[Weather] Acoustic rain gauge

[Dale Chatham]

I suspect that the "more complex theories" probably account for 
distortions of the drop due to it's being liquid.  What you mention is 
appropriate as mass is a cubic and cross sectional area is a square, 
ergo r^3 / r^2 = r.

Dave Flory wrote:
> For what it is worth, when doing calculations of drop growth via 
> collision/coalescense (where it is vital to know the relative speed 
> between the collector drop and the drops being collected), drop terminal 
> fall speed is typically taken to be directly proportional to the drop 
> radius, or v = k*r.  More complex theories exist, but this is a good 
> starting point.
>
>
> On Thu, 26 Apr 2007, Frédéric wrote:
>
>   
>> Le jeudi 26 avril 2007 12:55, Matthew Smith a écrit :
>>
>>     
>>> Measurement of rainfall by this means is not - as far as I can see -
>>> feasible.  Rainfall is about volume (per area); an acoustic transducer
>>> could only really measure the product of the velocity of an incoming
>>> drop multiplied by its mass (will the real physics guys help me out
>>> here?)  These "hits" would not contain the data to measure the
>>> volume/area.
>>>       
>> Tell me if I am wrong, but because of the air, and because a drop is not 
>> solid, its speed only depends on its mass, no? I mean, falling from 100m or 
>> 1000m, it will reach the same speed (if a physician could confirm...) So, 
>> measuring the energy of the sound can give its mass, so its volume.
>>
>> Air temperature and pressure are parameters to take into account... ut if 
>> there is wind, the drop will not produce the same sound, as it will not hit 
>> the surface with the same angle.
>>
>> Not very easy, so :o/
>>
>>
>>     
>
>   
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