Gene Fuller wrote:

Antenna books point out that the current in a
short antenna decreases in a straight line, not a sine curve, from the
feed point to the tip. (E.g. Kraus, 2nd Ed. page 216)

Isn't that simply because the slope of a sine wave near the zero
crossing closely approximates that of a straight line?

73, Jim AC6XG

Isn't that simply because the slope of a sine wave near the zero
crossing closely approximates that of a straight line?

73, Jim AC6XG



Looks like it, back in those days it was simpler to draw the straight line
approximating end of sine wave curve than bother to draw it precisly. Close
enough for unbelievers :-)

Yuri, K3BU.us

Tom Donaly wrote:
We're all marching in lockstep right back to the middle ages.

Walter Johnson didn't live in the middle ages, Tom. I would be
careful about disagreeing with him.
--
73, Cecil http://www.qsl.net/w5dxp


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Gene Fuller wrote:
This experiment demonstrates what happens to the "remaining eight feet"
when confronted with the conflict between the "need" for 90 degrees and
the availability of only 45 degrees.

Once again, you misunderstand what I am saying which is: For an unloaded
vertical antenna to have a purely resistive feedpoint impedance, i.e.
resonant, it must be at least 1/4WL long fed against a counterpoise. A
45 degree unloaded antenna does NOT have a purely resistive feedpoint
impedance, i.e. it is NOT resonant. We are discussing only resonant antennas
here. A properly loaded short mobile antenna does indeed have a purely resistive
feedpoint impedance, i.e. at resonance, it must exhibit something in the ballpark
of 90 degrees of antenna. Hint: There is 90 degrees between the purely resistive
current maximum point and the open end of the antenna. The coil MUST occupy some
of that 90 degrees.

I'm not saying the coil occupies every degree not occupied by the
wires but it does NOT occupy zero degrees. The argument is whether it
occupies zero degrees or not.

Please stop misunderstanding what I am saying. :-)

My computer did not blow up, and I suspect yours will survive as well.

Since you misunderstood, the rest of your posting is proceeding under
false premises.
--
73, Cecil http://www.qsl.net/w5dxp


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"Tom Donaly" wrote in message
m...
snip
Roy, why do you bother?
73
H.



Hi H.,
this is interesting in a sick sort of way. You've got Cecil,
a cloud cuckoo land philosopher espousing his inchoate, hand waving
"theory," Yuri, the consummate empiric, who thinks he can understand
nature by experiment alone, and Richard Harrison, who understands the
natural world almost solely through the agency of written authority; and
all of these erstwhile fellows together can't get it right.
We're all marching in lockstep right back to the middle ages.
73,
Tom Donaly, KA6RUH
Not me, baby. I'm going kicking and screaming.
73, H.

BTW
I took graduate E&M out of J D Jackson's "Classical Electrodynamics" in
1980.
The second semester was taught by one of the two profs I presently work for
at U Texas Physics.
My main boss is an old friend of Dave Jackson and one of my colleagues is a
student of Kraus.

"Jim Kelley" wrote in message
...
Gene Fuller wrote:

Antenna books point out that the current in a
short antenna decreases in a straight line, not a sine curve, from the
feed point to the tip. (E.g. Kraus, 2nd Ed. page 216)

Isn't that simply because the slope of a sine wave near the zero
crossing closely approximates that of a straight line?

73, Jim AC6XG


yup

Jim Kelley wrote:
Gene Fuller wrote:

Antenna books point out that the current in a short antenna decreases
in a straight line, not a sine curve, from the feed point to the tip.
(E.g. Kraus, 2nd Ed. page 216)


Isn't that simply because the slope of a sine wave near the zero
crossing closely approximates that of a straight line?

73, Jim AC6XG


Actually, no one really knows whether it's really a sine curve
or not because no one has ever been able to solve the integral
equation that would give an exact answer. The sine approximation
works o.k. because the far field is relatively insensitive to
changes in the shape of the current curve back at the antenna.
The best thing to do is to approximate the curve with a
moment method program on a computer. That's what the moment
method does best.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:
Actually, no one really knows whether it's really a sine curve
or not because no one has ever been able to solve the integral
equation that would give an exact answer.

It would only be a sine curve if the reflected current was equal
to the forward current, i.e. the antenna was lossless (no I^2*R losses
and no radiation). We know that the reflected current is roughly about
90% of the value of the forward current at the feedpoint of a dipole.
So the total current distribution approximates a cosine wave. In the
textbooks you will find general assumption statements like Kraus':

"IT IS GENERALLY ASSUMED that the current distribution of an infinitesimally
thin antenna is sinusoidal, and that the phase is constant over a 1/2WL
interval, changing abruptly by 180 degrees between intervals."

For all real-world current waves, there is an attenuation factor. The
reflected current arriving back at the feedpoint is always less than
the forward current. That's why the feedpoint impedance,
(Vfor+Vref)/(Ifor+Iref) is low but never zero for a dipole. The net
current cosine function is a ballpark assumption, not actual reality.

This is interesting because (Vfor+Vref)/(Ifor+Iref) is 75 ohms for a
1/2WL dipole but is about 12 ohms for a 75m bugcatcher. That means
the reflected waves are closer in magnitude to the forward waves
in the 75m bugcatcher than they are for a 1/2WL dipole. That makes
sense since the tip of the antenna is closer to the feedpoint for
the 75m bugcatcher than for the 1/2WL dipole.
--
73, Cecil http://www.qsl.net/w5dxp


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