- Asked 2/31
- 2080 Baisakh · 6 marks
Mean direction method, traverse ABCDEA (2080)
Short answer
Corrected horizontal angles: A=110°24′52.5″, B=90°15′37.5″, C=89°07′37.5″, D=70°11′52.5″.
The following observations were recorded in a theodolite traverse ABCDEA. Rule out the proper field book and compute horizontal angles by mean direction method and adjust them if necessary.
| Instrument station | Target Station | HCR Direct | HCR Reversed |
|---|---|---|---|
| A | D | 90°00′20″ | 269°59′40″ |
| B | 200°25′40″ | 20°25′30″ | |
| B | A | 89°59′30″ | 270°00′10″ |
| C | 180°16′10″ | 00°16′10″ | |
| C | B | 90°00′10″ | 270°00′10″ |
| D | 179°08′40″ | 359°08′20″ | |
| D | C | 89°59′50″ | 270°00′10″ |
| A | 160°12′40″ | 340°12′30″ |
Assumption: the title says ABCDEA, but only four instrument stations (A, B, C, D) are given, so the traverse is taken as the closed quadrilateral ABCDA with angle sum 360°.
Method. For each target the mean direction is found from the two faces:
The angle at a station = mean direction of the forward station - mean direction of the back station. For a closed traverse of sides, the sum of the interior angles must be ; 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°00′20″ | 269°59′40″ | 90°00′00″ | |
| B | 200°25′40″ | 20°25′30″ | 200°25′35″ | 110°25′35″ | |
| B | A | 89°59′30″ | 270°00′10″ | 89°59′50″ | |
| C | 180°16′10″ | 00°16′10″ | 180°16′10″ | 90°16′20″ | |
| C | B | 90°00′10″ | 270°00′10″ | 90°00′10″ | |
| D | 179°08′40″ | 359°08′20″ | 179°08′30″ | 89°08′20″ | |
| D | C | 89°59′50″ | 270°00′10″ | 90°00′00″ | |
| A | 160°12′40″ | 340°12′30″ | 160°12′35″ | 70°12′35″ |
Check and adjustment
Required sum = = 360°00′00″. Observed sum = 360°02′50″. Angular error = 360°02′50″ - 360°00′00″ = +0°02′50″. Correction per angle = -0°00′42.5″ (equal for all 4 angles).
| Angle at | Mean angle | Correction | Corrected angle |
|---|---|---|---|
| A | 110°25′35″ | -0°00′42.5″ | 110°24′52.5″ |
| B | 90°16′20″ | -0°00′42.5″ | 90°15′37.5″ |
| C | 89°08′20″ | -0°00′42.5″ | 89°07′37.5″ |
| D | 70°12′35″ | -0°00′42.5″ | 70°11′52.5″ |
| Sum | 360°02′50″ | -0°02′50″ | 360°00′00″ |
Answer: corrected horizontal angles: A = 110°24′52.5″, B = 90°15′37.5″, C = 89°07′37.5″, D = 70°11′52.5″.
✓ Answer checked by Claude Opus 5.5 (high effort)
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