Curvature and refraction in levelling

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.