John Smith wrote:
"The "major efficiency" you are stating is?"

Q is the usual statement of inductor efficiency.

Libear loading has its advocates. My edition of John Devoldere, ON4UN`s
"Low-Band DXing" is from 1994. On page 9-36 is Fig 9-46, Two-Band (80
and 160-m) vertical system using linear loading to bring the antenna to
resonance on 160 meters.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:
John Smith wrote:
"The "major efficiency" you are stating is?"

Q is the usual statement of inductor efficiency.

Libear loading has its advocates. My edition of John Devoldere, ON4UN`s
"Low-Band DXing" is from 1994. On page 9-36 is Fig 9-46, Two-Band (80
and 160-m) vertical system using linear loading to bring the antenna to
resonance on 160 meters.

Best regards, Richard Harrison, KB5WZI


Thanks Richard, I will see if I can locate an issue that old.

Regards,
JS

(Richard Harrison) wrote in news:24417-
:

John Smith wrote:
"The "major efficiency" you are stating is?"

Q is the usual statement of inductor efficiency.

Not to imply that Efficiency=Q.

In very simple terms, in the case of a coil, the inductance is
proportional to the cross sectional area of the coil. A circular coil
uses the least wire and hence the least resistance to achieve that area.
Techniques that increase the conductor length for the same inductance are
likely to produce a lossier (ie less efficient) inductor.


Libear loading has its advocates. My edition of John Devoldere, ON4UN`s
"Low-Band DXing" is from 1994. On page 9-36 is Fig 9-46, Two-Band (80
and 160-m) vertical system using linear loading to bring the antenna to
resonance on 160 meters.

Linear loading does have its advocates. The claims sometimes made
regarding efficiency are perhaps unjustified. My article at
http://www.vk1od.net/cobra/index.htm explores NEC based system models of
the Cobra antenna system (a linear loaded multiband antenna system) and
the loading mechanism is not inherently lossless.

The case of a non-inductive wire wound resistor is an extreme case of
inefficiency as a result of "linear loading". The objective is to arrange
the conductor so that incremental magnetising forces cancel and just the
loss is retained.

Owen

Owen wrote:
"Not to imply that Efficiency = Q."

It does.

Q = energy stored per cycle / energy lost per cycle. Q = XsubL / R

Efficiency = output / input.

Output is the energy given back by the coil when its field collapses.

Input is the energy required to charge the inductor`s field plus the
energy required to supply the losses.

Net output is equivalent to coil reactance and net input is equivalent
to the effective series resistance loss. Therefo Q = Efficiency.

Best regards, Richard Harrison, KB5WZI

(Richard Harrison) wrote in news:259-475487A3-140
@storefull-3258.bay.webtv.net:

Owen wrote:
....
Efficiency = output / input.
....
to the effective series resistance loss. Therefo Q = Efficiency.

So, you are saying that if a coil has a Q of 500, its efficiency is 500,
and that there is more output (power) than input (power)?

Owen

Owen wrote:
"So, you are saying that if a coil has a Q of 500, its efficiency is
500, and that there is more output (power) than input (power)?"

Don`t I wish! Our energy problems would be solved. Owen caught me not
thinking everything through and making a bone-headed mistake. I`ll take
some time to rethink the relation between Q and efficiency.

Best regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

...
Best regards, Richard Harrison, KB5WZI


Richard:

I'll break down and be honest, I cannot "dust you off."

However, how much "Q" is there in a straight wire radiator?

And, in "linear loading" that much of an "hinderence?"

You know, I have no horse here either--come to think of it!

Warm regards,
JS

John Smith wrote:
"However, how much "Q" is there in a straight wire radiator?"

That`s easy. Arnold B. Bailey has already done all the work for us in
"Antennas and Other Receiving Antennas", but it depends on how fat the
wire is.

For a thin-wire dipole, the 3 dB down bandwidth is 34%, so
F2-F1/Fo=0.34. For Bailey`s 200 MHz antenna, BW = F/Q and Q = 200/68=
2.94.

Best Regards, Richard Harrison, KB5WZI

Richard Harrison wrote:

...
For a thin-wire dipole, the 3 dB down bandwidth is 34%, so
F2-F1/Fo=0.34. For Bailey`s 200 MHz antenna, BW = F/Q and Q = 200/68=
2.94.

Best Regards, Richard Harrison, KB5WZI


Thanks Richard, good math! So many won't take the time ...

Warm regards,
JS