Wentworth’s Method of Slope Analysis
This practical demonstrates how to calculate the average slope of an area using contour crossings.
- Scale: 1 cm represents 1 km
- Representative fraction: 1 : 100,000
- Assumed contour interval: 100 metres
Step 1: Draw the Grid
Draw equally spaced horizontal and vertical lines across the contour map.
- Use a ruler to draw horizontal lines.
- Draw vertical lines at equal intervals.
- Label horizontal lines H1, H2, H3, etc.
- Label vertical lines V1, V2, V3, etc.
Step 2: Count Contour Crossings
Count the number of times each grid line intersects the contour lines.
Count repeated intersections with the same closed contour separately.
Step 3: Measure Ground Distance
Measure each grid line using a ruler.
Ground distance (km) = Map distance (cm) × 1
For example, a line measuring 5 cm on the correctly sized map represents 5 km.
Step 4: Straight-Line Observation Table
Enter the measurements for all horizontal and vertical grid lines.
| Grid Line | Ground Length (km) | Contour Crossings |
|---|
Step 5: Calculate Average Crossings
Average contour crossings per km =
Total Contour Crossings / Total Ground DistanceStep 6: Calculate Average Slope
Since the contour interval is assumed to be 100 metres and the ground distance is in kilometres:
tan θ = (Average Crossings × 100) / 1000
θ = tan⁻¹(tan θ)
Step 7: Verification Using Diagonal Lines
Draw diagonal lines across the same contour map.
- Measure the length of every diagonal line.
- Convert each length into ground distance.
- Count contour crossings.
- Enter the observations below.
| Diagonal Line | Ground Length (km) | Contour Crossings |
|---|
Step 8: Compare Results
A small difference indicates that the calculated slope is less sensitive to sampling direction.
Worked Numerical Example
The following values are hypothetical and are not measurements from the contour map.
| Measurement | Value |
|---|---|
| Total contour crossings | 80 |
| Total ground distance | 40 km |
| Average crossings | 80 / 40 = 2 |
| Contour interval | 100 m |
| tan θ | (2 × 100) / 1000 = 0.2 |
| Average slope | 11.31° |
tan θ = 0.2
θ = tan⁻¹(0.2)