Basically, efficiency is calculated by the ratio of radiation resistance to the sum of radiation resistance and system losses. If system losses are zero, efficiency will always be 100% by definition
Here are some "complete system" numbers drawn from EZNEC, VK1OD's line loss calculator, and W9CF's tuner simulator: Half-wave 80m dipole at 30ft made from #14 wire: Antenna copper losses: 0.19dB 100f
Hi Steve - You 've made a good point , however: an 80m dipole at 30 feet isn't going to be 1:1 on 50 ohm line and, how does it look off resonance out towards the band edges ? Gene / W2LU ____________
1. If the antenna is to be used for multiple bands, keep it nonresonant to If "feed-point impedance" means the impedance at the center of the dipole with open-wire feedline, it doesn't matter--it's o
The "cloudwarmer" function is more likely a result of the height rather than the length. There is nothing wrong with a "cloudwarmer" unless you are a DXer. A cloud warmer will probably outperform a d
I have trouble accepting that. We will have to agree to disagree. Ken WA8JXM _______________________________________________ _______________________________________________ TowerTalk mailing list Tow
Now put that matching network at the feedpoint and see what losses you get. What feedpoint Z did you get from EZNEC?... over what kind of soil? (because 30 feet is pretty low for 3.5MHz) ____________
Paul, THERE I agree with you because the high current portion (e.g. center portion of a dipole, or the "effective" quarter wave portion of a vertical) is used as a radiator. End loading works well be
What difference does it make, we don't put the matching network (lump) at the feed point (except people with remote tuners like the Icom AH-4). Most of us have a linear "matching network" of 30-100'
Gene, EZNEC predicts the feedpoint impedance of an 80m half-wave at 30ft over average ground as 51 Ohms at resonance - that was close enough for me to call the RG213 "matched" ;) But you raise a very
that's just basic electromagnetic theory. If you have accelerating charges, some energy is radiated away. All the antenna does is provide a place for those charges to move in a desirable fashion. You
Ken, If by "efficiency" we mean the power radiated by the dipole as EM energy compared to the power applied, what you say simply isn't true. If we apply 100W to a 70ft dipole made of #14 copper on 80
If you're building a 80m single band antenna, and it's short, and you want to feed with coax, then putting a lumped network at the feedpoint that gets you "close" to 50 or 75 ohms (depending on your
Not exactly. What end loading does is make the feedpoint impedance closer to your desired resistive value. Think of this... a 1/4 wavelength long dipole is going to look very much like a capacitor. I
Then you don't understand antennas very well. The bandwidth will be very narrow and the currents extremely high, but if the matching network had zero loss the radiating efficiency would be equivalent
Hi All, So...! How's DX been lately, anyway...? : >) ~73~ de Eddy VE3CUI - VE3XZ ** On 2012-01-17, at 2:00 PM, David Gilbert wrote: _______________________________________________ ___________________
Any rational to justify that superstition? Dave AB7E _______________________________________________ _______________________________________________ TowerTalk mailing list TowerTalk@contesting.com ht
Are you proposing we actually USE the antenna? Heretic! :-D 73, Mike NF4L _______________________________________________ _______________________________________________ TowerTalk mailing list TowerT
Hi SteveYes, but.... There are so many variables The best answer all depends............. switching to open wire and a tuner then adds the question of tuner efficiency which is quite variable and som
Niels Bohr said it well: "The opposite of a correct statement is indeed an incorrect statement. But the opposite of a profound TRUTH may well be ANOTHER profound truth..." If you wad each full-length