BE Civil Engineering (IOE, TU) Surveying II (IOE, CE 503b) Question Paper 2077 Nepal

Principle of Bowditch's rule. The compass (Bowditch) rule assumes that errors in linear and angular measurements are of the same order, so the closing error is distributed to each line in proportion to its length. Correction to latitude (or departure) of a line $=$ total error in latitude (or departure) $\times \dfrac{\text{length of line}}{\text{perimeter}}$.

Step 1 - Compute latitudes and departures. $L = \ell\cos\theta,\quad D = \ell\sin\theta$ (WCB convention: N is +lat, E is +dep).

• AB: $\theta=45$, $L=200\cos45=+141.421$, $D=200\sin45=+141.421$
• BC: $\theta=135$, $L=150\cos135=-106.066$, $D=150\sin135=+106.066$
• CD: $\theta=225$, $L=200\cos225=-141.421$, $D=200\sin225=-141.421$
• DA: $\theta=315$, $L=150\cos315=+106.066$, $D=150\sin315=-106.066$

Step 2 - Closing error.

$$\Sigma L = 141.421-106.066-141.421+106.066 = 0.000\text{ m}$$ $$\Sigma D = 141.421+106.066-141.421-106.066 = 0.000\text{ m}$$

The sums of latitudes and departures are each zero, so this traverse is already geometrically balanced; the closing error $e=\sqrt{(\Sigma L)^2+(\Sigma D)^2}=0$. No Bowditch correction is required (each correction $= 0 \times \ell/P = 0$).

Step 3 - Independent coordinates (Easting $E$ from departures, Northing $N$ from latitudes), starting at $A(1000,1000)$:

The traverse returns exactly to $A(1000.000, 1000.000)$.

Adjusted coordinates:

• A (1000.000, 1000.000) m
• B (1141.421, 1141.421) m
• C (1035.355, 1247.487) m
• D (893.934, 1106.066) m

Derivation (inclined sight, vertical staff). Let the line of sight make a vertical angle $\alpha$ with the horizontal, staff intercept $s$ = (top - bottom) hair reading, multiplying constant $k$, additive constant $C$. The line-of-sight (slope) distance to the staff equivalent is reduced for the obliquity of the staff. Standard reductions:

$$D = k\,s\cos^2\alpha + C\cos\alpha \qquad\text{(horizontal distance)}$$ $$V = k\,s\cdot\tfrac{1}{2}\sin 2\alpha + C\sin\alpha \qquad\text{(vertical component)}$$

The vertical intercept correction arises because the staff is vertical, not normal to the line of sight; resolving the slope distance $L=ks+C$ along and perpendicular gives the $\cos^2\alpha$ and $\tfrac12\sin2\alpha$ terms.

Given: $k=100$, $C=0.30$ m, $\alpha=632'=6.5333$, top $=2.600$, mid $=1.875$, bottom $=1.150$.

Step 1 - Staff intercept.

$$s = 2.600 - 1.150 = 1.450\text{ m}$$

Step 2 - Horizontal distance. $\cos\alpha = \cos 6.5333 = 0.993505$, $\cos^2\alpha = 0.987052$.

$$D = 100(1.450)(0.987052) + 0.30(0.993505)$$ $$D = 143.123 + 0.298 = 143.421\text{ m}$$

Horizontal distance $PQ \approx 143.42$ m.

Step 3 - Vertical component. $\sin 2\alpha = \sin 13.0667 = 0.226056$.

$$V = 100(1.450)(0.5)(0.226056) + 0.30\sin 6.5333$$ $$V = 16.389 + 0.30(0.113817) = 16.389 + 0.034 = 16.423\text{ m}$$

Step 4 - RL of Q. With instrument axis RL $=1248.55$ m, mid-hair reading $=1.875$ m, and a rising (+) sight:

$$RL_Q = RL_{axis} + V - (\text{mid reading}) = 1248.55 + 16.423 - 1.875$$ $$RL_Q = 1263.098\text{ m}$$

RL of $Q \approx 1263.10$ m.

Given: Deflection angle $\Delta = 40$, $R=300$ m, chainage of intersection point (PI) $=1245.00$ m, peg (chord) interval $=20$ m.

Step 1 - Tangent length.

$$T = R\tan\frac{\Delta}{2} = 300\tan 20 = 300(0.363970) = 109.191\text{ m}$$

Step 2 - Length of curve.

$$L = \frac{\pi R\Delta}{180} = \frac{\pi(300)(40)}{180} = 209.440\text{ m}$$

Step 3 - Chainages.

$$\text{Chainage of PC} = 1245.00 - T = 1245.00 - 109.191 = 1135.809\text{ m}$$ $$\text{Chainage of PT} = \text{PC} + L = 1135.809 + 209.440 = 1345.249\text{ m}$$

Step 4 - First sub-chord. The first peg after PC is at the next round 20 m multiple. PC $=1135.809$; next multiple of 20 is $1140.00$.

$$\text{First sub-chord} = 1140.00 - 1135.809 = 4.191\text{ m}$$

Step 5 - Deflection angle for first sub-chord. Using $\delta = \dfrac{1718.873'\, C}{R}$ (minutes), or $\delta=\dfrac{C}{2R}$ rad:

$$\delta_1 = \frac{1718.873 \times 4.191}{300} = 24.01' = 0\,24'\,01''$$

(Check via radians: $\delta_1=\dfrac{4.191}{2\times300}=0.0069850$ rad $=0.40021=024'$.)

Results:

• Tangent length $T = 109.191$ m
• Length of curve $L = 209.440$ m
• Chainage of PC $= 1135.809$ m
• Chainage of PT $= 1345.249$ m
• First sub-chord $= 4.191$ m, deflection angle $\delta_1 = 0 24' 01''$

Given: $v = 80$ km/h $= 80/3.6 = 22.222$ m/s, $R=400$ m, road width $B=7.5$ m, $\alpha$ (rate of change of radial acceleration) $=0.3$ m/s$^3$, $g=9.81$ m/s$^2$.

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

$$L = \frac{v^3}{\alpha R} = \frac{(22.222)^3}{0.3 \times 400}$$ $$v^3 = 22.222^3 = 10973.9\text{ m}^3/\text{s}^3$$ $$L = \frac{10973.9}{120} = 91.449\text{ m}$$

Length of transition curve $L \approx 91.45$ m.

(b) Superelevation.

$$e = \frac{v^2}{gR} \quad(\text{as a ratio}),\qquad \text{raising } = \frac{Bv^2}{gR}$$ $$\frac{v^2}{gR} = \frac{(22.222)^2}{9.81\times400} = \frac{493.83}{3924} = 0.12585$$

Superelevation amount over road width:

$$e_{height} = B\times0.12585 = 7.5\times0.12585 = 0.944\text{ m}$$

Superelevation rate $\approx 0.126$ (12.6%); raise of outer edge $\approx 0.944$ m.

(c) Shift of the circular curve.

$$S = \frac{L^2}{24R} = \frac{(91.449)^2}{24\times400} = \frac{8362.9}{9600} = 0.871\text{ m}$$

Shift $S \approx 0.871$ m.

Principle of triangulation. Triangulation is a method of control surveying in which the area is covered by a framework of connected triangles. Only one line (the base line) is measured very precisely; all other distances are computed trigonometrically from the measured angles using the sine rule:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$$

The horizontal angles at every station are measured with high-precision theodolites, and the lengths and coordinates of all stations are derived. It provides rigorous horizontal control over large areas with minimal linear measurement.

Well-conditioned triangle. A triangle is well-conditioned when an error in the measured angles produces the least error in the computed sides. From the sine rule, the fractional error in a computed side is minimised when the triangle is near-equilateral. Strictly, no angle should be less than $30$ nor greater than $120$; the ideal is the equilateral ($60$ each) or an isosceles triangle with base angles of about $56^{\circ}14'$.

Why near-equilateral is preferred. The sensitivity of a side to angular error is proportional to $\cot$ of the relevant angle. Very small or very large angles have large $\cot$ values (and the sine becomes ill-conditioned near $0$/$180$), magnifying the propagated error. Near-equilateral triangles keep all angles in the well-conditioned range, giving balanced, minimum error propagation.

Signals and towers.

• Signals (mark the station, give a target to sight):
  • Opaque (non-luminous) signals - pole signal, target signal, pole-and-brush, beacon, stone cairn (used for short to medium distances).
  • Luminous signals - sun signals (heliotrope, reflecting sunlight) and night signals (lamps), used for long sights/poor visibility.
• Towers: raised structures carrying the instrument and/or the observer above obstructions; types include timber (Bilby) towers and steel/masonry towers, often a double (inner instrument + outer observer) tower so the observer's movement does not disturb the instrument.

Given: $g_1=+2.5\%=+0.025$, $g_2=-1.5\%=-0.015$, $L=160$ m, apex (PI) at chainage $2400.00$ m, RL$_{PI}=325.000$ m.

Step 1 - Locate the start of the curve (BVC). Curve is symmetrical, so BVC is $L/2=80$ m before PI.

$$\text{Chainage of BVC} = 2400.00 - 80 = 2320.00\text{ m}$$

RL of BVC (going back up the $+2.5\%$ grade from PI):

$$RL_{BVC} = 325.000 - 0.025\times80 = 325.000 - 2.000 = 323.000\text{ m}$$

Step 2 - Rate of change of grade.

$$r = \frac{g_2-g_1}{L} = \frac{-0.015-0.025}{160} = \frac{-0.040}{160} = -0.00025\ \text{per m}$$

Step 3 - Distance to highest point from BVC. At the highest point the gradient is zero:

$$x = \frac{-g_1}{r} = \frac{-0.025}{-0.00025} = 100\text{ m from BVC}$$

Chainage of highest point $= 2320.00 + 100 =$ 2420.00 m.

Step 4 - RL of highest point. Curve elevation: $y = RL_{BVC} + g_1 x + \tfrac12 r x^2$.

$$y = 323.000 + 0.025(100) + \tfrac12(-0.00025)(100)^2$$ $$y = 323.000 + 2.500 - 1.250 = 324.250\text{ m}$$

Results:

• Chainage of highest point $= 2420.00$ m
• RL of highest point $= 324.250$ m

Tangential method (both angles positive). Let $s$ = vane separation, $\alpha_1$ = angle to lower vane, $\alpha_2$ = angle to upper vane, $D$ = horizontal distance.

$$D = \frac{s}{\tan\alpha_2 - \tan\alpha_1}, \qquad V = D\tan\alpha_1\ (\text{to lower vane})$$

Given: lower vane reading $=1.000$ m, $\alpha_1=330'=3.5$; upper vane reading $=3.000$ m, $\alpha_2=545'=5.75$. $s=3.000-1.000=2.000$ m.

Step 1 - Tangents.

$$\tan\alpha_1=\tan3.5=0.0611626,\quad \tan\alpha_2=\tan5.75=0.1006567$$ $$\tan\alpha_2-\tan\alpha_1 = 0.0394941$$

Step 2 - Horizontal distance.

$$D = \frac{2.000}{0.0394941} = 50.640\text{ m}$$

Horizontal distance $PQ \approx 50.64$ m.

Step 3 - Vertical height to lower vane.

$$V_1 = D\tan\alpha_1 = 50.640\times0.0611626 = 3.097\text{ m}$$

Step 4 - RL of Q. RL$_Q$ = RL of axis $+ V_1 -$ lower vane reading:

$$RL_Q = 1500.000 + 3.097 - 1.000 = 1502.097\text{ m}$$

(Check with upper vane: $V_2=D\tan\alpha_2=50.640\times0.1006567=5.097$ m; $RL_Q=1500.000+5.097-3.000=1502.097$ m. Consistent.)

Results:

• Horizontal distance $PQ = 50.64$ m
• RL of $Q = 1502.097$ m

Given: total deflection $\Delta=80$, $R_1=250$ m, $\Delta_1=35$, $R_2=400$ m.

Step 1 - Second deflection angle.

$$\Delta_2 = \Delta - \Delta_1 = 80 - 35 = 45$$

Step 2 - Tangent lengths of individual arcs.

$$t_1 = R_1\tan\frac{\Delta_1}{2} = 250\tan17.5 = 250(0.315299) = 78.825\text{ m}$$ $$t_2 = R_2\tan\frac{\Delta_2}{2} = 400\tan22.5 = 400(0.414214) = 165.685\text{ m}$$

The common tangent length $= t_1 + t_2 = 78.825 + 165.685 = 244.510$ m.

Step 3 - Total tangent lengths (PI to PC, PI to PT). Using the sine rule on the triangle formed by the two straights and the common tangent (angles opposite: at the apex the interior angle is $180-\Delta=100$):

$$T_1 = t_1 + \frac{(t_1+t_2)\sin\Delta_2}{\sin\Delta}$$ $$T_1 = 78.825 + \frac{244.510\sin45}{\sin80} = 78.825 + \frac{244.510(0.707107)}{0.984808}$$ $$T_1 = 78.825 + \frac{172.894}{0.984808} = 78.825 + 175.561 = 254.386\text{ m}$$ $$T_2 = t_2 + \frac{(t_1+t_2)\sin\Delta_1}{\sin\Delta}$$ $$T_2 = 165.685 + \frac{244.510\sin35}{\sin80} = 165.685 + \frac{244.510(0.573576)}{0.984808}$$ $$T_2 = 165.685 + \frac{140.244}{0.984808} = 165.685 + 142.407 = 308.092\text{ m}$$

Results:

• $\Delta_2 = 45$
• $T_1 = 254.386$ m (PI to PC)
• $T_2 = 308.092$ m (PI to PT)

Total station. A total station is an electronic/optical surveying instrument that integrates an electronic theodolite (for measuring horizontal and vertical angles), an electronic distance measurement (EDM) unit (for measuring slope distances), and an on-board microprocessor with data storage. It measures angles and distances simultaneously and computes coordinates, reduced levels, and horizontal/vertical distances on the fly.

Main components.

• Electronic theodolite with digital circles for horizontal and vertical angle readout.
• EDM unit (infrared or laser) with a coaxial telescope.
• Microprocessor / control panel for computation.
• Internal memory or data collector for storing observations.
• Reflecting prism (target) and tribrach/tripod mounting; many modern units are also reflectorless.

Distance measurement (EDM principle). The instrument emits a modulated infrared/laser beam toward a prism (or directly to a surface). The returning signal's phase difference (phase-comparison method) or the time of flight of a pulse is measured; knowing the speed of light and the modulation wavelength, the microprocessor computes the slope distance, then reduces it to horizontal and vertical components using the measured vertical angle.

Four advantages over theodolite-and-tape.

1. Speed and efficiency - angles and distances measured in one observation; large numbers of points fixed quickly.
2. Higher accuracy and no taping errors - electronic distance measurement avoids sag, temperature, and tension errors of a tape.
3. Automatic computation and recording - coordinates, RLs, areas, and setting-out data computed instantly; digital data eliminates booking and transcription errors.
4. Direct data transfer - observations download to CAD/GIS software, reducing office work and supporting setting-out, remote elevation, and missing-line measurement functions.

Three segments of GPS.

1. Space segment - the constellation of GPS satellites (nominally 24+) orbiting at about $20{,}200$ km altitude in 6 orbital planes, each continuously transmitting coded radio signals (carrier $L1$/$L2$, with C/A and P codes) and navigation data.
2. Control segment - the ground network of a master control station, monitor stations, and ground antennas that track the satellites, compute and update their orbital (ephemeris) and clock parameters, and upload corrections.
3. User segment - the GPS receivers (and antennas) that capture the satellite signals and compute position, velocity, and time.

Principle of position fixing (trilateration / ranging). Each satellite transmits a signal stamped with the time of transmission. The receiver measures the travel time $\Delta t$ of the signal and computes the pseudo-range $\rho = c\,\Delta t$ (with $c$ the speed of light). The true range from each satellite defines a sphere centred on that satellite; the receiver's position lies at the intersection of these spheres. Knowing each satellite's coordinates (from the ephemeris), the receiver solves for its own 3-D position.

Why four satellites. Position fixing involves four unknowns: the three position coordinates $(X, Y, Z)$ plus the receiver clock error $\delta t$ (the inexpensive receiver clock is not synchronised with the precise satellite clocks). Each satellite gives one pseudo-range equation:

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

Three satellites would suffice if the clock were perfect, but the clock bias adds a fourth unknown, so a minimum of four satellites (four range equations) is needed to solve for all four unknowns and obtain a reliable 3-D fix. Additional satellites improve geometry (lower DOP) and allow redundancy/adjustment.

Errors and accuracy. GPS positioning is degraded by satellite clock/ephemeris errors, ionospheric and tropospheric signal delays, multipath, and receiver noise. These are reduced by differential GPS (DGPS) and relative/static and RTK techniques, in which a known base station broadcasts corrections to the rover, achieving centimetre-level accuracy suitable for control and setting-out work.

Polar setting-out method. Compute the join (bearing and distance) from the instrument station $A$ to the design point $P$ from their coordinates.

Step 1 - Coordinate differences.

$$\Delta E = E_P - E_A = 1562.500 - 1500.000 = +62.500\text{ m}$$ $$\Delta N = N_P - N_A = 2042.000 - 2000.000 = +42.000\text{ m}$$

Both positive $\Rightarrow$ $P$ lies in the NE quadrant of $A$.

Step 2 - Horizontal distance.

$$D = \sqrt{\Delta E^2 + \Delta N^2} = \sqrt{62.500^2 + 42.000^2}$$ $$= \sqrt{3906.25 + 1764.00} = \sqrt{5670.25} = 75.301\text{ m}$$

Step 3 - Whole circle bearing $A\to P$.

$$\theta = \tan^{-1}\!\frac{\Delta E}{\Delta N} = \tan^{-1}\!\frac{62.500}{42.000} = \tan^{-1}(1.488095)$$ $$\theta = 56.0779 = 56 04' 41''$$

Since the RO is due north (bearing $0000'00''$), this WCB is also the horizontal angle to be turned clockwise from RO.

Setting out procedure. Set the theodolite/total station at $A$, sight $RO$ and set the horizontal circle to $0000'00''$; turn clockwise through $56 04' 41''$; along this line measure a horizontal distance of $75.301$ m to fix $P$.

Polar vs intersection method.

• Polar (radiation) method: a single instrument station fixes the point by one angle and one distance (bearing + distance). Fast and ideal for a total station/EDM; needs accurate distance measurement and clear line of sight from one station.
• Intersection method: the point is fixed by angles from two (or more) known stations, with no distance measurement. Useful where distances cannot be measured directly (across rivers/obstacles) but needs two set-ups and a good intersection geometry (well-conditioned, near $90$ intersection angle).

Practical check after fixing $P$. Re-observe $P$ from a second known station (independent check by intersection or by measuring the distance $P$ to an adjacent already-set corner) and confirm the measured value agrees with the design dimension within tolerance; alternatively check diagonals of the building rectangle.

Results:

• Bearing $A\to P = 56 04' 41''$
• Horizontal distance $A\to P = 75.301$ m