---
title: "Collimation: pegs 60 m apart, Q near B"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:two-peg-pegs-ab-60m-q-6m-from-b"
source: https://www.bctengineer.com/q/semester_3/surveying_i/two-peg-pegs-ab-60m-q-6m-from-b
site: BCT Engineering
---

# Collimation: pegs 60 m apart, Q near B

Surveying I · Chapter 5: Leveling

- Asked in: 2078 Bhadra, 5 marks
- Video: [Finding Collimation Error of a Level using Two Peg Test Method|Solved Problem](https://www.youtube.com/watch?v=-QssAt7Kx64) — Master Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/two-peg-pegs-ab-60m-q-6m-from-b
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

A level instrument was set up exactly mid-way between two pegs A and B 60 m apart and found the true difference of level = 0.320 m. The level instrument was then set up at a point Q on the line AB, 6 m from B and inside of AB, and the following readings were taken at A and B.

| Instrument at | Sighted to | Staff reading S (m), Middle |
|---|---|---|
| Q | A | 1.387 |
| Q | B | 1.069 |

Compute the correct staff readings on A and B when the line of collimation is exactly horizontal. Also compute the collimation precision ratio.

## Short answer

Collimation error -0.002 m over 48 m, precision 1 in 24000, line inclined downward. Correct staff readings for horizontal line: A = 1.389 m, B = 1.069 m.

## Answer

**Data:** A and B are 60 m apart; with the level at the mid-point the true difference of level is 0.320 m (reading on A is greater than on B by 0.320 m, since the apparent difference is also positive). Setup 2: level at Q on AB, 6 m from B (so 54 m from A): readings A = 1.387 m, B = 1.069 m.

#### Step 1: True difference of level (instrument at the mid-point)

With equal sight lengths the collimation error cancels, so

$$
\text{True difference of readings (A} - \text{B)} = 0.320\ \text{m}
$$

#### Step 2: Apparent difference from the second setup

$$
\text{Apparent difference} = 1.387 - 1.069 = 0.318\ \text{m}
$$

#### Step 3: Error and collimation precision

The reading error of a staff is proportional to its distance from the instrument. In the second setup the distances are A = 54 m and B = 6 m, so the difference of sight lengths is 48 m.

$$
e = \text{apparent} - \text{true} = 0.318 - 0.320 = -0.002\ \text{m}
$$

$$
\text{Error per metre of sight} = \frac{0.002}{48} = 0.000042\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{24000}
$$

This is 4.2 mm per 100 m of sight, which is within the usual permissible limit of about 10 mm per 100 m (1 in 10,000) for ordinary levelling, so the instrument needs no adjustment.

The readings are too small and the error grows with distance (the far staff reads too little), so the line of collimation is **inclined downward**.

#### Step 4: Correct readings for a horizontal line of sight (second setup)

Error in reading on A $= -0.000042 \times 54 = -0.00225$ m; error in reading on B $= -0.000042 \times 6 = -0.00025$ m.

$$
A = 1.387 - (-0.00225) = 1.38925\ \text{m}, \qquad B = 1.069 - (-0.00025) = 1.06925\ \text{m}
$$

Check: A $-$ B $= 1.38925 - 1.06925 = 0.320$ m, equal to the true difference.

**Answer:** correct readings A = 1.389 m (1.38925), B = 1.069 m (1.06925); collimation precision 1 in 24,000 (the line of collimation is inclined downward, but the error is small, 4.2 mm per 100 m).

## Check yourself

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

1. Apparent difference at Q is:
   A. 0.318 m
   B. 0.320 m
   C. 0.002 m
   D. 2.456 m
   Answer: A. 1.387 - 1.069 = 0.318 m. (Key point: Apparent difference)

2. e = 0.318 - 0.320 = -0.002 m means:
   A. The staff was inverted
   B. Line of collimation inclined downward
   C. Inclined upward
   D. No error
   Answer: B. The far staff reads too little, so the sight is downward. (Key point: Error sign)

3. The precision ratio is about:
   A. 1 in 2,400
   B. 1 in 600
   C. 1 in 24,000 (0.002/48)
   D. 1 in 30,000
   Answer: C. Sight difference 54 - 6 = 48 m; 0.002/48 = 1/24000. (Key point: Precision)

4. Corrected reading on A is:
   A. 1.38475 m
   B. 1.387 m
   C. 1.320 m
   D. 1.38925 m
   Answer: D. Error on A = -0.00225 m, so subtract it: 1.387 + 0.00225. (Key point: Corrected A)

## Related in Surveying I

- [Collimation error with level mid-way and near B](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-60m-staff-1-855-1-625) — Surveying I, chapter 5
- [Two peg, AB = 60 m, near A](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-ab-60m-near-a-6m-inside) — Surveying I, chapter 5
- [Collimation: level 5 m from B on AB produced](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-50m-staff-1-875-1-790-b-5m) — Surveying I, chapter 5
- [Two peg test: collimation inclined up or down](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-dumpy-cd-df-120m) — Surveying I, chapter 5
- [Two peg test, AB = 40 m, near B](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-ab-40m-inner-side-b) — Surveying I, chapter 5
- [Collimation error from 5 m and 45 m setup](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-collimation-ab-50m-true-diff-0-250) — Surveying I, chapter 5

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