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Thiessen polygon method with rainfall data

Short answer

The Thiessen polygon method weights each rain gauge's rainfall by the area of its polygon, formed by perpendicular bisectors between stations. For the given data, average rainfall = (Σ P_i A_i) / Σ A_i = 33500 / 450 = 74.44 mm ≈ 74.4 mm.

a) Explain the Thiessen polygon method for estimating the average rainfall over a catchment area. b) The following data were obtained from a rainfall event. Determine the average rainfall over the catchment using the Thiessen polygon method.

Rainfall StationABCD
Area (km²)10015080120
Rainfall (mm)80709065

a) Thiessen polygon method

The Thiessen polygon method gives the average rainfall by weighting each station's rainfall by the area it represents. It assumes the rainfall at any point equals that at the nearest gauge.

Procedure:

  1. Plot all rain gauge stations (inside and near the catchment) on a map of the catchment.
  2. Join adjacent stations by straight lines to form a network of triangles.
  3. Draw perpendicular bisectors of each side; they meet to form a polygon around each station.
  4. Measure (by planimeter or GIS) the area AiA_i of each polygon inside the catchment boundary.
  5. Compute:
Pˉ=P1A1+P2A2+⋯+PnAnA1+A2+⋯+An=∑PiAiA\bar{P} = \frac{P_1A_1 + P_2A_2 + \dots + P_nA_n}{A_1 + A_2 + \dots + A_n} = \sum P_i\frac{A_i}{A}
   +-------+--------+
   |   A   |    B   |   * = rain gauge
   |   *   |    *   |   lines = perpendicular
   +---+---+----+---+   bisectors
       |  C *   |
       +--------+

Merits: allows for uneven gauge spacing. Demerits: ignores orographic effects; polygons must be redrawn when the network changes.

b) Numerical

StationArea AiA_i (km²)Rainfall PiP_i (mm)PiAiP_iA_i
A100808000
B1507010500
C80907200
D120657800
Total45033500
Pˉ=∑PiAi∑Ai=33500450=74.44 mm\bar{P} = \frac{\sum P_iA_i}{\sum A_i} = \frac{33500}{450} = 74.44\ \text{mm}

Answer: Average rainfall over the catchment ≈ 74.4 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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