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From: "Talbot Andrew" <ACTALBOT@dera.gov.uk>
To: rsgb_lf_group@blacksheep.org
Subject: LF: RE: Formula
Date: Wed, 7 Jun 2000 11:42:27 +0100
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Walter  et al :- 

Doesn't specifying a dipole as the radiator automatically take antenna
aperture into account, and cancels out frequency from the equation ?  So
if you take just the spreading out of the total power over the area of a
sphere :

Watts per m^2  =  P . / (D ^ 2  x  4.pi)	
and  E ( V/m )  = SQTR( Power x. Resistance of free space )
so   Volts / m   =   SQRT(W/m^2  *   Z free space)   =   SQRT(P x 377 /
4.pi / D^2)

= SQRT(P) * 30 / D  

A dipole has a gain of 1.64 (power) so its "voltage gain" could be said
to be 1.28
so I reckon the formula should read    E  (V/m)  =    38.4 . SQRT(P) / D
(all in SI units of course)

Some way off a value of 7, but in the right region.   Can anyone spot if
I've slipped a digit or something ?

Andy  G4JNT

> ----------
> From: 	Walter Blanchard[SMTP:blanch@pncl.co.uk]
> Reply To: 	rsgb_lf_group@blacksheep.org
> Sent: 	2000-06-07 10:25
> To: 	rsgb_lf_group@blacksheep.org
> Subject: 	LF: Formula
> 
> I found the following in an article recently.
> 
> Quote:
> 
> "The intensity E (known as the field strength) of a transmission
> at a distance D from a source transmitting P watts of RF power via
> a half-wave dipole in a free, unobstructed space, can be estimated
> using the formula:
> 
> E=(7*sqrt (P))/D.
> 
> Thus, for a 2 watt transmitter, the theoretical field strength in free
> space at one metre distance is approximately 10 V/m and at 100 metres
> distance 0.1 V/m"
> 
> This formula is new to me and I can't find it in any
> textbooks I have.  Since it ignores frequency shouldn't
> it be power flux per unit area and not field strength?
> Can anyone tell me where it comes from?
> 
> Walter G3JKV.
> 
> 
> 


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