL 100.000 m), which is also the first change point.

- Set-up 1: BS 0.685 on station 1; IS 1.315 on station 2; FS $-1.825$ on the BM (point 3).
- Set-up 2: BS $-0.635$ on the BM; IS 1.205 on station 4; FS 1.235 on CP2 (point 5).
- Set-up 3: BS 2.631 on CP2; IS 1.355 on station 6; FS $-2.015$ on station 7.

A negative reading means the staff was held inverted against an overhead object: the point is above the line of sight, so RL $=$ HI $-$ (reading) $=$ HI $+$ the numerical value. Because the BM (3rd point) has the known RL, the HI of set-up 1 is $100.000 + (-1.825) = 98.175$ m, and the RLs of points 1 and 2 are found from it. The HI of set-up 2 is $100.000 + (-0.635) = 99.365$ m.

#### Level field book (height of instrument method)

| Station | BS | IS | FS | HI (m) | RL (m) |
|---|---|---|---|---|---|
| 1 | 0.685 |  |  | 98.175 | 97.490 |
| 2 |  | 1.315 |  |  | 96.860 |
| 3 (BM, CP1) | -0.635 |  | -1.825 | 99.365 | 100.000 |
| 4 |  | 1.205 |  |  | 98.160 |
| 5 (CP2) | 2.631 |  | 1.235 | 100.761 | 98.130 |
| 6 |  | 1.355 |  |  | 99.406 |
| 7 |  |  | -2.015 |  | 102.776 |

#### Arithmetic check

$\Sigma BS - \Sigma FS = 2.681 - (-2.605) = 5.286$ ; last RL $-$ first RL $= 102.776 - 97.490 = 5.286$. Both are equal, so the reduction is correct.


**Answer:** RLs of the stations 1 to 7 = 97.490, 96.860, 100.000, 98.160, 98.130, 99.406, 102.776 m (the BM, point 3, being 100.000 m).

## Check yourself

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

1. The instrument shifts after the 3rd and 6th readings. How are readings 3, 4 and 6 treated?
   A. 3rd is FS on the BM, 4th is BS on the BM from the new set-up, 6th is FS on CP2
   B. 3rd is BS, 4th is FS, 6th is IS
   C. 3rd and 4th are IS, 6th is BS
   D. All three are IS readings
   Answer: A. The reading before a shift is the FS; the first reading after is the BS on the same point. (Key point: Set-ups)

2. The BM has RL 100.000 and FS -1.825. What is the HI of set-up 1?
   A. 101.825 m
   B. 98.175 m
   C. 100.000 m
   D. 98.360 m
   Answer: B. HI = RL + reading = 100.000 + (-1.825) = 98.175 m. (Key point: HI of set-up 1)

3. What does a negative reading imply, and how is RL found?
   A. Point is below the line of sight; RL = HI + value
   B. It is ignored
   C. Point is above the line of sight; RL = HI + numerical value
   D. RL = HI - numerical value
   Answer: C. RL = HI - (-value) = HI + value. (Key point: Inverted staff)

4. What is the arithmetic check for this HI-method book?
   A. ΣBS + ΣFS = 5.286
   B. ΣIS = last RL
   C. ΣBS - ΣIS = 0
   D. ΣBS - ΣFS = last RL - first RL = 5.286 m
   Answer: D. 2.681 - (-2.605) = 5.286 and 102.776 - 97.490 = 5.286 m. (Key point: Check)

## Related in Surveying I

- [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 TBM1-TBM2 with (-)1.685 reading](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-tbm1-tbm2-with-minus-1-685) — Surveying I, chapter 5, asked 4×
- [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×
- [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×
- [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×

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