- 2081 Baisakh · 5+1 marks
Theodolite traverse ABCDA with missing VCR (68°30′10″)
Short answer
Corrected horizontal angles: A=70°11′50″, B=110°24′55″, C=89°07′55″, D=90°15′20″. Missing VCR: A to D face right=291°29′50″, B to A face left=44°14′20″, C to B face right=265°54′20″, D to C face left=89°45′40″.
The following observations were recorded in a theodolite traverse ABCDA. Compute the correct horizontal angles and find the missing readings from the following readings in the given table.
| Instrument station | Target station | HCR Face left | HCR Face right | VCR Face left | VCR Face right |
|---|---|---|---|---|---|
| A | D | 90°00′00″ | 269°59′50″ | 68°30′10″ | ? |
| B | 160°12′40″ | 340°12′30″ | |||
| B | A | 90°00′10″ | 270°00′20″ | ? | 315°45′40″ |
| C | 200°25′50″ | 20°26′10″ | |||
| C | B | 90°00′00″ | 269°59′40″ | 94°05′40″ | ? |
| D | 179°08′40″ | 359°08′30″ | |||
| D | C | 89°59′50″ | 270°00′10″ | ? | 270°14′20″ |
| A | 180°16′10″ | 00°16′10″ |
Assumption: the traverse ABCDA is a closed four-sided figure, so the sum of the interior angles must be 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′00″ | 269°59′50″ | 89°59′55″ | |
| B | 160°12′40″ | 340°12′30″ | 160°12′35″ | 70°12′40″ | |
| B | A | 90°00′10″ | 270°00′20″ | 90°00′15″ | |
| C | 200°25′50″ | 20°26′10″ | 200°26′00″ | 110°25′45″ | |
| C | B | 90°00′00″ | 269°59′40″ | 89°59′50″ | |
| D | 179°08′40″ | 359°08′30″ | 179°08′35″ | 89°08′45″ | |
| D | C | 89°59′50″ | 270°00′10″ | 90°00′00″ | |
| A | 180°16′10″ | 00°16′10″ | 180°16′10″ | 90°16′10″ |
Check and adjustment
Required sum = = 360°00′00″. Observed sum = 360°03′20″. Angular error = 360°03′20″ - 360°00′00″ = +0°03′20″. Correction per angle = -0°00′50″ (equal for all 4 angles).
| Angle at | Mean angle | Correction | Corrected angle |
|---|---|---|---|
| A | 70°12′40″ | -0°00′50″ | 70°11′50″ |
| B | 110°25′45″ | -0°00′50″ | 110°24′55″ |
| C | 89°08′45″ | -0°00′50″ | 89°07′55″ |
| D | 90°16′10″ | -0°00′50″ | 90°15′20″ |
| Sum | 360°03′20″ | -0°03′20″ | 360°00′00″ |
Answer: corrected horizontal angles: A = 70°11′50″, B = 110°24′55″, C = 89°07′55″, D = 90°15′20″.
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 (+ elevation, - depression).
| Observation | VCR Face left | VCR Face right | Vertical angle (90° - Z) |
|---|---|---|---|
| A to D | 68°30′10″ | 291°29′50″ | +21°29′50″ |
| B to A | 44°14′20″ | 315°45′40″ | +45°45′40″ |
| C to B | 94°05′40″ | 265°54′20″ | -4°05′40″ |
| D to C | 89°45′40″ | 270°14′20″ | +0°14′20″ |
✓ Answer checked by Claude Opus 5.5 (high effort)
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