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From: "hamilton mal" <g3kev.ham@virgin.net>
To: <rsgb_lf_group@blacksheep.org>
References: <000701c79ef8$58d0b7a0$42e8fc3e@g3kev> <000701c79f80$2a56a460$c4227ad5@w4o8m9>
Date: Sat, 26 May 2007 15:57:21 -0000
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Subject: LF: Re: Re: 500 khz ERP
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Jim.
I did say to keep the maths SIMPLE and as a rough guide to make a point. I
was not discussing a specific case nor preparing a white paper for OFCOM.
de Mal/G3KEV

----- Original Message -----
From: "James Moritz" <james.moritz@btopenworld.com>
To: <rsgb_lf_group@blacksheep.org>
Sent: Saturday, May 26, 2007 10:25 AM
Subject: LF: Re: 500 khz ERP


> Dear Mal, LF Group,
>
>
> > To justify my ERP argument and keep the maths simple, look at examples
> > below.
> >
> > 1. 10W rf output to the antenna and an efficiency of the average antenna
> 1%
> > the ERP would be 0.1W erp...
>
> The maths is not quite as simple as this - to calculate ERP in this way,
the
> directivity of the antenna compared to the reference 'half-wave dipole in
> free space' must also be included:
>
> ERP = (TX power) x (Antenna Efficiency) x (Directivity)
>
> Assuming the antenna is a vertical over a ground plane that is a small
> fraction of a wavelength long (or a small loop over a ground plane), its
> directivity can be assumed to be 2.6dB over a dipole, i.e. a factor of
1.8.
> So the ERP in your example would be 0.18W ERP. It is important to bear
this
> in mind in correspondence with Ofcom!
>
> Cheers, Jim Moritz
> 73 de M0BMU
>
>