Mark, Tom,
let me try to put some numbers to your
question...
Loop resistance: Wire diameter #8 AWG = 3.26 mm, skin
depth at 8.3 kHz ~ 0.66 mm, AC cross section 6.7
mm^2, AC resistance 2.6 mohm/m,
loop perimeter 244m, => wire resistance 0.62
ohm.
Depending on ground
conductivity, there will be additional eddy current losses in the ground,
maybe on the order of an ohm. Plus some capacitor ESR and connection
losses. Assume 2 ohms total resistance, giving i = 10 A from 200
watts.
Loop area: a=3720
m^2, magnetic field at distance r towards the side of the loop:
H = i*a /
(4*pi*r^3) ~ 3000 Am^2 / r^3
Noise background at 9 kHz:
Approximately 3 to 10
uV/m/sqrt(Hz), but quite variable depending on static background, time
of day, noise blanking etc. Assuming 5 uV/m/sqrtHz.
CW would be readable with 9
dB SNR in 50 Hz bandwidth => required signal
Emin = 14 + 9 + 17 dB
uV/m = 40 dBuV/m = 100 uV/m
Assuming a magnetic receive
antenna, this is equivalent to Hmin = Emin / 377 ohm = 0.26 uA/m
Solving for
distance:
rmax = (3000/0.26e-6)^0.33
m = 2.24 km
... Stefan, nice estimate indeed!
In reality, the receive loop
would also need to be horizontal (ie vertical axis), and may thus benefit from
picking up much less of the atmospheric noise. 20 dB lower noise
would extend the range by a factor of 2.15.
Furthermore, G3XBM and others
have demonstrated that in inhabited areas, the field of the TX loop
may be picked up by infrastructure like buried cables or water
conduits, inducing currents which may carry the signal to much further
distances in "utility aided earth mode".
But...
Though a horizontal loop
(vertical axis) may have significant inductive nearfields, it will not
radiate much into the far field because it has the wrong
polarization. The associated horizontal E-fields are simply being shunted
by the ground. To "get out", you'd need to
put the loop in upright orientation (ie horizontal axis). Assuming the
same loop area as before, the radiation resistance (including the loop
"image" below conducting ground) would be
Rrad = 2 * 31171 ohms
* a^2 / lambda^4 = 0.5 microohms
and radiated power
EMRP = i^2 * Rrad = 50
uW
But that would require a much
larger mechanical construction, using either very high supports or a much
lengthier loop.
Employing the existing supports
for an upright loop, the area would be only eg. 610 m^2, giving
Rrad = 0.013
microohms
EMRP = 1.3
uW.
A second option would be an
earth antenna, grounded at the far end. You might feed one corner of your
rectangle against ground, and connect the opposite corner to ground as
well. This would be an earth loop of length 61m*sqrt(2), and height 10 m above
ground plus say 48 m below ground (depending on conductivity) => a= 5000 m^2. But here the ground reflection is already
"included" so there won't be the extra factor of two for the image.
Thus
Rrad = 31171 ohms * a^2 / lambda^4 = 0.45
microohms
However the total resistance of
the two ground connections will be much higher than the 2 ohms of the
copper loop, say 200 ohms. So 200 watts would radiate only
EMRP = 0.45
uW.
You will be much better off
using the existing horizontal loop as a top load for a vertical "Marconi".
Assuming there are only few and small trees around it, effective height may be
perhaps 80% of 10m, leading to
Rrad = 1579 ohms *
8m^2/lambda^2 = 76.8 microohms
Optimistically assuming around
200 ohms for coil and ground losses, you coud get 1 A, giving
EMRP = 77 uW,
some two orders of magnitude
stronger than the magnetic and earth loop configurations. With around 1.4 nF capacitance and 30 kV voltage
withstand, you could then even QRO up to 800 Watts
for 308 uW EMRP. Which will easily put you in the ballpark of Dex' and Paul's
successful TA crossing.
Bottom line: Winding a large
coil will be well worth the effort.
And... if you want to go far, just forget about aural CW! This is only
for the biggies like SAQ, radiating about 30 kW.
Best 73,
Markus
(DF6NM)
Sent: Wednesday, June 25, 2014 11:27 PM
Subject: Re: LF: VLF in Canada - earth loop
Hello Mark,
200W in such a loop and 8.97 or 8.27 kHz, CW, my guess is that the
distance is less than 2 km. But it depends on the type of the RX antenna (E or H
field).
73, Stefan/DK7FC
Am 25.06.2014 18:47, schrieb mbdittmar@comcast.net:
Hi -
I've been following the discussion on 9 khz
transmit antennas with some interest, as a friend of mine, Tom, KD0VBR, has
done some preliminary experimentation in this area ( using large loading coils
w/ vertical ). Tom asked me to post a query regarding some antenna ideas
he has, and it would be great to hear the group's thoughts on them and whether
they might work or not. 73, Mark AB0CW
Below are his questions:
From:
"Pouliot, Tom"
<tomandmarti@hotmail.com>To:
"Dittmar, Mark"
<mbdittmar@comcast.net>Sent:
Wednesday, June 25, 2014 8:51:48 AM
Subject: RE: LF: VLF in
Canada - earth loop
Hi Mark,
That
would be great if you could post some questions about transmitting loops for
VLF.
Thinking about a square loop 200' per side
10' off the ground #8 wire with 200 watts into the antenna. How far away
could one expect to be able to detect a CW signal ? ( not QRSS)
Another
possibility would be a triangular loop 200' per side with a 50' high
apex.
