BE Civil Engineering (IOE, TU) Surveying II (IOE, CE 503b) Question Paper 2080 Nepal

Step 1 — Latitudes and Departures ($L=\ell\cos\theta,\ D=\ell\sin\theta$)

Sums:

• $\Sigma L = +100.000-129.904-191.511+137.888 = -83.527$ m
• $\Sigma D = +173.205+75.000-160.697-115.702 = -28.194$ m

Step 2 — Closing error

$$e = \sqrt{(\Sigma L)^2+(\Sigma D)^2}=\sqrt{(-83.527)^2+(-28.194)^2}=\sqrt{6976.8+794.9}=\sqrt{7771.7}=88.157\text{ m}$$

Bearing of closing error (the error vector points along $(\Sigma L,\Sigma D)$; the correction is opposite):

$$\theta_e=\tan^{-1}\!\left|\frac{\Sigma D}{\Sigma L}\right|=\tan^{-1}\frac{28.194}{83.527}=18.65^\circ.$$

Both $\Sigma L$ and $\Sigma D$ are negative ⇒ third quadrant ⇒ WCB $=180^\circ+18.65^\circ=198^\circ39'$.

Closing error $\approx \mathbf{88.16\ m}$ at WCB $\mathbf{198^\circ39'}$. (Deliberately large to exercise the full method; relative precision $=88.16/780\approx1/8.8$, which in practice signals a gross field blunder.)

Step 3 — Bowditch corrections. Correction to latitude/departure of a line $\propto$ its length:

$$C_{L,i}=-\Sigma L\cdot\frac{\ell_i}{\Sigma\ell},\qquad C_{D,i}=-\Sigma D\cdot\frac{\ell_i}{\Sigma\ell}.$$

Here $-\Sigma L=+83.527$, $-\Sigma D=+28.194$, $\Sigma\ell=780$.

Corrected latitudes & departures:

Check: $121.417-113.841-164.740+157.164=0.000$ ✓; $180.434+80.422-151.660-109.196=0.000$ ✓.

The adjusted traverse now closes exactly.

Step 1 — Derivation. For a staff held vertical and line of sight inclined at angle $\theta$, the staff intercept $s$ is not perpendicular to the line of sight. With $K$ the multiplying and $C$ the additive constant, the sloped distance along the line of collimation to the staff is $L=Ks\cos\theta+C$. Resolving:

$$\boxed{D=Ks\cos^2\theta+C\cos\theta},\qquad \boxed{V=\tfrac{1}{2}Ks\sin2\theta+C\sin\theta}.$$

With $C=0$: $D=Ks\cos^2\theta$, $V=\tfrac12 Ks\sin2\theta$.

Step 2 — Data. Staff intercept $s = 2.600-1.150 = 1.450$ m; $\theta=8^\circ30'=8.5^\circ$; $Ks=100\times1.450=145.000$ m.

Step 3 — Horizontal distance.

$$D=Ks\cos^2\theta=145.000\times\cos^2 8.5^\circ=145.000\times(0.989016)^2=145.000\times0.978153=141.832\text{ m}.$$

Horizontal distance PQ $=\mathbf{141.83\ m}$.

Step 4 — Vertical component.

$$V=\tfrac12 Ks\sin2\theta=\tfrac12\times145.000\times\sin17^\circ=72.500\times0.292372=21.197\text{ m}.$$

Step 5 — Reduced level of Q. Using RL of P, instrument height $h_i$, vertical component $V$, and middle hair reading $m$:

$$RL_Q = RL_P + h_i + V - m = 1245.600 + 1.420 + 21.197 - 1.875.$$ $$RL_Q = 1266.342\text{ m}.$$

Reduced level of Q $=\mathbf{1266.342\ m}$.

Step 1 — Curve elements. Deflection angle $\Delta=42^\circ$, $R=300$ m.

Tangent length: $T=R\tan(\Delta/2)=300\tan21^\circ=300\times0.383864=115.159$ m.

Length of curve: $L=\dfrac{\pi R\Delta}{180}=\dfrac{\pi\times300\times42}{180}=219.911$ m.

Step 2 — Chainages.

• Chainage of T1 $=$ Chainage of I $- T = 1840.000 - 115.159 = 1724.841$ m.
• Chainage of T2 $=$ Chainage of T1 $+ L = 1724.841 + 219.911 = 1944.752$ m.

T1 chainage $=\mathbf{1724.841\ m}$, T2 chainage $=\mathbf{1944.752\ m}$, $L=\mathbf{219.911\ m}$.

Step 3 — Chord layout. First sub-chord brings chainage from 1724.841 to the next full 20 m station 1740.000:

$$c_1 = 1740.000-1724.841 = 15.159\text{ m (initial sub-chord)}.$$

Thereafter full chords of 20 m. Last chord: $1944.752-1940.000 = 4.752$ m.

Step 4 — Deflection per chord. Rankine: $\delta = \dfrac{1718.873\,c}{R}$ minutes (since $\delta=\frac{c}{2R}$ rad $=\frac{1718.873\,c}{R}$ min).

• Full 20 m chord: $\delta=\dfrac{1718.873\times20}{300}=114.592'=1^\circ54'35.5''$.
• Initial sub-chord 15.159 m: $\delta_1=\dfrac{1718.873\times15.159}{300}=86.857'=1^\circ26'51.4''$.

Step 5 — Cumulative deflection table (first three pegs):

Check (final): total deflection at T2 should equal $\Delta/2 = 21^\circ00'00''$. Summing all chords' tangential angles over the full curve length gives $\delta_{tot}=\frac{1718.873\times219.911}{300}=1260.0'=21^\circ00'$ ✓.

Step 1 — Convert speed. $v=65\ \text{km/h}=65\times\frac{1000}{3600}=18.056\ \text{m/s}$.

Step 2 — Length of transition (rate of change of radial acceleration).

$$L=\frac{v^3}{c\,R}=\frac{(18.056)^3}{0.3\times250}=\frac{5886.5}{75}=78.487\ \text{m}.$$

Length of transition $L=\mathbf{78.49\ m}$.

Step 3 — Shift.

$$S=\frac{L^2}{24R}=\frac{(78.487)^2}{24\times250}=\frac{6160.2}{6000}=1.027\ \text{m}.$$

Shift $S=\mathbf{1.027\ m}$.

Step 4 — Superelevation.

$$e=\frac{v^2}{gR}=\frac{(18.056)^2}{9.81\times250}=\frac{326.0}{2452.5}=0.1329.$$

As a rate (tangent of camber angle), $e\approx0.133$, i.e. about $1$ in $7.5$.

Step 5 — Raising of outer edge. With full carriageway width $W=7.0$ m, the outer edge is raised relative to the inner by:

$$h = e\times W = 0.1329\times7.0 = 0.930\ \text{m}.$$

If superelevation is applied about the centre line, the outer edge rises $h/2=0.465$ m and the inner edge falls $0.465$ m.

Theoretical superelevation $e=\mathbf{0.133}$ (1 in 7.5); raising of outer edge $=\mathbf{0.930\ m}$ across full width.

Note: This theoretical $e$ exceeds the usual design maximum (~0.07); in practice $e$ is capped and part of the centripetal demand is met by side friction. The question asks for the full theoretical value.

Part (a) — Principle and classification.

Principle: Triangulation is a control-survey method in which a network of well-conditioned triangles is established; one side (the base line) is measured precisely, all angles are measured with a theodolite, and the lengths of all other sides are computed using the sine rule. Stations are chosen to be intervisible and form strong (well-conditioned) figures so that angular errors propagate minimally into computed lengths.

Classification:

Part (b) — Intervisibility.

Derivation of combined curvature+refraction coefficient: $\dfrac{1-2m}{2R}=\dfrac{1-0.14}{2\times6370}=\dfrac{0.86}{12740}=6.75\times10^{-5}\ \text{per km}^2$. Hence the correction at a point between the two stations is $C=0.0675\,d_1 d_2$ m, with $d_1,d_2$ in km.

Step 1 — Elevation of the straight line AB at the obstacle (15 km from A). Straight line from A(420 m) to B(530 m) over 40 km. At 15 km:

$$h_{LOS}=420+(530-420)\times\frac{15}{40}=420+110\times0.375=420+41.25=461.25\ \text{m}.$$

Step 2 — Curvature+refraction effect (earth bulge) at the obstacle, $d_1=15$ km, $d_2=25$ km:

$$C=0.0675\,d_1 d_2 = 0.0675\times15\times25 = 0.0675\times375 = 25.31\ \text{m}.$$

The bulge lifts the intervening ground toward the chord, i.e. the effective obstacle top relative to the straight chord becomes $475 + 25.31 = 500.31$ m.

Step 3 — Compare.

• Line-of-sight elevation at that point $=461.25$ m.
• Effective obstacle top $=500.31$ m.
• Clearance $=461.25 - 500.31 = -39.06\ \text{m}$ (negative).

The line of sight does NOT clear the obstacle; it is blocked by about $\mathbf{39.1\ m}$. Intervisibility would require raised signals/towers (about 39 m of elevation gain distributed at A and/or B).

Step 1 — Grades. $g_1=+2.5\%=+0.025$, $g_2=-1.5\%=-0.015$. Total change $A=g_1-g_2=0.025-(-0.015)=0.040$ (4%). Curve length $L=160$ m.

Step 2 — Highest point. On a summit parabola the highest point occurs where the gradient is zero. Measured from the first tangent point:

$$x=\frac{g_1\,L}{g_1-g_2}=\frac{0.025\times160}{0.040}=\frac{4.0}{0.040}=100.0\ \text{m}.$$

Highest point is $\mathbf{100\ m}$ from the start of the curve.

Step 3 — Rate of change of gradient per 20 m chain. Total grade change over the curve $=A=4\%$ over $L/20 = 160/20 = 8$ chains:

$$r=\frac{A}{L/20}=\frac{4\%}{8}=0.5\%\ \text{per }20\text{ m chain}.$$

Rate of change $=\mathbf{0.5\%}$ per 20 m chain (the grade falls 0.5% every 20 m).

Step 1 — Total deflection angle. For a compound curve the total deflection between the two main tangents equals the sum of the individual central angles:

$$\Delta=\Delta_1+\Delta_2=30^\circ+40^\circ=70^\circ.$$

Total deflection angle $=\mathbf{70^\circ}$.

Step 2 — Length of each arc. $L=\dfrac{\pi R\Delta}{180}$.

• Arc 1: $L_1=\dfrac{\pi\times200\times30}{180}=\dfrac{18849.6}{180}=104.720$ m.
• Arc 2: $L_2=\dfrac{\pi\times300\times40}{180}=\dfrac{37699.1}{180}=209.440$ m.

Step 3 — Total length.

$$L=L_1+L_2=104.720+209.440=314.159\ \text{m}.$$

Total length of compound curve $=\mathbf{314.16\ m}$.

Part (a) — Total station. A total station is an electronic/optical surveying instrument that integrates an electronic theodolite (horizontal and vertical angles) with an EDM (distances) and an on-board microprocessor with data storage. Functions integrated: angle measurement, distance measurement, computation of horizontal distance/coordinates/elevations, data recording, and (in most models) setting-out routines and digital data transfer.

Part (b) — Phase-comparison EDM principle. The instrument transmits a continuous, intensity-modulated carrier (light/IR) toward a reflector; the returned signal's modulation phase is compared with the outgoing reference. The double path equals an integer number of full modulation wavelengths plus a fractional part from the phase reading:

$$2D = n\lambda + \Delta\lambda,$$

where $n$ = integer ambiguity (resolved using several modulation frequencies) and $\Delta\lambda$ = part-wavelength.

Part (c) — Computation. $\lambda=20$ m, $n=43$, $\Delta\lambda=7.4$ m (double path):

$$2D = n\lambda + \Delta\lambda = 43\times20 + 7.4 = 860 + 7.4 = 867.4\ \text{m}.$$ $$D = \frac{867.4}{2} = 433.7\ \text{m}.$$

Slope distance $=\mathbf{433.70\ m}$.

Part (a) — Three segments.

1. Space segment — the constellation of satellites (nominally 24+) in ~20,200 km orbits, transmitting timed, coded radio signals on L-band carriers.
2. Control segment — the master control station and a network of monitor stations that track satellites, compute orbital (ephemeris) and clock corrections, and upload them to the satellites.
3. User segment — the GPS receivers and antennas that pick up satellite signals and compute position, velocity, and time.

Part (b) — Why four satellites. GPS positioning is trilateration using measured ranges (pseudoranges) to satellites of known position. Three unknowns ($X,Y,Z$) would in principle need three ranges; however, the receiver clock is not synchronized with system time, adding a fourth unknown — the receiver clock bias $\delta t$. Each pseudorange equation

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

contains four unknowns ($X,Y,Z,\delta t$), so four satellites (four equations) are required for a 3D fix.

Part (c) — Absolute vs. relative positioning.

Step 1 — Set up. In the tangential method, with both targets above the horizontal, the horizontal distance $D$ satisfies:

$$V_1 = D\tan\alpha_1,\qquad V_2 = D\tan\alpha_2,$$

where $V_1,V_2$ are the heights of the lower and upper vanes above the instrument axis, and the staff intercept is $s=V_2-V_1$ (upper vane higher).

Given: upper vane $\alpha_2=5^\circ30'$, lower vane $\alpha_1=3^\circ00'$, staff intercept $s = 3.000-1.000 = 2.000$ m.

Step 2 — Distance formula (both angles of elevation):

$$s = D\tan\alpha_2 - D\tan\alpha_1 = D(\tan\alpha_2-\tan\alpha_1)\ \Rightarrow\ D=\frac{s}{\tan\alpha_2-\tan\alpha_1}.$$

Step 3 — Compute.

• $\tan5^\circ30'=\tan5.5^\circ=0.096289$
• $\tan3^\circ00'=\tan3.0^\circ=0.052408$
• Difference $=0.096289-0.052408=0.043881$

$$D=\frac{2.000}{0.043881}=455.78\ \text{m}.$$

Horizontal distance $=\mathbf{455.78\ m}$.

Part (a) — Polar setting-out procedure.

1. Set up and level the total station over control station P; sight the reference object R and set the horizontal circle to the known bearing of PR (or zero).
2. From the coordinates, compute the bearing and distance of line PA (inverse computation).
3. Compute the angle to be turned from PR to PA.
4. Turn the instrument by that angle from PR and clamp the horizontal circle, fixing the direction PA.
5. Direct the prism along the line and, using the EDM, position it so the measured horizontal distance equals the computed PA; mark point A.
6. Check by re-observation or by setting out from a second control station.

Part (b) — Computation.

Deltas P→A: $\Delta E = 1060.000-1000.000 = +60.000$; $\Delta N = 2080.000-2000.000 = +80.000$.

Distance PA:

$$PA=\sqrt{\Delta E^2+\Delta N^2}=\sqrt{60^2+80^2}=\sqrt{3600+6400}=\sqrt{10000}=100.000\ \text{m}.$$

Horizontal distance PA $=\mathbf{100.000\ m}$.

Bearing of PA: $\theta_{PA}=\tan^{-1}\dfrac{\Delta E}{\Delta N}=\tan^{-1}\dfrac{60}{80}=\tan^{-1}0.75=36^\circ52'12''$ (both positive ⇒ NE quadrant).

Bearing of PR: $\Delta E = +100$, $\Delta N = 0$ ⇒ due East ⇒ WCB of PR $=90^\circ00'00''$.

Clockwise angle from PR to PA:

$$\beta = \theta_{PA}-\theta_{PR}=36^\circ52'12''-90^\circ00'00''=-53^\circ07'48''.$$

Adding $360^\circ$ for a clockwise turn: $\beta = 360^\circ-53^\circ07'48'' = 306^\circ52'12''$.

Set out A at horizontal distance $\mathbf{100.000\ m}$, turning $\mathbf{306^\circ52'12''}$ clockwise from PR (equivalently $53^\circ07'48''$ anticlockwise from PR).