Reciprocal levelling AB 250 m, RL 1450.500

Notation: instrument near A: staff on A (near) $a_1 = 1.252$, staff on B (far) $b_1 = 1.052$. Instrument near B: staff on A (far) $a_2 = 1.419$, staff on B (near) $b_2 = 1.253$. Rise of B above A $=$ BS $-$ FS. AB = 250 m, R = 6370 km (assumed).

RL of B

$$H_1 = a_1 - b_1 = 1.252 - 1.052 = +0.200\ \text{m}, \qquad H_2 = a_2 - b_2 = 1.419 - 1.253 = +0.166\ \text{m}$$ $$H = \frac{H_1 + H_2}{2} = \frac{0.200 + 0.166}{2} = +0.183\ \text{m}\ (\text{B higher than A})$$ $$\text{RL of B} = 1450.500 + 0.183 = 1450.683\ \text{m}$$

Total error in the far reading

$$e = \frac{H_2 - H_1}{2} = \frac{0.166 - 0.200}{2} = -0.017\ \text{m}$$

The negative sign means the far readings are too small by 0.017 m.

Combined curvature and refraction correction (250 m)

$$\frac{d^2}{2R} - \frac{1}{7}\frac{d^2}{2R} = \frac{6}{7} \times \frac{0.25^2}{2 \times 6370} \times 1000 = 0.00420\ \text{m} \approx 0.0042\ \text{m}$$

(Curvature and refraction together make the far readings too large, by 0.0042 m.)

Collimation error

$$\text{Collimation error} = e - (\text{C \& R}) = -0.0170 - 0.0042 = -0.0212\ \text{m in 250 m}$$

That is 8.5 mm per 100 m (about 1 in 11,800); the sign is negative, so the line of collimation is inclined downward.

Correct reading on A in the second set-up

The far reading on A (1.419 m) is too small by the total error of 0.017 m. The reading free of collimation, curvature and refraction errors is

$$a_2 - e = 1.419 - (-0.017) = 1.436\ \text{m}$$

(If only the collimation error is removed, with curvature and refraction left in, the reading is $1.419 + 0.0212 = 1.4402$ m.)

Answer: RL of B = 1450.683 m; combined correction = 0.0042 m; collimation error = −0.0212 m in 250 m (downward); correct reading on A in the second set-up = 1.436 m.