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R: R: phase shift with coax




----- Original Message -----
From: i8cvs <domenico.i8cvs@tin.it>
To: Stan <stan@capeonramp.com>
Sent: Thursday, July 12, 2001 4:47 AM
Subject: R: R: [amsat-bb] phase shift with coax


>
> ----- Original Message -----
> From: Stan <stan@capeonramp.com>
> To: i8cvs <domenico.i8cvs@tin.it>
> Sent: Wednesday, July 11, 2001 9:32 PM
> Subject: Re: R: [amsat-bb] phase shift with coax
>
>
> > Hello Dom,
> >
> > The question of the day:
> >
> > if the far field boundary is at (2D**2)/lambda,
> >
> > is D the length of the boom for a single yagi antenna ?
> >
> > And is it four times for 4 identical yagis of a 2 x 2 array ?
> >
> > Most of my calculations have been for a dish where D is clearly
> > the diameter.
> >
> > Stan, WA1ECF
> >
>
> Hi  Stan,
>
> The area of  far-field begins at a distance R from the antenna greater
than
> (2D ^2) / lambda
> where D is the diameter in meters of the antenna aperture or the antenna
> capture area and so the far field boundary is directly proportional to the
> antenna gain.
>
> In other words,greater the antenna gain,greater is D and greater is the
> distance R from the antenna at wich the far field begins.
>
> In case of a parabolic dish D is the diameter of the dish,or the diameter
of
> his mouth in meters but in case of a yagi or any other type of  antenna,D
is
> the diameter of his capture area A that can be computed using the
following
> formula.
> A= (G x lambda ^2 )/ 4 x 3,14    [ square meters]
>
> We can assume that the antenna capture area A is a circle wich area in
> square meters increases with the antenna gain G
>
> By geometry from the circle area A we can derive its radius
> r= SQR ( A/3,14 )  [meters]
> and so the diameter of the area is D=2 x r   [meters]
>
> As an example we can use a 2 meters array made by 4  yagi  in wich the
gain
> of a single yagi is 10 dBi
> The gain of two yagi is than in theory  13 dBi and  the gain for 4 yagi is
> 16 dBi or 40 time in power over the isotropic antenna in theory.
>
> The capture area of this array is:
>
> A= ( 40 x 2 ^2 )/ 4 x 3,14 = 12,73    [square meters]
>
> from wich D= 4   meters
>
> If we call R the distance from the last director of the array at wich the
> far field begins,than
>
> R= ( 2 x D^2 )/ lambda  = (2 x 4^2 )/ 2 = 16 meters
>
> Conclusion:
>
> Greater the antenna gain and greater is the distance from the antenna at
> wich the transiction from the near field and the far field begins.
>
> 73 de i8CVS Domenico
>
>
> >
> > > > Via the amsat-bb mailing list at AMSAT.ORG courtesy of AMSAT-NA.
> > > > To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org
> >


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