--- title: "Reciprocal levelling across river, 250 m" subject: "Surveying I" chapter: 5 key: "semester_3/surveying_i:reciprocal-leveling-river-250m-eyepiece-height" source: https://www.bctengineer.com/q/semester_3/surveying_i/reciprocal-leveling-river-250m-eyepiece-height site: BCT Engineering --- # Reciprocal levelling across river, 250 m Surveying I · Chapter 5: Leveling - Asked in: 2079 Bhadra, 6 marks - Video: [reciprocal levelling numerical | survey](https://www.youtube.com/watch?v=2u5QBEY1c1I) — CE (simplified engineering) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/reciprocal-leveling-river-250m-eyepiece-height - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-5 ## Question The results of reciprocal levelling between stations A and B 250 m apart on opposite sides of a wide river were as follows: | Level at | Height of eye piece (m) | Staff readings | |---|---|---| | A | 1.339 | 2.518 on B | | B | 1.332 | 0.524 on A | Find: a) The difference of level between A and B. b) Curvature and refraction correction if mean radius of earth = 6365 km. c) The error due to imperfect adjustment of the diaphragm wires. d) If RL of B = 1460.605 m, find RL of A. ## Short answer True difference of level: mean of two apparent differences gives 0.993 m, A higher than B. Combined curvature and refraction correction is 0.0042 m. Collimation error is 0.181 m in 250 m (upward). RL of A = 1461.599 m. ## Answer **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the true difference of level taken as the mean of the two apparent differences? A. The far-reading error (collimation, curvature, refraction) is the same in both sets and cancels in the mean B. The mean of two values is always more accurate than one C. The near readings contain all the error, so averaging removes it D. Curvature and refraction act only on the near staff Answer: A. The error in the far reading has the same size and effect on the apparent difference with opposite sign in the two sets, so it cancels when the two are averaged. (Key point: Mean of two sets) 2. At A the eyepiece height is 1.339 m and the reading on B is 2.518 m. What apparent difference of level does this give? A. A is higher than B by 3.857 m B. A is higher than B by 1.179 m C. B is higher than A by 1.179 m D. A is higher than B by 0.808 m Answer: B. A is higher than B by (reading on B) - (eyepiece height) = 2.518 - 1.339 = 1.179 m. (Key point: Apparent differences) 3. What is the final difference of level between A and B? A. 0.371 m, A higher B. 1.9870 m, A higher C. 0.9935 m (about 0.993 m), A higher D. 0.9935 m, B higher Answer: C. H = (1.179 + 0.808)/2 = 0.9935 m, and A is above B in both sets. (Key point: True difference) 4. For 250 m and R = 6365 km, what combined curvature and refraction correction is obtained? A. 0.0049 m (curvature only) B. 0.0056 m (curvature plus refraction) C. 0.1813 m D. 0.0042 m (curvature 0.00491 minus refraction 0.00070) Answer: D. Curvature = d^2/2R = 0.00491 m; refraction is 1/7 of it, 0.00070 m; the combined effect is the difference, 0.0042 m. 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