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Re: [TowerTalk] Takeoff Angles and Non-Reciprocal Propagation

To: Michael Keane K1MK <k1mk@alum.mit.edu>,TowerTalk <towertalk@contesting.com>
Subject: Re: [TowerTalk] Takeoff Angles and Non-Reciprocal Propagation
From: David Gilbert <xdavid@cis-broadband.com>
Date: Sat, 03 Nov 2007 11:18:30 -0700
List-post: <mailto:towertalk@contesting.com>
Ahh ... a very thoughtful and relevant reply.   Thanks!

I'll respond to your last comment first.  Yes, using HFTA to look at an 
irregular terrain feature from both sides is not representative of what 
really happens to a particular ray.  At first I thought I might be able 
to choose a particular angle (let's say 4 degrees) and simulate a 
returning signal by placing an antenna the proper distance and height 
(based on the tangent of 4 degrees) on the "other side" of the terrain 
feature.  Unfortunately, and for obvious reasons, HFTA won't compute 
signals at negative angles.  But HFTA did, at least, confirm that 
differently shaped terrain features can have a significantly different 
diffraction affect on the signal even if the heights are the same.

As for your other comment, diffraction of a ray around an edge (in any 
direction) is a function of shape as seen by the ray.  Any ray of energy 
will be affected by an object within a distance determined by the 
wavelength of the ray ... the ray of RF (or light) is not an infinitely 
thin beam in terms of its interaction with surroundings.  If it were, 
very few antennas (certainly no parasitic antenna) would work worth a 
darn.  Any one of those rays you describe will be affected by an object 
within a distance that is determined by its wavelength, so if the shape 
of that object appears different from one direction than from the other, 
different refraction will occur.  It seems to me that if, for example, 
the radius of curvature of an object as seen by a ray traveling from one 
direction is two wavelengths and the radius of curvature of that same 
object is 10 wavelengths as seen by a second ray trying to retrace the 
path of the first ray but going in the opposite direction, the second 
ray is not very likely to follow the path of the first ray once it gets 
past the object.

I'll try to draw a diagram to illustrate what I mean, and I'll post a 
link to it later today.

73 and thanks again,
Dave   AB7E



Michael Keane K1MK wrote:
> At 02:02 AM 11/3/2007, David Gilbert wrote:
>
>   
>> HFTA
>> uses ray tracing calculations for edge diffractions, surface
>> reflections, and combinations thereof to compile a net elevation profile
>> of the energy passing over the terrain after leaving the antenna.  To
>> say that a single ray coming from the far end of the path and "seeing"
>> an entirely different terrain profile from the return side (assuming
>> asymmetric features) would follow the same path to the antenna just
>> doesn't make sense.
>>     
>
> The assumption that an individual ray will follow a different path 
> when moving left-to-right than right-to-left is what shouldn't make sense.
>
> As you shoot individual rays outward from the antenna and the ray 
> encounters an obstacle, that single ray diffracts into multiple 
> diffracted rays travelling in a range of directions with a 
> well-defined distribution of amplitude with direction.
>
> All reciprocity says is that for any one of the multiple diffracted 
> rays, the amplitude of that diffracted ray depends only on the 
> included angle between the incident ray and the diffracted ray and 
> not on the direction of the incident ray.
>
>   
>> Even HFTA says (per the very simple test profiles I
>> generated) that an asymmetrical terrain feature results in asymmetrical
>> diffraction when viewed from both sides.  Unless you want to challenge
>> the algorithms and calculations built into HFTA, I don't see how how you
>> and N6RK can challenge that result.
>>     
>
> No one is challenging that result. That's because N6BV's comments in 
> the HFTA documentation and your tests both address a point that is 
> quite different from simple reciprocity.
>
> HFTA evaluates the outward travelling signal at infinity (more 
> precisely as a plane wave after the last refraction/diffraction) by 
> shooting rays from the antenna through the obstacle to infinity. 
> Placing the antenna on the opposite side of the obstacle and 
> re-running the HFTA ray trace is not the same as reversing the 
> direction of the rays in the original configuration and tracing from 
> infinity through the obstacle to the antenna.
>
> 73,
> Mike K1MK
>
> Michael Keane K1MK
> k1mk@alum.mit.edu
>
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