ALAN MELIA wrote:
Hi Piotr, Is it possible that Andy is radiating a much stronger ground-wave at that distance. I know the paths should be reciprocal, but if the weaker station ground wave was weaker, swamped in the noise?? Possibly the antennas for the others are radiating much more upward and the ground wave is being attenuated. The addition.subtraction of a ground wave at the same strength as the skywave could add or subtract 6dB (about the difference in the figures)....I know that the Lowfers in the States see a dead zone at their low power.....but that as with QRSS not a sophisticated system like WSPR.
A slant range calculation formula is given on my web site. The apparent reflection height at 136kHz
at night was lower of course 90-100km. At extreme range the apogee was calculated by raytracing at
about 86km (from memory) At short range and high angles the signal will penerate further before
being returned. The path calculated this was is in error because the refraction
"shortens" the path over a "mirror reflection" model.
http://www.alan.melia.btinternet.co.uk/simple.htm.
http://www.alan.melia.btinternet.co.uk/fading.htm Dont take too much notice
of the text it is quite old now but the equations work and can be solved at
fading nulls. You need two receivers at different distances to resolve the
ambiguities.
An interesting point is that if you can calculate the ground-wave strength, you
can estimate the skywave strength by the depth of the null(which asumes they
are out of phase of course).
I used the thumbnail figure that a watt ERP at 1km generates 300uV/m, for my
calculations. These kind of simple models can be very useful in trying to
understand what is happening and calculating whether any particular path is
possible...... and its good fun !!
Best Wishes
Alan G3NYK
Dear Alan,
Thank you for your response , thank you for your comments and bringing
my attention to your earlier work upon those issues. Had I known your
papers before, my initial post would have been different :but few
points are , say, equivalent. . according to slant calculations formula
.. sure, i did the same sketch ( well, in some way, they had to be the
same ) and i came up with equivalent eqns but in a more
straightforward way. i was using a "cosine theorem' which holds for an
arbitrary triangle: we have its a,b,c sides and between a and b we
have angle alpha then we have that c*c = a*a +b*b - 2*a*b*cos(alpha)
. if you make it a right triangle i.e. alpha = pi/2 then you have its
cosine = 0 and "classical" Pythagoras shows up. The reason that i 've
put simplifications was due to the numerical example i had in mind (
using Rik, on7yd, data ) After reading your post i immediately added
few lines to the program. so now it has one approximation less :)
assuming mean earth radius as 6370 km and using the notation from
your sketch ( dcf39,porto) the path AD+DB = 483.7 km (approximate) and
487.1 km ( slant)
with h = 120 km and AB distance 420 km .
now the first sentence from your "simple path-loss calculation" paper..
"the book say Fs at 1km from 1kW ERP stn is 300 mV/m"...
In my post it is the E = sqrt(60*P)/R as P is isotropic, erp is
mutiplied by 1.5 ( short dipole) we have 1000*1.5*60 = 90000 and its
sqrt is 300 [mV/m because R is 1 and expressed in km) so we know
exactly the 'origin' of 300 mV/m .. :)
Alan, it is just a 'hot' response to your post. I have to 'digest' the
content of your papers and this will take some time but i already like
very much your idea about
the 'oscillating layer' ( again, dcf39,porto paper and 'further
thoughts' chapter) will come back later to this ...
Yours, Piotr,
sq7mpj
qth: Lodz /jo91rs/
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