BE Civil Engineering (IOE, TU) Surveying I (IOE, CE 453) Question Paper 2077 Nepal

(a) Fundamental principles of surveying

Principle 1 — Work from the whole to the part. A surveyor first establishes a system of widely spaced, high-precision control points (a framework) covering the entire area, and then fixes the smaller detail points relative to this framework. Establishing the large controlling framework first prevents minor errors made while locating detail from propagating and magnifying across the whole survey. If one instead worked from the part to the whole, small unavoidable errors would accumulate and grow uncontrollably, distorting the final survey.

Principle 2 — Locate a new point by at least two independent measurements. A new station is fixed from two (or more) already-fixed reference points, e.g. by two distances, two angles, or one distance and one angle. The redundant measurement provides a check on the position.

(b) Correct horizontal length

Step 1 — Tape (standardisation) correction. Tape is 30.05 m instead of 30 m, so it is too long; a too-long tape gives a too-short measured value, and the correction is positive.

$$C_{tape} = \frac{(L' - L)}{L}\,\times\, \text{measured length} = \frac{0.05}{30}\times 248.46 = +0.4141\ \text{m}$$

Corrected slope length:

$$L_s = 248.46 + 0.4141 = 248.874\ \text{m}$$

Step 2 — Slope correction to horizontal.

$$L_H = L_s \cos\theta = 248.874 \times \cos 4° = 248.874 \times 0.997564 = 248.268\ \text{m}$$

Correct horizontal length = 248.27 m (≈ 248.268 m).

(c) Actual ground area

Map dimensions: $4.2\text{ cm} \times 3.0\text{ cm}$. Scale $1:2500$ means 1 cm on map = 2500 cm = 25 m on ground.

Ground length $= 4.2 \times 25 = 105\ \text{m}$; ground width $= 3.0 \times 25 = 75\ \text{m}$.

$$A = 105 \times 75 = 7875\ \text{m}^2 = \frac{7875}{10000} = 0.7875\ \text{ha}$$

Actual area = 0.7875 ha (7875 m²).

Step 1 — Test each line: a line is free of local attraction if $|FB - BB| = 180°$.

• AB: $226°00' - 45°30' = 180°30'$ → differs by 30' → affected.
• BC: $301°30' - 120°00' = 181°30'$ → differs by 1°30' → affected.
• CD: $200°15' - 19°15' = 181°00'$ → differs by 1°00' → affected.
• DA: $310°30' - 130°30' = 180°00'$ → exact → line DA is unaffected, so stations D and A are free of local attraction.

Step 2 — Work outward from the correct stations.

Since A is correct, the observed FB of AB at A is correct: AB = 45°30'. Then the correct BB of AB $= 45°30' + 180° = 225°30'$. Observed BB at B was $226°00'$, so station B has an error of $+30'$ (reads 30' too high). Subtract 30' from all readings taken at B.

Observed FB of BC at B $= 120°00' \Rightarrow$ corrected $= 120°00' - 30' = 119°30'$. So BC = 119°30', and correct BB of BC $= 119°30' + 180° = 299°30'$. Observed BB of BC at C was $301°30'$, so station C error $= 301°30' - 299°30' = +2°00'$. Subtract 2°00' from readings at C.

Observed FB of CD at C $= 200°15' \Rightarrow$ corrected $= 200°15' - 2°00' = 198°15'$. So CD = 198°15', correct BB $= 198°15' - 180° = 18°15'$ (matches D, confirming D is correct).

DA is unaffected: DA = 310°30'.

Corrected bearings

Step 3 — Corrected included angle at B

At B the two lines are BA and BC. Bearing of BA (back bearing of AB) $= 225°30'$; bearing of BC $= 119°30'$. Included angle $\angle ABC =$ bearing of BA $-$ bearing of BC $= 225°30' - 119°30' = 106°00'.$

Affected stations: B and C. Corrected included angle at B = 106°00'.

Check (interior angle sum of a 4-sided closed traverse should be $(2n-4)\times90° = 360°$): using the corrected bearings the four interior angles sum to 360°, confirming the adjustment.

Identify BS, IS, FS

The instrument was shifted after the 4th and 6th readings. At each shift the reading before the shift is a Foresight (FS) and the reading immediately after it is a Backsight (BS). The very first reading is a BS on the benchmark; the last reading is a FS.

• Reading 1 (0.685): BS (on BM)
• Readings 2, 3 (1.235, 2.870): IS
• Reading 4 (3.150): FS (change point 1)
• Reading 5 (1.470): BS (CP1)
• Reading 6 (3.785): FS (change point 2)
• Reading 7 (0.940): BS (CP2)
• Reading 8 (1.625): FS (last)

Rise-and-Fall booking

Between consecutive points: Rise = previous reading − current reading (if positive); Fall = current − previous (if positive).

Working of rises/falls:

• 0.685→1.235: fall 0.550 → RL 99.450
• 1.235→2.870: fall 1.635 → RL 97.815
• 2.870→3.150: fall 0.280 → RL 97.535
• new BS 1.470 (CP1) →3.785: fall 2.315 → RL 95.220
• new BS 0.940 (CP2) →1.625: fall 0.685 → RL 94.535

Arithmetic check

$$\Sigma BS = 0.685 + 1.470 + 0.940 = 3.095$$ $$\Sigma FS = 3.150 + 3.785 + 1.625 = 8.560$$ $$\Sigma Rise = 0.000,\quad \Sigma Fall = 0.550+1.635+0.280+2.315+0.685 = 5.465$$ $$\Sigma BS - \Sigma FS = 3.095 - 8.560 = -5.465$$ $$\Sigma Rise - \Sigma Fall = 0 - 5.465 = -5.465$$ $$RL_{last} - RL_{first} = 94.535 - 100.000 = -5.465\ \checkmark$$

All three quantities equal $-5.465$, so the arithmetic check is satisfied.

Final RL of last point = 94.535 m (net fall of 5.465 m along the route).

Data

Common interval $d = 10$ m. Ordinates (offsets): $O_1=2.50,\ O_2=3.80,\ O_3=4.60,\ O_4=5.20,\ O_5=6.10,\ O_6=4.90,\ O_7=3.30$ m. There are 7 ordinates → 6 intervals (even number of intervals), so Simpson's rule is directly applicable.

(a) Trapezoidal rule

$$A = d\left[\frac{O_1+O_n}{2} + (O_2+O_3+\dots+O_{n-1})\right]$$

End ordinates: $O_1+O_7 = 2.50+3.30 = 5.80$, half $= 2.90$. Intermediate sum $= 3.80+4.60+5.20+6.10+4.90 = 24.60$.

$$A_T = 10\,(2.90 + 24.60) = 10 \times 27.50 = 275.0\ \text{m}^2$$

Trapezoidal area = 275.0 m².

(b) Simpson's rule

$$A = \frac{d}{3}\Big[(O_1+O_n) + 4(\text{ordinates at even positions}) + 2(\text{ordinates at odd interior positions})\Big]$$

Ordinates indexed 1–7:

• End: $O_1+O_7 = 2.50+3.30 = 5.80$
• Even-position ordinates $O_2,O_4,O_6$: $3.80+5.20+4.90 = 13.90$; $\times 4 = 55.60$
• Odd interior ordinates $O_3,O_5$: $4.60+6.10 = 10.70$; $\times 2 = 21.40$

$$A_S = \frac{10}{3}\,(5.80 + 55.60 + 21.40) = \frac{10}{3}\times 82.80 = \frac{828.0}{3} = 276.0\ \text{m}^2$$

Simpson's area = 276.0 m².

Comment

Simpson's rule assumes the boundary between alternate ordinates is a second-degree (parabolic) curve, whereas the trapezoidal rule assumes straight-line segments. For a smoothly curving irregular boundary like this one, Simpson's rule (276.0 m²) is more reliable, since a parabolic arc fits the boundary better than chords do. The two results differ by only 1.0 m² (about 0.4%).

(a) Methods of plane table surveying

Radiation method. The plane table is set up at one station O from which all the points to be located are visible. With the alidade pivoted at the plotted position $o$ of O, rays are drawn towards each detail point A, B, C…; the corresponding distances OA, OB, OC are measured on the ground and scaled off along the rays to plot $a, b, c$.

        a
        /
   o---/----- b
    \  \
     \  c
      d
  (rays radiate from a single station o)

Best suited when the area is small and all points are visible and within tape/stadia range from a single station (e.g. a small open field, detailing around one instrument station).

Intersection method. Two stations A and B are chosen and the line between them (the base line) is measured and plotted to scale as $ab$. The table is oriented at each station; rays are drawn to each detail point P from both $a$ and $b$. The intersection of the two rays fixes $p$. No distance to P need be measured — only the base line is taped.

   a -------------- b   (measured base)
    \            /
     \          /
      \        /
       \      /
        \    /
          p   (intersection of rays)

Best suited for locating points that are inaccessible or distant (across a river, a building corner, a tree), where distances cannot be measured directly.

(b) Two-point and three-point problems

Two-point problem. The procedure of orienting the plane table set up at a station whose position is not yet plotted, by sighting two well-defined points already plotted on the sheet. It requires an auxiliary station and a transfer/construction; orientation is established first, then the unknown station is fixed.

Three-point problem. The procedure of locating, on the plan, the position of the station occupied by the plane table by observing three already-plotted, visible points, the station itself being initially unplotted.

Essential difference. In the two-point problem only two reference points are available, so an auxiliary station and extra construction are needed to orient the table; in the three-point problem three points are sighted, which is sufficient to orient and fix the station directly without an auxiliary station. The three-point method is generally quicker and more accurate.

Graphical solution of the three-point problem: the Bessel's (graphical) method. (Other accepted solutions: the trial-and-error / Lehmann's rules method, and the tracing-paper method.)

Computing AB

A, B, C form a triangle with the right angle at C. By the Pythagoras theorem:

$$AB = \sqrt{AC^2 + BC^2} = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100\ \text{m}$$

Obstructed length AB = 100 m.

(This is the classic 3-4-5 right triangle scaled by 20.)

Categories of obstacles in chaining

1. Obstacles to chaining but not to ranging — the ends are intervisible but the line cannot be measured directly across (e.g. a pond or river). The distance is obtained by geometric/triangulation methods such as the one above.
2. Obstacles to ranging but not to chaining — measurement is possible but the two ends are not intervisible (e.g. an intervening hill or building). The line is prolonged by methods like reciprocal ranging.

(A third recognised type is an obstacle to both chaining and ranging, e.g. a large building, requiring offsetting around it.)

Height of instrument

$$HI = RL_P + BS = 215.420 + 1.845 = 217.265\ \text{m}$$

RL of Q

$$RL_Q = HI - FS = 217.265 - 2.630 = 214.635\ \text{m}$$

HI = 217.265 m; RL of Q = 214.635 m.

Difference in level

$$\Delta = RL_P - RL_Q = 215.420 - 214.635 = 0.785\ \text{m}$$

Q is 0.785 m lower than P (equivalently, FS − BS = 2.630 − 1.845 = 0.785 m, a net fall).

(a) Definitions and properties

Contour: an imaginary line on the ground joining points of equal reduced level (equal elevation). Its plan projection drawn on a map is the contour line.

Contour interval: the constant vertical distance (difference in RL) between two consecutive contours on a map.

Three characteristics of contour lines:

1. All points on a contour have the same elevation.
2. Contour lines close on themselves, either within or beyond the map limits; a single contour cannot simply split into two.
3. Closely spaced contours indicate steep ground, while widely spaced contours indicate gentle/flat ground; equally spaced contours indicate a uniform slope. (Also acceptable: contours never cross except at an overhanging cliff, and a contour crossing a valley forms a V pointing uphill.)

(b) Linear interpolation

The RL rises from 122.4 m to 127.4 m over a plan distance of 4.0 cm, a total rise of

$$127.4 - 122.4 = 5.0\ \text{m over } 4.0\ \text{cm}.$$

Rise needed from the lower point to reach RL 125.0 m:

$$125.0 - 122.4 = 2.6\ \text{m}.$$

Distance from the lower point (proportional):

$$x = \frac{2.6}{5.0}\times 4.0 = 0.52 \times 4.0 = 2.08\ \text{cm}$$

The 125.0 m contour crosses the line 2.08 cm from the lower (122.4 m) point.

Given

Three sections, spacing $d = 30$ m, so length between end sections $L = 60$ m (two equal intervals — the prismoidal rule applies).

(a) Prismoidal rule

$$V = \frac{d}{3}\big[A_1 + 4A_2 + A_3\big] = \frac{30}{3}\big[12 + 4(20) + 16\big]$$ $$= 10\,[12 + 80 + 16] = 10 \times 108 = 1080\ \text{m}^3$$

Prismoidal volume = 1080 m³.

(b) Trapezoidal (average end area) rule

Apply over the two intervals:

$$V = d\left[\frac{A_1 + A_3}{2} + A_2\right] = 30\left[\frac{12+16}{2} + 20\right] = 30\,[14 + 20] = 30 \times 34 = 1020\ \text{m}^3$$

Trapezoidal volume = 1020 m³.

The prismoidal rule gives the more accurate value; the difference of 60 m³ is the prismoidal correction.

Weighted mean (most probable value)

Work with the seconds part relative to the base $48°24'00"$: deviations are 20\", 26\", 14\" with weights $2, 3, 1$.

\bar{s} = \frac{\sum w_i s_i}{\sum w_i} = \frac{2(20) + 3(26) + 1(14)}{2+3+1} = \frac{40 + 78 + 14}{6} = \frac{132}{6} = 22\"

Most probable value:

$$= 48°24'00" + 22" = \mathbf{48°24'22"}$$

Most probable value of the angle = 48°24'22".

Systematic vs accidental error

• Systematic error: follows a definite physical law, has a known sign/magnitude and is cumulative — it can be detected and eliminated/corrected (e.g. error due to a tape being too long, or non-adjustment of an instrument). Under the same conditions it always acts in the same direction.
• Accidental (random) error: arises from a multitude of small, uncontrollable causes; it is equally likely to be positive or negative, tends to compensate, cannot be eliminated, and is treated by the theory of probability/least squares (e.g. small errors in bisection or staff reading).

(a) Latitudes and departures

Latitude $= L\cos\theta$ (N positive, S negative); Departure $= L\sin\theta$ (E positive, W negative), where $\theta$ is the reduced (quadrantal) bearing angle.

Working:

• PQ: $120\cos30°=120(0.86603)=+103.923$; $120\sin30°=120(0.5)=+60.000$
• QR: $150\cos60°=150(0.5)=−75.000$ (S); $150\sin60°=150(0.86603)=+129.904$ (E)
• RS: $130\cos45°=130(0.70711)=−91.924$ (S); $130\sin45°=130(0.70711)=−91.924$ (W)
• SP: $110\cos50°=110(0.64279)=+70.707$ (N); $110\sin50°=110(0.76604)=−84.265$ (W)

$$\Sigma\text{Latitude} = +103.923 - 75.000 - 91.924 + 70.707 = +7.706\ \text{m}$$ $$\Sigma\text{Departure} = +60.000 + 129.904 - 91.924 - 84.265 = +13.715\ \text{m}$$

(b) Closing error

$$e = \sqrt{(\Sigma\text{Lat})^2 + (\Sigma\text{Dep})^2} = \sqrt{(7.706)^2 + (13.715)^2}$$ $$= \sqrt{59.38 + 188.10} = \sqrt{247.48} = 15.73\ \text{m}$$

Bearing of the closing error (direction of the error vector, measured from north). Since both $\Sigma$Lat and $\Sigma$Dep are positive, the error lies in the NE quadrant:

$$\tan\alpha = \frac{\Sigma\text{Dep}}{\Sigma\text{Lat}} = \frac{13.715}{7.706} = 1.7798 \Rightarrow \alpha = 60.66° = 60°40'$$

Closing error $e = 15.73$ m, bearing ≈ N 60°40' E (the correction to close the traverse acts in the opposite direction, S 60°40' W).

(c) Relative accuracy

Perimeter $= 120.0 + 150.0 + 130.0 + 110.0 = 510.0$ m.

$$\text{Relative accuracy} = \frac{e}{\text{perimeter}} = \frac{15.73}{510.0} = \frac{1}{32.4} \approx \frac{1}{32}$$

Relative accuracy ≈ 1 : 32. (This is a low precision, indicating the traverse would require re-observation in practice.)