# Question about a dual end fed antenna - possible?

Discussion in 'Antennas, Feedlines, Towers & Rotors' started by W4LLZ, Aug 5, 2020.

1. ### WA7ARKHam MemberQRZ Page

No, no horse, Trojan or otherwise.

If you set out to build a center-fed dipole that is exactly one wavelength long, then NEC says that the feedpoint impedance is 5245 Ohms and you say it is 10,000 Ohms. It doesn't matter which number you believe because either feed impedance makes it an impractical antenna.

It takes ~2300Vrms at the antenna feedpoint to produce 1000W into 5245 Ohms. However, arcing is related to peak voltage, not rms voltage, so that is what you would have to be concerned with when choosing a feedline, or it trying to reduce that impedance to something more useful using a transformer or LC matching network. 2300Vrms is ~3250Vpp.

About the only hope for matching this antenna to a modern radio would be to feed it with an odd multiple of 15.62ft of 4 to 6 inch-spaced 600 Ohm parallel ladder (OWL) line. At the design frequency of 14.463MHz, 1/4wl (4.76m) , 3/4wl (14.29m) , 5/4wl (23.82m), ... of OWL would transform 5245 Ohms to ~70 Ohms, requiring only a 1:1 balun to get from an unbalanced rig/tuner to the balanced line... Even then, if the spacers on the ladder line get wet, the impedance will change, and the losses will increase. Even when dry, 33m of OWL will loose 77W out of 1000W.

WA7ARK: It takes ~2300Vrms at the antenna feedpoint to produce 1000W into 5245 Ohms.

K9AXN: Well of course it does ---- on one .5λ wire. We are talking about a pair of .5λ wires combined to be a center fed 1λ dipole. All of the calculations found on post #19 are for a pair of .5λ wires combined to be a 1λ dipole. Check to see what NEC says about 500 watts to each of the .5λ halves. You may be surprised to find your same error result because NEC configures it as a .5λ radiating wire and a .5λ non radiating counterpoise.

You don't get to write off writers of theoretical material simply because they are not well known or do not boast their credentials. I may very well be a country boy that walks barefoot through the fields chewing on a wheat stem smoking corn silk in my corncob pipe, but you have a responsibility to at least acknowledge the logic of the past. There was a time when we could design an antenna knowing how it worked. There are a number of wayward theories tossed about by people that use todays tool set but are not qualified to interpret the results. Unverified opinions are not science, just gossip.

Next, a logical experiment used as a teaching tool for transmission lines and antennas from 1964: It can be done in any ham shack.

Regards Jim K9AXN

3. ### AK5BHam MemberQRZ Page

Perhaps---just perhaps---the OP (last heard from on the first page) has run away as fast as he could and decided to try making a good old resonant dipole out of his two end feds---and is now living happily ever after on his particular band of choice making plenty of Qs while the sun spots begin to once again shine our way---and the titan cartoonists fight tooth, nail ad infinitum.

I wouldn't be a bit surprised.

4. ### W9XMTHam MemberQRZ Page

K9AXN: Please post the proven Physics (even from 1964) supporting your belief that with 500 watts applied to both ".5λ halves" as you describe above, one half radiates, and the other half is a counterpoise that doesn't radiate.

5. ### WA7ARKHam MemberQRZ Page

What I want to know is how the antenna decides which of the two wires is the "radiator" and which is the "counterpoise"?

6. ### W9XMTHam MemberQRZ Page

First things first.

7. ### KB7WGHam MemberQRZ Page

Perhaps the purpose of the counterpoise, is to keep the feedpoint at a 0 electrical potential. This allows the element current to be always controlled by the open end pole potential. Antenna current is controlled by the open end potential. So, as far as one element is concerned, it doesn't care if counterpoise is radiating or not. As long as it is at 0 potential. The open ended potential goes to 2 times the input potential, after 180 degrees input.......on a 1/4 wave length element. This happens on all open wires, only the length of wire, sets the required input degrees necessary......for the double of potential.

The antenna current is 180 degrees out of phase with the pole voltage, because it is one pole driven.

So, after 180 degrees of excitation, the open pole voltage is double, and the current is maximum and out of phase.....and the current disassociates in an instant. Emitting the established current field from the element.

8. ### W9XMTHam MemberQRZ Page

However K9AXN stated in Reply 22 that both ".5λ halves" of the antenna system under discussion are separately driven with 500 watts.

Therefore, neither half is functioning as a "counterpoise." Both halves are active radiators.

9. ### KB7WGHam MemberQRZ Page

They counterpoise each other. The feed point is held at 0 or neutral potential. Whether the counterpoise radiates or not, it doesn't matter. The dipole elements are synced, but still independent radiators. They emit opposite fields. In a vertical ground radiator, the counterpoise(ground) does not radiate. Only one field is emitted at a time. The ground, neutralizes the feedpoint. If you elevate that vertical.....you must give it a 0 potential at the feedpoint. And there are several ways to do it.

Do elevated vertical ground elements radiate? Or just a counterpoise cap. for 0 potential at feedpoint?

It seems to me that all older hams have their own ideas about radiation, probably because no one truly knows.

Once we realized how the measurement adds to the result that we are looking for........maybe someone will figure it out.

When a charge is induced, it reacts. A measurement has an included reaction.

To see the inducement separately, we need to remove the reaction of the measurement.

10. ### W9XMTHam MemberQRZ Page

Agree, and that is unfortunate in view of the accurate and powerful software available for the last decade+ to analyze and understand the performance of antenna systems.

Some of it is even free.