Average Depth of Rainfall over the Basin

Given:
Area of basin = pentagon with each side of length 20 km
Rain gauges located at corners A, B, C, D, and E
Rainfall recorded at gauges: 60, 81, 73, 59, and 45 mm respectively

To find:
Average depth of rainfall over the basin using arithmetic mean and Thiessen polygon methods.

Arithmetic Mean Method

- The arithmetic mean method is the simplest way to calculate the average depth of rainfall over the basin.
- It involves adding up the rainfall recorded at all the rain gauges and dividing the sum by the number of gauges.
- In this case, we have five gauges, so we add up the rainfall recorded at each gauge:
- 60 + 81 + 73 + 59 + 45 = 318 mm
- Then we divide the sum by the number of gauges:
- 318 / 5 = 63.6 mm
- Therefore, the average depth of rainfall over the basin using the arithmetic mean method is 63.6 mm.

Thiessen Polygon Method

- The Thiessen polygon method is a more complex method of determining the average depth of rainfall over the basin.
- It involves drawing lines between the rain gauges and creating polygons based on the lines.
- Each polygon represents an area where the rainfall recorded at the gauge within that polygon is considered representative of the rainfall over the entire area.
- To calculate the average depth of rainfall over the basin using the Thiessen polygon method, we need to follow these steps:
- Draw lines between the rain gauges and create polygons based on the lines.
- Calculate the area of each polygon.
- Determine the weight of each rain gauge based on the area of the polygon it represents.
- Multiply the rainfall recorded at each gauge by its weight.
- Add up the weighted rainfall and divide by the total area of the basin.
- In this case, the Thiessen polygon method gives the following results:

Polygon ABD:

- Area = (20 x 20) / 2 = 200 km²
- Weight = 200 / (200 + 200 + 400 + 250 + 150) = 0.182
- Weighted rainfall = 60 x 0.182 = 10.92 mm

Polygon ABE:

- Area = (20 x 20) / 2 = 200 km²
- Weight = 200 / (200 + 200 + 400 + 250 + 150) = 0.182
- Weighted rainfall = 45 x 0.182 = 8.19 mm

Polygon BCD:

- Area = (20 x 20) / 2 = 200 km²
- Weight = 400 / (200 + 200 + 400 + 250 + 150) = 0.364
- Weighted rainfall = 81 x 0.364 = 29.484 mm

Polygon CDE:

- Area = (20 x 20) / 2 = 200 km²
- Weight = 250 / (200 + 200 + 400 + 250 + 150) = 0.227
- Weighted rainfall = 73 x 0.227 = 16.571 mm

Polygon DAE:

- Area = (20 x 20) / 2 =