Surveying – Contouring
Sub Engineer Civil Complete Notes + 50 MCQ Test
1. Introduction & Definitions
Primary Objectives
- To study the general topography and nature of the ground terrain.
- To select suitable alignment for roads, railways, canals, and pipelines.
- To calculate capacity of reservoirs and earthwork quantities (cutting and filling).
- To determine intervisibility between two stations.
2. Fundamental Terms in Contouring
| Term | Definition / Core Concept |
|---|---|
| Contour Interval (CI) | The constant vertical distance between two consecutive contour lines. Remains constant for a map. |
| Horizontal Equivalent (HE) | The horizontal distance between any two adjacent contour lines. Varies according to ground slope. |
| Contour Gradient | A line lying on the ground surface maintaining a constant inclination to the horizontal. |
| Direct Contouring | Location of contour points directly on the ground by staff readings (Accurate but Slow). |
| Indirect Contouring | Spot levels taken at grid/radial points and contour lines interpolated later (Fast and Practical). |
- Nature of Ground: Small for flat ground, large for hilly terrain.
- Scale of Map: Inversely proportional to scale (CI ∝ 1 / Scale).
- Purpose & Time: Small for detailed work; large when time/budget is limited.
3. Characteristics of Contours (Crucial for PYQs)
- Equal Elevation: All points on a single contour line have the same Reduced Level (RL).
- Uniform Slope: Equally spaced contour lines represent a uniform slope.
- Flat Surface: Straight, parallel, and widely spaced contour lines indicate flat terrain.
- Steep vs Gentle Slope: Closely spaced = Steep slope; Widely spaced = Gentle slope.
- Hill vs Depression:
- Hill: Closed contours with elevation values INCREASING towards the center.
- Pond / Depression: Closed contours with elevation values DECREASING towards the center.
- Overhanging Cliff: Contour lines CROSS each other at a point.
- Vertical Cliff: Contour lines UNITE/COINCIDE to form a single line.
- Ridge Line (Watershed) vs Valley Line:
- Ridge Line: U-shaped/V-shaped contours with convex curve towards HIGHER elevation (Crosses ridge at 90°).
- Valley Line: U-shaped/V-shaped contours with convex curve towards LOWER elevation (Crosses valley at 90°).
4. Methods of Contouring
| Method | Procedure / Suitability |
|---|---|
| Direct Method | Vertical control via level instrument & staff. Very precise; used for small areas and high accuracy requirements. |
| Grid Method (Squares) | Indirect method. Area divided into grids (5m to 20m). Used for flat/gentle terrain, reservoir sites. |
| Cross-Section Method | Indirect method. Cross-lines drawn perpendicular to centerline. Best for roads, railways, and canals. |
| Radial Lines Method | Indirect method. Lines radiate from central point using tacheometer. Best for hilly areas/hollows. |
5. Interpolation & Computation of Earthwork
Methods of Interpolation
1. Estimation: Rough judgement by eye.
2. Arithmetic Calculation: Precise, based on linear proportion.
3. Graphical Method: Fast, using tracing paper and proportional lines.
Earthwork Volume Formulas
- Trapezoidal Rule (Prismoidal Approximation): $V = H \left[ \frac{A_1 + A_n}{2} + A_2 + A_3 + \dots + A_{n-1} \right]$
- Prismoidal Rule (Simpson's Rule for Volume): $V = \frac{H}{3} \left[ (A_1 + A_n) + 4(\text{Sum of Even Areas}) + 2(\text{Sum of Odd Areas}) \right]$ (Requires odd number of cross-sections)
Contouring Formula Master Sheet
Usage: Quick estimation of contour interval based on map scale.
Example: Scale = 1 cm = 50 m → $CI = 25 / 50 = 0.5\text{ m}$.
Symbols: $\theta$ = Ground angle of slope.
Example: $CI = 2\text{ m}, HE = 40\text{ m} \rightarrow \tan(\theta) = 2/40 = 1/20$ (5%).
Symbols: $L$ = distance between two grid points, $\Delta H$ = level diff.
Example: $L=10\text{m}, \Delta H_{tot}=2\text{m}, \Delta H_{tar}=0.5\text{m} \rightarrow x = (0.5/2)\times 10 = 2.5\text{m}$.
Symbols: $H$ = Contour interval, $A_i$ = Area enclosed by $i^{\text{th}}$ contour line.
Condition: Number of area sections ($n$) MUST be odd.
Symbols: $K$ = Multiplying constant (100), $C$ = Additive constant (0), $S$ = Staff intercept.
1. Arithmetic Interpolation Numerical
Problem: Two points A and B are 20 m apart on a map. The RL of point A is 101.20 m and point B is 104.80 m. Find the map distance of 103.00 m contour line from point A.
Given:
Distance $L = 20\text{ m}$
$RL_A = 101.20\text{ m}, RL_B = 104.80\text{ m}$
Target Contour Level = $103.00\text{ m}$
Formula:
$x = \frac{RL_{\text{target}} - RL_A}{RL_B - RL_A} \times L$
Calculation:
$\Delta H_{\text{total}} = 104.80 - 101.20 = 3.60\text{ m}$
$\Delta H_{\text{target}} = 103.00 - 101.20 = 1.80\text{ m}$
$x = \frac{1.80}{3.60} \times 20 = 0.5 \times 20 = 10.00\text{ m}$
Final Answer: Distance of 103.00 m contour from point A = 10.00 m.
2. Reservoir Storage Volume (Prismoidal Rule)
Problem: Areas enclosed by contours in a lake site at 2 m interval are: 100 m², 150 m², 200 m², 280 m², and 350 m². Calculate the capacity of the reservoir using Prismoidal Rule.
Given:
Contour Interval $H = 2\text{ m}$
Areas: $A_1 = 100\text{ m}^2, A_2 = 150\text{ m}^2, A_3 = 200\text{ m}^2, A_4 = 280\text{ m}^2, A_5 = 350\text{ m}^2$
Total Areas $n = 5$ (Odd, so Prismoidal rule applies directly).
Formula:
$V = \frac{H}{3} \left[ (A_1 + A_5) + 4(A_2 + A_4) + 2(A_3) \right]$
Calculation:
$A_1 + A_5 = 100 + 350 = 450$
$4(A_2 + A_4) = 4(150 + 280) = 4(430) = 1720$
$2(A_3) = 2(200) = 400$
$V = \frac{2}{3} \times [450 + 1720 + 400] = \frac{2}{3} \times 2570 = 1713.33\text{ m}^3$
Final Answer: Reservoir Capacity = 1713.33 m³.
3. Slope and Contour Gradient Determination
Problem: On a map drawn to scale 1:1000, two consecutive contours of 100 m and 102 m are 4 cm apart. Determine the slope of the ground in percentage.
Given:
$CI = 102 - 100 = 2\text{ m}$
Map distance = $4\text{ cm}$, Scale = 1:1000
Ground distance (HE) = $4\text{ cm} \times 1000 = 4000\text{ cm} = 40\text{ m}$
Formula:
$\text{Slope (\%)} = \frac{\text{Contour Interval (CI)}}{\text{Horizontal Equivalent (HE)}} \times 100$
Calculation:
$\text{Slope (\%)} = \frac{2}{40} \times 100 = \frac{1}{20} \times 100 = 5\%$
Final Answer: Ground Slope = 5% (1 in 20).
MP Sub Engineer Civil - Contouring Test
Test your mastery of Contouring with 50 Sub Engineer PYQ-style and conceptual questions.
⏱️ Total Questions: 50
📊 Topics Covered: 17 Contour Sub-Modules
🧮 Numerical Questions: 20+
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