14-12-2018, 11:13
Hi,
Yes, it is the area of the opening which is most relevant, although the "walls" may have some effect (particularly with the FO sensor) in preventing "splash out". If both openings are circular then the calculation is very easy because the area is proportional to the square of the diameter. Thus the increased area is 2.15 ^ 2 = 4.6225 and the sensitivity (or resolution) improved to 1 / 4.6... = 0.2163.. of the original tip sensitivity.
I was also surprised by the suggested figure of 1mm / tip. My understanding of German/Spanish/Dutch (???) is very poor, but the data sheet linked in #6 appears to describe the "resolution" as 1.0 mm per hour and perhaps refers to the (wireless) transmitted data, not to the raw tip count. So I think a measurement of the real sensor (with synthetic rain) is necessary. For example, drip > 78.65 cc of water (0.0786 litre) slowly through the original gauge and count the number of tips (which should correspond to 10mm of rain).
Cheers, Alan.
Yes, it is the area of the opening which is most relevant, although the "walls" may have some effect (particularly with the FO sensor) in preventing "splash out". If both openings are circular then the calculation is very easy because the area is proportional to the square of the diameter. Thus the increased area is 2.15 ^ 2 = 4.6225 and the sensitivity (or resolution) improved to 1 / 4.6... = 0.2163.. of the original tip sensitivity.
I was also surprised by the suggested figure of 1mm / tip. My understanding of German/Spanish/Dutch (???) is very poor, but the data sheet linked in #6 appears to describe the "resolution" as 1.0 mm per hour and perhaps refers to the (wireless) transmitted data, not to the raw tip count. So I think a measurement of the real sensor (with synthetic rain) is necessary. For example, drip > 78.65 cc of water (0.0786 litre) slowly through the original gauge and count the number of tips (which should correspond to 10mm of rain).
Cheers, Alan.