BE Civil Engineering (IOE, TU) Surveying II (IOE, CE 503b) Question Paper 2078 Nepal

(a) Closing error and relative accuracy

For a closed traverse, $\sum L = 0$ and $\sum D = 0$ in theory. The residuals are the closing error components.

$$\sum L = 120.50 - 85.30 - 60.20 + 25.10 = +0.10\ \text{m}$$ $$\sum D = 80.40 + 95.20 - 110.60 - 64.80 = +0.20\ \text{m}$$

Closing error:

$$e = \sqrt{(\textstyle\sum L)^2 + (\textstyle\sum D)^2} = \sqrt{0.10^2 + 0.20^2} = \sqrt{0.05} = \mathbf{0.2236\ m}$$

Perimeter $P = 144.86 + 127.82 + 125.93 + 69.49 = 468.10\ \text{m}$.

Relative accuracy:

$$\frac{e}{P} = \frac{0.2236}{468.10} = \frac{1}{2093} \approx \mathbf{1\ in\ 2090}$$

(b) Bowditch's (compass) rule

Statement: The closing error is distributed to the latitudes and departures of each leg in proportion to the length of that leg. It assumes errors in linear and angular measurement are equally probable, i.e. the error in any leg is proportional to its length.

$$\text{Correction to latitude of a leg} = -\left(\sum L\right)\times\frac{\ell}{P}, \qquad \text{Correction to departure of a leg} = -\left(\sum D\right)\times\frac{\ell}{P}$$

With $\sum L/P = 0.10/468.10 = 0.0002136$ per metre, and $\sum D/P = 0.20/468.10 = 0.0004272$ per metre, the corrections (signs negative) are:

(The 0.001 residual is rounding; corrected sums close to zero.)

Sample working for AB: latitude correction $= -0.10\times144.86/468.10 = -0.031\ \text{m}$; departure correction $= -0.20\times144.86/468.10 = -0.062\ \text{m}$.

(c) Acceptability

The achieved accuracy is 1 in 2090, which is finer (better) than the permissible 1 in 2000 for a third-order traverse. The traverse is acceptable.

(a) Curve elements

Tangent length:

$$T = R\tan\frac{\Delta}{2} = 300\tan 24^\circ = 300\times0.44523 = \mathbf{133.569\ m}$$

Curve (arc) length:

$$L = \frac{\pi R \Delta}{180^\circ} = R\,\Delta_{\text{rad}} = 300\times0.837758 = \mathbf{251.327\ m}$$

Long chord:

$$LC = 2R\sin\frac{\Delta}{2} = 2\times300\times\sin 24^\circ = 600\times0.40674 = \mathbf{244.042\ m}$$

Mid-ordinate:

$$M = R\left(1-\cos\frac{\Delta}{2}\right) = 300(1-\cos 24^\circ) = 300(1-0.91355) = \mathbf{25.936\ m}$$

External distance:

$$E = R\left(\sec\frac{\Delta}{2}-1\right) = 300\left(\frac{1}{\cos24^\circ}-1\right) = 300(1.09464-1) = \mathbf{28.391\ m}$$

(b) Chainages of tangent points

$$\text{Chainage of T1} = \text{Chainage of PI} - T = 1245.50 - 133.569 = \mathbf{1111.931\ m}$$ $$\text{Chainage of T2} = \text{Chainage of T1} + L = 1111.931 + 251.327 = \mathbf{1363.259\ m}$$

(c) Deflection-angle table

The deflection angle for a chord $c$ is $\delta = \dfrac{1718.873\,c}{R}\ \text{(minutes)}$.

First sub-chord brings chainage from 1111.931 to the next 20 m peg (1120.000): $c_1 = 8.069\ \text{m}$. Full chords of 20 m run to chainage 1360.000. Last sub-chord = $1363.259 - 1360.000 = 3.259\ \text{m}$.

Per-chord deflection: full chord $\delta = 1718.873\times20/300 = 114.592' = 1^\circ54.59'$.

Check: the final cumulative deflection equals $\Delta/2 = 24^\circ$, confirming the table closes correctly.

(a) Derivation (inclined sight, vertical staff)

Let the staff intercept be $s$ (difference of top and bottom stadia readings) and the inclination $\theta$. If the staff were normal to the line of sight, the inclined distance along the line of collimation would be $D' = ks + c$. Because the staff is held vertical, the effective intercept normal to the sight line is $s\cos\theta$, so the inclined distance is

$$L = ks\cos\theta + c.$$

Resolving along the horizontal and vertical:

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

where $D$ is the horizontal distance and $V$ the vertical component between the instrument axis and the mid (central) staff reading.

(b) Horizontal distance

Staff intercept: $s = 2.345 - 0.985 = 1.360\ \text{m}$. (Check: mid reading $1.665 = (2.345+0.985)/2$ ✓)

With $k=100$, $c=0$, $\theta = 8^\circ12'$, $\cos\theta = 0.98977$:

$$D = 100\times1.360\times\cos^2 8^\circ12' = 136.0\times0.97964 = \mathbf{133.233\ m}$$

(c) Reduced level of Q

Vertical component:

$$V = 100\times1.360\times\tfrac{1}{2}\sin 16^\circ24' = 68.0\times0.28232 = \mathbf{19.198\ m}$$

With an elevation sight, the RL of Q is obtained from:

$$RL_Q = RL_P + h_i + V - r$$

where $h_i = 1.450\ \text{m}$ (height of instrument) and $r = 1.665\ \text{m}$ (central/mid staff reading).

$$RL_Q = 1024.600 + 1.450 + 19.198 - 1.665 = \mathbf{1043.583\ m}$$

(a) Functions of a transition curve

1. To gradually introduce centrifugal force between the straight and the circular arc, avoiding a sudden jerk.
2. To enable the driver to turn the steering gradually for comfort and safety.
3. To gradually introduce the designed superelevation (cant) and the extra widening.
4. To improve aesthetics and provide a smooth, pleasing alignment.

Common forms: (i) the clothoid / ideal transition spiral (length proportional to curvature) and (ii) the cubic parabola; the cubic spiral and the lemniscate are also used.

(b) Length of transition curve

Design speed $v = 50\ \text{km/h} = 50\times\dfrac{1000}{3600} = 13.889\ \text{m/s}$.

By the rate of change of radial acceleration method:

$$L_s = \frac{v^3}{C\,R} = \frac{(13.889)^3}{0.3\times250} = \frac{2679.0}{75} = \mathbf{35.72\ m}$$

(Adopt $L_s = 36\ \text{m}$ in practice.)

(c) Shift and superelevation

Shift of the curve:

$$S = \frac{L_s^2}{24R} = \frac{35.72^2}{24\times250} = \frac{1276.0}{6000} = \mathbf{0.213\ m}$$

Superelevation (full, ignoring friction):

$$e = \frac{v^2}{gR} = \frac{13.889^2}{9.81\times250} = \frac{192.90}{2452.5} = 0.0787$$

Superelevation as a height across road width $B = 7.5\ \text{m}$:

$$E = e\times B = 0.0787\times7.5 = \mathbf{0.590\ m}$$

Thus the outer edge is raised by about 0.59 m; in practice this would be limited to the design maximum cant (commonly 7 %, giving $E \approx 0.525\ \text{m}$).

(a) Triangulation

Triangulation is a method of horizontal control surveying in which the relative positions of stations are fixed by measuring the angles of a network of connected triangles, with at least one side (the baseline) measured precisely; all other sides are computed trigonometrically using the sine rule.

Systems/figures used: (i) simple triangles / chain of triangles, (ii) braced quadrilaterals, and (iii) polygons with a central station (centred figures). (Any two suffice.)

(b) Computation

Third angle:

$$C = 180^\circ - A - B = 180^\circ - 58^\circ12'30'' - 64^\circ48'15'' = \mathbf{56^\circ59'15''}$$

By the sine rule, with each side opposite its angle:

$$\frac{AB}{\sin C} = \frac{BC}{\sin A} = \frac{AC}{\sin B}$$

$\sin C = \sin 56^\circ59'15'' = 0.83853$, $\sin A = \sin 58^\circ12'30'' = 0.84979$, $\sin B = \sin 64^\circ48'15'' = 0.90466$.

Side BC (opposite angle A):

$$BC = AB\cdot\frac{\sin A}{\sin C} = 2500.00\times\frac{0.84979}{0.83853} = \mathbf{2534.04\ m}$$

Side AC (opposite angle B):

$$AC = AB\cdot\frac{\sin B}{\sin C} = 2500.00\times\frac{0.90466}{0.83853} = \mathbf{2697.68\ m}$$

(c) Triangular (figure) adjustment

Rule: the three angles of a plane triangle must sum to exactly $180^\circ$; the misclosure is distributed equally among the three observed angles (assuming equal weight/reliability).

Misclosure $= 180^\circ00'30'' - 180^\circ = +30''$. Correction to each angle $= -30''/3 = -10''$.

Corrected angles:

$$A = 58^\circ12'30'' - 10'' = \mathbf{58^\circ12'20''}$$ $$B = 64^\circ48'15'' - 10'' = \mathbf{64^\circ48'05''}$$ $$C = 56^\circ59'15'' - 10'' = \mathbf{56^\circ59'05''}$$

(Check: $58^\circ12'20'' + 64^\circ48'05'' + 56^\circ59'05'' = 180^\circ00'00''$ ✓)

Setup

$g_1 = +0.03$, $g_2 = -0.02$, length $L = 160\ \text{m}$. The BVC is at $L/2 = 80\ \text{m}$ before the PVI.

Chainage of BVC $= 1880.000 - 80.000 = 1800.000\ \text{m}$. RL of BVC $= 452.000 - g_1\times80 = 452.000 - 0.03\times80 = 452.000 - 2.400 = 449.600\ \text{m}$.

(a) Location of highest point

For a summit curve the highest point is where the slope is zero. Distance from BVC:

$$x = \frac{g_1\,L}{g_1 - g_2} = \frac{0.03\times160}{0.03-(-0.02)} = \frac{4.80}{0.05} = \mathbf{96.0\ m\ from\ BVC}$$

(Chainage $= 1800.000 + 96.000 = 1896.000\ \text{m}$.)

(b) RL of highest point

The elevation on the parabola at distance $x$ from BVC:

$$y = RL_{BVC} + g_1 x + \frac{(g_2-g_1)}{2L}x^2$$ $$y = 449.600 + 0.03(96) + \frac{(-0.02-0.03)}{2\times160}(96)^2$$ $$y = 449.600 + 2.880 + \left(\frac{-0.05}{320}\right)(9216) = 449.600 + 2.880 - 1.440 = \mathbf{451.040\ m}$$

(a) Sketch

            PI
            /\
           /  \
   T1 ___ /    \ ___ T2
      \   o     o   /
  arc1 \  R1   R2  / arc2
        \ ( PCC ) /
         \__   __/
            \_/
T1 = first tangent point   PCC = point of compound curvature
T2 = second tangent point  O1,O2 = centres of the two arcs

A compound curve has two (or more) arcs of different radii curving in the same direction, meeting at the point of compound curvature (PCC), with a common tangent there.

(b) Arc lengths and total deflection

$$L_1 = \frac{\pi R_1 \Delta_1}{180} = R_1\,\Delta_{1,\text{rad}} = 200\times\frac{30\pi}{180} = 200\times0.523599 = \mathbf{104.72\ m}$$ $$L_2 = \frac{\pi R_2 \Delta_2}{180} = R_2\,\Delta_{2,\text{rad}} = 300\times\frac{40\pi}{180} = 300\times0.698132 = \mathbf{209.44\ m}$$

Total deflection angle of the compound curve:

$$\Delta = \Delta_1 + \Delta_2 = 30^\circ + 40^\circ = \mathbf{70^\circ}$$

Total curve length $L = L_1 + L_2 = 104.72 + 209.44 = \mathbf{314.16\ m}$.

(a) Advantages of a total station

1. Simultaneous electronic measurement of angles and distances, with instant on-board reduction to horizontal distance, level difference and coordinates.
2. High accuracy and fast EDM ranging, eliminating chaining/taping errors.
3. Automatic data recording/storage and direct download to a computer (no manual booking errors).
4. Built-in software for area, remote elevation, missing line, setting-out and traverse computations.

(b) Reduction of slope distance

Let $S = 458.620\ \text{m}$ and vertical angle $\alpha = +6^\circ30'00''$.

Horizontal distance:

$$H = S\cos\alpha = 458.620\times\cos 6^\circ30' = 458.620\times0.99357 = \mathbf{455.672\ m}$$

Difference in elevation (instrument axis to reflector centre; equal heights so it equals the ground level difference):

$$V = S\sin\alpha = 458.620\times\sin 6^\circ30' = 458.620\times0.11320 = \mathbf{51.917\ m}$$

(a) Principle and four-satellite requirement

Principle (trilateration by ranging): A GPS receiver measures the travel time of coded signals from satellites whose positions are known from the broadcast ephemeris. Multiplying travel time by the speed of light gives the range (pseudo-range) to each satellite. The receiver position lies on a sphere about each satellite; the intersection of these spheres fixes the position.

Why four satellites: There are four unknowns — three position coordinates $(X, Y, Z)$ and the receiver-clock bias $\Delta t$ (the inexpensive receiver clock is not synchronised with the precise satellite atomic clocks). Three satellites give three range equations for the three coordinates; a fourth is needed to solve for the clock error as well, so a minimum of four satellites is required for a 3-D fix.

(b) Error sources and differential technique

Four error sources:

1. Ionospheric and tropospheric (atmospheric) refraction delays.
2. Satellite and receiver clock errors.
3. Ephemeris (satellite orbit) errors.
4. Multipath (signal reflection) and receiver noise.

The technique used to remove or reduce most of the common (correlated) errors is Differential GPS (DGPS) / relative positioning, in which corrections from a base station on a known point are applied to the rover observations.

(a) Setting out vs ordinary surveying

(b) Right-angle check by the 3-4-5 (diagonal) method

Lay out the two sides $24\ \text{m}$ and $18\ \text{m}$ from a corner. The corner is a true right angle if the measured diagonal equals the value from Pythagoras. The 3-4-5 rule is the practical small-scale version (sides 3 and 4 give a hypotenuse of 5).

Numerical check of the diagonal:

$$d = \sqrt{24^2 + 18^2} = \sqrt{576 + 324} = \sqrt{900} = \mathbf{30.000\ m}$$

If the tape between the far corners reads $30.000\ \text{m}$, the corner is square. Repeat for both diagonals (both should read 30.000 m); equal diagonals confirm a true rectangle.

(a) Rule for forward bearing

Fore bearing of the next line = (back bearing of the previous line) + (clockwise included angle at the forward station).

The back bearing of a line = its fore bearing $\pm 180^\circ$.

Final-value rule: if the computed bearing exceeds $360^\circ$, subtract $360^\circ$ so that the result lies between $0^\circ$ and $360^\circ$.

(b) Bearing of QR

Back bearing of PQ (i.e. bearing of QP):

$$\text{B.B. of PQ} = 112^\circ30' + 180^\circ = 292^\circ30'$$

Fore bearing of QR = back bearing of PQ + included angle at Q:

$$\text{F.B. of QR} = 292^\circ30' + 98^\circ15' = 390^\circ45'$$

Since this exceeds $360^\circ$, subtract $360^\circ$:

$$\text{WCB of QR} = 390^\circ45' - 360^\circ = \mathbf{30^\circ45'}$$

Reduced (quadrantal) bearing: $30^\circ45'$ lies in the first quadrant (0°–90°), so:

$$\text{R.B. of QR} = \mathbf{N\ 30^\circ45'\ E}$$