Of course, Bart's math is fine. Seems like there's a bit more to the
story. The current may be less because the actual antenna impedance
may be a bit higher than 50 ohms, but 50 is probably a reasonable
figure for calculations, to be on the safe side. And Chuck's right
that the heating may be much less because of the mode of operation.

But...what current would be too much? The RF resistance of the 26AWG
wire at 4MHz is about 125 ohms/1000 feet, or about 3.06 times the DC
resistance, at 20C. The DC current to get the same heating as a 4MHz
RF current would then be about sqrt(3.06)=1.75 times the RF
current--close to 8 amps, using Bart's calculated RF current. (It's
actually a bit less as the wire gets hot and its resistance goes up,
because the skin depth increases with increased resistance, but use
1.75 times for a worst-case.) Since the fusing current (current at
which a wire in nom. 20C ambient will melt) is 20.5A for 26AWG copper,
you're very unlikely to melt the wire, but it very well might get
pretty hot, depending on actual operating conditions.

Suggestion: suspend a foot or so of 26AWG like you used in your
antenna, with a weight on the bottom end to put similar tension in the
wire to what you have in your antenna. Run a DC current through it,
and see if anything bad happens (like getting too hot, stretching too
much, or whatever). Use a DC current so that the heating will be
similar to what you'd get worst-case with your antenna setup. Or use
RF, if it's easy for you to do that...DC would be easy for me. Expect
to get very similar heating if you use about 1.75 times as much DC
current as what the RF current will be. So try about 8 amps for a
worst-case test.

Cheers,
Tom

Bart Rowlett wrote in message ...
Hal Rosser wrote:
"gibberdill" wrote in message
...

I plan to use my 26 ga. 75 meter dipole (stealth antenna)-- now with 100

W--

with my new amplifier, with 800-1000 W output. Is that too small a gauge?
How much current can I expect it to take? Even if it doesn't get too hot,
will it be significantly less efficient?



Ohms law is gonna put 20 amps on that wire (1000 watts / 50 ohms)

P = I*I * R
I = sqrt(P / R) = 4.47 A

bart