Average Rainfall by Thiessen Polygons

Easyrainfallthiessenaveraging

A watershed is divided into two Thiessen polygons that cover it completely with no overlap. Gauge 1 represents 2 ha and recorded 20 mm of rainfall; gauge 2 represents 3 ha and recorded 40 mm during the same storm. What is the average storm rainfall depth over the watershed?

A30 mm
B32 mm
C28 mm
D160 mm

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Answer: B. 32 mm

Weight each depth by the area it represents: P_avg = Σ(A_i P_i)/ΣA_i. Total area = 2 + 3 = 5 ha, so P_avg = (2·20 + 3·40)/5 = (40 + 120)/5 = 32 mm (equivalently 0.4·20 + 0.6·40 = 8 + 24). 30 mm is the unweighted mean, which ignores that gauge 2 represents more area. 28 mm swaps the weights (0.6·20 + 0.4·40). 160 mm is the sum of area-depth products never divided by the total area.

P_{a v g}

= \frac{\sum A _{i} P _{i}}{\sum A _{i}}

= \frac{2 ( 20 ) + 3 ( 40 )}{5}

= \frac{160}{5}

= 32

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Average Rainfall by Thiessen Polygons

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