In a message dated 11/4/01 4:40:12 PM Pacific Standard Time, K3BU@aol.com
writes:<<
In a message dated 11/2/01 11:08:56 AM Eastern Standard Time, K3BU@aol.com
writes:
> aa4lr writes:
> > Now, pi/2 wl divided by 50 is pi/100 -- which is 0.0314159..., and
pi/2
> > wl divided by 60 is pi/120 -- which is 0.02618. Close, but not quite
> > 0.025.
> Simple,
> Tom is (always) right, you are using wrong Pi(e :-)
> Yuri
>
Actually it is a little less.
Just to be more precise and not to accumulate errors while lying out
radials
:-)
AA4LR example is valid for the length of the arc with radius of the radial
length, being .25 wavelength and that is 0.02618 wl.
Proper formula for the straight distance A between the tips of the radials
(length R) and number of radials N is
A = 2R sin 180deg/N
so for 60 radials .25 wl. long
A = 2 x 0.25 x sin 3deg
A= 0.5 x 0.052335956
A= 0.026167978 wl.
For those who want to be precise, as they say, hope it helps.
Yuri, K3BU >>
Yuri: I duplicated your figure A=.026167978 WL by creating 2 "right
triangles" with a line at 3 degrees bisecting the line bewteen the .25 WL
tips (for further clarification). Without a right triangle the sine factor
is not correct. Yuri I didn't know you were good on math also. hi
Actually most all this radial math nonsence goes away with 1/2 wave
verticals. With the finances of the average DXer today, I cannot see why any
DXer screws around with 1/4 waves. A 1/2 wave is already resonant and
doesn't need any more help in that department. I've had great success with
1/2 waves with just a ground rod to cool the coax shield feeding an L
network. How much E-field loss is there at the base without any radials?
I'm not sure. The higher the base of a 1/2 wave this problem reduces itself.
I will run a comparison with a side by side 1/2 waves--one with a ground rod
and one over a good radial system. WWVH has some radials out a ways. I
don't think they have to be a 1/4 wave although it sure can't hurt.
Has anyone else any 1/2 wave vertical experience? K7GCO
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