--- title: "Mean angles and VCR, station O (A-D)" subject: "Surveying I" chapter: 7 key: "semester_3/surveying_i:mean-direction-o-abcd-vcr-120-15-10-ls10" source: https://www.bctengineer.com/q/semester_3/surveying_i/mean-direction-o-abcd-vcr-120-15-10-ls10 site: BCT Engineering --- # Mean angles and VCR, station O (A-D) Surveying I · Chapter 7: Transit and Theodolite - Asked in: 2081 Bhadra, 6 marks; 2075 Asoj, 6 marks (asked 2 times) - Video: [Compute mean horizontal method by Mean direction method. #theodolite ](https://www.youtube.com/watch?v=D_hF_bbdk6U) — Er Dipak Sir - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/surveying_i/mean-direction-o-abcd-vcr-120-15-10-ls10 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7 ## Question A theodolite is set over station O to measure direction to stations A, B, C and D. The observed circle readings are as follows. Compute the mean horizontal angle and adjust them if necessary. Also calculate the missing data of the vertical circle reading. | Instrument | Target Station | Telescope | Horizontal Circle Readings | Vertical Circle Readings | |---|---|---|---|---| | O | A | Direct | 00°00′10″ | 120°15′10″ | | | | Reversed | 180°00′20″ | ? | | | B | Direct | 60°55′10″ | ? | | | | Reversed | 240°55′20″ | 308°51′40″ | | | C | Direct | 140°50′50″ | ? | | | | Reversed | 320°51′20″ | 269°15′10″ | | | D | Direct | 270°20′10″ | 177°20′10″ | | | | Reversed | 90°20′20″ | ? | | | A | Direct | 00°00′20″ | 89°00′10″ | | | | Reversed | 180°00′30″ | ? | ## Short answer Mean directions: (Direct + (Reversed−180°))/2, reduced to A=0°00′00″. Closing error 10″ distributed equally gives adjusted angles: AOB=60°54′57.5″, BOC=79°55′47.5″, COD=129°29′07.5″, DOA=89°40′07.5″. Missing vertical readings from FL+FR=360°: A rev=239°44′50″, B dir=51°08′20″, C dir=90°44′50″, D rev=182°39′50″, A clos rev=270°59′50″. ## Answer **Method.** The mean direction of each target is found from the two faces: mean = (Direct + (Reversed - 180°)) / 2. The directions are referred to the first reading on A (taken as 0°00′00″). The second reading on A closes the horizon, so the difference from 360° is the closing error, which is distributed equally. #### Mean horizontal directions | Station | Face | Reading L (D) | Reading R (Rev) | R - 180 | Mean direction | Direction from A | Correction | Corrected direction | |---|---|---|---|---|---|---|---|---| | A | | 00°00′10″ | 180°00′20″ | 0°00′20″ | 0°00′15.0″ | 0°00′00.0″ | - | 0°00′00.00″ | | B | | 60°55′10″ | 240°55′20″ | 60°55′20″ | 60°55′15.0″ | 60°55′00.0″ | -0°00′02.50″ | 60°54′57.50″ | | C | | 140°50′50″ | 320°51′20″ | 140°51′20″ | 140°51′05.0″ | 140°50′50.0″ | -0°00′05.00″ | 140°50′45.00″ | | D | | 270°20′10″ | 90°20′20″ | 270°20′20″ | 270°20′15.0″ | 270°20′00.0″ | -0°00′07.50″ | 270°19′52.50″ | | A (closing) | | 00°00′20″ | 180°00′30″ | 0°00′30″ | 0°00′25.0″ | 0°00′10.0″ | -0°00′10.00″ | 360°00′00″ | #### Mean angles and adjustment | Angle | Mean angle (before adjustment) | Correction | Adjusted angle | |---|---|---|---| | AOB | 60°55′00.0″ | -0°00′02.50″ | **60°54′57.50″** | | BOC | 79°55′50.0″ | -0°00′02.50″ | **79°55′47.50″** | | COD | 129°29′10.0″ | -0°00′02.50″ | **129°29′07.50″** | | DOA | 89°40′10.0″ | -0°00′02.50″ | **89°40′07.50″** | | Sum | 360°00′10.0″ | -0°00′10.0″ | 360°00′00″ | Closing error = 0°00′10.0″ (second reading on A larger than the first). The sum of the four angles at O must be 360°, and it is 360°00′10.0″; the error of 0°00′10.0″ is small and is therefore distributed equally: correction = -0°00′02.50″ per angle. #### Missing vertical circle readings For a theodolite with a zenith-angle vertical circle, FL + FR = 360° for a correct instrument. So the missing reading = 360° - the other face. The vertical angle is $90° - Z$ (positive = elevation, negative = depression). | Target | VCR Direct (FL) | VCR Reversed (FR) | Mean zenith distance | Vertical angle (90° - Z) | |---|---|---|---|---| | A | 120°15′10″ | **239°44′50″** | 120°15′10″ | -30°15′10″ | | B | **51°08′20″** | 308°51′40″ | 51°08′20″ | +38°51′40″ | | C | **90°44′50″** | 269°15′10″ | 90°44′50″ | -0°44′50″ | | D | 177°20′10″ | **182°39′50″** | 177°20′10″ | -87°20′10″ | | A (closing) | 89°00′10″ | **270°59′50″** | 89°00′10″ | +0°59′50″ | **Answer:** adjusted angles AOB = 60°54′57.50″, BOC = 79°55′47.50″, COD = 129°29′07.50″, DOA = 89°40′07.50″; missing VCR: A (reversed) 239°44′50″, B (direct) 51°08′20″, C (direct) 90°44′50″, D (reversed) 182°39′50″; A closing (reversed) 270°59′50″. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the mean direction of a target found from the two faces? A. (Direct + Reversed) / 2 B. (Direct + (Reversed - 180°)) / 2 C. Direct - Reversed D. (Reversed - Direct) / 2 Answer: B. The reversed reading differs by 180°, so it is reduced by 180° before averaging with the direct reading. (Key point: Mean direction formula) 2. The sum of the four angles at O is 360°00'10''. What is the correction per angle? A. -0°00'10'' B. -0°00'05'' C. +0°00'02.5'' D. -0°00'02.5'' Answer: D. The sum is 10'' too large, so subtract 10''/4 = 2.5'' from each angle. (Key point: Closing error distribution) 3. For a zenith-angle vertical circle with no index error, how is a missing face reading found? A. Face L + Face R = 180°, so subtract from 180° B. Face L = Face R C. Face L + Face R = 360°, so subtract from 360° D. Add 90° to the other face Answer: C. FL + FR = 360° for a correct instrument, so the missing reading = 360° - the other face. (Key point: Missing VCR rule) 4. Target A has zenith distance 120°15'10''. What is its vertical angle? A. +30°15'10'' B. -30°15'10'' C. +120°15'10'' D. -239°44'50'' Answer: B. Vertical angle = 90° - Z = 90° - 120°15'10'' = -30°15'10'', a depression. (Key point: Vertical angle from zenith) ## Close variants set in other exams - [Mean angles and VCR, station O (reading 179°59′40″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-o-abcd-first-reversed-179-59-40) — Surveying I, chapter 7, asked 3× ## Related in Surveying I - [Direction theodolite, least count 6″, station D](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Adirection-theodolite-ls-6-sec-station-d) — Surveying I, chapter 7 - [Mean direction angles with VCR (125°25′20″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-traverse-abcda-vcr-125-25-20) — Surveying I, chapter 7 - [Theodolite traverse ABCDA with missing VCR (68°30′10″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atheodolite-traverse-abcda-hcr-vcr-68-30-10) — Surveying I, chapter 7 - [Mean angles; VCR and VA from A](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atraverse-abcdea-vcr-va-from-a-to-e-and-b) — Surveying I, chapter 7 - [Theodolite traverse ABCDA, booking format with VCR](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atraverse-abcda-booking-format-vcr-120-14-20) — Surveying I, chapter 7 - [Mean horizontal angles, station O (P, Q, R)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-horizontal-angles-o-pqrp) — Surveying I, chapter 7