Here's a paper where some folks measured corona currents, charged raindrops
and E field in Yellowstone:
http://docs.lib.noaa.gov/rescue/mwr/095/mwr-095-12-0912.pdf
From one of the charts, it looks like the rain drops were typically
charged with around 0.1-0.2 esu. (I think 1 esu = 3.3356E-10 C ), but
there's a lot of variability. They propose a relationship between the E
field and the charge on the drops due to them charging by conduction with
ions resulting from the corona discharge off the grass.
Another paper suggests charges in the area of 1-2pC (1E-12 C) which is
consistent with the data in the above paper.
Other sources suggest that raindrops are 1-4 mm in diameter. This
corresponds to a falling speed of around 6 m/second. The volume of one
drop is then, about 1 mm^3 (V=4/3 pi r^2)
If we consider a rain rate of 1 inch/hr (25 mm/hr), that would be .0069
mm/sec. Aa hypothetical rain gauge of 1 square meter will have traversed a
path some 6 meters long and swept up all the drops, resulting in a volume
of 1mx1mx.0069mm collected (6.9E-3 * 1E6 = 6.9E3 drops). At 1 pC per drop,
we're looking at a current of about 7 nA for a cross section of 1 square
meter (about 10 square feet)
Now look at a plane going 100 mi/hr (48 m/sec). It's going to pick up 8
times the number of drops in a second, so the charging current will be even
higher
Now, lets say we have an object which is going to discharge at 1kV. What's
the capacitance if it's going to discharge 500 times a second. V=I * T/C
. I'll use 10 nA as the charging current. 1E3 = 10E-9 * 2E-3/C >>> C =
20E-12/1E3 >> C = .02 pF
That's a pretty small capacitance, and not a heck of a lot of stored energy
(0.02 nC*1kV = .02 microJoule), but, in terms of power being dissipated in
the discharges, we're looking at 10 microWatts.. -20 dBm
There's some unrealistic numbers in here (something that intercepts 1
square meter of rain isn't going to have a capacitance as low as 0.02 pF..
a 1m diameter sphere has a C of about 50 pF), but I think if you fool
around with the relative sizes, you can find an object size that collects
enough charge, but is small enough to discharge frequently enough to create
the effects that are observed.
For one thing, we probably don't need to get to 1kV. Corona inception
voltages can be much lower, and, we could actually have a glow discharge
occuring from a very sharp point at a quite low voltage (perhaps 100V or
lower?). You could also have some other nonlinear effects occuring...
charging current flowing unidirectionally through a rusty joint,
etc. forming a sort of charge pump so that the collecting area can be
larger than the capacitance that discharges. (this would be like the fast
strobe in Fruengel.... a big resistor from a HV source formed a constant
current source)
The radio receiver only needs to see -100 dBm or so in a 3kHz BW. Those
little sparks have their power spread over at least 30MHz, so the spark
power probably needs to be on the order of -60 dBm, which is still quite
low. Not all the power winds up in the antenna either...
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Stations", and lot's more. Call Toll Free, 1-800-333-9041 with any questions
and ask for Sherman, W2FLA.
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