On May 11, 2:26*am, wrote:
Hi Tom,

I should have been more explicit.

I took the "Axial Propagation Factor" (4.372 rad/m) figure which was
given by the HamWaves calculator and multiplied it by the coil length
(155mm) to find the effective electrical length of the coil (38.83
degrees). Then I took cos(38.83)=0.779 as the fall-off in current
across the coil.

73,
Steve G3TXQ

Hi Steve,

OK, so I suppose you are assuming that the current distribution will
follow a cosine along electrical degrees of your antenna, with a
maximum at the base/feedpoint. If that's the case, then would you not
account for the bottom 10 feet of wire, about 20.5 electrical
degrees? If I do that and assume 1 amp at the feedpoint, I should see
about .9367 amps at 20.5 degrees and 0.5101 amps at (20.5+38.83)
electrical degrees. 0.5101/.9367 would then be the ratio of currents
between the ends of the coil, and that's 0.5446, only a 45.54 percent
fall-off.

In fact, it seems to me that the idea of cos(38.83 degrees) = .779
would imply a fall-off of 22.1%... and that tells me that perhaps I'm
still not understanding your model very well. Maybe you are NOT
assuming the current along the electrical degrees of the antenna, up
from the feedpoint, will have a cosine distribution. At this point, I
have to say that I'm just not at all sure what your model really is.
Perhaps you are making different assumptions about the current
distribution...

Also, if you still have the model around, try adding a top hat to the
upper wire. For simplicity, you can just use a simple "T" structure,
where the top horizontal wire is, say, five feet long total. With
such a configuration, what's the current distribution along the
radiating element going to be?

Of course, what I'm suggesting here is that one must be careful to
test ones models at corner cases before putting too much faith in
them, and even then, one must always be wary of cases where the model
may go awry.

Cheers,
Tom