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Message-ID: <3.0.1.16.19990624180651.2f0fc7fc@mail.cc.kuleuven.ac.be>
Date: Thu, 24 Jun 1999 18:06:51
To: rsgb_lf_group@blacksheep.org
From: "Rik Strobbe" <rik.strobbe@fys.kuleuven.ac.be>
Subject: Re: LF: ERP Calculations
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At 14:53 24/06/99 +0100, you wrote:
>Whilst browsing the fascinating information, I came across a really 
>easy way of calculating ERP - at least I thought I had.
>
>The book gives:
>
>Radiation resistance = 160 x pi squared x antenna height squared, 
>all divided by wavelength squared (height and wavelength in same 
>units).
>
>By multiplying this by the square of your antenna current you have 
>the ERP - simple.
>
>BUT
>
>The Admiralty Handbook - and many derivatives - uses the factor 
>160 at the start of the formula. Many other books (and we have a 
>very large collection in the RSGB Library) including the definitive 
>Terman, give 60 instead. Now this is almost three times less!!!
>
>Which is right - or have I missed something vital?

Pure mathematical the radiation resistance of a short (in wavelength)
vertical monopole above a perfect ground is :

   Ra = 40 . Pi^2 . l^2 / L^2

where Pi = 3.1415... , l = antenna-length , L = wavelength and ^2 means
squared.

simplified for 136.75kHz this means that Ra (milli-Ohm) = 0.082 x l(meter)

The same vertical with a infinite top-capacitance has a radiation
resistance of

   Ra  = 160 . Pi^2 . l^2 / L^2

so 4 times the radiation resistance of the same vertical without tophat and
with the same antennacurrent it will have a 6dB higher ERP.

Any 'real-world' vertical with tophead wil have a radiation resistance
somewhere inbetween.

An easy approach to understand this increase of radiation resistance due to
the tophat is to look at the current distribution over the antenna :
- For a 'pure' vertical the current at the feedingpoint is maximum and
lineary decreases to 0 at the top, so the average current is 0.5 times the
feeding-current.
- For a infinite tophat the current all over the antenna will be constant
and equal to the feeding-current, so also the average current will be equal
to the feeding current (so it is double compared to a pure vertical).
Double current means quadruple power, so 6dB gain.

A 'quick and dirty' method to guestimate the radiation resistance of a
topheaded vertical is to try to 'reconstruct' (or measure) the current
distribution along the antenna.
With an inverted-L antenna (with single topload wire) this is rather easy,
assuming a linear current decrease from feeding point to end (where the
current is 0). That way you can determine the current at the top of the
vertical section and so the average current in this vertical section.
An example :
Assume we have an inverted-L antenna of 10m height and 30m topwire. Total
antenna-length is 30m, so the current at the top will be 75% of the feeding
current and the average current will be 0.875 times the feeding current.
A pure 10m vertical (no tophat) would have an average current of 0.5 times
the feeding current and a radiation resistance of 8.2 milliOhm. The
topheaded vertical has a 1.75 times higher average current (in the vertical
section) and so the radiation resistance will be 1.75^2 = 3.06 times higher
(= 25 milliOhm).

With more complex tophats calculations are less 'straightforward' bot not
so much harder if we assume that the voltage over the antenna is constant
and as a result of this via each 'pF' of antenna capacitance the same
amount of current 'disapears'. The capacitance of the complete antenna can
be measure in various ways and the capacitance of the vertical section can
be estimated as 6pF per meter.
An example :
Assume we have a 10m high vertical with a number of tophat wires, the total
capacitance of the antenna is measured as 300pF. Taking 6pF/m the 10m
vertical section will have a capacitance of 60pF, the remaining 240pF is in
the tophat.
Since the 60pF of the vertical section is 20% of the total antenna
capacitance also 20% of the feeding current will 'disapear' in the vertical
section and so the current at the top of the vertical section will be 0.8
times the feeding current and the average current will be 0.9 times the
feeding current.
This is 1.8 times the average current that a 10m pure vertical would have,
so the radiation resistance of this antenna will be 1.8^2 . 8.2 = 26.6
milliOhm.

So far my 'view' on this item. But I have some other question on the same
topic, I will put in in another mail to makes it not too complicated.

73, Rik


Rik Strobbe  ON7YD
rik.strobbe@fys.kuleuven.ac.be
Villadreef 14  B-3128 Baal  BELGIUM   (JO20IX)