[TowerTalk] Good vacation antenna

[Wes]

Then the essence of this of this is what effect does a base loading inductor 
have on the current profile on the radiator.  My answer is none.

It's too bad we can't have attachments here so I'll try to describe it with 
words.  I "built" a model of a vertical in AutoEZ and used EZNEC Pro/2+ V 7.0 
running the NEC 5 engine to do the computations.  I used 7.25 MHz with a height 
of 1/8 wavelength at that frequency (16.55 feet), diameter of 0.1 inch and 33 
segments.  I used perfect ground and conductor loss of zero.  The current source 
was 1.0A.

The feedpoint Z without any loading is 6.15 -j444.6 (SWR = 651). Using the 
current plotting feature of AutoEZ, the current profile v. segment is a nearly 
linear, negative slope line from (almost) 1.0A down to (almost) zero.  (Almost 
because current is reported along the length of a segment, not at the ends)  I 
used the snapshot feature to capture this plot.

Next I added a series inductor at the feedpoint of 1.48 +j444.6 (Q=300).  The 
calculated feedpoint Z=7.64 j0 (SWR = 6.55).  The current profile is nearly 
identical as before.  The snapshot completely overlays the newly calculated plot.

My conclusion: A non-radiating, high Q inductor at the feedpoint of an 
electrically short monopole affects the feedpoint Z but has no effect on the 
current distribution on the radiator, which is determined entirely by its 
geometry.  A load placed anywhere on the radiator other than the feedpoint is a 
different story.

Wes  N7WS



On 7/3/2025 4:19 PM, David Gilbert wrote:
>
> The loss is the same (probably), but the current profile at the antenna is 
> different (definitely).
>
> Dave AB7E
>
>
> On 7/3/2025 4:10 PM, Wes wrote:
>> I'll confess some difficulty (probably my failing) in understanding your 
>> point(s).  To try and simplify the situation to something my pea brain may 
>> understand, let's use a toroid at either location. Now tell me what the 
>> difference is.
>>
>> Wes  N7WS
>>
>>