---
title: "Reciprocal levelling: RL of A, B at 1160.18 m"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:reciprocal-leveling-rl-b-1160-18"
source: https://www.bctengineer.com/q/semester_3/surveying_i/reciprocal-leveling-rl-b-1160-18
site: BCT Engineering
---

# Reciprocal levelling: RL of A, B at 1160.18 m

Surveying I · Chapter 5: Leveling

- Asked in: 2076 Chaitra, 2+2+2 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-1160-18
- 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 |

a) If the reduced level of B is 1160.18 m above the datum, what is the reduced level of A?
b) Assuming that the atmospheric conditions remain unchanged during the two sets of observations, calculate the collimation error, precision ratio, and the combined curvature and refraction correction if the distance AB is 300 m.

## Short answer

RL of A = 1160.413 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.

#### a) Reduced level of A

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

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

$$
\text{RL of A} = \text{RL of B} - H = 1160.18 + 0.2325 = 1160.4125\ \text{m} \approx 1160.413\ \text{m}
$$

#### b) Collimation error, precision ratio and combined correction

**Total error in the far reading:**

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

**Combined curvature and refraction correction** (AB = 300 m, R = 6370 km):

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

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

**Collimation error and precision ratio:** 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 = 1160.413 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. In the second set (instrument near B), the staff on A (far) reads 1.191 m and on B (near) 1.416 m. What is H2 = a2 - b2?
   A. -0.225 m
   B. +0.225 m
   C. -0.240 m
   D. +2.607 m
   Answer: A. H2 = 1.191 - 1.416 = -0.225 m, meaning B is lower than A in that set. (Key point: Apparent differences)

2. With H = -0.2325 m and RL of B = 1160.18 m, why is RL of A = 1160.413 m?
   A. B is higher than A, so A = RL of B - 0.2325
   B. B is lower than A by 0.2325 m, so A = RL of B + 0.2325
   C. The error e is added to RL of B
   D. A and B have equal RL after averaging
   Answer: B. RL of A = RL of B - H = 1160.18 + 0.2325 = 1160.4125 m. (Key point: RL of A)

3. How is the total error in the far reading found, and what is it here?
   A. e = (H1 + H2)/2 = -0.2325 m
   B. e = H1 - H2 = -0.015 m
   C. e = (H2 - H1)/2 = +0.0075 m
   D. e = H2 + H1 = -0.465 m
   Answer: C. The half difference of the two apparent differences gives the far-reading error: (-0.225 + 0.240)/2 = +0.0075 m. (Key point: Total far-reading error)

4. How is the collimation error obtained from e, and what is its value for 300 m?
   A. e plus combined correction = 0.0136 m
   B. e alone = 0.0075 m
   C. combined correction minus e = -0.00144 m, upward
   D. e minus combined curvature and refraction = 0.0075 - 0.00606 = 0.00144 m
   Answer: D. e contains both the collimation error and the curvature-refraction error, so the latter (0.00606 m) is subtracted. (Key point: Collimation error)

## Close variants set in other exams

- [Reciprocal levelling: RL of A, B at 1260.18 m](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Areciprocal-leveling-rl-b-1260-18) — 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