BE Civil Engineering (IOE, TU) Surveying II (IOE, CE 503b) Question Paper 2079 Nepal

Theory. For a line of length $L$ and WCB $\theta$:

$$\text{Latitude} = L\cos\theta, \qquad \text{Departure} = L\sin\theta$$

For a closed loop the algebraic sums of latitudes and of departures should each be zero. The residuals are the closing errors $\sum L$ (in latitude) and $\sum D$ (in departure).

(a) Consecutive coordinates

Closing error in latitude $e_L = +1.023$ m, in departure $e_D = +0.372$ m.

$$e = \sqrt{e_L^2 + e_D^2} = \sqrt{1.023^2 + 0.372^2} = \sqrt{1.0465+0.1384}=\mathbf{1.089\ m}$$

Direction of closing error (measured as a WCB of the error vector $e_L,e_D$):

$$\theta_e = \tan^{-1}\!\frac{e_D}{e_L} = \tan^{-1}\frac{0.372}{1.023} = 20°02′ \text{ (NE)}.$$

The correction is applied in the opposite sense.

(b) Relative precision

$$\text{Precision} = \frac{e}{\sum L} = \frac{1.089}{778.0} = \frac{1}{715}.$$

Since $1\text{ in }715$ is finer than the permissible $1\text{ in }500$, the traverse is acceptable for third-order work.

(c) Bowditch adjustment Correction to latitude of a line $= -e_L\dfrac{L}{\sum L}$; correction to departure $= -e_D\dfrac{L}{\sum L}$.

The adjusted latitudes and departures now sum to zero, so the traverse is geometrically closed. Closing error 1.089 m, precision 1 in 715, adjusted by Bowditch rule as tabulated.

(a) Derivation (inclined sight, vertical staff) Let the staff intercept be $s$ (top − bottom reading) and the angle of elevation $\theta$. The rays strike the inclined staff; resolving the inclined stadia distance $ks+c$ onto the horizontal and vertical axes:

$$\boxed{D = ks\cos^2\theta + c\cos\theta} \qquad \boxed{V = \tfrac{1}{2}ks\sin 2\theta + c\sin\theta}$$

With an anallactic lens $c=0$, so $D = ks\cos^2\theta$ and $V = \tfrac{1}{2}ks\sin 2\theta$.

(b) Computation Staff intercept: $s = 2.630 - 1.150 = 1.480$ m. Mid (axial) reading $m = 1.890$ m, $\theta = 6°00′$.

Horizontal distance:

$$D = 100\times1.480\times\cos^2 6° = 148.0\times(0.99452)^2 = 148.0\times0.98904 = \mathbf{146.383\ m}.$$

Vertical component:

$$V = \tfrac12\times100\times1.480\times\sin 12° = 74.0\times0.20791 = \mathbf{+15.385\ m}.$$

Reduced level of the line-of-sight (instrument) axis:

$$\text{RL}_{axis} = \text{RL}_P + h_i = 100.000 + 1.450 = 101.450\ \text{m}.$$

Reduced level of Q (rising sight, so add $V$, subtract the axial staff reading):

$$\text{RL}_Q = \text{RL}_{axis} + V - m = 101.450 + 15.385 - 1.890 = \mathbf{114.945\ m}.$$

Horizontal distance PQ = 146.383 m; RL of Q = 114.945 m.

Given: $R = 300$ m, deflection $\Delta = 40°00′$, chainage PI = 1245.000 m, peg interval 20 m.

(a) Curve elements

$$T = R\tan\tfrac{\Delta}{2} = 300\tan 20° = 300\times0.36397 = \mathbf{109.191\ m}$$ $$L = \frac{\pi R\Delta}{180} = \frac{\pi\times300\times40}{180} = \mathbf{209.440\ m}$$ $$\text{Long chord } C = 2R\sin\tfrac{\Delta}{2} = 600\sin20° = 600\times0.34202 = \mathbf{205.212\ m}$$ $$\text{Mid-ordinate } M = R\!\left(1-\cos\tfrac{\Delta}{2}\right)=300(1-\cos20°)=300(1-0.93969)=\mathbf{18.092\ m}$$ $$\text{Apex distance } E = R\!\left(\sec\tfrac{\Delta}{2}-1\right)=300(1.06418-1)=\mathbf{19.253\ m}$$

(b) Chainages of tangent points

$$\text{Chainage }T_1 = \text{Chainage PI} - T = 1245.000 - 109.191 = \mathbf{1135.809\ m}$$ $$\text{Chainage }T_2 = \text{Chainage }T_1 + L = 1135.809 + 209.440 = \mathbf{1345.249\ m}$$

(c) Deflection-angle setting-out table First sub-chord = first round 20 m chainage after T1 = 1140.000 − 1135.809 = 4.191 m; then full 20 m chords. Deflection angle for a chord $\ell$: $\delta = \dfrac{1718.873\,\ell}{R}$ minutes.

• Peg 1 ($\ell_1 = 4.191$ m): $\delta_1 = 1718.873\times4.191/300 = 24.012' = 0°24′01″$ → cumulative $0°24′01″$
• Peg 2 ($\ell = 20$ m): $\delta = 1718.873\times20/300 = 114.592' = 1°54′36″$ → cumulative $2°18′37″$
• Peg 3 ($\ell = 20$ m): $\delta = 1°54′36″$ → cumulative $4°13′12″$

(Check: the final cumulative deflection at T2 must equal $\Delta/2 = 20°00′$.)

Tangent length 109.191 m, curve length 209.440 m, T1 at 1135.809 m, T2 at 1345.249 m.

Given: $g_1 = +3\%$, $g_2 = -2\%$, length $L = 180$ m, grade-intersection chainage 1500.000 m, RL of intersection 250.000 m. The curve is symmetrical, so it extends $L/2 = 90$ m each side of the intersection.

(a) RL of tangent points BVC chainage $= 1500 - 90 = 1410.000$ m, EVC chainage $= 1500 + 90 = 1590.000$ m.

$$\text{RL}_{BVC} = 250.000 - g_1\cdot\tfrac{L}{2} = 250.000 - 0.03\times90 = \mathbf{247.300\ m}$$ $$\text{RL}_{EVC} = 250.000 + g_2\cdot\tfrac{L}{2} = 250.000 - 0.02\times90 = \mathbf{248.200\ m}$$

(b) Highest point RL on a parabolic curve at distance $x$ from BVC:

$$y(x) = \text{RL}_{BVC} + \frac{g_1}{100}x - \frac{(g_1-g_2)}{200L}x^2.$$

The highest point occurs where $dy/dx = 0$:

$$x_{high} = \frac{g_1 L}{g_1 - g_2} = \frac{3\times180}{3-(-2)} = \frac{540}{5} = 108.0\ \text{m from BVC}.$$

Chainage $= 1410 + 108 = \mathbf{1518.000\ m}$.

$$y_{high} = 247.300 + 0.03\times108 - \frac{5}{200\times180}\times108^2 = 247.300 + 3.240 - 1.620 = \mathbf{248.920\ m}.$$

(c) RLs at 30 m intervals (offset from grade $= \frac{g_1-g_2}{200L}x^2 = \frac{5}{36000}x^2$):

BVC RL 247.300 m, EVC RL 248.200 m, highest point at chainage 1518.000 m, RL 248.920 m.

(a) Triangulation vs trilateration Triangulation is a control-survey method in which a network of well-shaped triangles is formed; one side (the base line) is measured precisely and all the angles are observed, the remaining sides being computed by the sine rule. Trilateration measures all the sides (with EDM) and computes the angles. Triangulation is preferred (i) in hilly/undulating terrain where long inter-visible lines are easier than chaining, and (ii) when high angular precision is available but direct distance measurement is difficult, e.g. across rivers or gorges.

(b) Well-conditioned triangle A triangle is well-conditioned when its computed sides are least affected by small errors in the observed angles — conventionally no angle smaller than 30° or larger than 120°. By the sine rule a computed side $b = a\dfrac{\sin B}{\sin A}$. Taking the differential, the fractional error in $b$ due to an error $\delta A$ in angle $A$ is proportional to $\cot A\,\delta A$, and similarly $\cot B\,\delta B$. The total propagated error is minimised when $\cot^2 A + \cot^2 B$ is minimum subject to $A+B = 120°$ (for the third angle 60°). Minimising gives $A = B = 60°$, i.e. the equilateral triangle, where each $\cot 60° = 0.577$ is small and equal — hence the most reliable computed sides.

(c) Height of signal at B The combined curvature-and-refraction correction is $C = 0.0673\,d^2$ m.

Distance of visible horizon from the 4 m instrument at A:

$$d_1 = \sqrt{\frac{h_A}{0.0673}} = \sqrt{\frac{4}{0.0673}} = \sqrt{59.435} = 7.709\ \text{km}.$$

Remaining distance from horizon to B:

$$d_2 = 18 - 7.709 = 10.291\ \text{km}.$$

Required height of signal above the ground at B for the line of sight grazing the horizon:

$$h_B' = 0.0673\,d_2^2 = 0.0673\times10.291^2 = 0.0673\times105.90 = 7.127\ \text{m}.$$

This is the height the signal must rise above B's reduced level relative to the horizon line. Since B already stands $980 - 240 = 740$ m above A, the line of sight clears B's ground comfortably; the governing requirement here is the visibility (grazing) height:

$$\boxed{h_{signal} \approx 7.13\ \text{m above the station mark at B}.}$$

(If the elevations are level, the same formula gives the minimum tower; the large height difference of B over A means a 7.13 m signal is more than sufficient for inter-visibility.)

Given: $R = 400$ m, $V = 65$ km/h, rate of change of radial acceleration $C = 0.3\ \text{m/s}^3$.

(a) Length of transition curve (by rate of change of radial acceleration):

$$L = \frac{0.0215\,V^3}{C\,R} = \frac{0.0215\times65^3}{0.3\times400} = \frac{0.0215\times274625}{120} = \frac{5904.44}{120} = \mathbf{49.20\ m}.$$

(b) Shift of the circular curve:

$$S = \frac{L^2}{24R} = \frac{49.20^2}{24\times400} = \frac{2420.6}{9600} = \mathbf{0.252\ m}.$$

(c) Superelevation:

$$e = \frac{V^2}{127R} = \frac{65^2}{127\times400} = \frac{4225}{50800} = 0.0832 = \mathbf{8.32\%}.$$

For a 7.0 m carriageway this corresponds to a raising of the outer edge by $e\times b = 0.0832\times7.0 = 0.582$ m.

L = 49.20 m, shift = 0.252 m, superelevation = 8.32%.

Given: $R_s = 250$ m, $\Delta_s = 35°$; $R_l = 400$ m, $\Delta_l = 25°$.

Total deflection $\Delta = \Delta_s + \Delta_l = 35° + 25° = \mathbf{60°}$.

Arc tangent lengths:

$$t_s = R_s\tan\tfrac{\Delta_s}{2} = 250\tan17.5° = 250\times0.31530 = 78.825\ \text{m}$$ $$t_l = R_l\tan\tfrac{\Delta_l}{2} = 400\tan12.5° = 400\times0.22169 = 88.678\ \text{m}$$

Total tangent lengths from the point of intersection (using the sine relation):

$$T_1 = t_s + (t_s+t_l)\frac{\sin\Delta_l}{\sin\Delta} = 78.825 + 167.503\times\frac{\sin25°}{\sin60°} = 78.825 + 167.503\times0.48803 = \mathbf{160.566\ m}$$ $$T_2 = t_l + (t_s+t_l)\frac{\sin\Delta_s}{\sin\Delta} = 88.678 + 167.503\times\frac{\sin35°}{\sin60°} = 88.678 + 167.503\times0.66222 = \mathbf{199.616\ m}$$

Lengths of arcs:

$$L_s = \frac{\pi R_s\Delta_s}{180} = \frac{\pi\times250\times35}{180} = 152.716\ \text{m}, \quad L_l = \frac{\pi\times400\times25}{180} = 174.533\ \text{m}.$$

Total length $= 152.716 + 174.533 = \mathbf{327.249\ m}$.

Δ = 60°, T1 = 160.566 m, T2 = 199.616 m, total curve length = 327.249 m.

(a) Components of a total station A total station is an integrated electronic instrument combining an electronic theodolite, an electronic distance measuring (EDM) unit and an on-board microprocessor with data storage.

        +-----------------------------+
        |        TOTAL STATION        |
        |                             |
  Target| EDM unit  -->  Microprocessor  --> Display / Keyboard
  prism |  (IR/laser)        |              |
        | Electronic       Memory /        On-board
        | theodolite       data card        software
        | (Hz & V circles)                  (COGO, area, etc.)
        +-----------------------------+

• Electronic theodolite: measures horizontal and vertical angles using electronic (digital) circles with photo-detectors.
• EDM unit: measures slope distance to a reflecting prism (or reflectorless) by phase or pulse comparison of a modulated infra-red/laser beam.
• Microprocessor: reduces slope distance and angles to horizontal distance, vertical distance, coordinates; performs COGO, traverse, area and setting-out computations.
• Display and keyboard: for input of station data and read-out of results.
• Data storage (internal memory / card / Bluetooth): records observations for transfer to a computer.

(b) Operations done more efficiently than theodolite + tape

1. Simultaneous measurement of angle and distance with automatic reduction to horizontal distance and reduced level.
2. Direct computation and storage of three-dimensional coordinates (E, N, RL) of observed points.
3. Stake-out / setting-out of points from stored design coordinates, with real-time left/right and in/out guidance.
4. On-board area, remote-elevation (REM) and missing-line (MLM) computations, plus paperless data logging that eliminates booking errors.

(a) Principle of GPS positioning GPS determines a receiver's position by trilateration in space. Each satellite continuously broadcasts its position (from the navigation message/ephemeris) and a precise time signal. The receiver measures the travel time of the signal and computes the pseudo-range $\rho = c\,\Delta t$ to each satellite, where $c$ is the speed of light. Geometrically each range places the receiver on the surface of a sphere centred on that satellite; the intersection of the spheres fixes the position.

For a 3-D fix four unknowns must be solved: the three coordinates $(X, Y, Z)$ of the receiver and the receiver-clock bias $\delta t$ (inexpensive receiver clocks are not synchronised with the satellite atomic clocks). Each satellite provides one pseudo-range equation:

$$\rho_i = \sqrt{(X_i-X)^2 + (Y_i-Y)^2 + (Z_i-Z)^2} + c\,\delta t.$$

Four unknowns require four independent equations, hence a minimum of four satellites must be tracked simultaneously. (Three suffice only if height is known, e.g. 2-D navigation.)

(b) Absolute vs differential GPS

• Absolute (point) positioning: a single receiver computes its position autonomously from satellite ranges. It carries the full satellite-clock, ephemeris, ionospheric/tropospheric and multipath errors, giving accuracy of a few metres (3–10 m).
• Differential GPS (DGPS): a reference (base) receiver on a known point computes range corrections and transmits them to a roving receiver. Because errors common to both receivers (atmospheric delay, ephemeris, satellite-clock) largely cancel, accuracy improves to sub-metre, and to centimetre level with carrier-phase RTK.

DGPS gives the better accuracy because the spatially correlated errors are eliminated by differencing the observations of the base and rover stations.

Procedure (rectangular building 20 m × 12 m)

1. Establish the first corner A on the ground from the reference (building) line, fixed by the architectural setting-out plan (offset from the road boundary).
2. Set the theodolite at A, level it and sight along the reference line; this gives the direction of the long (20 m) wall. Tape 20.000 m along this line and drive peg B.
3. Turn off 90°00′00″ at A (using the horizontal circle, face left then face right and mean to remove instrument error). Along this perpendicular tape 12.000 m and fix peg D.
4. Shift the theodolite to B, turn 90° from line BA and tape 12.000 m to fix peg C (the fourth corner).
5. Drive batter (profile) boards about 1–1.5 m clear of each corner and stretch strings between them so the corner pegs can be re-established after excavation. Mark foundation widths on the boards.

Checks

1. Diagonal check: both diagonals AC and BD must equal $\sqrt{20^2+12^2}=\sqrt{400+144}=\sqrt{544}=\mathbf{23.324\ m}$; equal diagonals confirm the rectangle is true.
2. Closing/perimeter check: independently measure side CD (should be 20.000 m) and DA (should be 12.000 m); any misclosure indicates a layout error to be corrected before excavation.

(a) Field determination of $k$ and $c$ For a horizontal line of sight the stadia equation is $D = k\,s + c$, where $s$ is the staff intercept. To find the constants:

1. On level ground drive pegs at several known horizontal distances $D_1, D_2, \dots$ (e.g. 30, 60, 90, 120 m) measured from the instrument's vertical axis with a tape.
2. With the line of sight horizontal, read the staff intercept $s_1, s_2, \dots$ at each peg.
3. Each observation gives a linear equation $D_i = k\,s_i + c$. Solving any two simultaneously (or fitting a straight line to all pairs) yields $k$ (the slope) and $c$ (the intercept). Using more than two points and least squares improves reliability.

(b) Computation Observation 1: $50.000 = k(0.495) + c$ Observation 2: $100.000 = k(0.998) + c$

Subtracting (1) from (2):

$$100.000 - 50.000 = k(0.998 - 0.495) \Rightarrow 50.000 = k(0.503)$$ $$k = \frac{50.000}{0.503} = \mathbf{99.40}.$$

Substitute into (1):

$$c = 50.000 - 99.40\times0.495 = 50.000 - 49.207 = \mathbf{0.793\ m}.$$

(These are close to the nominal $k=100,\ c\approx0.8$ for an external-focusing instrument.)

For a staff intercept of 1.350 m (horizontal sight):

$$D = k\,s + c = 99.40\times1.350 + 0.793 = 134.190 + 0.793 = \mathbf{134.983\ m}.$$

k = 99.40, c = 0.793 m; horizontal distance = 134.983 m ≈ 134.98 m.