Tom Ring wrote:

Sounds reasonable, Reg. To put it for simple people like me, it would
mean it's a transmission line of diameter x with an infinite diameter
shield. Then we feed 2 of them, balanced, colinear, and that's our R
sub r.

Did I misunderstand?


Minus the radiation, of course.

Now, the question is, how much does that change things.

"Tom Ring" wrote in message
. ..
Tom Ring wrote:

Sounds reasonable, Reg. To put it for simple people like me, it
would
mean it's a transmission line of diameter x with an infinite
diameter
shield. Then we feed 2 of them, balanced, colinear, and that's
our R
sub r.

Did I misunderstand?


Minus the radiation, of course.

Now, the question is, how much does that change things.

==================================
Tom,

Twas not I who posed the original "input impedance of an infinitely
long dipole" question.

The radiation resistance does not enter very much into the solution.
But in any case the radiation resistance is not the one we are
accustomed to because the current distribution along the wire is not
of sinewaveform but decays rather slowly exponentially.

I've just remembered the name of the real antenna which best fits the
problem. It is the terminated Beverage which is just a very long
horizontal wire some distance above ground.

The Zo of the Beverage is 60 * Ln( 4 * Height / d ) ohms, where d is
wire diameter. When terminated its input resistance at LF is Zo (see
the Bible, the ARRL Antenna Book). A typical value of Zo is 550 ohms.

Note that radiation resistance does not enter the formula although it
cannot be denied radiation does occur. The formula is a close
approximation which serves present purposes.

Putting two Beverages back-to-back to make a dipole we get an input
impedance of 1100 ohms. The infinite dipole is in the same high
impedance ballpark.

To calculate Zo of an isolated infinite dipole we shall have to change
dimensions Height above ground disappears and is replaced by
wavelength (or frequency).

Zo = 60 * ( Ln( Wavelength / 2 / d ) - 1 ) ohms, approximately.

A more exact formula involves inverse hyperbolic functions and
wavelength, height, wire length, and wire diameter, but nobody ever
uses it. You won't find it in Terman.

On the favourite American 40-meter band with a 14 AWG infinitely long
wire Zo = 505 ohms.

Which makes the dipole input impedance = 1010 ohms but a nice, round
1000 ohms is near enough for me.

Your 'infinite shield' is a fair description for the return path but
the end-effect is fairly large. I prefer 'the rest of the Universe'.
But the nearest point is still the Earth's surface.

I leave the people, who attempted to use Smith Charts and Eznec to
solve the interesting problem, to fathom out where they (or the charts
or Eznec) went wrong.
----
Reg.

Reg Edwards wrote:
The Zo of the Beverage is 60 * Ln( 4 * Height / d ) ohms, where d is
wire diameter. When terminated its input resistance at LF is Zo (see
the Bible, the ARRL Antenna Book). A typical value of Zo is 550 ohms.

Note that radiation resistance does not enter the formula although it
cannot be denied radiation does occur. The formula is a close
approximation which serves present purposes.

Putting two Beverages back-to-back to make a dipole we get an input
impedance of 1100 ohms. The infinite dipole is in the same high
impedance ballpark.

To calculate Zo of an isolated infinite dipole we shall have to change
dimensions Height above ground disappears and is replaced by
wavelength (or frequency).

Zo = 60 * ( Ln( Wavelength / 2 / d ) - 1 ) ohms, approximately.

A more exact formula involves inverse hyperbolic functions and
wavelength, height, wire length, and wire diameter, but nobody ever
uses it. You won't find it in Terman.

On the favourite American 40-meter band with a 14 AWG infinitely long
wire Zo = 505 ohms.

Which makes the dipole input impedance = 1010 ohms but a nice, round
1000 ohms is near enough for me.

Your 'infinite shield' is a fair description for the return path but
the end-effect is fairly large. I prefer 'the rest of the Universe'.
But the nearest point is still the Earth's surface.

I leave the people, who attempted to use Smith Charts and Eznec to
solve the interesting problem, to fathom out where they (or the charts
or Eznec) went wrong.
----
Reg.

In message , Reg
Edwards writes

"Tom Ring" wrote in message
...
Tom Ring wrote:

Sounds reasonable, Reg. To put it for simple people like me, it
would
mean it's a transmission line of diameter x with an infinite
diameter
shield. Then we feed 2 of them, balanced, colinear, and that's
our R
sub r.

Did I misunderstand?


Minus the radiation, of course.

Now, the question is, how much does that change things.

==================================
Tom,

Twas not I who posed the original "input impedance of an infinitely
long dipole" question.

The radiation resistance does not enter very much into the solution.
But in any case the radiation resistance is not the one we are
accustomed to because the current distribution along the wire is not
of sinewaveform but decays rather slowly exponentially.

I've just remembered the name of the real antenna which best fits the
problem. It is the terminated Beverage which is just a very long
horizontal wire some distance above ground.

The Zo of the Beverage is 60 * Ln( 4 * Height / d ) ohms, where d is
wire diameter. When terminated its input resistance at LF is Zo (see
the Bible, the ARRL Antenna Book). A typical value of Zo is 550 ohms.

Note that radiation resistance does not enter the formula although it
cannot be denied radiation does occur. The formula is a close
approximation which serves present purposes.

Putting two Beverages back-to-back to make a dipole we get an input
impedance of 1100 ohms. The infinite dipole is in the same high
impedance ballpark.

To calculate Zo of an isolated infinite dipole we shall have to change
dimensions Height above ground disappears and is replaced by
wavelength (or frequency).

Zo = 60 * ( Ln( Wavelength / 2 / d ) - 1 ) ohms, approximately.

A more exact formula involves inverse hyperbolic functions and
wavelength, height, wire length, and wire diameter, but nobody ever
uses it. You won't find it in Terman.

On the favourite American 40-meter band with a 14 AWG infinitely long
wire Zo = 505 ohms.

Which makes the dipole input impedance = 1010 ohms but a nice, round
1000 ohms is near enough for me.

Your 'infinite shield' is a fair description for the return path but
the end-effect is fairly large. I prefer 'the rest of the Universe'.
But the nearest point is still the Earth's surface.

I leave the people, who attempted to use Smith Charts and Eznec to
solve the interesting problem, to fathom out where they (or the charts
or Eznec) went wrong.
----
Reg.



I'd like to say thanks to all who responded to my original question.
There have been a variety of answers.

The only answers I feel might be wrong are those where there is a
reactive component. Surely, this can be removed by lengthening the
antenna by the appropriate amount - but as it is already infinite, this
might be difficult.

Anyway, all very interesting.
Cheers,
Ian.

--

"Cecil Moore" bravely wrote to "All" (19 Sep 05 09:27:25)
--- on the heady topic of " 73 Ohms, How do you get it?"

CM Subject: 73 Ohms, How do you get it?
CM Xref: core-easynews rec.radio.amateur.antenna:216868



CM pezSV7BAXdag wrote:
The limit for Z does not exist
or is (in general) the complex infinity.

CM As the length of a dipole is increased, for the same
CM power input, more energy is radiated during the first
CM transcient cycle and less is available for reflection
CM from the ends of the dipole. Reflected energy is what
CM is causing the feedpoint impedance to change. As the
CM length of the dipole is incrementally increased, the
CM magnitude of the reflected energy is incrementally
CM decreased. I believe Balanis alludes to this characteristic
CM of standing-wave antennas.

CM The feedpoint impedance is Zfp =3d (Vfor+Vref)/(Ifor+Iref)
CM using phasor addition.

CM The limit of that equation as Vref and Iref go to zero
CM is Vfor/Ifor. That's what happens for an infinitely
CM long dipole. That's also what happens during the transient
CM phase of a finite dipole. Thus, Vfor/Ifor can be thought
CM of as the characteristic impedance of the dipole. Seems
CM to me, Vfor/Ifor could actually be measured during the
CM transient phase of a long finite dipole. Will a TDR
CM report the ratio of V/I for an RF pulse?


Cecil, an infinitely long antenna is simply an impedance transformation
between different mediums. i.e. wire to free space.

A*s*i*m*o*v

Asimov wrote:
Cecil, an infinitely long antenna is simply an impedance transformation
between different mediums. i.e. wire to free space.

We know one of the impedances to be 377 ohms.
Question is, what is the other impedance?
--
73, Cecil, http://www.qsl.net/w5dxp

It's so simple to derive the value of 73 ohms.
Just divide 146 by 2.
Or multiply 36.5 by 2.
There are other ways of doing it.

Cec, if you divide the well-known 377 ohms by 5.164, which is the
square-root of 26.67, you get precisely 73 ohms.

What's the problem?
---
Reg

Reg Edwards wrote:
Cec, if you divide the well-known 377 ohms by 5.164, which is the
square-root of 26.67, you get precisely 73 ohms.

But, but, but, Reg, you said the feedpoint impedance of
an infinite dipole is 1200 ohms, or some such. That's
very close to 377 ohms times pi.
--
73, Cecil http://www.qsl.net/w5dxp

But, but, but, Reg, you said the feedpoint impedance of
an infinite dipole is 1200 ohms, or some such. That's
very close to 377 ohms times pi.
================================

Cec, for once, you ageed with me.

But, but, but, coincidences in arithmetic are certain to occur.
----
Reg.