---
title: "Level book TBM1 to TBM2 (4th, 7th)"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:level-book-tbm1-tbm2-1449-870"
source: https://www.bctengineer.com/q/semester_3/surveying_i/level-book-tbm1-tbm2-1449-870
site: BCT Engineering
---

# Level book TBM1 to TBM2 (4th, 7th)

Surveying I · Chapter 5: Leveling

- Asked in: 2078 Bhadra, 6 marks; 2080 Baisakh, 6 marks; 2072 Chaitra, 6 marks (asked 3 times)
- 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-tbm1-tbm2-1449-870
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

The following readings were taken during a levelling work from TBM1 to TBM2: 2.191, 2.505, 2.325, 1.496, 3.019, 2.513, 2.811, 1.752 and 3.824 m. The level instrument was changed after the 4th and 7th readings. Enter the above readings in a level field book format and compute RLs of all the points and adjust the RLs if error arises. RLs of TBM1 and TBM2 are 1449.870 and 1448.710 m respectively. (Distance between TBM1 and TBM2 was 200 m.)

## Short answer

Computed RL of TBM2 is 1448.701 m, misclosure –0.009 m. Correct proportionally by distance (equal spacing assumed). Adjusted RLs: TBM1 1449.870, Pt2 1449.557, Pt3 1449.739, CP1 1450.569, Pt6 1451.077, CP2 1450.780, TBM2 1448.710 m.

## Answer

**Entering the readings.** The level was set up three times (it was changed after the 4th and 7th readings), so:

- Setup 1: 1st reading 2.191 = BS on TBM1; 2nd (2.505) and 3rd (2.325) = IS; 4th (1.496) = FS on change point CP1.
- Setup 2: 5th (3.019) = BS on CP1; 6th (2.513) = IS; 7th (2.811) = FS on change point CP2.
- Setup 3: 8th (1.752) = BS on CP2; 9th (3.824) = FS on TBM2.

HI $=$ RL of point $+$ BS, and RL of a point $=$ HI $-$ staff reading.

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

| Station | BS | IS | FS | HI (m) | RL (m) |
|---|---|---|---|---|---|
| TBM1 | 2.191 |  |  | 1452.061 | 1449.870 |
| 2 |  | 2.505 |  |  | 1449.556 |
| 3 |  | 2.325 |  |  | 1449.736 |
| CP1 (4th/5th) | 3.019 |  | 1.496 | 1453.584 | 1450.565 |
| 6 |  | 2.513 |  |  | 1451.071 |
| CP2 (7th/8th) | 1.752 |  | 2.811 | 1452.525 | 1450.773 |
| TBM2 |  |  | 3.824 |  | 1448.701 |

#### Arithmetic check

$\Sigma BS - \Sigma FS = 6.962 - (8.131) = -1.169$ ; last RL $-$ first RL $= 1448.701 - 1449.870 = -1.169$. Both are equal, so the reduction is correct.

#### Adjustment of misclosure

Computed RL of the last point = 1448.701 m; known RL = 1448.710 m.

$$
\text{Closing error} = 1448.701 - 1448.710 = -0.009\ \text{m}
$$

The correction at a point is $-(-0.009) \times d / 200$, where $d$ is the distance of the point from TBM1. The 7 staff points are taken at equal spacing, so the 200 m is divided into 6 equal parts of 33.33 m.

| Station | Distance (m) | Correction (m) | Adjusted RL (m) |
|---|---|---|---|
| TBM1 | 0 | +0.0000 | 1449.870 |
| 2 | 33.3333 | +0.0015 | 1449.557 |
| 3 | 66.6667 | +0.0030 | 1449.739 |
| CP1 (4th/5th) | 100 | +0.0045 | 1450.569 |
| 6 | 133.333 | +0.0060 | 1451.077 |
| CP2 (7th/8th) | 166.667 | +0.0075 | 1450.780 |
| TBM2 | 200 | +0.0090 | 1448.710 |


**Answer:** the computed RL of TBM2 is 1448.701 m against the known 1448.710 m, a closing error of −0.009 m (9 mm low). Adjusted RLs: TBM1 1449.870, point 2 1449.557, point 3 1449.739, CP1 1450.569, point 6 1451.077, CP2 1450.780, TBM2 1448.710 m.

*Assumption:* the distances to the intermediate points are not given, so equal spacing between successive staff points is assumed for the proportional adjustment.

## Check yourself

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

1. Level changed after the 4th and 7th readings. How are the 4th and 5th readings classified?
   A. 4th is FS on CP1; 5th is BS on CP1
   B. Both are IS
   C. 4th is BS; 5th is FS
   D. Both are FS
   Answer: A. The 4th reading is the last at setup 1 (FS on change point CP1) and the 5th is the first at setup 2 (BS on the same point). (Key point: Classifying readings)

2. ΣBS = 6.962 and ΣFS = 8.131. What does the arithmetic check give?
   A. −1.169, equal to last RL − first RL
   B. +1.169
   C. 0
   D. −0.009
   Answer: A. ΣBS − ΣFS = −1.169 and 1448.701 − 1449.870 = −1.169, so the reduction is correct. (Key point: Arithmetic check)

3. Computed RL of TBM2 is 1448.701 m against the known 1448.710 m. What is the closing error?
   A. −0.009 m
   B. +0.009 m
   C. +1.169 m
   D. 0.0045 m
   Answer: A. Computed minus known = 1448.701 − 1448.710 = −0.009 m. (Key point: Closing error)

4. How is the correction applied to a point at distance d from TBM1?
   A. Equally to all points
   B. In proportion to d: 0.009 × d/200 added
   C. Only to the last point
   D. Subtracted from the RL in proportion to d
   Answer: B. The error is distributed in proportion to distance; for CP1 at 100 m the correction is +0.0045 m. (Key point: Adjustment)

## Close variants set in other exams

- [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×

## 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