With H = -0.2325 m and RL of B = 1160.18 m, why is RL of A = 1160.413 m?

B is lower than A by 0.2325 m, so A = RL of B + 0.2325

B is higher than A, so A = RL of B - 0.2325

The error e is added to RL of B

A and B have equal RL after averaging

How is the collimation error obtained from e, and what is its value for 300 m?

e minus combined curvature and refraction = 0.0075 - 0.00606 = 0.00144 m

e plus combined correction = 0.0136 m

combined correction minus e = -0.00144 m, upward

Asked 2/31
2076 Chaitra · 2+2+2 marks

Reciprocal levelling: RL of A, B at 1160.18 m

Short answer

RL of A = 1160.413 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 at	Staff reading A (m)	Staff reading B (m)
1	1.486	1.726
2	1.191	1.416

a) If the reduced level of B is 1160.18 m above the datum, what is the reduced level of A? b) Assuming that the atmospheric conditions remain unchanged during the two sets of observations, calculate the collimation error, precision ratio, and the combined curvature and refraction correction if the distance AB is 300 m.

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

a) Reduced level of A

H_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 = \frac{H_1 + H_2}{2} = -0.2325\ \text{m}\ (\text{B is lower than A by } 0.2325\ \text{m})

\text{RL of A} = \text{RL of B} - H = 1160.18 + 0.2325 = 1160.4125\ \text{m} \approx 1160.413\ \text{m}

b) Collimation error, precision ratio and combined correction

Total error in the far reading:

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

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

\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}

\text{Combined correction} = 0.00706 - 0.00101 = 0.00606\ \text{m} \approx 0.0061\ \text{m}

Collimation error and precision ratio: the error in the far reading is made up of the collimation error and the combined curvature and refraction error, so

\text{Collimation error} = e - (\text{C \& R}) = 0.0075 - 0.00606 = 0.00144\ \text{m in 300 m}

\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 = 1160.413 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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Reciprocal levelling: RL of A, B at 1160.18 m

a) Reduced level of A

b) Collimation error, precision ratio and combined correction

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