Reciprocal levelling across river, 250 m

Data: AB = 250 m. Level at A: height of the eyepiece above A = 1.339 m, staff reading on B = 2.518 m. Level at B: height of the eyepiece above B = 1.332 m, staff reading on A = 0.524 m. The height of the eyepiece above the station acts as the (near) staff reading on that station.

a) Difference of level between A and B

Level at A: A is higher than B by (reading on B) $-$ (eyepiece height) $= 2.518 - 1.339 = 1.179$ m.

Level at B: A is higher than B by (eyepiece height) $-$ (reading on A) $= 1.332 - 0.524 = 0.808$ m.

The true difference of level is the mean of the two (the error in the far reading cancels):

$$H = \frac{1.179 + 0.808}{2} = 0.9935\ \text{m} \approx 0.993\ \text{m}\ (\text{A higher than B})$$

The total error in each far reading:

$$e = \frac{1.179 - 0.808}{2} = 0.1855\ \text{m}$$

b) Curvature and refraction correction

With $d = 0.250$ km and $R = 6365$ km:

$$\text{Curvature} = \frac{d^2}{2R} = \frac{0.25^2}{2 \times 6365} = 0.00491\ \text{m}$$ $$\text{Refraction} = \frac{1}{7} \times 0.00491 = 0.00070\ \text{m}$$ $$\text{Combined correction} = 0.00491 - 0.00070 = 0.00421\ \text{m} \approx 0.0042\ \text{m}\ (\text{to be subtracted from the far reading})$$

c) Error due to imperfect adjustment of the diaphragm wires (collimation error)

The total error in the far reading $e$ is the sum of the collimation error and the combined curvature-refraction error, so

$$\text{Collimation error} = e - 0.0042 = 0.1855 - 0.0042 = 0.1813\ \text{m in 250 m}$$

The readings are too large, so the line of collimation is inclined upward.

d) RL of A

$$\text{RL of A} = \text{RL of B} + H = 1460.605 + 0.9935 = 1461.599\ \text{m}$$

Answer: (a) difference of level = 0.993 m (A above B); (b) combined curvature and refraction correction = 0.0042 m; (c) collimation error = 0.181 m in 250 m (inclined upward); (d) RL of A = 1461.599 m.

Note: the two apparent differences (1.179 m and 0.808 m) disagree by 0.371 m, which is unusually large for a 250 m crossing; the figures are used as printed. The method does not change if a printed value differs.