CE-303 Surveying - Dec 2023 Question Paper

Under each question, How to answer lists the points an answer should cover, outlined by CampusPrep from the syllabus. Check the details against your textbook.

1. 7 Marks
   aExplain the different types of surveying in details. What is the fundamental principle of surveying?
   How to answer

   • Classification based on nature of field: Land, Marine, Astronomical
   • Classification based on object/purpose: Engineering, Military, Mine, Geological, Archeological
   • Classification based on instruments: Chain, Compass, Plane Table, Theodolite, Tacheometric, Photogrammetric
   • Fundamental principles: Working from whole to part, Location of a point with respect to at least two reference points
   • Importance of whole to part in preventing accumulation of errors
2. 7 Marks
   bThe following consecutive readings were taken with a dumpy level along a chain line at a common interval of 15 meter. Reduced Level (RL) of the first point was $98.085\text{ m}$. the instrument was shifted after fourth and ninth readings. Calculate RL of other points by height of instruments method. 3.150, 2.245, 1.125, 0.860, 3.125, 2.760, 1.835, 1.470, 1.965, 1.225, 2.390, 3.035
   How to answer

   • Construct standard leveling field book table: Station, BS, IS, FS, HI, RL, Remarks
   • Assign readings: BS1=3.150, IS2=2.245, IS3=1.125, FS4=0.860; BS5=3.125, IS6=2.760, IS7=1.835, IS8=1.470, FS9=1.965; BS10=1.225, IS11=2.390, FS12=3.035
   • Calculate HI for each instrument setup
   • Compute RLs for all stations
   • Apply arithmetic check: $\sum \text{BS} - \sum \text{FS} = \text{Last RL} - \text{First RL}$
3. 7 Marks
   aDescribe temporary adjustment of theodolite. What are the fundamental lines of theodolite?
   How to answer

   • Temporary adjustments steps: Setting over the station, Centering (using plumb bob or optical plummet), Leveling up (using plate bubble and foot screws), Elimination of parallax (focusing eyepiece and objective)
   • List fundamental lines: Line of collimation, Axis of the plate levels, Horizontal axis (trunnion axis), Vertical axis, Axis of the altitude level bubble
   • State desired geometric relationships among these lines
4. 7 Marks
   bThe following table is related to a closed traverse PQRS. Find out the missing coordinates marked X and Y. Also calculate the independent coordinates of the points P, Q and S, if the independent coordinate of R is 300, 1000. $$\begin{array}{|c|c|c|c|c|}\hline \text{Point} & \multicolumn{4}{c|}{\text{Corrected consecutive coordinates}} \\ \cline{2-5} & \multicolumn{2}{c|}{\text{Latitude}} & \multicolumn{2}{c|}{\text{Departure}} \\ \hline \text{P} & 562.40 & & \text{Nil} & \text{Nil} \\ \hline \text{Q} & 180.30 & & \text{Y} & \\ \hline \text{R} & & 595.86 & 58.78 & \\ \hline \text{S} & & \text{X} & & 572.72 \\ \hline \end{array}$$
   How to answer

   • Identify Northing/Southing from Latitude columns and Easting/Westing from Departure columns
   • Apply traverse closing conditions: $\sum L = 0$ to solve for $X$
   • Apply $\sum D = 0$ to solve for $Y$
   • Use known independent coordinates of R (300, 1000) and consecutive coordinates to compute independent coordinates of S, P, and Q successively
5. 7 Marks
   aExplain the principle of stadia method in details.
   How to answer

   • Basic optical principle of stadia tacheometry
   • Definition of staff intercept $s$ and stadia wire interval $i$
   • Derivation/statement of distance equation: $D = (f/i)s + (f + d)$
   • Explanation of multiplying constant $k = f/i = 100$ and additive constant $c = f + d \approx 0$ with anallatic lens
   • Neat ray diagram showing stadia diaphragm and staff intercept
6. 7 Marks
   bDifferentiate between the fixed-hair method and movable hair method. Discuss the advantages and disadvantages of each method.
   How to answer

   • Contrast principle: In fixed hair, interval $i$ is constant, staff intercept $s$ varies; in movable hair, staff intercept $s$ is constant (fixed vanes), hair interval $i$ varies
   • Comparison table based on speed, accuracy, mechanical complexity, and instrumentation
   • Advantages and disadvantages of fixed-hair method (simple, fast, but difficult for very long sights)
   • Advantages and disadvantages of movable-hair method (accurate over long sights, but slow and micrometer errors possible)
7. 7 Marks
   aExplain the elements of simple circular curve with neat sketch.
   How to answer

   • Neat diagram of a simple circular curve connecting two straights with PI, PC ($T_1$), PT ($T_2$), intersection angle, deflection angle $\Delta$, radius $R$
   • Formulas: Tangent length $T = R \tan(\Delta/2)$
   • Length of curve $L = R \Delta = \pi R \Delta / 180^\circ$
   • Length of long chord $L_c = 2R \sin(\Delta/2)$
   • Mid-ordinate / Versine $M = R(1 - \cos(\Delta/2))$
   • Apex distance $E = R(\sec(\Delta/2) - 1)$
8. 7 Marks
   bWhat do you mean by transition curves? What are the different types and uses of transition curves?
   How to answer

   • Definition: Non-circular curve with continuously changing radius introduced between a straight and circular curve
   • Objectives/Uses: Gradual introduction of centrifugal force, gradual introduction of superelevation, gradual increase of extra-widening, comfortable ride, reduced wear and tear
   • Types: Euler's spiral / clothoid, Cubic parabola, Lemniscate of Bernoulli
   • Characteristics and ideal conditions of a transition curve
9. 7 Marks
   aWrite short notes on the following terms: i) Sounding equipments ii) Nautical sextant iii) Tide gauge
   How to answer

   • Sounding equipments: Sounding pole/rod, lead line, echo sounder / fathometer; used to measure depth of water below surface
   • Nautical sextant: Reflecting instrument based on principle of double reflection to measure horizontal and vertical angles from a moving boat
   • Tide gauge: Non-registering (staff gauge, float gauge) and self-registering automatic tide gauges; used to record water level variations for determining Mean Sea Level (MSL) and datum
10. 7 Marks
    bDetermine the number of photographs required to cover an area $40\text{ km} \times 25\text{ km}$. If the scale of the photograph is $1\text{ in } 10,000$ and the format is $230\text{ mm} \times 230\text{ mm}$. Take longitudinal and side overlap as $60\%$ and $30\%$ respectively.
    How to answer

    • Calculate ground dimensions covered by one photograph: $S_g = 0.230 \times 10,000 = 2300\text{ m} = 2.3\text{ km}$
    • Effective advance per photograph along flight line: $L_1 = 2.3 \times (1 - 0.60) = 0.92\text{ km}$
    • Effective spacing between flight lines: $W_1 = 2.3 \times (1 - 0.30) = 1.61\text{ km}$
    • Number of photographs per strip: $n_1 = (40 / 0.92) + 1 = 43.48 + 1 \approx 45$
    • Number of flight strips: $n_2 = (25 / 1.61) + 1 = 15.53 + 1 \approx 17$
    • Total number of photographs: $N = n_1 \times n_2 = 45 \times 17 = 765$
11. 7 Marks
    aExplain prismatic compass used in surveying? What are the temporary adjustments of prismatic compass?
    How to answer

    • Description of prismatic compass: Non-magnetic metal box, magnetic needle attached to circular graduated ring (0° to 360°), prism with sighting slit, sight vane with horsehair, lifting lever, mirror/sunglasses
    • Working principle: Sighting and reading done simultaneously, reads Whole Circle Bearings (WCB) from South as 0° (inverting prism shows North as 0°)
    • Temporary adjustments: 1. Centering over the station using plumb bob, 2. Leveling using ball and socket arrangement, 3. Focusing of the prism by moving slide
12. 7 Marks
    bThe following are the bearings of a closed traverse ABCDEA, taken by a prismatic compass, in an area where local attraction was suspected. Find out the included angles of the traverse and correct them if necessary? $$\begin{array}{|c|c|c|}\hline \text{Line} & \text{Fore Bearing} & \text{Back bearing} \\ \hline \text{AB} & 165^\circ 30' & 345^\circ 30' \\ \hline \text{BC} & 40^\circ 45' & 221^\circ 45' \\ \hline \text{CD} & 315^\circ 45' & 135^\circ 45' \\ \hline \text{DE} & 250^\circ 30' & 69^\circ 30' \\ \hline \text{EA} & 120^\circ 45' & 300^\circ 45' \\ \hline \end{array}$$
    How to answer

    • Calculate included angle at each station using: $\text{Angle} = \text{FB of forward line} - \text{BB of previous line}$ (or by drawing sketches at each station)
    • Station angles: Calculate $\angle A, \angle B, \angle C, \angle D, \angle E$
    • Sum of interior angles for 5-sided polygon must equal $(2 \times 5 - 4) \times 90^\circ = 540^\circ$
    • Identify station free from local attraction (lines AB, CD, EA have $|\text{FB} - \text{BB}| = 180^\circ$, e.g., line AB and line CD)
    • Check total angular error, distribute correction equally if needed, and write final corrected angles
13. 7 Marks
    aA series of offset taken from chain line at $10\text{ m}$ intervals in the following orders: 2.7, 3.22, 5.64, 2.88, 7.29, 6.13, 8.66, 3.24, 2.11 Calculate the area enclosed between the survey lines by Trapezoidal and Simpsons rule.
    How to answer

    • Identify given data: $d = 10\text{ m}$, $n = 9$ ordinates ($O_1$ to $O_9$)
    • Trapezoidal rule: Area = $10 \times [ (2.7 + 2.11)/2 + (3.22 + 5.64 + 2.88 + 7.29 + 6.13 + 8.66 + 3.24) ] = 10 \times [2.405 + 37.06] = 394.65\text{ m}^2$
    • Simpson's rule ($n = 9$ is odd, so applicable directly): Area = $\frac{10}{3} [ (2.7 + 2.11) + 4(3.22 + 2.88 + 6.13 + 3.24) + 2(5.64 + 7.29 + 8.66) ]$
    • Sum of even ordinates = $15.47$, sum of remaining odd ordinates = $21.59$
    • Area = $\frac{10}{3} [ 4.81 + 4(15.47) + 2(21.59) ] = \frac{10}{3} [ 4.81 + 61.88 + 43.18 ] = \frac{10}{3} [109.87] = 366.23\text{ m}^2$
14. 7 Marks
    bWrite the short notes on the following terms: i) Latitude and departure ii) Trigonometric leveling iii) Compound curves
    How to answer

    • Latitude: Projection on N-S axis ($L = l \cos \theta$), Departure: Projection on E-W axis ($D = l \sin \theta$), signs in four quadrants
    • Trigonometric leveling: Determining difference in elevation by measuring vertical angles and horizontal distances, cases of accessible and inaccessible base
    • Compound curves: Combination of two or more circular curves of different radii curving in the same direction with a common tangent at the junction point (PCC)
15. 7 Marks
    aWhat is meant by relief displacement? Derive the expression of relief displacement.
    How to answer

    • Definition: Displacement of the photographic image of a ground feature from its true datum position due to elevation/height above or below datum
    • Characteristics: Radial from principal point / nadir point; positive for points above datum, negative below datum
    • Neat diagram showing camera lens O, focal length f, flying height H, height of object h, radial distances r and r0
    • Derive from similar triangles: $r = \frac{f R}{H - h}$ and $r_0 = \frac{f R}{H}$
    • Result: $d = r - r_0 = \frac{r h}{H}$
16. 7 Marks
    bA circular curve has $350\text{ m}$ radius and deflection angle of $45^\circ$. Calculate: i) Length of curve ii) Tangent length iii) Length of long chord iv) Mid-ordinate, and v) Chainage of tangent points If chainage of intersection point is $900\text{ m}$.
    How to answer

    • i) Tangent length: $T = 350 \times \tan(22.5^\circ) = 350 \times 0.4142 = 144.97\text{ m}$
    • ii) Length of curve: $l = \frac{\pi \times 350 \times 45^\circ}{180^\circ} = 274.89\text{ m}$
    • iii) Length of long chord: $L = 2 \times 350 \times \sin(22.5^\circ) = 700 \times 0.3827 = 267.88\text{ m}$
    • iv) Mid-ordinate: $M = 350 \times (1 - \cos 22.5^\circ) = 350 \times (1 - 0.9239) = 26.64\text{ m}$
    • v) Chainage of $T_1 = 900 - 144.97 = 755.03\text{ m}$
    • Chainage of $T_2 = 755.03 + 274.89 = 1029.92\text{ m}$