Re: Length-in-feet Formula

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From: Stephen Modena (ab4el@cybernetics.net)
Date: Tue Mar 14 1995 - 07:39:41 EST


I received this...which makes me think I should have worded my
posting better.
 
> Steve, I saw your post concerning using a 40M dipole for a 2M
> antenna. Just have one question, what is the "N" in the formula,
> and where did you find this referenced. I am always searching for
> additional antenna ideas and would like to understand this concept.
>
> ------- Length-in-feet = ( 0.95 + N ) * 492 / Freq-in-MHz --------
>
> 73 KC5DVT Wayne email...ehayes@vnet.ibm.com

The reference is "The ARRL Antenna Book" ...harmonic antennas. Or look
in the ARRL Handbook.

Allow me to correct my formula ("formula" means an empirical calculation
method).

        Length(feet) = (N-0.05) * 492 / Freq-in-MHz

This is the formula for "harmonic" antennas. Now I will illustrate the
role of "N". N is the harmonic number.

I want to make an antenna for 40 M...specifically for 7.00 MHz. Because
this is the fundamental frequency (1st harmonic), N=1.

        L = (1 - 0.05) * 492 / 7.00 (because N=1)
        L = 0.95 * 492 / 7.00
        L = 468 / 7.00 (Is "468" familiar?)
        L = 66.86 ft

I cut the wire and erect the dipole.

I decide to use this 40 M dipole on 15 M because at 3-halfwaves operation
it's feed point will be a current null. Let's calculate the resonant freq
of this 40 M dipole. N=3 because this is 3rd harmonic. (7 x 3 = 21)

        L = (3 - 0.05) * 492 / Freq (because N=3, 7 MHz * 3 = 21 MHz)
        Freq = (3 - 0.05) * 492 / Length (did you follow the rearrangement?)
        Freq = 2.95 * 492 / 66.86 (wire antenna already exists)
        Freq = 21.71 MHz

Ooops...the 40 M dipole appears to be "too short." :^) Not to worry
if you are using a "good" TransMatch.

Let's see how long a three halfwaves antenna would be on 21 MHz.

        L = 2.95 * 492 / 21.00 = 69.11 ft

And what is the "40 M" resonant frequency of that antenna?

        69.11 = 0.95 * 492 / Length
        L = 468/69.11 = 6.76 MHz

Will it help to make the original antenna shorter on 40 M? Let cut one
for the other end of the band: 7.3 MHz.

        L = (1-0.05) * 492 / 7.3 = 64.03 ft
        Freq = (3-0.05) * 492 / 64.03 = 22.67 MHz

Twice we have proved that when we intend to operate a 40 M antenna
intensively on both 40 M and 15 M, we perhaps should make a compromise
is the direction of cutting to the low end of 40 M...or even below.

Why?

Because I use a MFJ cheapy "fix-taps" TransMatch...and it is problematic
to tame the "SWR" on 15 M, the shorter the antenna is on 40 M.

If you have a *nice* roller inductor-type TransMatch, it really doesn't
matter what you cut your antenna to...you will always be able to find
a setting to achive 1:1 "SWR". :^)

Hint: I put "SWR" in quotes, so no flame wars please. ;^) A transmatch
device TRANSforms a presented impedance to achieve a MATCH....which
we usually determine via an SWR meter at the transmitter coaxial plug.

Now, does one need to be concerned with whether the
*particular* 40 M antenna is actually even-harmonic or odd-harmonic
length on 2 M? I think not. I believe *other* factors come into
dominant play and make it moot.

--
73/Steve/AB4EL ab4el@Cybernetics.NET in Raleigh, NC  35.81245N, 78.65849W


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