Volume of rainfall over an area of land

Volume is the measure of three-dimensional space. Three length measurements are needed to calculate a volume. In the case of rainfall, two of the length measurements are used in obtaining the area of the land and the third measurement is the depth of rainwater that has fallen.

Usually, a rain gauge is the measuring device that gives the depth measurement. If it is reported that, say, $15$15 mm of rain fell at some location, we understand this to mean the rainwater would cover the ground everywhere to a depth of $15$15 mm if none of it soaked away.

We calculate the volume of rainfall over a given area by multiplying the area measurement by the depth measurement, making sure that the units are the same. For example, if we want to express the volume of water in cubic metres, we need to express the area in square metres and the depth in metres.

It is useful to note that it is the cross-sectional area of the land that matters. This is the area as seen when looking directly down from above. As a consequence, the slope of the terrain or its roughness does not affect the volume of rainwater that would be calculated. The following example illustrates this principle in the case of a sloping roof. 

Example 1

The parallelogram on the left represents an end-view of a sloping roof onto which rain has fallen to a depth $d$d. The rectangle on the right has the same area as the parallelogram. (We invite you to explain why this assertion is true.)

The sloping roof has a cross-section of $w$w, the same as the rectangle on the right. Therefore,  If the length of the roof is $L$L, both shapes have an area looking down from above of $w\times L.$w×L. We conclude that the volume of rain that has fallen on the roof is $w\times L\times d$w×L×d.

 

 

Generally speaking, the difficulty of calculating a volume of rainfall or of water flowing in a pipe or river, is due to the problem of finding the cross-sectional area involved. In many cases, it is necessary to estimate the cross-sectional area by dividing the shape into smaller shapes that are approximately rectangles or trapezoids whose areas are easy to calculate.

Example 2

Suppose $30$30 mm of rainfall occurs in the space of $4$4 hours over the area shown in the following radial survey diagram. The area is surrounded on all sides by a drainage system. How much water, in cubic metres per minute must the system cope with?

 

Using the sine formula for area, we find that the total area is

$\frac{1}{2}\times41\times55\times\sin45^\circ+\frac{1}{2}\times55\times48\times\sin34^\circ$12​×41×55×sin45°+12​×55×48×sin34°

This is $1535.4$1535.4 m$^2$2

The area is covered by rainwater to a depth of $30$30 mm. This is $0.03$0.03 m. Therefore, the volume of water is $1535.4\times0.03=46.06$1535.4×0.03=46.06 m$^3$3.

This amount of water flows through the drain in $4$4 hours, which is $4\times60=240$4×60=240 minutes. Therefore the flow rate is $\frac{46.06}{240}=0.19$46.06240​=0.19 m$^3/$3/minute.

Worked Examples

Question 1

Question 2

Question 3