How to calibrate a gauge sensor

[zboblamont]

@pierrot10 Flawed mathematics, 0.01m2 is not 10mm2 but 10,000mm2....

[pierrot10]

@mfalkvidd
Thank for your reply, but something is interreting to me. If the funnel area is twice bigger and the bucket is the same, what will it change? the bucket will toggle when it will get 0.2794mm of water, specially if the hole the funnel has the same diameter. Am I wrong?
If the diamter is bigger, the buncket will toggle faster.

[mfalkvidd]

@pierrot10 the bucket will tip based on a volume (or technically weight, but that is almost the same thing) of water, not based on a distance.

Lest's say it is raining 1mm/h.

With the current funnel, the bucket will tip x times.
If you had a funnel that was twice as big (area, as seen from above), the bucket would tip x * 2 times, since the funnel would be collecting water twice as fast.

[pierrot10]

@mfalkvidd Ok, but if the diameter of the hole of the funnel does not change, dot it really make a difflrence?

What is unclear for me,

• I which to display the result in mm/h.
• If I full up 1 liter of water into a bottle, I got about 353 toggles.
• If divide 1000ml by 353 I get 2.832mm

Let's say, I wrongly counted the tips of the buncket (I may missed 2or 4 tips over 353) and there were 358 tips then

1000/358=2.793 => 2.794

Are you agree with my reasoning?

Should I conclude that bucket is 2.794 and not 0.2794?

BUT, I oft read this

The rain gauge is a self-emptying 
tipping bucket type.  Each 0.011” 
(0.2794 mm) of rain causes one mo
mentary contact closure that can 
be recorded with a digital counter or
 microcontroller interrupt input

I am a bit frustrated because I oft read the number of 0.2794 and I have the same gauge meter as the above picture.

Then if within one hour, my bucket tips 358 time because I poured 1 liter of water, the calcul is

358*0.2794mm=100.02mm/h

BUT, 1mm/h is egal to 1liter of water (for 1m2)

So I guess my problem is a mathematical issue :sweat_smile: , or conversion. Isn't? I missed something :fearful:

If I make an error of conversion and specially, if I forget to take in consideration the area of the funnel surface.

I am bit confuse :)

I am not at home yet, and I can not calculate the surface of my funnel yet but I will do as soon as possible (it should be around 50mm*110mm=5500mm2)

[pierrot10]

@pierrot10 Other stupid question, does 1liter of water from a bottle, will cover 1m2 of surface with a high of 1mm?

I red, 1mm/h is egal to 1liter over 1m2 with a high of 1mm

For sure, I missundertand something :confounded:

[mfalkvidd]

@pierrot10 yes, one litre spread evenly over 1m2 will equal 1mm height.

But you're not spreading that litre equally over 1m2. You are pouring everything into the funnel, right?

If you take a regular drinking glass and pour 1 litre into it, will the height of the water in the glass be 1mm?

[bgunnarb]

@pierrot10
Since I am using almost the same tipping bucket rain gauge, let me try to explain how it works:

The top of my funnel measures 109 x 49 mm = 5341 mm2 (square millimeters). For now, let us not worry about the rounded corners of the funnel.
If it rains 1 mm the funnel will collect 5341 mm3 (cubic millimeters) which is 5,341 cubic centimeters equal to 5,34 milliliters. This is difficult to use for calibration so let's try with 10 mm of rain which then equals 53,4 milliliters or 0,53 deciliters of water.

So, you pour 53,4 milliliters of water into the funnel, not spilling any and you count the number of bucket tips. In my example, when I do this, the bucket tips 30 times. So if the bucket tips 30 times during a time period, it has rained 10 mm during that time. Or, the other way around: One tip of the bucket = 10/30 = 0,33 mm of rain.

Do not worry about the size of the bucket inside. What matters is how many times the bucket tips for a given amount of water. It is the area of the funnel which is important.
When it rains 1 mm an area of one square meter will receive 1 liter of water but the little funnel will only receive 5,34 milliliters as explained above.

Now, if you really want to be accurate, you should take away the area of the rounded corners. But that will be less than 0,5% and there are other sources of inaccuracy.

[pierrot10]

@mfalkvidd said in How to calibrate a gauge sensor:

You are pouring everything into the funnel, right?

Hello, yes that right, I poured the litre from my bottle into the funnel.

If you take a regular drinking glass and pour 1 litre into it, will the height of the water in the glass be 1mm?

Yes, of course, that make sense!

I forgetten, but my 3 gauge are in the filed and I can neasure the area surface, unfortunately, but let's say, it 0.05m*0.11m = 0.0055m2 or 55000mm2. I do have the same as below

So in my case, each tips will be egal to
0.2794mm of high over 0.0055m2 of surface

If I know I need 358 tips for 1 liter of water (is egal of 1mm over 1m2) within 1 hour. How can I make sure that 0.2794mm is the correct value of my bucket?

0.2794*358=100.02mm/h

But 100.02mm/h is over 0.0055m2 of surface.

But how can I calculate the high if I poured the same amount of water in 1m2
It should be 1mm because we know that 1 liter over 1m2 has a hight of 1mm

But how calculate it in order to make sure that 0.2794 is the correct value of the bucket?

[pierrot10]

@pierrot10
I make easier. If one liter make 358 tips. One bucket has a value of 2.7932

rain fall [mm/h] = 2.7932 * nb of tips

[bgunnarb]

@pierrot10
1 mm of rain over 1 square meter is 1 liter of water. Correct!
One liter makes 358 tips so one tip is then 2,7932 milliliters that is correct.
But rainfall mm/hr is 2,7932 * nbr of tips only if the area of the funnel is 1 square meter.

Your funnel is only 55 square centimeters so you will only collect 55/10000 of the rain that falls on the whole square meter.

The example you are using at the top of your post does not have the size of the funnel anywhere in the code.