CE-303 Surveying - Jun 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
   aThe following staff readings were obtained during a levelling work with the instrument being shifted after the $3^{\text{rd}}$, $6^{\text{th}}$ and $9^{\text{th}}$. Readings: 2.215, 0.955, 0.985, 2.345, 2.105, 1.295, 1.325, 2.327, 1.498, 0.975 and 1.625. Find the reduced levels of the remaining points if the RL is 200.00 for the first point using Height of instrument method.
   How to answer

   • Set up standard levelling field book table with columns: Station, BS, IS, FS, HI, RL, Remarks.
   • Group readings into setups: Shift after 3rd reading (2.215 BS, 0.955 IS, 0.985 FS), shift after 6th reading (2.345 BS, 2.105 IS, 1.295 FS), shift after 9th reading (1.325 BS, 2.327 IS, 1.498 FS), last setup (0.975 BS, 1.625 FS).
   • Calculate Height of Instrument (HI) and RL of each station sequentially.
   • Apply arithmetic check: Sum of BS - Sum of FS = Last RL - First RL.
2. 7 Marks
   bDefine Whole Circle Bearing and Reduced circle Bearing. Convert WCB to RCB and RCB to WCB of the following: i) $125^\circ 45'$ ii) $325^\circ 30'$ iii) $\text{S} 25^\circ 45' \text{W}$ iv) $\text{N} 40^\circ 15' \text{W}$
   How to answer

   • Define Whole Circle Bearing (measured clockwise from North, 0° to 360°).
   • Define Reduced Bearing (measured from North or South towards East or West, 0° to 90°).
   • Convert (i) 125°45' to Quadrantal Bearing: S (180° - 125°45') E = S 54°15' E.
   • Convert (ii) 325°30' to Quadrantal Bearing: N (360° - 325°30') W = N 34°30' W.
   • Convert (iii) S 25°45' W to WCB: 180° + 25°45' = 205°45'.
   • Convert (iv) N 40°15' W to WCB: 360° - 40°15' = 319°45'.
3. 7 Marks
   aDiscuss about the Obstacles in Chaining with a neat sketch.
   How to answer

   • Classify obstacles: (1) Chaining free, ranging obstructed (e.g., hill), (2) Ranging free, chaining obstructed (e.g., pond, river), (3) Both chaining and ranging obstructed (e.g., building).
   • Explain geometrical solutions for obstacles like ponds and rivers (perpendicular offsets, triangles).
   • Provide neat explanatory sketches for each case.
4. 7 Marks
   bA steel tape was exactly $20\text{ m}$ long at $20^\circ\text{C}$ when supported throughout its length under a pull of $5\text{ Kg}$. A line measured with this tape under a pull of $16\text{ kg}$ and at a mean temperature of $32^\circ\text{C}$ was found to be $680\text{ m}$ long. Assuming the tape is supported at every $20\text{ m}$, find the true length of the line. Given data area of $\text{C/s} = 0.03\text{ cm}^2$, $E = 2.1 \times 10^6\text{ kg/cm}^2$, $\alpha = 11 \times 10^{-6}\text{ per } ^\circ\text{C}$ and weight of tape $= 10\text{ g/cc}$.
   How to answer

   • Calculate weight of one tape length ($20\text{ m}$): $W_1 = A \times l \times d = 0.03 \times 2000 \times 10 = 600\text{ g} = 0.6\text{ kg}$.
   • Number of spans: $n = 680 / 20 = 34$.
   • Calculate temperature correction: $C_t = \alpha (T - T_0) L = 11 \times 10^{-6} \times (32 - 20) \times 680 = +0.08976\text{ m}$.
   • Calculate pull/tension correction: $C_p = \frac{(P - P_0) L}{A \cdot E} = \frac{(16 - 5) \times 680}{0.03 \times 2.1 \times 10^6} = +0.11873\text{ m}$.
   • Calculate sag correction for 34 spans: $C_s = \frac{n \cdot l_0 \cdot W_1^2}{24 P^2} = \frac{680 \times (0.6)^2}{24 \times 16^2} = 0.03984\text{ m}$.
   • Compute total correction: $C_{total} = C_t + C_p - C_s$ and True length $= 680 + C_{total}$.
5. 7 Marks
   aDefine True meridian, Magnetic meridian, Arbitrary meridian and Grid meridian.
   How to answer

   • True meridian: Line passing through the geographical north and south poles and the observer's position; constant over time.
   • Magnetic meridian: Direction indicated by a freely suspended, balanced magnetic needle influenced only by Earth's magnetic field.
   • Arbitrary meridian: Any convenient, fixed direction arbitrarily assumed for small local surveys.
   • Grid meridian: Meridian formed by the central meridian of a national or map grid system adopted for an entire region.
6. 7 Marks
   b

   The following are the bearings of a closed traverse. Calculate the interior angles of the traverse and also check whether the bearing is free from local attraction.

   Side	FB	BB
   AB	$\text{N}45^\circ 30'\text{E}$	$\text{S}45^\circ 30'\text{W}$
   BC	$\text{S}60^\circ 0'\text{E}$	$\text{N}60^\circ 0'\text{W}$
   CD	$\text{S}10^\circ 30'\text{W}$	$\text{N}10^\circ 30'\text{E}$
   DA	$\text{N}75^\circ 45'\text{W}$	$\text{S}75^\circ 45'\text{E}$

   How to answer

   • Check local attraction: For all lines AB, BC, CD, and DA, the numerical values of FB and BB are identical with opposite quadrant designations, meaning $|FB - BB| = 180^\circ$; therefore, all stations are free from local attraction.
   • Convert bearings to WCB: AB = 45°30', BC = 120°00', CD = 190°30', DA = 284°15'.
   • Calculate interior angles: $\angle A$, $\angle B$, $\angle C$, and $\angle D$ using back bearing and fore bearing differences.
   • Verify theoretical sum of interior angles: $(2 \times 4 - 4) \times 90^\circ = 360^\circ$.
7. 7 Marks
   aWhat do you know about repeating theodolite, direction theodolite and Gale's table.
   How to answer

   • Repeating theodolite: Features lower and upper clamps allowing angle accumulation/repetition to increase precision.
   • Direction theodolite: Has a single vertical axis and reads angles directly around the horizon relative to an initial direction/line.
   • Gale's Table: Standard tabular form used for computing and adjusting the coordinates of a closed traverse (bearings, consecutive coordinates, corrections, independent coordinates).
8. 7 Marks
   b

   The following records are obtained in a traverse survey, where the length and bearing of the last line were not recorded. Calculate the length of DA and bearing of AB.

   Side	Length	Bearing
   AB	100.0	?
   BC	80.5	$140^\circ 30'$
   CD	60.0	$220^\circ 30'$
   DA	?	$310^\circ 15'$

   How to answer

   • Formulate closed traverse equations: $\sum L = 0$ and $\sum D = 0$.
   • Calculate latitudes and departures of known lines BC and CD.
   • Express equations in terms of unknown bearing $\theta_{AB}$ and length $l_{DA}$.
   • Eliminate $\theta_{AB}$ using $\sin^2\theta_{AB} + \cos^2\theta_{AB} = 1$ to obtain a quadratic equation in $l_{DA}$.
   • Solve for positive length $l_{DA}$ and subsequently determine bearing $\theta_{AB}$.
9. 7 Marks
   aDescribe the process of permanent adjustment in theodolite.
   How to answer

   • State the purpose of permanent adjustments (maintaining fundamental line relationships).
   • Adjustment of plate level tube (axis perpendicular to vertical axis).
   • Adjustment of horizontal axis / trunnion axis (perpendicular to vertical axis via spire test).
   • Adjustment of line of sight / collimation test (line of collimation perpendicular to horizontal axis).
   • Adjustment of altitude bubble and vertical circle index zero.
10. 7 Marks
    bWhat are the methods of traversing by theodolite.
    How to answer

    • Traversing by included angles (closed loop traverse).
    • Traversing by deflection angles (route surveys, highways, railways).
    • Traversing by direct observation of angles to right.
    • Fast needle method / azimuth method.
    • Loose needle method.
11. 7 Marks
    aTwo tangents AB and BC intersects a B. Another line DE intersects AB and BC at D and E such that $\angle \text{ADE} = 150^\circ$ and $\angle \text{DEC} = 140^\circ$. The radius of the first curve is $200\text{ m}$ and that of the second is $300\text{ m}$. The chainage of B is $950\text{ m}$. Calculate all data necessary for the setting out the compound curve.
    How to answer

    • Determine deflection angles: $\Delta_1 = 180^\circ - 150^\circ = 30^\circ$ and $\Delta_2 = 180^\circ - 140^\circ = 40^\circ$, total deflection $\Delta = \Delta_1 + \Delta_2 = 70^\circ$.
    • Calculate tangent lengths: $t_1 = R_1 \tan(\Delta_1 / 2) = 200 \tan 15^\circ = 53.59\text{ m}$; $t_2 = R_2 \tan(\Delta_2 / 2) = 300 \tan 20^\circ = 109.19\text{ m}$.
    • Calculate common tangent length: $DE = t_1 + t_2 = 162.78\text{ m}$.
    • Solve $\triangle BDE$ using sine rule to compute distances $BD$ and $BE$.
    • Determine main tangent lengths $T_1 = BD + t_1$ and $T_2 = BE + t_2$.
    • Compute chainages: Chainage of $T_1 = \text{Chainage of } B - T_1$; Chainage of PCC $= \text{Chainage of } T_1 + L_1$; Chainage of $T_2 = \text{Chainage of PCC} + L_2$.
12. 7 Marks
    bWhat are the initial and final sub-chords and what is super elevation and why it is provided
    How to answer

    • Define initial sub-chord: Chord between point of curve ($T_1$) and first full chainage peg.
    • Define final sub-chord: Chord between last full chainage peg and point of tangency ($T_2$).
    • Define super-elevation (cant): Raising of the outer edge/rail above the inner edge/rail along a curved path.
    • Explain purpose: Counteract centrifugal force, prevent overturning/skidding, ensure passenger comfort and uniform wheel load distribution.
    • Give formula: $e = \frac{v^2}{gR}$.
13. 7 Marks
    aWhat is the versed sine of curve and what are the different types of the Curve settings.
    How to answer

    • Define versed sine (mid-ordinate of curve): The distance from the midpoint of the long chord to the apex of the curve ($M = R(1 - \cos(\Delta/2))$).
    • Classify curve setting methods into Linear methods (Offsets from long chord, Successive bisection of arcs, Offsets from tangents, Offsets from chords produced).
    • Classify Angular methods (Rankine's method of tangential angles, Two-theodolite method, Tacheometric method).
14. 7 Marks
    bTwo tangents AB and BC intersects at a point B at chainage $150.5\text{ m}$. Calculate all the necessary data for setting out a circular curve of radius $100\text{ m}$ and deflection angle is $30^\circ$ by the method of offset from the long chord.
    How to answer

    • Compute tangent distance: $T = 100 \tan 15^\circ = 26.79\text{ m}$.
    • Compute length of curve: $L = \frac{\pi \times 100 \times 30}{180} = 52.36\text{ m}$.
    • Compute chainages: $T_1 = 150.5 - 26.79 = 123.71\text{ m}$, $T_2 = 123.71 + 52.36 = 176.07\text{ m}$.
    • Compute length of long chord: $L_c = 2 \times 100 \sin 15^\circ = 51.76\text{ m}$, so half-chord $= 25.88\text{ m}$.
    • Calculate mid-ordinate: $O_0 = 100(1 - \cos 15^\circ) = 3.41\text{ m}$.
    • Tabulate offsets $O_x = \sqrt{R^2 - x^2} - (R - O_0)$ at $5\text{ m}$ intervals: $x = 5, 10, 15, 20, 25\text{ m}$.
15. 7 Marks
    a

    To find the RL of station B two observation, are taken by a theodolite from station A one to a BM and other to the station B. The records are a follow;

    Inst. Station	Staff Station	Target Staff	Vertical angle	Staff Reading	Remark
    A	BM	LOWER	$-10^\circ 0'$	0.655	RL OF BM = 510.500 m
    		UPPER	$-7^\circ 0'$	2.655
    A	B	LOWER	$-5^\circ 0'$	1.250
    		UPPER	$+4^\circ 0'$	3.200

    How to answer

    • Identify system as Tangential Tacheometry.
    • Calculate vane spacing for BM: $s_1 = 2.655 - 0.655 = 2.000\text{ m}$.
    • Since both angles to BM are angles of depression ($\theta_1 = 10^\circ$, $\theta_2 = 7^\circ$): $D_1 = \frac{s_1}{\tan 10^\circ - \tan 7^\circ}$.
    • Compute height of instrument (HI) relative to BM.
    • Calculate vane spacing for B: $s_2 = 3.200 - 1.250 = 1.950\text{ m}$.
    • One angle of depression ($\alpha_1 = 5^\circ$) and one angle of elevation ($\alpha_2 = 4^\circ$): $D_2 = \frac{s_2}{\tan 5^\circ + \tan 4^\circ}$.
    • Compute $V_2 = D_2 \tan 5^\circ$ and derive the RL of station B.
16. 7 Marks
    bWhat do you mean by subtense bar and what is the difference between a fixed hair tachometer and subtense theodolite?
    How to answer

    • Define Subtense Bar: A horizontal bar of fixed length (usually 2 m or 3 m) with targets at both ends mounted horizontally on a tripod at the target station.
    • Fixed hair tachometer: Uses fixed stadia wires in diaphragm; angle is constant, intercept on vertical staff varies with distance ($D = ks + c$).
    • Subtense theodolite: Measures variable small horizontal angle subtended by a base of fixed length (subtense bar); distance $D = \frac{s}{2} \cot(\theta / 2)$.
    • Compare advantages: Subtense method eliminates refraction errors common in vertical staff readings over uneven ground.