BE Civil Engineering (IOE, TU) Surveying II (IOE, CE 503b) Question Paper 2076 Nepal

Concept. For a line of length $L$ and whole-circle bearing $\theta$:

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

For a closed traverse $\sum L = 0$ and $\sum D = 0$ ideally; the residuals give the closing error.

(a) Latitudes and departures

$\sum \text{Latitude} = 141.421 - 106.066 - 176.777 + 106.066 = \mathbf{-35.356\ m}$

$\sum \text{Departure} = 141.421 + 106.066 - 176.777 - 106.066 = \mathbf{-35.356\ m}$

(b) Closing error

Error in latitude $e_L = \sum\text{Lat} = -35.356$ m (so the correction is $+35.356$). Error in departure $e_D = \sum\text{Dep} = -35.356$ m (correction $+35.356$).

Magnitude of closing error:

$$e = \sqrt{e_L^2 + e_D^2} = \sqrt{35.356^2 + 35.356^2} = 35.356\sqrt{2} = \mathbf{50.0\ m}.$$

Bearing of the closing error (direction of the residual $e_L,e_D$, both negative → third quadrant):

$$\tan\alpha = \frac{|e_D|}{|e_L|} = 1 \Rightarrow \alpha = 45°,\ \text{WCB} = 180°+45° = \mathbf{225°}.$$

(The correction is applied in the opposite direction, WCB 45°.)

Perimeter $\sum L = 200+150+250+150 = 750$ m, so relative precision $= 50.0/750 \approx 1/15$ (poor — illustrative figures).

(c) Bowditch adjustment. Correction to a line is proportional to its length:

$$C_{L,i} = -e_L\cdot\frac{L_i}{\sum L} = +35.356\cdot\frac{L_i}{750}, \qquad C_{D,i} = +35.356\cdot\frac{L_i}{750}.$$

Per-metre factor $= 35.356/750 = 0.047141$ m/m.

Check: $\sum C_{lat}=9.428+7.071+11.785+7.071=35.356$ ✓; corrected $\sum\text{Lat}=150.849-98.995-164.992+113.137=-0.001\approx 0$ ✓ (rounding). Similarly corrected $\sum\text{Dep}\approx 0$ ✓.

Adjusted traverse closes; corrected latitudes and departures are tabulated above.

(a) Derivation (vertical staff, inclined line of sight). Let the staff intercept be $s$ (top − bottom reading) and the angle of inclination of the line of sight be $\theta$. The inclined (slope) distance along the line of sight to the centre of the staff is $L = Ks\cos\theta + C$. Resolving:

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

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

(b) Horizontal distance. Staff intercept $s = 2.900 - 1.150 = 1.750$ m. Middle (axial) reading $m = 2.025$ m. $\theta = 8°20' = 8.3333°$. $\cos\theta = 0.98944$, so $\cos^2\theta = 0.97900$.

$$D = 100\times 1.750\times 0.97900 = \mathbf{171.32\ m}.$$

(c) Reduced level of Q.

$$V = \tfrac12 \times 100 \times 1.750 \times \sin(16°40') = 87.5 \times 0.28680 = 25.095\ \text{m}.$$

($\sin 16°40' = \sin 16.6667° = 0.28680$.)

RL of line of sight at instrument axis $=$ RL of P $+$ height of instrument $= 1250.00 + 1.40 = 1251.40$ m.

For an elevation sight:

$$\text{RL of Q} = \text{RL of P} + h_i + V - m = 1250.00 + 1.40 + 25.095 - 2.025.$$ $$\text{RL of Q} = \mathbf{1274.47\ m}.$$

(a) Curve geometry. Deflection angle $\Delta = 40°$, $R = 300$ m.

Tangent length:

$$T = R\tan\frac{\Delta}{2} = 300\tan 20° = 300\times 0.36397 = \mathbf{109.19\ m}.$$

Length of curve:

$$L = \frac{\pi R\Delta}{180} = \frac{\pi\times 300\times 40}{180} = \mathbf{209.44\ m}.$$

Chainage of T1 $=$ Chainage of PI $- T = 1850.00 - 109.19 = \mathbf{1740.81\ m}.$ Chainage of T2 $=$ Chainage of T1 $+ L = 1740.81 + 209.44 = \mathbf{1950.25\ m}.$

First sub-chord (to bring chainage to a 20 m multiple): first peg at chainage 1760.00 m, so initial chord $c_1 = 1760.00 - 1740.81 = 19.19$ m. Final sub-chord $= 1950.25 - 1940.00 = 10.25$ m. The intermediate chords are full 20 m chords.

(b) Deflection angle for a full 20 m chord. Rankine: $\delta = \dfrac{1718.873\,c}{R}$ minutes (since $\delta = \tfrac{c}{2R}$ rad $= \tfrac{90c}{\pi R}$ deg $= \tfrac{1718.873\,c}{R}$ min).

$$\delta_{20} = \frac{1718.873\times 20}{300} = 114.591' = 1°54'35''.$$

Rounded to the $20''$ least count: $\mathbf{1°54'40''}$ (for tabulation we keep computed value and round each cumulative angle).

(c) Deflection-angle table (cumulative angles set out from T1; round each total to nearest $20''$).

First (sub) chord $c_1 = 19.19$ m: $\delta_1 = \dfrac{1718.873\times 19.19}{300} = 109.951' = 1°49'57''.$

Check (final peg). Total deflection at T2 should equal $\Delta/2 = 20°$. Sum of all tangential angles from $c_1$, the full chords, and the final sub-chord equals $20°00'00''$, confirming the setting-out is consistent.

(The instrument is set at T1, sighting along the rear tangent for $0°00'$, then each cumulative deflection is turned and the chord measured to fix each peg.)

Given. $R = 250$ m, $v = 60$ km/h $= 60\times\tfrac{1000}{3600} = 16.667$ m/s, $\alpha = 0.3$ m/s³, width $b = 7.5$ m, $g = 9.81$ m/s².

(a) Length of transition (rate of change of radial acceleration).

$$L = \frac{v^3}{\alpha R} = \frac{16.667^3}{0.3\times 250} = \frac{4629.6}{75} = \mathbf{61.73\ m}.$$

($16.667^3 = 4629.6$.)

(b) Superelevation (fully balanced).

$$e = \frac{v^2}{gR} \quad(\text{as a ratio}), \qquad \text{raise of outer edge } E = \frac{b\,v^2}{gR}.$$ $$\frac{v^2}{gR} = \frac{16.667^2}{9.81\times 250} = \frac{277.78}{2452.5} = 0.11327.$$

Superelevation as a slope $\approx 1$ in $8.83$. Raise of the outer edge across the carriageway:

$$E = b\times 0.11327 = 7.5\times 0.11327 = \mathbf{0.850\ m}.$$

(c) Shift of the circular curve.

$$S = \frac{L^2}{24R} = \frac{61.73^2}{24\times 250} = \frac{3810.6}{6000} = \mathbf{0.635\ m}.$$

Given. $g_1 = +2.5\% = +0.025$, $g_2 = -1.5\% = -0.015$, length $L = 120$ m, PI chainage $= 2000.00$ m, RL of PI $= 315.00$ m. Total change of grade $A = g_1 - g_2 = 0.025-(-0.015) = 0.040$.

(a) RL of BVC and EVC. The curve is symmetrical, so BVC and EVC are $L/2 = 60$ m each side of the PI.

Chainage BVC $= 2000.00 - 60 = 1940.00$ m; Chainage EVC $= 2000.00 + 60 = 2060.00$ m.

RL of BVC $=$ RL of PI $- g_1\times(L/2) = 315.00 - 0.025\times 60 = 315.00 - 1.50 = \mathbf{313.50\ m}.$

RL of EVC $=$ RL of PI $+ g_2\times(L/2) = 315.00 + (-0.015)\times 60 = 315.00 - 0.90 = \mathbf{314.10\ m}.$

(b) Highest point. Measure $x$ from BVC. The grade-line RL is $313.50 + g_1 x$, and the parabolic offset (below grade line, summit) is $\dfrac{A x^2}{2L}$. Curve RL:

$$y = 313.50 + 0.025x - \frac{0.040\,x^2}{2\times 120} = 313.50 + 0.025x - \frac{x^2}{6000}.$$

Highest point where $\dfrac{dy}{dx}=0$:

$$0.025 - \frac{2x}{6000} = 0 \Rightarrow x = \frac{0.025\times 6000}{2} = 75.0\ \text{m from BVC}.$$

Chainage of highest point $= 1940.00 + 75.00 = \mathbf{2015.00\ m}.$

RL at $x = 75$ m:

$$y = 313.50 + 0.025\times 75 - \frac{75^2}{6000} = 313.50 + 1.875 - 0.9375 = \mathbf{314.44\ m}.$$

(a) Triangulation. Triangulation is a method of horizontal control in which the area is covered by a network (chain) of connected triangles; one line (the base line) is measured precisely and all the angles of the triangles are measured, so that the lengths of all other sides are computed trigonometrically using the sine rule. Purposes: (i) to establish accurately located horizontal control points over a large area for subsequent detailed surveys and mapping; (ii) to fix the positions of important features (towers, boundaries) and to provide a framework for engineering works such as bridges, tunnels and dams.

(b) Well-conditioned triangle. A well-conditioned triangle is one whose shape gives the least error in the computed sides for a given error in the measured angles — i.e. the computed length is least sensitive to angular error. This occurs when no angle is too small. The ideal is the equilateral triangle (each angle 60°); in practice angles should lie between about 30° and 120°. Equilateral (or near-equilateral) triangles are preferred because the sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}$ amplifies angular error when an angle is very small (since $\frac{d a}{a}\propto \cot A\,dA$ becomes large as $A\to 0$), so balanced angles minimise propagated error.

(c) Signal vs station mark.

• A station mark is the permanent ground mark (a stone, peg, or embedded mark) that physically defines the exact horizontal position of the triangulation station; it is buried/protected for future recovery.
• A signal is a temporary device (pole, target, beacon or heliotrope) erected vertically over the station so that it is visible from other stations during angle observation. The signal marks where to point the theodolite; the station mark defines the true point.

(a) Total station. A total station is an electronic surveying instrument that integrates an electronic theodolite (for measuring horizontal and vertical angles) with an electronic distance measuring (EDM) unit and an on-board microprocessor with data storage. Quantities measured/computed directly: (i) horizontal angle, (ii) vertical (zenith) angle, (iii) slope distance, (iv) horizontal distance, plus derived values such as vertical height difference and 3D coordinates (E, N, RL).

(b) Principle of EDM. EDM measures distance by transmitting a modulated electromagnetic (light/infrared/microwave) wave to a reflector and measuring the phase difference (or the travel time) between the transmitted and returned signal. Knowing the wavelength/modulation frequency and the velocity of the wave, the instrument resolves the number of whole wavelengths plus the fractional phase to obtain the double path distance, and reports half of it: $D = \tfrac{1}{2} (n\lambda + \Delta\lambda)$.

(c) Computation. Slope distance $S = 245.380$ m, zenith angle $Z = 86°30'$ (measured from the vertical).

Horizontal distance:

$$H = S\sin Z = 245.380\times \sin 86°30' = 245.380\times 0.99813 = \mathbf{244.92\ m}.$$

($\sin 86.5° = 0.99813$.)

Vertical (elevation) difference:

$$V = S\cos Z = 245.380\times \cos 86°30' = 245.380\times 0.06105 = \mathbf{14.98\ m}.$$

($\cos 86.5° = 0.06105$.) Since the zenith angle is less than 90°, the prism is above the instrument's horizontal line of sight by 14.98 m.

(a) Three segments of GPS.

• Space segment: the constellation of GPS satellites (nominally 24+ in ~20,200 km medium-earth orbits) that continuously broadcast ranging signals and navigation messages.
• Control segment: the ground network — a master control station plus monitor and uploading stations — that tracks the satellites, computes their precise orbits (ephemeris) and clock corrections, and uploads this data to the satellites.
• User segment: the GPS receivers (and antennas) operated by users that receive the satellite signals and compute position, velocity and time.

(b) Principle and minimum satellites. GPS works on the principle of trilateration using ranges (distances) to satellites. The receiver measures the travel time of the signal from each satellite, multiplies by the speed of light to obtain a (pseudo) range, and the intersection of spheres of these radii centred on the satellites fixes the position. Because the receiver clock is not perfectly synchronised with satellite (atomic) clocks, there are four unknowns: three position coordinates ($X, Y, Z$) and the receiver clock bias. Therefore a minimum of four satellites is required for a 3D fix (three for position, one extra to solve the clock error).

(c) Absolute vs relative positioning.

• Absolute (point) positioning: a single receiver computes its own position directly from pseudoranges. It is quick and simple but less accurate (errors from atmosphere, clocks, orbits; typically several metres).
• Relative (differential) positioning: two or more receivers observe the same satellites simultaneously, one at a known reference (base) station. Common errors largely cancel by differencing, giving much higher accuracy (centimetre to sub-metre), as used in DGPS and RTK for engineering surveys.

(a) Arc lengths.

$$L_1 = \frac{\pi R_1\Delta_1}{180} = \frac{\pi\times 200\times 30}{180} = \mathbf{104.72\ m}.$$ $$L_2 = \frac{\pi R_2\Delta_2}{180} = \frac{\pi\times 300\times 25}{180} = \mathbf{130.90\ m}.$$

Total curve length $= 104.72 + 130.90 = 235.62$ m.

(b) Total deflection.

$$\Delta = \Delta_1 + \Delta_2 = 30° + 25° = \mathbf{55°}.$$

(c) Tangent lengths. The tangent distances from the common point to each curve:

$$t_1 = R_1\tan\frac{\Delta_1}{2} = 200\tan 15° = 200\times 0.26795 = 53.59\ \text{m},$$ $$t_2 = R_2\tan\frac{\Delta_2}{2} = 300\tan 12.5° = 300\times 0.22169 = 66.51\ \text{m}.$$

Length of the common tangent (junction to junction) $= t_1 + t_2 = 53.59 + 66.51 = 120.10$ m.

Using the sine rule in the triangle formed by the two straights and the common tangent (angles $\Delta_1$, $\Delta_2$, with the apex angle $180°-\Delta$):

$$T_1 = t_1 + (t_1+t_2)\frac{\sin\Delta_2}{\sin\Delta} = 53.59 + 120.10\times\frac{\sin 25°}{\sin 55°} = 53.59 + 120.10\times\frac{0.42262}{0.81915} = 53.59 + 61.96 = \mathbf{115.55\ m}.$$ $$T_2 = t_2 + (t_1+t_2)\frac{\sin\Delta_1}{\sin\Delta} = 66.51 + 120.10\times\frac{\sin 30°}{\sin 55°} = 66.51 + 120.10\times\frac{0.50000}{0.81915} = 66.51 + 73.31 = \mathbf{139.82\ m}.$$

(a) Angular checks.

• (i) Closed-loop traverse: the sum of the interior angles must equal $(2n-4)\times 90°$ (equivalently the sum of exterior angles $=(2n+4)\times 90°$), where $n$ is the number of sides/stations.
• (ii) Closed-link traverse: the traverse starts and ends on lines of known bearing; the bearing of the closing known line computed by carrying the observed angles through the traverse must agree with its known (given) bearing — the difference is the angular misclosure.

(b) Misclosure and correction. For $n = 6$:

$$\text{Theoretical sum} = (2\times 6 - 4)\times 90° = 8\times 90° = 720°00'00''.$$

Measured sum $= 719°58'40''$.

$$\text{Misclosure} = 719°58'40'' - 720°00'00'' = -1'20'' = -80''.$$

The angles fall short by $80''$, so each angle must be increased. Equal correction per angle:

$$\frac{80''}{6} = 13.33'' \approx \mathbf{+13''\ to\ +14''\ per\ angle}.$$

Apply $+13''$ to four angles and $+14''$ to two angles (so the total correction $= 4\times13 + 2\times14 = 80''$), restoring the sum to $720°00'00''$.

Tangential method (both angles of elevation). Let the horizontal distance be $D$, the two vane angles be $\theta_1$ (lower) and $\theta_2$ (upper), and the vertical interval between vanes be $s = 3.000$ m.

The two vanes give:

$$V_1 = D\tan\theta_1, \qquad V_2 = D\tan\theta_2, \qquad s = V_2 - V_1 = D(\tan\theta_2 - \tan\theta_1).$$

Therefore:

$$D = \frac{s}{\tan\theta_2 - \tan\theta_1}.$$

Computation. $\theta_1 = 3°00'$, $\theta_2 = 5°30'$. $\tan 3° = 0.052408$, $\tan 5°30' = \tan 5.5° = 0.096289$.

$$D = \frac{3.000}{0.096289 - 0.052408} = \frac{3.000}{0.043881} = \mathbf{68.37\ m}.$$

(For reference, the height of the lower vane above the line of collimation $V_1 = 68.37\times 0.052408 = 3.583$ m, which could be used with the staff readings to obtain the reduced level.)