Surveying-I Chapter Six: Contouring Study Notes

Course Information and Content Outline

• Institution: University of Hargeisa (Jaamacadda Hargeysa), College of Engineering.

• Course: Surveying-I.

• Instructor: Engr. Abdirahman Dayr.

• Date: 01 September 2024.

• Chapter: Chapter Six - Contouring.

• Chapter Outline:

  • Definition of contouring terms.

  • Objects of preparing contour maps.

  • Uses of contour maps.

  • Characteristics of contours.

  • Methods of contouring.

  • Methods of interpolation of contours.

Definitions and Fundamental Concepts

• Contour Line: The line of intersection of a level surface with the ground surface is known as the contour line or simply the contour. It is also defined as a line passing through points of equal reduced levels. For example, a contour of $100\,m$ indicates that all points on that line have a Reduced Level ($RL$) of $100\,m$. Similarly, a contour of $99\,m$ includes only points with an $RL$ of $99\,m$.

• Contour Map: A map that shows only the contour lines of a specific area.

• Contour Interval ($CI$): The vertical distance between any two consecutive contours. For instance, if a map contains contours at $100\,m$, $98\,m$, and $96\,m$, the contour interval is $2\,m$.

  • The contour interval for a particular map must remain constant.

  • The choice of interval depends on:

    • The nature of the ground (flat or steep).

    • The scale of the map.

    • The purpose of the survey.

  • Interval Standards:

    • Flat Country: Generally smaller intervals (e.g., $0.25\,m$, $0.50\,m$, $0.75\,m$).

    • Steep Slope/Hilly Area: Generally greater intervals (e.g., $5\,m$, $10\,m$, $15\,m$).

    • Small-scale Map: Larger intervals (e.g., $1\,m$, $2\,m$, $3\,m$) suitable for large areas with less detail.

    • Large-scale Map: Smaller intervals (e.g., $0.25\,m$, $0.5\,m$, $0.75\,m$) suitable for small areas with more detail.

• Horizontal Equivalent ($HE$): The horizontal distance between any two consecutive contours. Unlike the contour interval, the horizontal equivalent is not constant and varies according to the steepness of the ground.

  • Steep slopes: Contour lines run close together.

  • Flatter slopes: Contour lines are widely spaced.

Objects and Uses of Contour Maps

• Object of Preparation: While general maps show locations of roads, rivers, and towns, they do not illustrate the nature of the ground surface. For engineering projects like roads and railways, understanding the ground surface is essential for locating suitable alignments and estimating earthwork volumes. Therefore, a contour map is essential for all engineering projects.

• Specific Uses:

  1. Understanding the nature of the ground surface of a country.

  2. Selecting suitable sites or economical alignments for engineering projects.

  3. Computing the approximate capacity of a reservoir or the area of a catchment.

  4. Marking suitable routes for a given gradient on the map.

  5. Portraying approximate quantities for earthwork computation.

Characteristics of Contours

1. Steep vs. Flat Ground: Contour lines are closer together the near the top of a hill or high ground and wider apart near the foot. This indicates a steep slope towards the peak and a flatter slope towards the base.

2. Depressions: Contour lines are closer near the bank of a pond or depression and wider apart towards the center, indicating a steep slope at the bank and flat slope at the center.

3. Uniform Slope: Uniformly spaced contour lines indicate a uniform slope.

4. Crossing Contours: Contour lines cannot cross one another, with the exception of an overhanging cliff. In the case of an overhanging cliff, the overlapping portion must be shown with a dotted line.

5. Ridge Lines: When higher values are inside a loop, it indicates a ridge line. Contour lines cross ridge lines at right angles ($90^\circ$).

6. Valley Lines: When lower values are inside a loop, it indicates a valley line. Contour lines cross valley lines at right angles.

Methods of Contouring

Direct Method

In this method, contours are directly traced out in the field by locating a number of points on each contour. These points are surveyed, plotted on a plan, and the contours are drawn through them. This method is slow, tedious, and used for small areas requiring great accuracy.

• Vertical Control: Locating points on the contours.

  • A Benchmark ($BM$) is required in the project area.

  • The level is set up, and a Back-sight ($BS$) is taken on the $BM$.

  • Example: If $RL$ of $BM = 500\,m$ and $BS = 1.523\,m$, then Height of Instrument ($HI$) is:       $HI = 500\,m + 1.523\,m = 501.523\,m$

  • To find the $500\,m$ contour, the staff reading must be:       $501.523\,m - 500\,m = 1.523\,m$

  • To locate the $501\,m$ contour, the staff reading must be:       $501.523\,m - 501\,m = 0.523\,m$

  • Note: A contour of $502\,m$ cannot be located with this setup because the $HI$ is only $501.523\,m$. However, lower contours ($499\,m$, $498\,m$) can be located depending on the staff length ($4\,m$, $4.5\,m$, or $5\,m$).

  • Shifting Instrument: To locate lower points (e.g., $497\,m$, $496\,m$), shift the level to a new station ($B$), take a back-sight on a forward station ($F$) of known $RL$, and calculate the new $HI$.

• Horizontal Control: Plotting points on the plan (e.g., using a plane table). Once vertical points are located, they must be surveyed based on the extent of the area.

Indirect Method

Guide points (spot levels) are selected and surveyed. These points do not need to be on the contour itself. These points are plotted and serve as the basis for the interpolation of contours. This method is faster and more economical.

1. Cross-Section Method:

   • Cross-sections are set out perpendicular to the center line of the area.

   • Spacing intervals: $20\,m$ for hilly country; $100\,m$ for flat country.

   • Salient features of the center line and cross-sections are located.

2. Square Method:

   • The area is divided into squares ($5\,m$ to $25\,m$ sides).

   • Levels of the corners and salient features are determined.

   • Contours are interpolated between the corners.

3. Tacheometric Method: Used as an indirect method for surveying.

Comparison: Direct vs. Indirect Methods

Methods of Interpolation of Contours

Interpolation is the process of finding the position of contour points between guide points.

1. By Estimation: A rough method used for small-scale maps. No calculations involved.

2. By Arithmetic Calculations: A very accurate but tedious method where positions are calculated mathematically.

3. By Graphical Method: Interpolation using tracing paper or tracing cloth.

Mathematical Examples and Logic

Example 1: Cross-Section Method (RL = 100m)

• Given: Horizontal distance between stations = $10\,m$. Required contour elevation = $100\,m$.

Technique 1: Arithmetic Proportion

Formula: $\text{Distance} = \frac{\text{Required RL} - \text{Station 1 RL}}{\text{Station 2 RL} - \text{Station 1 RL}} \times \text{Horizontal Distance}$

• Calculation 1 (Between RL 99 and RL 102):     $\frac{100 - 99}{102 - 99} \times 10 = \frac{1}{3} \times 10 = 3.33\,m$

• Calculation 2 (Between RL 99 and RL 101.5):     $\frac{100 - 99}{101.5 - 99} \times 10 = \frac{1}{2.5} \times 10 = 4\,m$

• Calculation 3 (Between RL 98 and RL 102):     $\frac{100 - 98}{102 - 98} \times 10 = \frac{2}{4} \times 10 = 5\,m$

• Calculation 4 (Between RL 99 and RL 101):     $\frac{100 - 99}{101 - 99} \times 10 = \frac{1}{2} \times 10 = 5\,m$

• Calculation 5 (Between RL 97 and RL 101):     $\frac{100 - 97}{101 - 97} \times 10 = \frac{3}{4} \times 10 = 7.5\,m$

Technique 2: Similar Triangles

Using the slope of the large triangle to find the horizontal offset ($x$) of the required contour.

• Formula logic: $\frac{\text{Total Rise}}{\text{Total Distance}} = \frac{\text{Partial Rise}}{x}$

• Example 1 (Between RL 99 and RL 102):     $\frac{102 - 99}{10} = 0.3$ (Slope of large triangle)     $\frac{100 - 99}{x} = \frac{1}{x}$ (Slope of small triangle)     $0.3 = \frac{1}{x} \rightarrow x = \frac{1}{0.3} = 3.33\,m$

• Example 2 (Between RL 99 and RL 101.5):     $\frac{101.5 - 99}{10} = 0.25$     $\frac{100 - 99}{x} = \frac{1}{x}$     $0.25 = \frac{1}{x} \rightarrow x = \frac{1}{0.25} = 4\,m$

• Example 3 (Between RL 98 and RL 102):     $\frac{102 - 98}{10} = 0.4$     $\frac{100 - 98}{x} = \frac{2}{x}$     $0.4 = \frac{2}{x} \rightarrow x = \frac{2}{0.4} = 5\,m$

Example 2: Square Method (RL 98.5m, 99.5m, and 100.5m)

• Given: Horizontal distance = $10\,m$.

• The estimation method is used to mark the points for contours at $98.5\,m$, $99.5\,m$, and $100.5\,m$ without specific calculation, based on visual interpolation between corner spot levels.

Exercises

• Task: Draw contour lines using the Square Method.

• Parameters: Horizontal distance between stations = $10\,m$. Required contours = $101.5\,m$ and $103\,m$.

• Requirement: Apply both the Estimation method and the Arithmetic Calculations method.