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#1
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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 |
#2
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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 |
#3
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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 |
#4
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