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Date: Sat, 30 Jan 2010 15:18:36 +0100
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Subject: Re: LF: Signal strengths, WSPR reports , propagation
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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/