---
title: "Two peg test: collimation inclined up or down"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:two-peg-dumpy-cd-df-120m"
source: https://www.bctengineer.com/q/semester_3/surveying_i/two-peg-dumpy-cd-df-120m
site: BCT Engineering
---

# Two peg test: collimation inclined up or down

Surveying I · Chapter 5: Leveling

- Asked in: 2081 Bhadra, 5 marks
- 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-dumpy-cd-df-120m
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

While carrying out permanent adjustment of a dumpy level by two peg method, the following observations were made.

| Instrument Station | Staff reading C | Staff reading D | Remarks |
|---|---|---|---|
| At mid-point F | 2.000 | 3.000 | Distance between C and D = 100 m |
| Near C (i.e. DF=120 m) | 1.500 | 2.750 | |

Check whether the line of collimation is inclined upward or downward. What is the reading in C and D with respect to the horizontal line of sight when the instrument is nearer to C?

## Short answer

True difference (mid-point): 3.000-2.000 = 1.000 m (D lower). Apparent near C: 2.750-1.500 = 1.250 m. Error = 0.250 m over 100 m sight difference; line of collimation inclined upward. Correct horizontal readings: C=1.450 m, D=2.450 m.

## Answer

**Given:** staffs at C and D, 100 m apart. Setup 1: instrument at the mid-point F, readings C = 2.000 m, D = 3.000 m. Setup 2: instrument on the line DC produced, near C, with DF = 120 m, so CF = 120 − 100 = 20 m; readings C = 1.500 m, D = 2.750 m.

#### Step 1: True difference of level (mid-point setup)

Equal sight lengths remove the collimation error, so

$$
\text{True difference (D below C)} = 3.000 - 2.000 = 1.000\ \text{m}
$$

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

$$
\text{Apparent difference} = 2.750 - 1.500 = 1.250\ \text{m}
$$

#### Step 3: Error and direction

$$
e = 1.250 - 1.000 = 0.250\ \text{m for a difference of sight lengths } (120 - 20) = 100\ \text{m}
$$

The error per metre of sight is $0.250/100 = 0.0025$ m/m (1 in 400; angle about 8′36″). The far staff D reads more than it should, so the line of collimation is **inclined upward**.

#### Step 4: Readings for a horizontal line of sight (instrument near C)

Error at C (20 m): $0.0025 \times 20 = 0.050$ m. Error at D (120 m): $0.0025 \times 120 = 0.300$ m.

$$
C = 1.500 - 0.050 = 1.450\ \text{m}, \qquad D = 2.750 - 0.300 = 2.450\ \text{m}
$$

Check: $2.450 - 1.450 = 1.000$ m, equal to the true difference.

**Answer: the line of collimation is inclined upward; the readings for a horizontal line of sight are C = 1.450 m and D = 2.450 m.** The cross-hair is then adjusted to read 1.450 on C (or 2.450 on D) with the bubble centred.

## Check yourself

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

1. Why is the mid-point setup reading used for the true difference?
   A. Equal sight lengths cancel the collimation error
   B. The staff is vertical there
   C. Curvature is larger there
   D. It is nearer to the BM
   Answer: A. So 3.000 - 2.000 = 1.000 m is true. (Key point: True difference)

2. The apparent difference from the second setup is:
   A. 1.500 m
   B. 1.250 m
   C. 1.000 m
   D. 0.250 m
   Answer: B. 2.750 - 1.500 = 1.250 m. (Key point: Apparent difference)

3. The error 0.250 m is spread over a sight difference of 100 m (120 - 20). The line of collimation is:
   A. Horizontal
   B. Cannot tell
   C. Inclined upward because the far staff D reads too much
   D. Inclined downward
   Answer: C. Apparent difference exceeds the true one, so the far staff reads high: upward. (Key point: Direction)

4. The corrected readings for a horizontal sight are:
   A. C = 1.500, D = 2.500
   B. C = 1.550, D = 3.050
   C. C = 1.450, D = 2.750
   D. C = 1.450, D = 2.450
   Answer: D. Errors 0.0025 x 20 = 0.050 and 0.0025 x 120 = 0.300 are subtracted. (Key point: Corrected readings)

## Related in Surveying I

- [Two peg test and permanent adjustment](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-test-procedure) — Surveying I, chapter 5, asked 4×
- [Two peg: closing error and precision of level](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-cd-50m-e-mid-f-5m) — Surveying I, chapter 5
- [Two peg test, AB 100 m, Q 10 m](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-ab-100m-q-10m-inside) — Surveying I, chapter 5
- [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
- [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

## Related in other subjects

- [Coordinates of instrument station O and station A](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aresection-pwd-instrument-station-a) — Surveying II, chapter 5
- [Pressure to drive oil up inclined pipe](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aviscous-fluid-upslope-pressure) — Hydraulics, chapter 1