On 5/23/2013 7:59 PM, David Ryeburn wrote:
In article , John S
wrote:

EZNEC says the feed-point impedance at the bottom "V" is:

80M 1639 - J 11890 ohms
40M 128.9 + J 0.4186 ohms (design center)
30M 3364 - J 409.1 ohms
20M 222.6 + J 171.4 ohms
15M 119.5 + J 301.5 ohms
10M 401.1 + J 412.7 ohms

Not including the effects of tree proximity, of course.

Thanks for doing this. Many Windows programs will run under CrossOver (a
Mac implementation of WINE) on my Mac, but unfortunately EZNEC is not
one of them.

When you say bottom "V" do you mean you are you feeding the thing down
where the two sloping legs come together near ground level? That wasn't
my plan; I would have an insulator up in the air halfway between the two
high tree branches, and attach a balun and the coax feed line at that
point. The low point at the back, where the two sloping legs meet, would
simply have them joined together there (or not, on 80 and 30) and I
wouldn't feed it there.

What frequencies in each of these bands did you use? I'm a bit surprised
at the relatively high reactance on 20, 15 and 10. I'd want to cut the
thing for the CW ends of the bands. Unless you cut the antenna right at
the bottom end of the band, the harmonic resonances march up into the
bands as you move to the higher frequency bands, so I'd expect the
antenna to be too short at the CW ends of the bands on the higher bands.
E.g. if design frequency is 7.05 MHz then we're looking at 14.1 MHz,
21.15 MHz, and 28.2 MHz whereas I'm going to want to operate around
14.05, 21.05, and 28.05.

Also the high resistance and high capacitive reacance for your figures
on 80 and 20 makes me suspect that your analysis is with the loop closed
on those bands. My idea was to walk around to the back and unclip the
jumper there, so that instead of a loop, on 80 and on 30 it would be a
doublet all bent into a sloping triangle, somwhat more than a half wave
long on 80 and a bit les than 3/2 waves long on 30. I'm taking into
account end effect only at the "ends", where the two sloping wires are
now detached from each other, on 30. So the 5% shortening would only
apply to the last quarter wave on each side (ending at that low-height
insulator), not to the two half wave sections either side of the center
insulator, and since 492*2 + 468 = 1452, I'd expect the resonant length
at 10.125 MHz to be 1452/10.125 = 143.4 ft, while 1005/7.05 = 142.55 ft
would be the calculated length for a full-wave 40 m loop. So I'd expect
relatively low resistance at the feedpoint on 80 and 40, with reactance
moderately high and inductive on 80 (where it's a good bit too long for
a half-wave) and pretty low and slightly capacitive on 30 (where it's
just a bit too short for 3/2 wave).

If you ran the analysis with the loop closed on those two bands, would
you mind doing it over with the loop open?

7.05MHz 117.3 - J 0.2996 ohms closed loop
3.525MHz 9.365 + J 111.4 ohms open loop
10.125MHz 253.9 + J 60.16 ohms open loop