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I'm trying to design something that will measure the amount of rainfall in a small container. And I was wondering if there was a function I could use to calculate the resistance of the metal probes when say we've had 5ml of rain and then 25ml of rain.

With the output of these probes I can put these in a potiential divider and then a have a voltage output proportional to the amount of rainfall.

Is this a sensible approach to measure the amount of rainfall electronically?

Dean
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  • We make people take on a simple version of this for our introductory EMF class at school. Capacitance was much easier for us to model in a simple classroom setting, but that does not mean with better equipment you could not measure resistance directly. – Kortuk Aug 28 '12 at 14:06
  • @Kortuk I was thinking of using 2 thin copper rods, in a beaker. Held together by a bit of wood, so nothing sophisticated. – Dean Aug 28 '12 at 14:12
  • @Dean Copper rods will corrode quickly and effect your measurements. Wood will soak up water and could also alter your readings. To make matters worse, depending on what's in the air at the time of the rain the water will have different stuff in it, which effects the conductivity of water. Pure water is non-conductive. –  Aug 28 '12 at 15:50

3 Answers3

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I'm not sure that will work well; you probably won't see a factor 5 difference between 5 ml and 25 ml.

I would suggest something capacitive. The relative permittivity of water is about 80 times that of air, so you should be able measure a difference in capacitance between two plates with water between them compared to air. So the capacitance will be a linear function of immersion level. The advantage of the (insulated) plates is that you won't get electrolytic effects.

stevenvh
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    You actually would see a lot of difference in resistance depending on the wetted fraction of the rod-length (ie, depth). The problem is you would also see a lot of difference depending on the contaminants in the water, and quite possibly the state of the rods (likely differing along their length based on typical exposure) though something like platinum might help. – Chris Stratton Aug 28 '12 at 14:34
  • @Chris - Yes, you'll see a difference, but it won't be 20 % for 25 ml compared to 5 ml. I'll rephrase that. (I just tested this in the bathroom, and it's roughly half the 5 ml value at 25 ml.) Don't you agree that the capacitive meter will be more linear? – stevenvh Aug 28 '12 at 14:50
  • @stevenh done right for example in a tall narrow rain gauage, resistance will be almost linear when the materials are consistent, so perhaps your setup was dominated by end effects. But capacitance may be a good idea for working around the very serious contamination/calibration problem with resistance. – Chris Stratton Aug 28 '12 at 15:03
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Look at the Pourbaix diagram of Cu to see how the oxidation state of copper changes with applied potential. Depending on the applied voltage, you might end up setting up a redox reaction. This will not only destroy your leads, but also give you wrong results (the redox reaction is like a battery in series; the dissolving probes' length is changing ruining linearity; the solution's ionic resistance is changing with dissolved ion concentration).

You can still use the voltage divider if you use a low voltage (< 0.2V? - see Pourbaix diagram) and high frequency signal to avoid the redox reaction. While you can still use your voltage divider idea, measuring the capacitance will probably give you better results.

(You could use a nobler metal than Cu and increase the applied V)

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    Could you show a Pourbaix diagram and how to read one in the answer. Assume voltage in the circuit to be 5V. – Dean Aug 28 '12 at 15:40
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As contributors have indicated, capacitance will be easier than resistance. If you must use resistance, I suggest a non-polarizing electrode, like silver/silver chloride, using AC stimuli that integrate to zero, and not DC, would help you avoid the effects of electrode potentials and would give you your best results-- but you still have many of the problems described above concerning contaminants and nonlinearities (and movement artifacts, if you're not careful).

For capacitance approaches on stuff like this, I like using the capacitive sensor as the capacitor (or one capacitor) in some sort of monostable vibrator like a one-shot or appropriate 555 timer circuit to generate a square wave. If you need a DC voltage, you can use a frequency-to-voltage conversion technique (maybe even just a low-pass filter, if you have things set up so your capacitance is linear with your duty cycle), or you can pass it to the CCP or timer interrupt of a microcontroller to measure the frequency or pulse width.

Scott Seidman
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