Topband
[Top] [All Lists]

Re: Topband: Term "cancel fields", apology

To: "Guy Olinger K2AV" <olinger@bellsouth.net>
Subject: Re: Topband: Term "cancel fields", apology
From: "Tom W8JI" <w8ji@w8ji.com>
Reply-to: Tom W8JI <w8ji@w8ji.com>
Date: Wed, 1 Aug 2012 20:48:30 -0400
List-post: <topband@contesting.com">mailto:topband@contesting.com>
Hi Guy,

> This is a non-issue, Tom.

I think it is a critical issue, because it demonstrates the difference 
between EM radiation and induction fields that only store and return energy 
to the system.

> It stems from what I meant when I said "cancel fields".
>
> Apparently some considerable number, even the majority, think this
> phrase means cancel fields entirely.  To others this means
> cancellation is taking place at some level.

Then the question arises, how much does it reduce fields?

As far as I know, all we can do is redistribute those fields as evenly as 
possible over the widest area possible in the lossy media to reduce losses.

> The FCP radiates at -30 dBi according to NEC4. That's really low for
> an antenna element. But, it DOES radiate.

But that has little to do with local losses in the soil.

> According to NEC4, using area integration of field squared data from
> the counterpoises only, excluding the vertical radiator, the FCP only
> has 8.2% of the power loss in the dirt as a pair of 1/4 wave raised
> radials.

What do you think of this example using a lossless half wave dipole antenna 
over the same soil:

Dipole with lossless wire:

freespace 100% eff  and 72.3 ohm
45 feet 46% eff and 37.6 ohms
10 feet 6% efficiency and 54 ohms

Comparing an FCP dipole:

Freespace 99.2% eff 0.03 ohms
45 feet 20% eff and 0.04 ohms
10 feet 0.4% 1 ohm

Here is my confusion:

The theory  (correct me if I misunderstood it)  is folding the element 
reduces fields in the earth compared to spread-out elements, and 
concentrating the field in a small area also reduces earth losses. I 
understand the article to say reducing the area exposed to fields reduces 
loss.

If that is the case, why is a full-size lossless-conductor dipole's 
efficiency higher (earth losses lower) than a dipole made from two right 
angle FCP's? Both have equal soils and both are lossless internally, so all 
losses are by fields in the earth.

What I understood, and always found true, is spreading fields out over a 
larger cross section of lossy media reduced losses. This is because current 
and electric field density is less.

Why does a 260 ft wide lossless dipole have less earth loss than a lossless 
compact folded antenna at the same height? What am I missing?

73 Tom


 

_______________________________________________
UR RST IS ... ... ..9 QSB QSB - hw? BK

<Prev in Thread] Current Thread [Next in Thread>