I haven't followed this thread completely, so this has probably already been
said:

The easiest was for me to think of the problem is with image theory.  The
source antenna has an "image" as far below the ground plane as it is above
the ground plane.  Solve for the hypotenuse of both triangles and take the
difference for phase, remembering that the horizontally polarized image is
180 degrees out of phase to start with while the vertical image is in phase.
For an approximation, use Friis equation to get the amplitude of each path
and add them up.  I say approximation because it assumes far field
propagation, which may not be the case depending on the range and frequency.

Hope that helps,

Brent DeWitt, AB1LF
Milford, MA

-----Original Message-----
From: CR [mailto:k...@earthlink.net] 
Sent: Saturday, August 01, 2015 8:01 AM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC

On 7/31/2015 10:59 AM, Pawson, James wrote:
> I can do this simply when the TX and RX antennae are the same height 
> above the reflecting surface as the point of reflection lies halfway 
> between the two antennae, Distance_tx = Distance_rx. The direct and 
> reflected paths can be calculated using simple geometry and the 
> wavelength is given by lambda = c / f.
> However when the height of the RX antenna is different to the height 
> of the TX antenna then the horizontal distance to the reflection point 
> is no longer equidistant. I can see that the ratio Height_tx / 
> Distance_tx = Height_rx / Distance_rx remains the same because the 
> angle of reflection is the same. But I'm left with two unknown 
> Distance terms in the equation.

-
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