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I have bought this cheap rain gauge (only the rain gauge, not the other parts) for measuring rainfall with a Raspberry Pi. It works as far as I can tell, every bucket swing will result in an interrupt that I can further process.

The datasheet is located here and this is where my questions begin:

I have calibrated the device and found that around 1.2-1.3 mL will tip the bucket. Now rainfall is usually measured in mm/m², so the datasheet also covers this: 0.2794mm per contact. Now how do I get this to a measurement in mm/m²? The area of the rain gauge is about ~0.0055m² (11x5cm). Is it safe to assume that the rainfall is the same in this m²? So if I measure 5mm of rainfall, then it's also 5mm per m²?

Hope that was not too complicated in my explanation.

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    $\begingroup$ I had a similar one several years ago. It was very accurate but the electronics failed after several months $\endgroup$ – blacksmith37 Aug 19 at 18:17
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mm per square meter is a measurement of volume, it's not really a depth divided by an area. You can tell as much because if you have 1 mm per square meter and you increase the coverage area to 2 square meters, the depth doesn't change---the volume increases. You just need to convert your volume measurement.

$$ 1.2 mL = 1.2 cm^{3} $$ 1.2 cubic centimeters is equivalent to a 1.2 cm depth over an area of 1 square cm. $$ 1.2 cm = 12 mm $$ $$ 1 cm^{2} * \frac{1 m^{2}}{100^{2} cm^{2}} = 0.0001m^{2} $$ 12mm per 0.0001 square meters is not a useful metric. You have to obviously make the square meters 1 square meter, but you have to be careful. When you increase the area, the depth will necessarily decrease (because the volume has to be equivalent). Multiplying the area by 10,000 square meters per square meter will give you 1 square meter; subsequently, the depth must be divided by the same amount. $$ \frac{12mm}{10,000} = 0.0012 mm $$ So you have 0.0012 mm per square meter with each tip of the rain gauge. We can check our work by converting it back to a normal volume. $$ 0.0012 mm * 1 m^{2} * \frac{1000^{2} mm^{2}}{1 m^{2}} = 1200 mm^{3} = 1.2 mL $$

As a side note, here is a useful tool provided by the United States Geological Survey for converting depth per area measurements to equivalent volumes: https://water.usgs.gov/edu/activity-howmuchrain-metric.html

Edit

Going back to the depth given by the data sheet, I ran the numbers and $$ (0.2794mm)/(1000mm/m)*(0.0055m^2) \approx 1.54 mL $$ which about matches the volume you have. So if you want the depth for one square meter, you just have to multiply the area by the factor of $\frac{1m^{2}}{0.0055m^{2}}$ and divide the depth by the same factor. So essentially you want to multiply the depth by the catchment area of the gauge. $$ (0.2794 mm) * (0.0055 m^2/m^2) = 0.00154 mm $$ as the depth for 1 square meter.

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  • $\begingroup$ Thanks for this answer. Where do the 0.2794mm factor in? Or is this just a random measurement that I should avoid? $\endgroup$ – fsp Aug 19 at 18:59
  • $\begingroup$ It seems like the 0.2794 mm is a measurement of the depth in the catchment cup which triggers it to tip over. So, my assumption is that if you know the area of the catchment cup (not the area of the catchment area of the gauge) you can do the same conversion as above. It's hard to know for sure because the data sheet isn't very explicit. $\endgroup$ – Reifier Aug 19 at 19:01
  • $\begingroup$ The problem with that approach is the catchment cup usually has an irregular shape, so the average cross-sectional area isn't easy to obtain. Measuring the volume is a far better approach! $\endgroup$ – Reifier Aug 19 at 19:10
  • $\begingroup$ Thanks. I still have a question regarding the normalized rainfall, would I take the 1 m² and divide by the catchment area of the gauge? That would give me a factor. Or how do I compare this measurements with other ones? $\endgroup$ – fsp Aug 19 at 19:19
  • $\begingroup$ I'm not so sure about what you're looking for; if you divide by the catchment area of the gauge the quotient will be the depth of the water that would be in a container with that cross-sectional area, since it's a volume measurement. What's really being normalized here is the volume per area of rainfall; let's say the rain delivers 10 cubic centimeters of rain in every square meter of space. Instead we can think of that as a depth of rain in that same square meter. So we start with a [volume]/[area] and we divide the volume by area again to get the depth, so we're left with [depth]/[area]. $\endgroup$ – Reifier Aug 19 at 20:43

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