---
title: "Curvature and refraction in levelling"
subject: "Surveying I"
chapter: 5
key: "semester_3/surveying_i:curvature-refraction-derivation"
source: https://www.bctengineer.com/q/semester_3/surveying_i/curvature-refraction-derivation
site: BCT Engineering
---

# Curvature and refraction in levelling

Surveying I · Chapter 5: Leveling

- Asked in: 2074 Chaitra, 4 marks
- Video: [6.8 Curvature and Refraction Correction](https://www.youtube.com/watch?v=aDlF3rvG5fI) — Civil Modules
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/surveying_i/curvature-refraction-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5

## Question

Discuss briefly the effect of curvature and refraction in levelling. Derive an expression for curvature correction and for combined curvature and refraction correction.

## Short answer

Curvature makes reading too high: hc = d²/(2R) = 0.0785 d² m (d in km), subtracted. Refraction bends sight down: correction hr = hc/7 = 0.0112 d² m, added. Combined correction = 0.0673 d² m, subtracted. Equalizing BS and FS distances cancels the error.

## Answer

#### Effect of curvature

The earth is a sphere, so a level surface curves away from the horizontal line of sight. A horizontal line of sight at the instrument departs from the level surface through the instrument, so the staff reading at a distance $d$ is **greater** than the true reading by the curvature correction.

Let O be the centre of the earth, R its radius, A the instrument position on the level surface, and AC the horizontal line of sight (tangent at A), of length $d$ to the staff. The staff point B lies on the level surface vertically below C, and $BC = h_c$ is the curvature error. In the triangle OAC, right-angled at A, $OA = R$ and $OC = R + h_c$:

$$
(R + h_c)^2 = R^2 + d^2 \;\Rightarrow\; 2Rh_c + h_c^2 = d^2
$$

$h_c^2$ is negligible, so the curvature correction is

$$
h_c = \frac{d^2}{2R}
$$

With $R = 6370$ km and d in km, $h_c = 0.0785\,d^2$ metres. This correction is **subtracted** from the staff reading.

#### Effect of refraction

Light rays from the staff pass through layers of air of different density and bend towards the earth, so the staff appears higher and the reading is **less** than the true one. The path is nearly a circular arc with a radius of about 7 R, so

$$
h_r = \frac{d^2}{2 \times 7R} = \frac{1}{7}\,h_c = 0.0112\,d^2\ \text{m}
$$

This correction is **added** to the reading.

#### Combined correction

$$
h = h_c - h_r = \frac{d^2}{2R} - \frac{d^2}{14R} = \frac{6}{7}\frac{d^2}{2R} = \frac{3d^2}{7R} = 0.0673\,d^2\ \text{m}
$$

(d in km). Thus the staff reading is made too high by $0.0673\,d^2$ metres; the combined correction is subtracted. For $d = 100$ m, $h = 0.0673 \times 0.01 = 0.0007$ m, which is negligible in ordinary levelling, and is eliminated by equal BS and FS distances.

## Check yourself

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

1. Due to curvature of the earth, the staff reading at distance d is:
   A. Greater than the true reading
   B. Less than the true reading
   C. Unchanged
   D. Zero
   Answer: A. The level surface falls away from the horizontal sight line, so the reading is too high. (Key point: Curvature effect)

2. From (R + hc)^2 = R^2 + d^2 with hc^2 neglected, the curvature correction is:
   A. d/(2R)
   B. d^2/(2R)
   C. d^2/R
   D. 2d^2/R
   Answer: B. 2R hc = d^2 gives hc = d^2/2R = 0.0785 d^2 m (d in km). (Key point: Curvature formula)

3. Refraction correction relative to curvature is:
   A. Equal to hc
   B. 1/7 of hc, giving a reading too high
   C. 1/7 of hc, giving a reading too low
   D. 7 times hc
   Answer: C. The ray path has radius about 7R so hr = d^2/14R = hc/7, bending toward earth so the reading is lower. (Key point: Refraction)

4. The combined correction h = hc - hr equals:
   A. 0.0785 d^2 metres, added
   B. 0.0112 d^2 metres, subtracted
   C. 0.0897 d^2 metres, subtracted
   D. 0.0673 d^2 metres, subtracted from the reading
   Answer: D. hc - hc/7 = 6/7 x 0.0785 d^2 = 0.0673 d^2; the reading is too high so it is subtracted. (Key point: Combined correction)

## Related in Surveying I

- [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×
- [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×
- [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 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
- [Reciprocal levelling: why and formula](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Areciprocal-leveling-derivation) — Surveying I, chapter 5, asked 3×

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