---
title: "Reciprocal levelling: RL of A, B at 1260.18 m"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:reciprocal-leveling-rl-b-1260-18"
source: https://www.bctengineer.com/q/semester_3/surveying_i/reciprocal-leveling-rl-b-1260-18
site: BCT Engineering
---

# Reciprocal levelling: RL of A, B at 1260.18 m

Surveying I · Chapter 5: Leveling

- Asked in: 2074 Chaitra, 6 marks
- Video: [1st Numerical of Reciprocal Levelling](https://www.youtube.com/watch?v=To9HwB1a0dQ) — Ekeeda GATE & ESE
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/reciprocal-leveling-rl-b-1260-18
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

Reciprocal levelling was conducted across a wide river to determine the difference in level of points A and B, A situated on one bank of the river and B situated on the other. The following results on the staff held vertically at A and B from level stations 1 and 2 respectively were obtained. The level station 1 was near to A and station 2 was near to B.

| Instrument at | Staff reading A (m) | Staff reading B (m) |
|---|---|---|
| 1 | 1.486 | 1.726 |
| 2 | 1.191 | 1.416 |

If the reduced level of B is 1260.18 m above the datum, what is the reduced level of A? Assuming that the atmospheric conditions remain unchanged during the two sets of observations, calculate a) the combined curvature and refraction correction if the distance AB is 300 m, and b) the collimation error.

## Short answer

RL of A = 1260.413 m. Combined curvature and refraction correction = 0.0061 m. Collimation error = 1.44 mm in 300 m, precision ratio about 1:208,000.

## Answer

**Notation:** station 1 is near A and station 2 is near B. At station 1: staff on A (near) $a_1 = 1.486$, staff on B (far) $b_1 = 1.726$. At station 2: staff on A (far) $a_2 = 1.191$, staff on B (near) $b_2 = 1.416$. The rise of B above A is BS $-$ FS.

#### Reduced level of A

$$
H_1 = a_1 - b_1 = 1.486 - 1.726 = -0.240\ \text{m}, \qquad H_2 = a_2 - b_2 = 1.191 - 1.416 = -0.225\ \text{m}
$$

$$
H = \frac{H_1 + H_2}{2} = -0.2325\ \text{m}\ (\text{B is lower than A by } 0.2325\ \text{m})
$$

$$
\text{RL of A} = \text{RL of B} - H = 1260.18 + 0.2325 = 1260.4125\ \text{m} \approx 1260.413\ \text{m}
$$

**Total error in the far reading:**

$$
e = \frac{H_2 - H_1}{2} = \frac{-0.225 - (-0.240)}{2} = +0.0075\ \text{m}
$$

#### a) Combined curvature and refraction correction (AB = 300 m, R = 6370 km)

$$
\frac{d^2}{2R} = \frac{0.3^2}{2 \times 6370} = 0.00706\ \text{m}, \qquad \text{refraction} = \frac{1}{7} \times 0.00706 = 0.00101\ \text{m}
$$

$$
\text{Combined correction} = 0.00706 - 0.00101 = 0.00606\ \text{m} \approx 0.0061\ \text{m}
$$

#### b) Collimation error

The error in the far reading is made up of the collimation error and the combined curvature and refraction error, so

$$
\text{Collimation error} = e - (\text{C \& R}) = 0.0075 - 0.00606 = 0.00144\ \text{m in 300 m}
$$

$$
\text{Precision ratio} = \frac{0.00144}{300} \approx \frac{1}{207\,600}
$$

The error is positive, so the far readings are too large (the line of collimation is inclined upward). The total error in the far sight, including curvature and refraction, is 0.0075 m in 300 m, or 1 in 40,000.

**Answer:** RL of A = 1260.413 m; combined curvature and refraction correction = 0.0061 m; collimation error = 1.44 mm in 300 m (about 0.48 mm per 100 m), precision ratio about 1 in 208,000.

## Check yourself

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

1. The staff readings from station 1 are A = 1.486 m, B = 1.726 m. What does this tell about B relative to A?
   A. B is lower than A by 0.240 m (apparent)
   B. B is higher than A by 0.240 m
   C. B is lower than A by 3.212 m
   D. B is at the same level as A
   Answer: A. Rise of B above A = BS - FS = 1.486 - 1.726 = -0.240 m, a fall. (Key point: Rise or fall)

2. With RL of B = 1260.18 m and H = -0.2325 m, what is RL of A?
   A. 1259.948 m
   B. 1260.413 m
   C. 1260.180 m
   D. 1260.4125 m minus 0.0075 m
   Answer: B. B is 0.2325 m below A, so RL of A = 1260.18 + 0.2325 = 1260.4125 m, about 1260.413 m. (Key point: RL of A)

3. Which expression gives the combined curvature and refraction correction for 300 m (R = 6370 km)?
   A. d^2/2R + (1/7)(d^2/2R) = 0.0081 m
   B. d^2/2R only = 0.0071 m
   C. d^2/2R - (1/7)(d^2/2R) = 0.00706 - 0.00101 = 0.0061 m
   D. (1/7)(d^2/2R) = 0.0010 m
   Answer: C. Refraction is 1/7 of curvature and acts opposite to it, so the combined correction is 6/7 of d^2/2R = 0.0061 m. (Key point: Combined correction)

4. The collimation error comes out +1.44 mm in 300 m. What does this mean?
   A. Far readings are too small; the line of collimation is inclined downward
   B. The instrument is perfectly adjusted
   C. Only the near readings are in error
   D. Far readings are too large; the line of collimation is inclined upward
   Answer: D. A positive error means the far readings are too large, which happens when the line of collimation tilts upward. (Key point: Sign of collimation error)

## Close variants set in other exams

- [Reciprocal levelling: RL of A, B at 1160.18 m](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Areciprocal-leveling-rl-b-1160-18) — Surveying I, chapter 5, asked 2×

## Related in Surveying I

- [Reciprocal levelling across river, 250 m](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Areciprocal-leveling-river-250m-eyepiece-height) — Surveying I, chapter 5
- [Reciprocal levelling; RL of closed traverse stations](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Areciprocal-leveling-closed-traverse-bcdefgb) — 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