--- title: "Mean direction angles with VCR (125°25′20″)" subject: "Surveying I" chapter: 7 key: "semester_3/surveying_i:mean-direction-traverse-abcda-vcr-125-25-20" source: https://www.bctengineer.com/q/semester_3/surveying_i/mean-direction-traverse-abcda-vcr-125-25-20 site: BCT Engineering --- # Mean direction angles with VCR (125°25′20″) Surveying I · Chapter 7: Transit and Theodolite - Asked in: 2080 Bhadra, 4+1+1 marks - 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-traverse-abcda-vcr-125-25-20 - Whole chapter: https://www.bctengineer.com/md/semester_3/surveying_i/chapter-7 ## Question Theodolite readings are taken during traversing. Compute the angles by mean direction method and correct them if necessary. Also find the missing data of VCR. | Inst. station | Sighted to | HCR Face left | HCR Face right | VCR Face left | VCR Face right | |---|---|---|---|---|---| | A | D | 90°0′10″ | 270°0′10″ | 125°25′20″ | ? | | | B | 186°24′40″ | 6°24′40″ | ? | 304°50′10″ | | B | A | 90°0′20″ | 270°0′0″ | | | | | C | 188°8′50″ | 8°8′50″ | | | | C | B | 90°0′0″ | 270°0′20″ | | | | | D | 171°4′0″ | 351°4′20″ | | | | D | C | 90°0′20″ | 270°0′0″ | | | | | A | 174°24′0″ | 354°24′20″ | | | ## Short answer Corrected horizontal angles: A=96°24′12.5″, B=98°08′22.5″, C=81°03′42.5″, D=84°23′42.5″. Missing VCR: A to D face right=234°34′40″, A to B face left=55°09′50″. ## Answer **Assumption:** a closed traverse ABCDA (four stations), so the interior angles must add up to 360°. **Method.** For each target the mean direction is found from the two faces: $$ \text{Mean direction} = \frac{\text{Face L} + (\text{Face R} \pm 180^\circ)}{2} $$ The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of $n$ sides, the sum of the interior angles must be $(n-2) \times 180^\circ$; the difference is the angular error, which is distributed equally among the angles (the angles are measured with the same instrument and care). #### Field book and mean angles | Inst. stn | Sighted to | Face L | Face R | Mean direction | Mean angle | |---|---|---|---|---|---| | A | D | 90°0′10″ | 270°0′10″ | 90°00′10″ | | | | B | 186°24′40″ | 6°24′40″ | 186°24′40″ | **96°24′30″** | | B | A | 90°0′20″ | 270°0′0″ | 90°00′10″ | | | | C | 188°8′50″ | 8°8′50″ | 188°08′50″ | **98°08′40″** | | C | B | 90°0′0″ | 270°0′20″ | 90°00′10″ | | | | D | 171°4′0″ | 351°4′20″ | 171°04′10″ | **81°04′00″** | | D | C | 90°0′20″ | 270°0′0″ | 90°00′10″ | | | | A | 174°24′0″ | 354°24′20″ | 174°24′10″ | **84°24′00″** | #### Check and adjustment Required sum = $(4-2) \times 180^\circ$ = 360°00′00″. Observed sum = 360°01′10″. Angular error = 360°01′10″ - 360°00′00″ = +0°01′10″. Correction per angle = -0°00′17.5″ (equal for all 4 angles). | Angle at | Mean angle | Correction | Corrected angle | |---|---|---|---| | A | 96°24′30″ | -0°00′17.5″ | **96°24′12.5″** | | B | 98°08′40″ | -0°00′17.5″ | **98°08′22.5″** | | C | 81°04′00″ | -0°00′17.5″ | **81°03′42.5″** | | D | 84°24′00″ | -0°00′17.5″ | **84°23′42.5″** | | Sum | 360°01′10″ | -0°01′10″ | 360°00′00″ | **Answer:** corrected horizontal angles: A = 96°24′12.5″, B = 98°08′22.5″, C = 81°03′42.5″, D = 84°23′42.5″. #### Missing VCR readings **Missing VCR.** For a vertical circle graduated as zenith angle, the sum of the face-left and face-right readings of the same target is 360° when the instrument has no index error. So the missing reading = 360° - the given reading. The vertical angle is $90^\circ - Z$ (+ elevation, - depression). | Observation | VCR Face left | VCR Face right | Vertical angle (90° - Z) | |---|---|---|---| | A to D | 125°25′20″ | **234°34′40″** | -35°25′20″ | | A to B | **55°09′50″** | 304°50′10″ | +34°50′10″ | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A to D reads 90°0'10'' (L) and 270°0'10'' (R). What is the mean direction? A. 180°00'10'' B. 90°00'10'' C. 90°00'20'' D. 270°00'10'' Answer: B. (90°0'10'' + (270°0'10'' - 180°))/2 = 90°00'10''. (Key point: Mean direction) 2. The mean angles sum to 360°01'10''. What is the correction applied to each angle? A. -0°00'17.5'' B. -0°00'35'' C. -0°01'10'' D. +0°00'17.5'' Answer: A. 70'' / 4 = 17.5'' is subtracted from each angle. (Key point: Angular error) 3. Angle at A is 96°24'30'' before correction. What is its corrected value? A. 96°24'47.5'' B. 96°24'12.5'' C. 96°24'30'' D. 96°23'20'' Answer: B. 96°24'30'' - 17.5'' = 96°24'12.5''. (Key point: Corrected angle) 4. A to B has FR VCR 304°50'10''. What is FL? A. 124°50'10'' B. 34°50'10'' C. 55°09'50'' D. 55°10'10'' Answer: C. FL = 360° - 304°50'10'' = 55°09'50''. 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