Dear LF,
on the phone, Stefan mentioned that he was running
about 0.4 A into the nearly vertical (70°), 100 m long kite
antenna. Thus radiated power would be
EMRP = 1579 ohm * (0.4 A * sin(70°)
* 50 m / 33 km)^2 = 0.5 mW = -3 dBm.
Using 1/distance field decay above lossless flat
earth, at 830 km we would expect a field strength
E = -3 + 49.5 - 20 log (830) dBuV/m = -12
dBuV/m.
However Paul's calibrated spectrum at 14:38 UT
showed a peak B = 3.1 fT at 8970 Hz, equivalent to an electrical
field
E = c * B = 0.93 uV/m = - 0.5
dBuV/m.
Taking into account a couple of dB's
for measurement uncertainties, we are still left with an observed signal
very significantly stronger than expected. This seems to support Alexander's statement about a slower
than inverse distance decay in the two-dimensional ionosphere-earth
waveguide.
On http://freenet-homepage.de/df6nm/vlf/vlf_DHO_dualfreq.htm
, I have another observation pointing out an ionospheric effect
on VLF. In 2006, the German naval transmitter DHO38, normally on 23.4
kHz, was simultaneously transmitting on 18.5 kHz as
well. Surprisingly the fieldstrengths and the diurnal patterns for the two
frequencies appeared to be completely different, presumably due to
different phaseshifts between ground and skywave component.
Best regards,
MArkus (DF6NM)
----- Original Message -----
Sent: Thursday, February 25, 2010 6:53
PM
Subject: Re: LF: AW: 9 Dreamers
Hellow, Stefan.
> If that
calculation is reasonably correct, what distance could be
> reached
with 1,4mW @ 8,9 kHz
With such a condition you'll get about 2.5 uV/m
at 100 km. Seems it can be
recievid. If there is no atmospheric and
industrial noise it should be
very strong signal. But all depends on noise
on 9 kHz. I have no ideas
about noise on 9 kHz. I neglet ionosphere in
estimations. But at D=100 km
it should be approximately right.
Anyway if you can use 100 m high antenna then few of 100's km you
should
reach. May be substantionally more. For large distances dependance
of E
on D changes. On the distances more then ~100 km field should
have
behaviour E ~ sqrt(1/D). This yelds only 10 dB attenuation if
distance
became 10 times more. Thus if you'll have 20 dB over noise at 100
km (if
noise is 0.25 uV/m for example) then you'll have ~10 dB over noise
at 1000
km.
Do it if you have such an oportunity! It is very
interesting.
Regards,
Alexander
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