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Sparks 
Posted: August 04, 2014 03:44 pm

Sr. Member Group: Trusted Members Posts: 401 Member No.: 750 Joined: April 08, 2004 
Hello,
Quick disclaimer to this post: I'm terrible at maths & understand very little of what I'm looking at Random wikipedia has led me to radio recently, specifically VHF. The antennae are lineofsight, but are affected by interference due noise within the 'fresnel zone'. Wiki  Fresnel Zone Looking at the calculation for the widest point of the zone (assuming midpoint between the antennae), I can fill in the bits of the equasion; d1 & d2 = 100m wavelength = 2m (approx 150mhz, I think) That gets me a radius (Fn) of 10. The next part 'simplifies' the equasion to find the greatest radius. My first problem is where has 11.819 come from? Additionally, I get: f = 299792458/2 D = 200 ..... r = 13.65 Shouldnt they be the same!?!?!? To try to find an answer I discovered a different page ( clicky ) which simplifies to the same formula but uses a constant value of 8.657. This gives me a radius of 9.999, which is close enough to 10 for me Anyway, question #1  where did the two contants come from, and is wikipedia right? Talking of constants & VHF  I have another question if you dont mind; The internet says ( here ) that the range in NM of VHF is; 2.25 * sqrt( antenna height) So, 2.25.... is that some kind of pythagorean fudge value? Thank you very much for your time. I'm sure it'll be easy once I know the answers Sparks. 
Sch3mat1c 
Posted: August 05, 2014 04:37 am

Forum Addict ++ Group: Moderators Posts: 20,553 Member No.: 73 Joined: July 24, 2002 
I don't have a clue what cutoff point those formulas were derived with. They're probably different, hence the similar (within a factor of 2) yet different coefficients.
Since we're talking fringes of a wave, it doesn't matter very much. It's not like something magically stops interacting with the field, just outside the zone. It's a matter of degree, a few dB here or there. Same thing with the range. You can do straightforward geometric reasoning, but waves don't just stop at the horizon, they diffract a little bit beyond (plus various subtle or dependent effects, like ground waves, multipath, atmospheric scattering and so on). So a more subtle argument has to ask, what attenuation? And even more subtle: what altitude is the receiver at, because that diffraction zone can have interference fringe effects as well, and how much of that is tolerable? Calculating any of these is usually somewhere between tedious and analytically impossible (i.e., no closed form solution, no simple expression with constants; the best you can do is chug it through a computer to get an approximate value). Tim  Answering questions is a tricky subject to practice. Not due to the difficulty of formulating or locating answers, but due to the human inability of asking the right questions; a skill that, were one to possess, would put them in the "answering" category.

Sparks 
Posted: August 05, 2014 09:17 am

Sr. Member Group: Trusted Members Posts: 401 Member No.: 750 Joined: April 08, 2004 
Thank you, I guess it makes sense.
On the other hand, I checked the wiki page history which shows that the fresnel zone constant of 8.675 was added in 2010, and was only changed to 11.819 six days ago! 
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