---
title: "Thiessen polygon method with rainfall data"
subject: "Hydro Power"
chapter: 2
key: "semester_6/hydro_power:thiessen-polygon-explain-numerical"
source: https://www.bctengineer.com/q/semester_6/hydro_power/thiessen-polygon-explain-numerical
site: BCT Engineering
---

# Thiessen polygon method with rainfall data

Hydro Power · Chapter 2: Introduction to Hydrology

- Asked in: 2081 Chaitra, 3+3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/thiessen-polygon-explain-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-2

## Question

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 Station | A | B | C | D |
|---|---|---|---|---|
| Area (km²) | 100 | 150 | 80 | 120 |
| Rainfall (mm) | 80 | 70 | 90 | 65 |

## 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.

## Answer

#### 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 $A_i$ of each polygon inside the catchment boundary.
5. Compute:
$$
\bar{P} = \frac{P_1A_1 + P_2A_2 + \dots + P_nA_n}{A_1 + A_2 + \dots + A_n} = \sum P_i\frac{A_i}{A}
$$

```text
   +-------+--------+
   |   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
| Station | Area $A_i$ (km²) | Rainfall $P_i$ (mm) | $P_iA_i$ |
|---|---|---|---|
| A | 100 | 80 | 8000 |
| B | 150 | 70 | 10500 |
| C | 80 | 90 | 7200 |
| D | 120 | 65 | 7800 |
| **Total** | **450** | | **33500** |

$$
\bar{P} = \frac{\sum P_iA_i}{\sum A_i} = \frac{33500}{450} = 74.44\ \text{mm}
$$

**Answer:** Average rainfall over the catchment ≈ 74.4 mm.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does the Thiessen polygon method assume about rainfall at any point?
   A. It equals the average of all gauges
   B. It increases with elevation
   C. It equals the rainfall at the nearest gauge
   D. It is uniform over the catchment
   Answer: C. Each point is assigned the rainfall of the closest gauge, giving a polygon of influence for each station. (Key point: Basic assumption)

2. How is the polygon around each station drawn?
   A. Draw perpendicular bisectors of the lines joining adjacent stations
   B. Draw circles of equal radius round each station
   C. Join points of equal rainfall
   D. Divide the catchment into equal squares
   Answer: A. Adjacent stations are joined to form triangles and the perpendicular bisectors of the sides form the polygons. (Key point: Constructing polygons)

3. For stations A-D with areas 100, 150, 80, 120 km2 and rainfall 80, 70, 90, 65 mm, what is sum(PiAi) and the total area?
   A. 33,500 and 450
   B. 33,500 and 350
   C. 7,800 and 450
   D. 30,500 and 450
   Answer: A. 8000 + 10500 + 7200 + 7800 = 33,500 mm km2 and the total area is 450 km2. (Key point: Weighted mean calculation)

4. What average rainfall results from the Thiessen method for this data?
   A. 75.0 mm
   B. 74.4 mm
   C. 77.9 mm
   D. 76.3 mm
   Answer: B. 33500/450 = 74.44 mm; the simple mean of 76.25 mm would ignore the different areas. (Key point: Final result)

## Related in Hydro Power

- [Arithmetic mean and Thiessen average precipitation](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aarithmetic-thiessen-numerical) — Hydro Power, chapter 2
- [Techniques for average depth of rainfall](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aaverage-rainfall-techniques) — Hydro Power, chapter 2
- [Firm and secondary energy from average monthly flows](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afirm-secondary-energy-2075) — Hydro Power, chapter 3

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- [Thiessen average rainfall, four gauges at rectangle corners](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Athiessen-rectangle-four-gauges) — Engineering Hydrology, chapter 2
- [Methods of average rainfall over catchment](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aavg-rainfall-methods) — Engineering Hydrology, chapter 2, asked 3×
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