Antonio Vernucci wrote:
Whether the current flows in the wire or in the rod depends on
inductance and inductive reactance.

A thin wire has greater inductance and impedance per unit length than
a thick rod.


True, the rod having a resistance by far higher than that of the copper wire, I would believe that current will anyway pass through the wire, despite its lower diameter.

73

Tony


Actually if the rod conducts it will share the current flow according to
the law or parallel resistance.

Dave WD9BDZ

Sorry I will not be able to do the test before some time.

I will report later.

Tony I0JX

"Antonio Vernucci" ha scritto nel messaggio ...
Hi Dan

thanks for your in-depth analysis on EZ-NEC. The only step that surpises me a bit is that, with the top and bottom shorting wires in place, the current through the rod gets quite high, almost 2/3 that of the wire. On the other hand is always difficult to predict reality when electromagnetic phenomena are involved.

The next test you suggested, i.e. with the wire fully insulated from the rod, is just what I had in mind to do. A spacing of about 1-inch would be easy to get, by sliding the copper wire through the rings where the fishing nylon wire is supposed to run. So I'll do that first. I have a good quantity of silver-plated teflon-coated wire, so will try with that as it will stand a fairly voltage. I should be able to make that test by next monday or tuesday afternoon, and I will report results here.

73

Tony I0JX

ha scritto nel messaggio ups.com...
Tony,

From an EZNEC model with essentially the situation you described:
Wire numbers 2 and 5 are both 7.9m high, frequency 10MHz, fed against a
radial system in free space.

Wire number 2 is 25.4mm diameter, 21 segments, a 10 ohm load in each
segment, total "rod resistance" 210 ohms + negligable copper loss. The
wire centers are 7cm apart. Wire number 5 is 2mm diameter copper.
Wires 2 and 5 are shorted together top and bottom (no contact along the
length, but maybe we can see why you had arcing)

As you can see below, the currents in the two "wires" with the top and
bottom shorted are not in phase. Moreover, the current in the rod
(Wire No. 2) is quite appreciable, almost 2/3 that of the wire.


Wire No. 2:
Segment Conn Magnitude (A.) Phase (Deg.)
te1 W4E1 .39744 -167.9
2 .40049 -168.9
3 .39918 -169.3
4 .39463 -169.2
5 .38725 -168.8
6 .3773 -168.1
7 .36499 -166.9
8 .35056 -165.4
9 .33425 -163.5
10 .31635 -161.0
11 .29723 -157.9
12 .27734 -154.0
13 .25727 -149.2
14 .23781 -143.2
15 .22002 -135.8
16 .20528 -126.8
17 .19529 -116.1
18 .19188 -104.1
19 .19661 -91.39
20 .21058 -78.89
21 W3E1 .23611 -66.69

Wire No. 5:
Segment Conn Magnitude (A.) Phase (Deg.)
1 W6E1 .64698 170.23
2 .66204 169.08
3 .67227 168.00
4 .67834 166.94
5 .68045 165.86
6 .67871 164.76
7 .6732 163.63
8 .66403 162.44
9 .6513 161.18
10 .63513 159.83
11 .61567 158.39
12 .59307 156.82
13 .56752 155.11
14 .53922 153.20
15 .50842 151.06
16 .47537 148.63
17 .44036 145.82
18 .40374 142.51
19 .36589 138.50
20 .32725 133.51
21 W3E2 .28776 126.90

So how's the loss? Well, here's the table of load data:


Frequency = 10 MHz

Load 1 Voltage = 14.71 V. at -167.86 deg.
Current = 1.471 A. at -167.86 deg.
Impedance = 10 + J 0 ohms
Power = 21.62 watts

Load 2 Voltage = 14.82 V. at -168.9 deg.
Current = 1.482 A. at -168.9 deg.
Impedance = 10 + J 0 ohms
Power = 21.96 watts

Load 3 Voltage = 14.77 V. at -169.29 deg.
Current = 1.477 A. at -169.29 deg.
Impedance = 10 + J 0 ohms
Power = 21.81 watts

Load 4 Voltage = 14.6 V. at -169.24 deg.
Current = 1.46 A. at -169.24 deg.
Impedance = 10 + J 0 ohms
Power = 21.32 watts

Load 5 Voltage = 14.33 V. at -168.83 deg.
Current = 1.433 A. at -168.83 deg.
Impedance = 10 + J 0 ohms
Power = 20.53 watts

Load 6 Voltage = 13.96 V. at -168.07 deg.
Current = 1.396 A. at -168.07 deg.
Impedance = 10 + J 0 ohms
Power = 19.49 watts

Load 7 Voltage = 13.5 V. at -166.95 deg.
Current = 1.35 A. at -166.95 deg.
Impedance = 10 + J 0 ohms
Power = 18.24 watts

Load 8 Voltage = 12.97 V. at -165.43 deg.
Current = 1.297 A. at -165.43 deg.
Impedance = 10 + J 0 ohms
Power = 16.82 watts

Load 9 Voltage = 12.37 V. at -163.46 deg.
Current = 1.237 A. at -163.46 deg.
Impedance = 10 + J 0 ohms
Power = 15.29 watts

Load 10 Voltage = 11.7 V. at -160.97 deg.
Current = 1.17 A. at -160.97 deg.
Impedance = 10 + J 0 ohms
Power = 13.7 watts

Load 11 Voltage = 11 V. at -157.86 deg.
Current = 1.1 A. at -157.86 deg.
Impedance = 10 + J 0 ohms
Power = 12.09 watts

Load 12 Voltage = 10.26 V. at -153.99 deg.
Current = 1.026 A. at -153.99 deg.
Impedance = 10 + J 0 ohms
Power = 10.53 watts

Load 13 Voltage = 9.519 V. at -149.17 deg.
Current = 0.9519 A. at -149.17 deg.
Impedance = 10 + J 0 ohms
Power = 9.061 watts

Load 14 Voltage = 8.799 V. at -143.19 deg.
Current = 0.8799 A. at -143.19 deg.
Impedance = 10 + J 0 ohms
Power = 7.742 watts

Load 15 Voltage = 8.141 V. at -135.8 deg.
Current = 0.8141 A. at -135.8 deg.
Impedance = 10 + J 0 ohms
Power = 6.627 watts

Load 16 Voltage = 7.595 V. at -126.79 deg.
Current = 0.7595 A. at -126.79 deg.
Impedance = 10 + J 0 ohms
Power = 5.769 watts

Load 17 Voltage = 7.226 V. at -116.11 deg.
Current = 0.7226 A. at -116.11 deg.
Impedance = 10 + J 0 ohms
Power = 5.221 watts

Load 18 Voltage = 7.099 V. at -104.08 deg.
Current = 0.7099 A. at -104.08 deg.
Impedance = 10 + J 0 ohms
Power = 5.04 watts

Load 19 Voltage = 7.274 V. at -91.39 deg.
Current = 0.7274 A. at -91.39 deg.
Impedance = 10 + J 0 ohms
Power = 5.292 watts

Load 20 Voltage = 7.791 V. at -78.88 deg.
Current = 0.7791 A. at -78.88 deg.
Impedance = 10 + J 0 ohms
Power = 6.071 watts

Load 21 Voltage = 8.736 V. at -66.69 deg.
Current = 0.8736 A. at -66.69 deg.
Impedance = 10 + J 0 ohms
Power = 7.632 watts

Total applied power = 1535 watts

Total load power = 271.9 watts
Total load loss = 0.846 dB

The situation is VERY much improved by simply removing the top and
bottom shorting wires (this changes the base impedance very much, by
the way. I adjusted the source to still give ~1500 watts applied)

Total applied power = 1576 watts

Total load power = 30.53 watts
Total load loss = 0.085 dB

The current in the two wires isn't in phase either, but the average
magnitude of the current in the rod is more than a factor of ten below
the average magnitude of current in the wire.

So, space the wire a couple of inches out from the rod with insulators
and you should be fine, except, as Reg said before, where the rod is
1/2 wavelength long.

If you still get heating, I suppose you'll just have to use the rod as
a center support for an inverted-vee antenna or something!

The arcing is coming from the fact that the currents are out-of-phase
by some amount (40 degrees) in the two conductors... and so are the
voltages... you get a potential difference between a point on the wire
and the corresponding point on the rod. I doubt it's sufficient to
jump any distance air gap, it's more that the carbon can't take the
current that wants to flow between the rod and wire and is burning, but
that's just a guess, and you know how good my guess was originally.

Dan