JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II SURVEYING Previous Question Paper (October - 2020 - R15)
Answer any five questions
All questions carry equal marks
Main
Examine the following notes on a compass survey for local attraction. Determine correct bearings. Also determine the included angles at A, B, C, D and E.
| Station | FB | BB |
|---|---|---|
| A | S 10° 0′ W | N 85° 0′ E |
| B | S 77° 0′ E | N 10° 0′ E |
| C | N 05° 0′ E | N 75° 0′ W |
| D | N 54° 0′ W | S 02° 0′ E |
| E | S 88° 0′ W | S 50° 0′ E |
Answer
To detect local attraction, we compare the difference between forward bearing (FB) and backward bearing (BB). The theoretical difference should be 180°. The observed differences are:
- A: |190° - 65°| = 125° (Affected)
- B: |103° - 190°| = 87° (Affected)
- C: |5° - 285°| = 280° (Affected)
- D: |306° - 182°| = 124° (Affected)
- E: |268° - 130°| = 138° (Affected)
Since all stations are affected, we identify the station with the least variation or use the interior angle method. Assuming station C is free from local attraction (as it shows the smallest discrepancy relative to adjacent lines), we calculate corrected bearings:
| Line | FB | BB |
|---|---|---|
| AB | S 10° 0' W | N 10° 0' E |
| BC | S 77° 0' E | N 77° 0' W |
| CD | N 05° 0' E | S 05° 0' W |
| DE | N 54° 0' W | S 54° 0' E |
| EA | S 88° 0' W | N 88° 0' E |
Included angles are calculated using the formula: Included Angle = FB of next line - BB of previous line. Adjusting for 360° where necessary:
- Angle A: (FB AB - BB EA) = 190° - 88° = 102°
- Angle B: (FB BC - BB AB) = 103° - 190° + 360° = 273° (Exterior) or 87° (Interior)
- Angle C: (FB CD - BB BC) = 5° - 283° + 360° = 82°
- Angle D: (FB DE - BB CD) = 306° - 185° = 121°
- Angle E: (FB EA - BB DE) = 268° - 126° = 142°
Explain the two–point problem and the different methods of solving it.
Answer
The two-point problem is a method of plane table surveying used to orient the table at a desired station point by observing two well-defined points whose positions are already plotted on the drawing sheet. Unlike the three-point problem, this method involves establishing the location of the instrument station using two known points and an auxiliary station.
The procedure involves the following steps:
- Select two points A and B on the ground, which are already plotted on the map as a and b.
- Choose an auxiliary station C at a convenient location.
- Set up the plane table at C and orient it by sighting A and B. Draw rays from c towards A and B.
- Shift the table to the required station P. Orient the table by sighting the auxiliary station C and the point A.
- Draw a ray from p towards B. The intersection of this ray with the ray drawn from A determines the position of B on the map.
- The discrepancy between the plotted position of B and the actual position indicates the error in orientation, which is corrected using the tracing paper method or the trial-and-error method.
There are two primary methods to solve the two-point problem:
- Tracing Paper Method: A piece of tracing paper is fixed on the board. The station P is marked, and rays are drawn to A, B, and C. The paper is then shifted over the board until the rays pass through the plotted points a, b, and c. The point p is then transferred to the drawing sheet.
- Trial and Error Method: This involves repeatedly adjusting the orientation of the table by sighting the points and checking the intersection of rays until the plotted position of the station coincides with the actual ground position. This method is more accurate but time-consuming compared to the tracing paper method.
The following staff readings are observed successively with a dumpy level, the instrument having been moved after third, sixth and eighth readings: 2.228, 1.606, 0.988, 2.090, 2.864, 1.262, 0.602, 1.982, 1.044, 2.684 m. Enter the above readings in a page of a level book and calculate the R.L. of points if the first reading is taken with a staff held on a bench mark of 432.384 m. Also perform the arithmetic check. Use height of instrument method.
Answer
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Explain the direct methods of contouring. Explain the advantages and disadvantages of these methods.
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Calculate the volume of earth work by prismoidal formula in a road embankment with the following data in meters:
| Chainage | 0 | 100 | 200 | 300 | 400 |
|---|---|---|---|---|---|
| Ground level | 201.70 | 202.90 | 202.40 | 204.70 | 206.90 |
Formation level at the chainage 0 is 202.30 m, top width is 2.0 m, side slopes are 2 to 1. The longitudinal gradient of the embankments is 1 in 100 rising. The ground is assumed to be level across the longitudinal section.
Answer
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The following are the areas at 100 ft interval found along a piece of ground. The areas computed are 0, 197.47 Sq. ft; 542.73 Sq. ft; 636.61 Sq. ft; 482.22 Sq. ft; 0. Find the volume of cutting by prismoidal and trapezoidal formulae.
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Explain in detail how permanent adjustments are carried using theodolite instrument.
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What do you understand by Global positioning system? Give an overview of GPS.
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