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falls as guides ? similar to the The are the isohye The avera over the i ... an-1 a catchmen The isohyet me.... tions are large in number. please illustrate this answer how this question solved Station Coordinates Precipitation (cm) Station EXAMPLE 2.5 In a catchment area, approximated by a circle of diameter 100 km, four rainfall stations are situated inside the catchment and one station is outside in its neigh- bourhood. The coordinates of the centre of the catchment and of the five stations are given below. Also given are the annual precipitation recorded by the five stations in 1980. Determine the average annual precipitation by the Thiessen-mean method. Diameter: 100 km. Centre: (100, 100) Distance are in km 1 2 3 4 5 Total SOLUTION: The catchment area is drawn to scale and the stations are marked on it (Fig. 2.15). The stations are joined to form a set of triangles and the perpendicular bisector of each side is then drawn. The Thiessen-polygon area enclosing each station is then identified. It may be noted that station 1 in 1. this problem does not have any area of influence in the catchment. The areas of various Thiessen poly- gons are determined either by a planimeter or by placing an overlay grid. Boundary of area 2 1 (30,80) (70, 100) 85.0 135.2 abcd dce ecbf fba 2141 1609 2141 1963 7854 Mean precipitation = 121.84 cm. area Fig. 2.14). The procedure is evels. ed with a planimeter. If d as the bounding line. assumed to be acting isohyets and if a₁, az, precipitation over the (2.9) ethods especially when the sta- 3 (100, 140) 95.3 Area Fraction of total (km²) 0.2726 0.2049 0.2726 0.2499 1.000 -1 + P 2 4 (130, 100) 146.4 d 2 a Rainfall 3 85.0 135.2 95.3 146.4 102.2 to ke C b 5 5 (100, 70) 102.2 Fig. 2.15 Thiessen Polygons- Example 2.5 e 4 Weighted P (cm) (col. 4 × col. 5) 36.86 19.53 39.91 25.54 121.84