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Re: [TowerTalk] 80m vertical size calculation

To: StellarCAT <rxdesign@ssvecnet.com>, "V. Sciucka" <vytenis.sciucka@gmail.com>
Subject: Re: [TowerTalk] 80m vertical size calculation
From: <w5prchuck@gmail.com>
Date: Wed, 13 Sep 2017 14:39:30 -0500
List-post: <mailto:towertalk@contesting.com>
I agree.  But, what was wanted is a formula where only one adjustment was 
needed. I think that will only work when the entire element is one diameter 
such as a wire.  The ratio will get close, but will probably need at least one 
more “tweak.”

Chuck W5PR

Sent from Mail for Windows 10

From: StellarCAT
Sent: Wednesday, September 13, 2017 2:24 PM
To: V. Sciucka; Chuck Dietz
Cc: Charles Morrison; towertalk@contesting.com
Subject: Re: [TowerTalk] 80m vertical size calculation

these are only starting points! Local conditions - the antenna itself, the 
surroundings, the ground, etc etc will all conspire to move that value up or 
down slightly. KNOWING what your particular antenna is in place - that is 
the perfect starting point.

g.



-----Original Message----- 
From: V. Sciucka
Sent: Wednesday, September 13, 2017 2:46 PM
To: Chuck Dietz
Cc: Charles Morrison ; towertalk@contesting.com
Subject: Re: [TowerTalk] 80m vertical size calculation

I just thought that it might explain why different formulas are used.
Charlie gave me formula which is widely available: 246/f(mhz) =  element
(feet), but I also found https://www.dxengineering.com/
techarticles/verticalantennainfo/dx-engineering-comtek-verti
cals-for-phased-arrays (see p.5) where 234/f(mhz) is used.


--------------------
Vytenis


2017-09-13 21:33 GMT+03:00 Chuck Dietz <w5prchuck@gmail.com>:

> Maybe I don't understand, but I would think that if you had a tower with
> an aluminum tube "stinger" on top, this might not work because of the
> change in percentage of the various diameters.
>
> Chuck W5PR
>
> On Wed, Sep 13, 2017 at 1:00 PM Charles Morrison <junkcmp@gmail.com>
> wrote:
>
>> Not length to diameter, It is independent of diameter.
>>
>> It is a simple method to determine a difference of length as a ratio 
>> based
>> on frequency.
>>
>>
>> On Wed, Sep 13, 2017 at 11:41 AM, V. Sciucka <vytenis.sciucka@gmail.com>
>> wrote:
>>
>> > Thanks Gary K9RX and Charlie N1RR, 2nd question is clear now.
>> > Charlie also gave formula for length 246/f(mhz) =  element (feet) which
>> I
>> > assume includes length to diameter ratio or this ratio is not so much
>> > important.
>> >
>> > --------------------
>> > Vytenis
>> > _______________________________________________
>> >
>> >
>> >
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>


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