Hi Bill,
That looks like a piece of cake to me also. I could make my lower run out of
1/2" copper pipe, but the vertical sections and top run would have to be
wire of whatever gauge that would support the current. When is your second
edition Loop book coming out??? Your first one was absolutely an awesome
read.
73's,
Mike>WE0H
http://www.geocities.com/we0h/lf.html
-----Original Message-----
From: [email protected] [mailto:[email protected]]On Behalf
Of Ashlock,William
Sent: Tuesday, May 28, 2002 6:16 PM
To: '[email protected]'
Subject: LF: Loop Amperage ~~~~~~
Hi 136-ers,
Everyone in EU is screaming to the ARRL and the FCC to permit you folks to
use 1W erp. I have
calculated the amperage to your loop to achieve this 1W erp and the figures
frighten me. You said your loop was suspended amongst trees and I would say
with this high amperage I will be able to see your smoke signals before I
hear you signal. Let us have your calculated antenna amperage to see if my
figures compare?
Well let's see...... I presently have ~1w input into the loop and the
current is ~1.5A. If I go to an input of 100w the current will be sq rt 100
x 1.5 = 15A. From a 'wasted power' standpoint the loop is a big 100w wire
wound resistor covering 50'x 50' = 2500 sq ft. Don't see a problem here.
Could even be a good perch for the birds in the winter?
The present ERP with a 1w input is about 1mw (according to the radiation
resistance formula that I like the best). I would therefore have to run 1KW
to produce a 1w ERP and a current of sq rt 1000 = 31.6 X 1.5A = 47.4A .
Well, yes, that frightens me too. BUT..... If I utilize my 450ft wide back
woods the situation improves. The radiation resistance goes up to the square
of the loop area gain to 81X. The ERP for 1w of input is now 81mw. To reach
a 1w ERP the current multiplier will need to be only 1/.081 = 12.3 X 1.5 =
18.5A. The actual input power for this loop will depend on the conductor
used. I calculate the total Rac (including soil loss) would have to be 2.92
ohms minimum for an input power of 1KW, and 18.5A, and this would be
equivalent to a 2.92/9 = .325 ohm Rac with a 50'x50' loop. (So my 4 x #12
Litz conductor or a 1/2" Cu pipe would be required...... not cheep!) So, it
appears that all I am doing, when power limited like this, is to spread out
the 1KW power loss over a larger portion of the back woods! But at least
it's doable.
Comments, please.
Bill A
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