Ken Bessler wrote:
I just put up an inverted V for 30 meters.
I started out with each leg being 24'0". This
gave me a low SWR at 9.5665 mhz which
works out to 229.6 instead of the usual 234/F.

As I trimmed, I decided to keep track of how
much I trimmed and what the nnn/F number
would be. As I got closer to my goal of 10.15,
the number went down, eventually ending up
at 227.28/10.1955=22.292' Also, the 2:1 swr
bandwidth went up - it started at 567 kc and
ended up at 655 kc.

Either way, I got the antenna up and it's working
fine - I'm just curious why the formula for length
and the bandwidth changed as the antenna got
shorter.

Ken KG0WX


234/f is just a starting point.
Dave WD9BDZ

David G. Nagel wrote:
Ken Bessler wrote:
I just put up an inverted V for 30 meters.
I started out with each leg being 24'0". This
gave me a low SWR at 9.5665 mhz which
works out to 229.6 instead of the usual 234/F.
As I trimmed, I decided to keep track of how
much I trimmed and what the nnn/F number
would be. As I got closer to my goal of 10.15,
the number went down, eventually ending up
at 227.28/10.1955=22.292' Also, the 2:1 swr
bandwidth went up - it started at 567 kc and
ended up at 655 kc.
Either way, I got the antenna up and it's working
fine - I'm just curious why the formula for length
and the bandwidth changed as the antenna got
shorter.
Ken KG0WX

234/f is just a starting point.

The "starting point" in question was a low 40m dipole, strung in the
back alley of the old ARRL HQ building.

For any other antenna, anywhere else in the universe, the "magic number
234" is going to be slightly different.

The difference in SWR bandwidth between 2:1 points is a bit more
complicated, and probably can't be explained in a one-liner. It will be
mostly determined by the interplay between two factors:
1. What the resonant impedance is (in relation to 50 ohms), which
determines the minimum SWR.
2. How quickly the reactive part of the feedpoint impedance changes with
frequency, for different dipole lengths.



--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

You've gotten a number of good answers, but maybe I can still add a
little helpful information.

The resonant length and the bandwidth of an antenna are determined by
some basic electromagnetic principles. Although simple in concept, the
exact solution for the impedance (and therefore the resonant frequency
and bandwidth) of even an elementary dipole is actually very complex.
The most common method involves solution of a triple integral equation,
which can't be done directly at all, but requires a computer to
numerically approximate the result.(*) The formulas you see in handbooks
are just a rough approximation that's more-or-less good over a limited
range of conditions. The actual resonant frequency and bandwidth are
affected by wire diameter, height above ground, and angle between the
wires, as well as just the wire length. And the relationships aren't
really simple at all.

So the bottom line is that the formulas work well enough to get you into
the ballpark, from which you've usually got to do some trimming -- just
as you did. You can't expect more than that from them.

Readily available, inexpensive or free, computer programs can do the
complex calculations from fundamental electromagnetic principles with
rather astounding accuracy, in a small fraction of a second for a simple
antenna. The computed results can still differ from reality, though, due
to differences between the model antenna and the real one, like nearby
objects or wire insulation not included in the model, wire sag,
capacitance of end insulators, common mode feedline current, and so
forth. But they'll still get you much closer than the simple handbook
formulas. However, the simple formulas and a bit of cut and try are
perfectly adequate for many simple antennas, and might easily be faster
in the long run for someone not familiar with the programs.

(*) Before the ready availability of computers, many different methods
were devised to approximate the solution, with varying degrees of
complexity and accuracy.

Roy Lewallen, W7EL

Ken Bessler wrote:
I just put up an inverted V for 30 meters.
I started out with each leg being 24'0". This
gave me a low SWR at 9.5665 mhz which
works out to 229.6 instead of the usual 234/F.

As I trimmed, I decided to keep track of how
much I trimmed and what the nnn/F number
would be. As I got closer to my goal of 10.15,
the number went down, eventually ending up
at 227.28/10.1955=22.292' Also, the 2:1 swr
bandwidth went up - it started at 567 kc and
ended up at 655 kc.

Either way, I got the antenna up and it's working
fine - I'm just curious why the formula for length
and the bandwidth changed as the antenna got
shorter.

Ken KG0WX

I'm just curious why the formula for length
and the bandwidth changed as the antenna got
shorter.

Ken KG0WX

===============================

The resonant length of an antenna depends on -

Length of wire.

But also to a far smaller extent on -

Increasing conductor diameter.
Increasing conductor insulation thickness.
Reducing height above ground.
Close proximity to antenna supports, buildings, trees and indoors.

All of which can reduce the resonant length to very slightly less than the
theoretical maximum value of 150/MHz metres for a halfwave dipole.

Or 492/MHz feet, pruned by a very few percent.

Behaviour versus the included angle of an inverted-V is slightly peculiar.
As the included angle approaches zero (which nobody ever uses) the resonant
length approaches that of an ordinary open-wire transmission line, 150/MHz
again.

"Very slightly less" is of the order of 1 or 2 or 3 percent unless you have
a VERY low antenna. Everybody's antenna is slightly different. Just keep a
pair of pruning shears handy. Or bend the wire back on itself.
----
Reg, G4FGQ

Ken, KG0WX wrote:
"I`m just curious why the formula for length and the bandwidth changed
as the antenna got shorter."

In an inverted V, the capacitance effect at its high-voltage ends is
enhanced by their nearness to earth.

Best regards, Richard Harrison, KB5WZI

On Thu, 2 Dec 2004 13:12:15 -0600, (Richard
Harrison) wrote:
In an inverted V, the capacitance effect at its high-voltage ends is
enhanced by their nearness to earth.

Hi Richard,

Then that is dashed by his having trimmed the ends (presumably
everything at the same height, and as such in a direction away from
earth) and the multiplier shrank more. More likely the capacitance
between the ends grew instead (corresponds to all factors observed).

73's
Richard Clark, KB7QHC

"Richard Clark" wrote in message
...
On Thu, 2 Dec 2004 13:12:15 -0600, (Richard
Harrison) wrote:
In an inverted V, the capacitance effect at its high-voltage ends is
enhanced by their nearness to earth.

Hi Richard,

Then that is dashed by his having trimmed the ends (presumably
everything at the same height, and as such in a direction away from
earth) and the multiplier shrank more. More likely the capacitance
between the ends grew instead (corresponds to all factors observed).

73's
Richard Clark, KB7QHC

Now THAT makes sense! I was worried that at the rate
I observed the nnn/f formula decline, your average 10m
dipole would be stupidly short!

And, for the record, I never touched the height of the feedpoint
or the angle of the legs. Instead of tying to the tips of the legs,
I attached the guy lines (heavy duty natural twine) about 8" up
from the ends. That way, I could trim the ends without having
to redo the supports. I wrapped the twine around the ends to
keep everything straight.

So, only 2 things changed - the height of the ends (10') and the
length of the legs.

Thanks everyone for the ideas.....

Ken KG0WX