Kilcullen Science and Engineering

Kilcullen Science and Engineering - Exploring Science, Engineering, and Technology

Tuesday, February 03, 2026

What Volume Does a 1 m Change in Water Level Represent in the Golden Falls Lake?

Google map showing a lake
Golden Falls lake. Image courtesy Google Maps, Airbus and Maxtar Techmologies
Distance can easily be measured on Google Maps on desktop by right clicking and selecting "Measure distance" from the context menu. Then simply click the start point and the end point, or draw a series of line segments by repeatedly left clicking and following an outline until the end point is reached. This is useful for measuring distances, but you may not have known that it's also possible to measure area by drawing a series of line segments around for instance a field, and then clicking on the start point again. You can read a complete article I wrote about how to do this here.

I'm curious about how much volume the Golden Falls compensatory lake can hold and what 1 m of level actually equates to in terms of volume. Calculating this is relatively simple, by determining the area of the lake in square metres and multiplying by 1 m. It's not totally accurate obviously because of the sloped sides of the lake, and area increases as water rises. Also the apparent area of the lake might  be somewhat larger than when this satellite image was taken sometime last year. However, I roughly measured the area of the lake as explained above, using the "beaches" as a guideline, to get a ball park figure. 

From the map above, area is approximately 256,000 square metres. 

A cubic metre is simply a cube with sides 1 m long.

So multiply area in square metres by height in metres to give volume:

Volume =  256,000 x 1 = 256,000 cubic metres

Cubic metres are usually written using the SI symbol for length, "m" with a superscript "3" meaning it's cubed.  (SI comes from the French Système international d'unités)

So volume = 256,000 m³ 

The ESB have been discharging 35 m³ of water per second from the dam in the last couple of days. Multiplying flow per second by the number of seconds in a day, this works out at:

35 x 60 x 60 x 24 =  3,024,000 m³

Under normal circumstances that would drop the lake level by:

3,0240,000 / 256,000 = 11.8 m approx

However, flow into the lake from the Poulaphouca dam is reducing this drop in level. Currently, stats aren't available for flow rates from the Poulaphouca dam. However, knowing average daily flow discharges from the Golden Falls dam from ESB stats and the discrepancy between drop in headrace level at Golden Falls and what it should actually be, it should be relatively straightforward to calculate a value for flow.

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