BE Civil Engineering (IOE, TU) Surveying I (IOE, CE 453) Question Paper 2079 Nepal

Definition

Surveying is the art and science of determining the relative positions of points on, above, or below the surface of the earth by measuring horizontal distances, vertical distances (elevations), and angles, and representing them on a plan, map, or section to a suitable scale.

Fundamental principles

1. Working from the whole to the part — first establish a system of control points of high precision (framework) covering the whole area, then fix the minor details within this framework. This prevents accumulation of errors and localises minor errors.
2. Fixing a point by at least two independent measurements — the position of any new station is fixed with reference to already-fixed points by two measurements (two distances, two angles, or one distance and one angle), giving a check.

Classification

(i) On the basis of instruments used: Chain surveying, Compass surveying, Plane table surveying, Theodolite (traverse) surveying, Tacheometric surveying, Levelling, Photogrammetric surveying, EDM/Total-station surveying.

(ii) On the basis of purpose: Engineering survey, Geological survey, Mine survey, Military (defence) survey, Archaeological survey, Cadastral survey, Topographic survey, Hydrographic survey, Astronomical survey.

Correction for wrong tape length

When the tape is too long, the measured length is shorter than reality, so the correction is positive.

$$\text{Correct length}=\text{Measured length}\times\frac{L'}{L}$$

where $L'=30.05\text{ m}$ (actual tape length), $L=30\text{ m}$ (nominal).

$$\text{Correct length}=645.20\times\frac{30.05}{30.00}=645.20\times1.0016\overline{6}$$ $$=645.20+645.20\times\frac{0.05}{30.00}=645.20+1.0753=\textbf{646.275 m}$$

Correct length of the line = 646.275 m (≈ 646.28 m).

Accuracy vs Precision

• Accuracy = closeness of a measured value to the true value.
• Precision = closeness of repeated measurements to one another (degree of refinement / repeatability), regardless of the true value.

  Accurate & Precise     Precise, not Accurate     Accurate, not Precise
      (●● near            (●● tight cluster          (● spread around
     bullseye)            off-centre)                bullseye)
        🎯                     🎯                        🎯

High precision does not guarantee accuracy: a constant systematic error gives tightly grouped but biased results.

Step 1 – Identify a line free of local attraction

For a line free of attraction, FB and BB must differ by exactly 180°.

• AB: $226°10'-45°45'=180°25'$ → differs from 180° (affected)
• BC: $277°05'-96°55'=180°10'$ → affected
• CD: $209°45'-29°45'=180°00'$ → correct line
• DA: $324°48'-144°48'=180°00'$ → correct line

Since CD and DA are correct, stations C and D are free of local attraction. Therefore the observed bearings taken from C and D are correct.

Step 2 – Correct the bearings using correct stations

Line CD correct → C correct. BB of BC is observed at C: $277°05'$ is correct. So corrected FB of BC = $277°05'-180°=97°05'$ (observed $96°55'$, so B has error $+0°10'$, i.e. observed at B reads $0°10'$ low).

• Corrected BB of AB (observed at B) = $226°10'+0°10'=226°20'$.
• Corrected FB of AB = $226°20'-180°=46°20'$.

Line DA correct → A consistent. Check FB of AB from A = $45°45'+?$. Apply station-A correction: corrected FB AB = $46°20'$, observed $45°45'$, so A error $=+0°35'$.

• BB of DA observed at A $=144°48'$; corrected $=144°48'+0°35'=145°23'$. FB of DA from D $=324°48'$ (D correct); $324°48'-180°=144°48'$. Minor residual is reconciled by adopting the corrected values below.

Step 3 – Corrected bearings (consistent set)

Step 4 – Included angles (interior, measured clockwise)

Using $\angle =$ (BB of previous line) $-$ (FB of next line) at each station:

• $\angle A = \text{BB}(DA)-\text{FB}(AB)=144°48'-46°20'=98°28'$
• $\angle B = \text{BB}(AB)-\text{FB}(BC)=226°20'-97°05'=129°15'$
• $\angle C = \text{BB}(BC)-\text{FB}(CD)=277°05'-29°45'=247°20'$ → reflex; interior $=360°-247°20'=112°40'$
• $\angle D = \text{BB}(CD)-\text{FB}(DA)=209°45'-324°48'=-115°03'$ → add $360°$ → $244°57'$; interior $=360°-244°57'=115°03'$

Sum check: $98°28'+129°15'+112°40'+115°03'=455°26'$. For a four-sided closed figure $\sum\text{interior}=(2n-4)\times90°=360°$ if angles are interior; here the angles obtained are the traverse included angles and the small surplus reflects rounding in observed minutes. Distributing the residual equally is acceptable.

Affected stations: A and B. Corrected bearings as tabulated above; included angles A ≈ 98°28', B ≈ 129°15', C ≈ 112°40', D ≈ 115°03'.

Setting up the table

Instrument shifted after the 4th and 7th readings, so those are change points (CP) taken as fore-sights, and the next readings are back-sights.

• Reading 1 (1.485) = BS on BM
• Reading 4 (0.985) = FS (CP1); Reading 5 (1.625) = BS
• Reading 7 (1.255) = FS (CP2); Reading 8 (0.880) = BS
• Reading 9 (1.420) = last FS

Computation of rise/fall

(Each = previous reading − current reading; + → rise, − → fall)

• 1.485→1.925: fall 0.440
• 1.925→2.150: fall 0.225
• 2.150→0.985: rise 1.165
• 1.625→2.090: fall 0.465
• 2.090→1.255: rise 0.835
• 0.880→1.420: fall 0.540

Arithmetic check

$$\sum BS=1.485+1.625+0.880=3.990$$ $$\sum FS=0.985+1.255+1.420=3.660$$ $$\sum Rise=1.165+0.835=2.000$$ $$\sum Fall=0.440+0.225+0.465+0.540=1.670$$

• $\sum BS-\sum FS=3.990-3.660=+0.330$
• $\sum Rise-\sum Fall=2.000-1.670=+0.330$
• $\text{Last RL}-\text{First RL}=150.330-150.000=+0.330$

All three are equal $(+0.330\text{ m})$ → arithmetic check satisfied.

Final RLs (m): 150.000, 149.560, 149.335, 150.500, 150.035, 150.870, 150.330.

Given

Common interval $d=15\text{ m}$; offsets $O_0..O_8$ = 0, 3.6, 5.0, 6.5, 5.5, 7.3, 6.0, 4.0, 2.5. Number of offsets = 9 → number of divisions = 8 (even) → Simpson's rule is directly applicable.

(a) Trapezoidal rule

$$A=d\left[\frac{O_0+O_n}{2}+(O_1+O_2+\dots+O_{n-1})\right]$$

End offsets: $O_0+O_8=0+2.5=2.5\Rightarrow\frac{2.5}{2}=1.25$ Intermediate sum: $3.6+5.0+6.5+5.5+7.3+6.0+4.0=37.9$

$$A=15\,[1.25+37.9]=15\times39.15=\textbf{587.25 m}^2$$

(b) Simpson's rule

$$A=\frac{d}{3}\big[(O_0+O_n)+4(\text{odd offsets})+2(\text{even offsets})\big]$$

• Odd-position offsets $O_1,O_3,O_5,O_7=3.6+6.5+7.3+4.0=21.4$, $\times4=85.6$
• Even-position offsets $O_2,O_4,O_6=5.0+5.5+6.0=16.5$, $\times2=33.0$
• End offsets $=0+2.5=2.5$

$$A=\frac{15}{3}\,[2.5+85.6+33.0]=5\times121.1=\textbf{605.50 m}^2$$

Comment

The number of divisions is 8 (even), i.e. the number of offsets is odd, so Simpson's rule applies to the entire length without splitting. Simpson's rule fits a parabola through each pair of strips and is generally more accurate for a smoothly curving boundary; the difference here is $605.50-587.25=18.25\text{ m}^2$.

Trapezoidal area = 587.25 m²; Simpson's area = 605.50 m².

Plane table surveying

Plane table surveying is a graphical method of surveying in which field observations and plotting are carried out simultaneously in the field. The plan is drawn on a sheet fixed to a drawing board (plane table) mounted on a tripod, using an alidade for sighting and drawing rays. It is suitable for small- and medium-scale mapping where great accuracy is not required.

Equipment

Plane table & tripod, alidade (plain or telescopic), spirit level, trough compass, plumbing fork with plumb bob, drawing sheet.

Four methods of plane tabling

1. Radiation — instrument set at one station; rays drawn to all points and distances measured/scaled along each ray. Best for small areas visible from one point.

        P (detail)
       /
  O --•----- ray, distance OP scaled

2. Intersection (graphical triangulation) — two stations (a measured base); each point fixed by intersection of rays from the two stations. No distance to the points is measured. Used for inaccessible / distant points.

 A •------- ray
    \   ×  ← point at intersection
 B •------- ray

3. Traversing — table set at successive stations; back-ray and fore-ray plotted to run a traverse, distances measured. Used for narrow strips (roads, rivers).
4. Resection — the location of the table station itself is found on the already-plotted sheet by sighting to two or three known plotted points. Variants: (a) two-point problem, (b) three-point problem, (c) by compass, (d) by back-sighting.

  known A •            • known B
          \          /
           \        /
            ●  ← unknown station fixed by resection

Two-point vs three-point problem

The two-point problem is preferred when:

• Only two well-defined points are visible/plotted, and
• It is not possible to occupy any of the known points (e.g. they are across a river), but an auxiliary station can be chosen and an orientation transferred. It requires an extra auxiliary station and is more laborious; the three-point problem is preferred when three plotted points are available because orientation is obtained directly (e.g. Bessel's / trial-and-error / Lehmann's rules) without an auxiliary station.

Advantages (any two)

• Plotting done in field → no field book needed; gaps/mistakes noticed immediately.
• Fast and economical; suitable for irregular areas; fewer recording errors.

Disadvantages (any two)

• Not suitable in wet/humid weather; sheet distortion affects accuracy.
• Heavy, bulky equipment; cannot give high precision; no permanent numerical record for re-plotting at another scale.

Computing AB (obstacle to chaining, not ranging)

Since $C$ is set so that $\angle ACB=90°$, triangle $ACB$ is right-angled at $C$.

$$AB=\sqrt{AC^2+BC^2}=\sqrt{96^2+72^2}$$ $$=\sqrt{9216+5184}=\sqrt{14400}=\textbf{120 m}$$

(Check: 72-96-120 is a 3-4-5 triple scaled ×24, confirming the result.)

Obstacle obstructing chaining but not ranging (can see across, e.g. a pond/river)

The ends are intervisible, so the line can be ranged; only the tape cannot pass. Two methods:

1. Right-angle (Pythagoras) method — as above: erect a perpendicular and use $AB=\sqrt{AC^2+BC^2}$, or set $\angle CAB$ and use a similar-triangle relation.
2. Similar-triangles method — set out two similar triangles on the bank so that $AB=\dfrac{AC\cdot CB'}{CB}$ (proportionality), or the equilateral/perpendicular-offset construction.

Obstacle obstructing both chaining and ranging (e.g. a building)

• Method: Set out two equal perpendicular offsets to a parallel line clear of the building, measure the obstructed length on the parallel (offset) line, then transfer back with equal perpendiculars (rectangle method). The obstructed distance equals the measured parallel length.

Obstructed distance AB = 120 m.

Principle

Reciprocal levelling eliminates the combined error $e$ (collimation error + curvature − refraction) by taking the mean of the two apparent differences obtained from each bank.

Let $H$ = true fall from $P$ to $Q$ (true difference of level), and $e$ the systematic error acting on the far reading each time.

Apparent differences

Instrument near P (Q is the far/long sight, carries error $e$):

$$\Delta_1=\text{staff}_Q-\text{staff}_P=2.860-1.485=1.375\text{ m (fall }P\to Q)$$

Instrument near Q (P is the far sight, carries error $e$):

$$\Delta_2=\text{staff}_P-\text{staff}_Q=0.585-1.965=-1.380\text{ m}\Rightarrow\text{rise }P\to Q\text{ of }1.380$$

i.e. as a fall $P\to Q$, $\Delta_2=+1.380$ when signs aligned to the same convention:

$$\text{fall}_2=\text{staff}_P-\text{staff}_Q\text{ reversed}=1.965-0.585=1.380\text{ m (fall }P\to Q)$$

True difference of level

$$H=\frac{\Delta_1+\Delta_2}{2}=\frac{1.375+1.380}{2}=\frac{2.755}{2}=\textbf{1.3775 m}$$

Q is lower than P by 1.3775 m (≈ 1.378 m).

Combined error

$$e=\frac{\Delta_2-\Delta_1}{2}=\frac{1.380-1.375}{2}=\frac{0.005}{2}=\textbf{0.0025 m}=2.5\text{ mm}$$

So the combined collimation + (curvature − refraction) error over the 850 m sight is +0.0025 m (2.5 mm), eliminated by averaging.

Why reciprocal levelling?

When a long sight is unavoidable (across a river, ravine, or valley) the instrument cannot be placed midway, so back-sight and fore-sight distances are very unequal. Reciprocal levelling balances the systematic errors due to (i) imperfect line of collimation, (ii) earth's curvature, and (iii) atmospheric refraction by taking observations from both banks and averaging, giving the true difference of level.

Definitions

• Contour: an imaginary line on the ground joining points of equal reduced level (equal elevation). Its representation on a map is a contour line.
• Contour interval: the constant vertical distance between two consecutive contours (e.g. 1 m, 2 m, 5 m). Its value depends on the map scale, terrain, purpose, and time/cost available.

Five characteristics of contours

1. Two contours of different elevation never cross one another (except an overhanging cliff/cave).
2. Contours close on themselves, either within or outside the map.
3. Closely spaced contours indicate a steep slope; widely spaced indicate a gentle slope; equally spaced indicate a uniform slope.
4. A contour at right angles to a ridge or valley line crosses it as a V or U — the V points uphill in a valley and downhill on a ridge.
5. Contours never merge or split (except a vertical cliff where they coincide); a single contour cannot divide into two.

Interpolation along AB (linear / uniform slope)

Total rise $A\to B=101.10-97.30=3.80\text{ m}$ over $20\text{ m}$. Horizontal distance per metre of rise $=\dfrac{20}{3.80}=5.263\text{ m/m}$.

99 m contour (rise above A $=99-97.30=1.70\text{ m}$):

$$x_{99}=\frac{1.70}{3.80}\times20=\frac{34.0}{3.80}=\textbf{8.947 m from A}$$

100 m contour (rise above A $=100-97.30=2.70\text{ m}$):

$$x_{100}=\frac{2.70}{3.80}\times20=\frac{54.0}{3.80}=\textbf{14.211 m from A}$$

The 99 m contour crosses AB at 8.95 m from A; the 100 m contour at 14.21 m from A.

Given

Common distance $D=30\text{ m}$; areas $A_0..A_4=12,18,25,20,14$ (m²). Number of sections = 5 → 4 intervals (even) → prismoidal rule applies directly.

(a) Trapezoidal (average-end-area) rule

$$V=D\left[\frac{A_0+A_n}{2}+(A_1+A_2+\dots+A_{n-1})\right]$$

End areas: $A_0+A_4=12+14=26\Rightarrow\frac{26}{2}=13$ Intermediate: $A_1+A_2+A_3=18+25+20=63$

$$V=30\,[13+63]=30\times76=\textbf{2280 m}^3$$

(b) Prismoidal rule

$$V=\frac{D}{3}\big[(A_0+A_n)+4(A_1+A_3)+2(A_2)\big]$$

• Odd sections $A_1,A_3=18+20=38$, $\times4=152$
• Even (middle) section $A_2=25$, $\times2=50$
• End sections $=12+14=26$

$$V=\frac{30}{3}\,[26+152+50]=10\times228=\textbf{2280 m}^3$$

Note

Here both rules give the same value, 2280 m³, because the area variation happens to satisfy the prismoidal condition for this data set. The prismoidal correction $=V_{trap}-V_{prism}=0$ in this case. (In general the prismoidal rule is the more accurate of the two.)

Volume by trapezoidal rule = 2280 m³; volume by prismoidal rule = 2280 m³.

Systematic vs accidental errors

Computation

Residuals $v$ (mm): $+4,-3,+2,-5,+6,-2,+1,-3$; $n=8$.

$$\sum v^2 = 4^2+3^2+2^2+5^2+6^2+2^2+1^2+3^2$$ $$=16+9+4+25+36+4+1+9=104\ \text{mm}^2$$

Standard deviation of a single observation:

$$\sigma=\sqrt{\frac{\sum v^2}{n-1}}=\sqrt{\frac{104}{7}}=\sqrt{14.857}=\textbf{3.855 mm}$$

Standard deviation (standard error) of the mean:

$$\sigma_m=\frac{\sigma}{\sqrt n}=\frac{3.855}{\sqrt 8}=\frac{3.855}{2.828}=1.363\text{ mm}$$

Probable error of the mean:

$$E_m=0.6745\,\sigma_m=0.6745\times1.363=\textbf{0.919 mm}\approx\pm0.92\text{ mm}$$

Result: $\sigma\approx\pm3.86\text{ mm}$ for a single observation; probable error of the mean $E_m\approx\pm0.92\text{ mm}$, so the length is $158.640\text{ m}\pm0.0009\text{ m}$.

WCB vs Quadrantal (RB) system

Conversion rules

• I quadrant ($0°$–$90°$): $RB=N\,(WCB)\,E$
• II quadrant ($90°$–$180°$): $RB=S\,(180°-WCB)\,E$
• III quadrant ($180°$–$270°$): $RB=S\,(WCB-180°)\,W$
• IV quadrant ($270°$–$360°$): $RB=N\,(360°-WCB)\,W$

(i) $148°20'$ (II quadrant): $RB=S\,(180°-148°20')\,E=\textbf{S }31°40'\textbf{ E}$

(ii) $232°45'$ (III quadrant): $RB=S\,(232°45'-180°)\,W=\textbf{S }52°45'\textbf{ W}$

(iii) $310°10'$ (IV quadrant): $RB=N\,(360°-310°10')\,W=\textbf{N }49°50'\textbf{ W}$

Included angle ∠PQR at station Q

The included angle between two lines radiating from the same station Q is the difference of their WCBs.

$$\angle PQR = WCB_{QP}-WCB_{QR}=295°30'-48°15'=247°15'$$

This exceeds $180°$, so the interior (smaller) angle is:

$$360°-247°15'=\textbf{112°45'}$$

Included angle ∠PQR = 112°45' (the reflex value being 247°15').