Fiberglass Flagpoles for Antenna Masts

Discussion in 'Technical' started by KF7JAF, Jul 1, 2014.

  1. KF7JAF

    KF7JAF Member

    Has anyone ever used a fiberglass flagpole for an antenna support? I found a supplier of fifty foot flagpoles and thought when I get the funds together I could put up two, one at each end of the small suburban lot I live on, and string a doublet between them. Has anyone tried this?
  2. W5HRO

    W5HRO Administrator

    That's actually not a bad idea, but it can be pricey. To get some heavy enough to support a wire in-between they run around $250 each minimum for 20' and around $350 each for 30' and anywhere from $500 to $1000+ each for 40' so I guess it depends on how high you want them to be.
  3. WD5JKO

    WD5JKO Member

    Realize too that the poles will need to be guyed on at least two levels. Putting the pole on the property line pretty much says at least one guy, and likely two guys will go over into adjacent property. One thought is to put the poles 10' within two diagonally opposite property line corners. The rather steep guy line angle will have more "down" force than sideway force, so when tightened, the fiberglass pole will bow. To minimize this, guys at more levels might be needed.

    Continuing with the w5pyt antenna stories, One of Bob's last major efforts was to build a multi element array for 80m. He put up eight 50' masts using dual staggered 2 X 4's bolted together and hinged from the bottom. The details are foggy in my mind now, but the topic came up about guying, how many guys, and how many levels. Before he got the antennas up and setup, a storm came by and demolished all the towers. Bob's final 80M antenna was a diagonal turnstile with the top-most point up about 300'. That antenna "Rocked".

  4. W5HRO

    W5HRO Administrator

    Not sure if you would need guy wires. If you buy the flag pole bases they should be self supporting. The only issue might be if they bowed toward each other a lttle due to the antenna wire being stung between them. It would depend on the amout of flexing if any.
  5. W5RKL

    W5RKL Member

    Why not use only one pole, centered in the back of the lot, to hold the center of an inverted vee? The antenna ends could serve as "guys" to keep the pole upright.

    Mike W5RKL
  6. W5HRO

    W5HRO Administrator

    It's because most of us have a bad habit of always overthinking things :icon_crazy:

    Anyway, that's actually a good idea. He did say he wanted a doublet, but there is no reason why it couldn't be in an inverted Vee configuration and it would solve most of the gain problem compared to using inverted Vees fed with coax. It would give him the best of both worlds for both short and long distance coverage.

  7. KF7JAF

    KF7JAF Member

    there's no good place to guy it to in the front yard. The two flagpoles would allow an 80 meter doublet with drooping ends, don't think an inverted vee would fit for 80.
  8. W5HRO

    W5HRO Administrator

    Flat horizontal doublet (aka Center Fed Zepp) at 3.600Mc, 130' total, 65' each leg.

    Inverted Vee with a 22 degree apex angle at 3.600Mc, 127.4' total, 63.7' each leg.


    Inverted Vee with a 45 degree apex angle at 3.600Mc, 123.5' total, 61.75' each leg.

    I think what he was saying is install the flagpole in the center of your back yard. Inverted Vees are a little shorter than horizontal ones. With a 45 degree apex angle the total length would only be 123.5' and you could always make it even shorter by cutting it for 3.700Mc or for 3.800Mc instead. At 3.800Mc the total length would only be 117' with each leg at 58.5'. That still might be too long for your property and only you would know that for certain, but if the flagpole is 50' tall then it would probably fit easily using a 45 degree apex angle. Is your back yard at least 45' to 47' wide? That's what it comes out to using the other 2 angles which would be around 67.5 degrees each. 45 + 67.5 + 67.5 = 180 degrees. The total length (width) of the base would only be somewhere around 44.8'. Probably just a little more after compensating for everything else.

    P.S. If you were to put up an Inverted Vee make the OWL length exactly the same as the total antenna length. That would be 117' for 3.800Mc. Halfwave dipoles fed with OWL are doublets, but when fed with coax they are not.
  9. KM1H

    KM1H Guest

    The simple approach is to sink a steel pipe tall enough so that the guy will be attached at a correct angle. The 150' Rohn 55 commercial tower across the street uses 12' H beams above ground level to support all 3 guy positions.

    A ham friend has a pair of 100' towers on a 60x100 suburban lot in MA, he uses galvanized pipes filled with concrete as support and they have been fine for over 30 years.

    Used TV antenna or CB vertical towers still show up in the 40-50' range in good condition in the boonies and surburbs. These are about half the width of Rohn 25.

  10. W5HRO

    W5HRO Administrator

    He's going to use a fiberglass flagpole or two. If you buy the base pieces for those and have them professionally installed you really don't need any guy wires. Now, that's only with the load of a flag of course, but if he is only supporting wire it shouldn't be that critical. The inverted Vee would be the best option if he had the room for it. As long as he didn't pull the wires really tight and left some slack it shouldn't be a problem. The other good thing since its fiberglass is he could just tape the OWL (like heavy 450-ohm ladder line) down the pole. You cant do that with metal telescopic masts or towers.

    My whole back yard is pool so I don't have a place to install one. I wish I did because a 75-meter inverted Vee would work great and give me that long distance coverage back to the Midwest and East Coast regions. The trick is to make sure the maximum voltage is at the right place on the wire. Usually inverted Vee's place it at the ends close to the ground with the max current at the feed-point which is bad. Making the OWL length the same half-wave length as the total length of the antenna instead of only a quarter-wave length will help solve that.