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Reciprocal levelling: RL of A, B at 1260.18 m

Short answer

RL of A = 1260.413 m. Combined curvature and refraction correction = 0.0061 m. Collimation error = 1.44 mm in 300 m, precision ratio about 1:208,000.

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 atStaff reading A (m)Staff reading B (m)
11.4861.726
21.1911.416

If the reduced level of B is 1260.18 m above the datum, what is the reduced level of A? Assuming that the atmospheric conditions remain unchanged during the two sets of observations, calculate a) the combined curvature and refraction correction if the distance AB is 300 m, and b) the collimation error.

Notation: station 1 is near A and station 2 is near B. At station 1: staff on A (near) a1=1.486a_1 = 1.486, staff on B (far) b1=1.726b_1 = 1.726. At station 2: staff on A (far) a2=1.191a_2 = 1.191, staff on B (near) b2=1.416b_2 = 1.416. The rise of B above A is BS −- FS.

Reduced level of A

H1=a1−b1=1.486−1.726=−0.240 m,H2=a2−b2=1.191−1.416=−0.225 mH_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=H1+H22=−0.2325 m (B is lower than A by 0.2325 m)H = \frac{H_1 + H_2}{2} = -0.2325\ \text{m}\ (\text{B is lower than A by } 0.2325\ \text{m}) RL of A=RL of B−H=1260.18+0.2325=1260.4125 m≈1260.413 m\text{RL of A} = \text{RL of B} - H = 1260.18 + 0.2325 = 1260.4125\ \text{m} \approx 1260.413\ \text{m}

Total error in the far reading:

e=H2−H12=−0.225−(−0.240)2=+0.0075 me = \frac{H_2 - H_1}{2} = \frac{-0.225 - (-0.240)}{2} = +0.0075\ \text{m}

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

d22R=0.322×6370=0.00706 m,refraction=17×0.00706=0.00101 m\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} Combined correction=0.00706−0.00101=0.00606 m≈0.0061 m\text{Combined correction} = 0.00706 - 0.00101 = 0.00606\ \text{m} \approx 0.0061\ \text{m}

b) Collimation error

The error in the far reading is made up of the collimation error and the combined curvature and refraction error, so

Collimation error=e−(C & R)=0.0075−0.00606=0.00144 m in 300 m\text{Collimation error} = e - (\text{C \& R}) = 0.0075 - 0.00606 = 0.00144\ \text{m in 300 m} Precision ratio=0.00144300≈1207 600\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 = 1260.413 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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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