At 07:15 AM 10/11/2006, Eugene Hertz wrote:
>Ok, this may be the simplest question I will ever ask here so here goes.
>
>During my research for OCF dipoles, I have read about some horror
>stories of dipoles breaking or coming apart due to the stress of
>wind, weight and other factors. Especially I've heard that the
>carolina windoms, with their extra weight in the center caused by
>the matching unit, the isolator, and the length of feedline can
>cause a big sag at the feedpoint. Several folks I've corresponded
>with indicated that they attempted to "lift up" the feedpoint as
>high as possible by putting more tension on the wires (legs) to
>hoist it higher. This was for people that did not have a convenient
>center mounting location like a tower or tree or house.
>
>In thinking about this, it seems that these antennas have lots of
>phyiscal stress on them from both gravity as well as tension on the
>ends of the wires in the horizontal plane to try to counteract gavity.
>
>My thought was this: By using one rope (dacron etc) to connect leg1
>to a tree and another rope to connect to leg2 to a tree (with
>pulleys and weights, of course), this is putting lots of
>"horizontal" stress on these wires and the center matching
>conductor/insulator thingy. This force is always trying to pull
>apart the antenna.
>
>Why not take one very long rope, thread it through the leg1
>insulator, the center insulator/matching unit and then the leg2
>insulator? Surely the two ends of the wires would have to be
>attached somewhat taught to the rope, but this could be done with
>some black wire-ties along the length of each wire. Or tie some
>knots in the rope somehow to hold the insulators in place.
>
>My thought was, this would all but eliminate any horizonatal stress
>normally put onto these wires and we would be left only with the
>stress of gravity vertically. I would imagine too, that this would
>help keep the center up a big better than nothing going through the
>center point.
>
>Any thoughts? Could this work? What am I missing?
>Eugene
Yes, doing what you say would put the load on the messenger cable
(the dacron line). However, the real problem is that if you try and
pull it up too high (reducing the sag) the loads get very high, very
quickly. It's a classic physics problem..
Better to just let the darn thing sag in the middle, and keep the
loads reasonable.
To a first order, the tension for a symmetric arrangement is
(weight/2)/sin(sag angle)
or
(weight/2) / (<sag distance>/<dist from support to weight>)
rearranging a bit:
Tension = weight * <dist from support to weight along wire>/(2*<sag distance>)
As the angle or sag distance gets small, you get towards a something
divided by zero situation.
Say your center insulator and isolator and coax hanging down weighs
something like 10 pounds, and you've got an 80 meter dipole (so the
distance from support to center is 66 ft). If you let it sag 10 ft,
the tension will be 10*66/(2*10) = 33 pounds. If you try to pull it
up so it only sags 1 ft, the tension will be 330 pounds.
Realistically speaking, the RF performance won't suffer all that much
even with sags of 20-30 degrees, which would be 20 odd feet of
sag. The real issue is how close is it getting to the ground. The
"effective height" of a sagging dipole is somewhere between the
feedpoint height and the support heights, but closer to the feedpoint
height than the support heights.
Jim
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