20m vertical dipole
 

Picture 2 fiberglass poles, each 16.5 feet long, mounted
horizontally on top of a 10 foot long wooden pole, with a dipole cut
for 20m strung between them. The coax feeding the dipole is taped to
the wooden pole. Picture this antenna on top of a building, then
position the antenna with the wooden pole parallel with the ground,
extending past the edge of the building, and the fiberglass poles
extended vertically. A vertical dipole! This is what I'm thinking of
putting together for a trip out to a friends lighthouse, and the end of
the lower element should be about 30 feet above the water. Being above
salt water, I'm expecting a takeoff angle of less than 10 degrees, and
hopefully closer to 3-5 (or less) for some DX work.
I've never built an HF antenna before, but I have had success with
a 4-element 2m quad I built years ago. Does this design look like it
will work? Any and all (well, most, at least) comments will be given
their appropriate attention. Thanks.

Dave

20m vertical dipole
 

On 29 Nov 2006 12:42:43 -0800, "Dave" wrote:

Any and all (well, most, at least) comments will be given
their appropriate attention. Thanks.

Hi Dave,

Sounds like a lot of work when gravity can do most of the fiberglass
poles' job. Try a sleeve dipole (vertical, of course):

Take a coax and strip off 16.5 feet of its jacket;
Pull back the coax shield over itself (like taking off a sock from the
top) back to the 16.5 foot mark and keep pulling it another 16.5 feet;
You now have an exposed inner conductor, and the coax shield rolled
over 16.5 feet of jacketed coax (you need at least 33 feet of coax to
do this). Feed this prepared coax with more coax (or simply use a
very long coax with one connector at the transmitter).

This is where gravity comes to work:
Hang the inner/outer prepared coax over the end of the horizontal
support, with the feed tracing along the horizontal support.

Same amount of wire, no need for the fiberglass and even if you may be
closer to the waves, it hardly matters (unless you have 20 foot surf
conditions).

73's
Richard Clark, KB7QHC

20m vertical dipole
 

Richard,

You forgot to tell him about trimming the folded back coax shield in
order to account for the reduction in velocity factor. The folded back
shield and the coax outer shield have a dielectric constant between
them that is based on the cable jacket material and a minimal air gap.

You covered this quite well in an earlier post you made:

http://www.archivum.info/rec.radio.a.../msg00373.html

The original posters idea was easy to build and will work quite well
horizontally or vertically. He should put an RF choke (coax coil or
ferrite beads) near the feedpoint in order to reduce the RF flowing on
the outside shield of the cable.

Roger





Richard Clark wrote:
On 29 Nov 2006 12:42:43 -0800, "Dave" wrote:

Any and all (well, most, at least) comments will be given
their appropriate attention. Thanks.

Hi Dave,

Sounds like a lot of work when gravity can do most of the fiberglass
poles' job. Try a sleeve dipole (vertical, of course):

Take a coax and strip off 16.5 feet of its jacket;
Pull back the coax shield over itself (like taking off a sock from the
top) back to the 16.5 foot mark and keep pulling it another 16.5 feet;
You now have an exposed inner conductor, and the coax shield rolled
over 16.5 feet of jacketed coax (you need at least 33 feet of coax to
do this). Feed this prepared coax with more coax (or simply use a
very long coax with one connector at the transmitter).

This is where gravity comes to work:
Hang the inner/outer prepared coax over the end of the horizontal
support, with the feed tracing along the horizontal support.

Same amount of wire, no need for the fiberglass and even if you may be
closer to the waves, it hardly matters (unless you have 20 foot surf
conditions).

73's
Richard Clark, KB7QHC

20m vertical dipole
 

On 29 Nov 2006 14:17:29 -0800, wrote:

You forgot to tell him about trimming the folded back coax shield in
order to account for the reduction in velocity factor. The folded back
shield and the coax outer shield have a dielectric constant between
them that is based on the cable jacket material and a minimal air gap.

You covered this quite well in an earlier post you made:

Hi Roger,

I took a quick glance at your link, but I don't think I paid much
attention to velocity factor (given all the variables). For all of a
life of precision I've been devoted to, I can summarily toss off those
issues (when data has already pounded me down with the weight of
imponderables).

One way to approach this is to take SWR readings with the progressive
rolling back of the shield over the lower jacketing. If someone could
do this with a stepper motor, then we would have a new antenna design
(new in the sense that it's new to/from me, and probably anticipated
looooong ago).

73's
Richard Clark, KB7QHC