---
title: "Thiessen weights for semicircular catchment"
subject: "Engineering Hydrology"
chapter: 2
key: "semester_5/engineering_hydrology:thiessen-weights-semicircle"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/thiessen-weights-semicircle
site: BCT Engineering
---

# Thiessen weights for semicircular catchment

Engineering Hydrology · Chapter 2: Precipitation

- Asked in: 2072 Chaitra, 12 marks
- Video: [Thiessen polygon method | calculation of mean rainfall | Engineering hydrology | Civil engineering](https://www.youtube.com/watch?v=xIuveNLwYm4) — CIVIL HUT
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/thiessen-weights-semicircle
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-2

## Question

The catchment area of a basin may be approximated as a semicircle of radius r km with respect to the coordinate axis set up with its origin at the center of the circle and the x-axis coincident with the diameter; the area lies in the first and second quadrants and the position coordinates of the rain gauge stations are $(0,0)$, $\left(\frac{r}{2},\frac{r}{2}\right)$ and $\left(\frac{-r}{2},\frac{r}{2}\right)$ km. Show that the Thiessen weights of the gauges are given by $\frac{0.5}{\pi}$, $(0.5 - 0.25/\pi)$ and $(0.5 - 0.25/\pi)$ respectively.

## Short answer

Perpendicular bisectors of stations at (0,0), (r/2,r/2), (-r/2,r/2) divide the semicircle. O's polygon is a triangle with area r²/4; total area πr²/2 gives weight 0.5/π. The rest splits symmetrically, giving each remaining gauge weight 0.5 - 0.25/π.

## Answer

**Setup:** Semicircle of radius $r$ in the upper half-plane, area $A = \pi r^2/2$. Gauges: $O(0,0)$, $P_1(r/2, r/2)$, $P_2(-r/2, r/2)$.

```text
        y
        |   P2 .  | . P1
        |   (-r/2,r/2) (r/2,r/2)
   -----+----O-----+---- x
       -r          r
```

#### Step 1: Perpendicular bisectors
- Between $P_1$ and $P_2$: they are at the same height, so the bisector is $x = 0$.
- Between $O$ and $P_1$: points equidistant from both satisfy $x^2+y^2 = (x-r/2)^2 + (y-r/2)^2$, which gives $x + y = r/2$.
- Between $O$ and $P_2$: similarly $-x + y = r/2$.

#### Step 2: Polygon of the gauge at O
The area closer to $O$ than to $P_1$ or $P_2$ is bounded by the x-axis, $y = r/2 - x$ and $y = r/2 + x$. This is a triangle with vertices $(-r/2, 0)$, $(r/2, 0)$ and $(0, r/2)$ (all inside the semicircle):

$$
A_O = \tfrac{1}{2} \times r \times \frac{r}{2} = \frac{r^2}{4}
$$

$$
W_O = \frac{A_O}{A} = \frac{r^2/4}{\pi r^2/2} = \frac{0.5}{\pi}
$$

#### Step 3: Remaining two gauges
The rest of the semicircle is divided by $x = 0$ into two equal halves (by symmetry):

$$
A_{P_1} = A_{P_2} = \frac{1}{2}\left(\frac{\pi r^2}{2} - \frac{r^2}{4}\right)
$$

$$
W_{P_1} = W_{P_2} = \frac{1}{2}\left(1 - \frac{0.5}{\pi}\right) = 0.5 - \frac{0.25}{\pi}
$$

Numerically: $W_O = 0.159$, $W_{P_1} = W_{P_2} = 0.4204$; the sum $= 1.000$. Hence proved.

## Check yourself

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

1. What is the perpendicular bisector of P1(r/2,r/2) and P2(-r/2,r/2)?
   A. x = 0
   B. y = 0
   C. x + y = r/2
   D. y = r/2
   Answer: A. They are at the same height, symmetric about the y-axis, so bisector is x = 0. (Key point: Bisector between P1 and P2)

2. What shape bounds the polygon of gauge O?
   A. Square of side r
   B. Triangle with vertices (-r/2,0), (r/2,0), (0,r/2)
   C. Semicircle
   D. Rectangle r by r/2
   Answer: B. The x-axis and lines y = r/2 -/+ x form this triangle of area r^2/4. (Key point: Polygon at origin)

3. What is W_O?
   A. 0.25/pi
   B. 0.5 - 0.25/pi
   C. 0.5/pi
   D. 1/3
   Answer: C. A_O/A = (r^2/4)/(pi r^2/2) = 0.5/pi = 0.159. (Key point: Weight of O)

4. How are the weights of P1 and P2 obtained?
   A. Equal to W_O
   B. Each is 1/3
   C. Each is 0.5/pi + 0.25
   D. Remaining area split equally by symmetry: 0.5(1 - 0.5/pi)
   Answer: D. Remaining semicircle is divided by x = 0 into equal halves giving 0.5 - 0.25/pi each. (Key point: Weights of P1 and P2)

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