Reg Edwards wrote:
Antenna conductors ARE transmission lines and the same sort of
calculations apply.

You're right, Reg. Without reflections from the ends of a
dipole, the feedpoint impedance would be hundreds of ohms.
A standing wave antenna is like a lossy transmission line
where the loss is to radiation. And the SWR on a 1/2WL dipole
standing wave antenna is quite high - in the neighborhood of
20:1.
--
73, Cecil http://www.qsl.net/w5dxp


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It is equally correct to say that the radiation resistance of a dipole
is 2*73 = 146 ohms, since this is the value when the radiation
resistance is assumed to be uniformly distributed along its length as
is the conductor resistance.

For calculating purposes the radiation and conductor resistances can
simply be added together.

And when referred to the end of a dipole, at HF the radiation
resistance is of the order of Q-squared * 73/2 = 3000 ohms. Although
this is somewhat indeterminate because of the difficulty of feeding an
isolated dipole at one end. The conductor diameter also plays a
significant part.
----
Reg.

Just about any antenna textbook will show you the calculation of a half
wavelength, infinitely thin dipole in free space. For that special case,
the answer ends up being simply 30 * Cin(2 * pi), where Cin is a
modification of the cosine integral Ci[*]. That's where the "magic
number" comes from. Kraus' _Antennas_ is just one of the many textbooks
which give the derivation for this. It takes about 3 pages and 23
equations for Kraus to derive.

One assumption made in calculation of this value is that the current
distribution is sinusoidal, an assumption that's true only for an
infinitely thin antenna. For finite thickness wire, the calculation
becomes much more difficult. The radiation resistance changes only
slowly with wire diameter, however, so sinusoidal distribution is a
reasonably good approximation provided that the antenna is thin.
Feedpoint reactance, though, varies much more dramatically with both
antenna length and diameter. Calculating its value exactly requires
solution of a triple integral equation which can't be solved in closed
form. That's why computer programs are used to solve it numerically.

To include the effect of ground, you need to calculate the mutual
impedance between the antenna and its "image". If the antenna is about
0.2 wavelength above ground or higher (for a half wave antenna -- the
height must be greater if the antenna is longer), you can assume that
the ground is perfect and get a pretty good result. Below that height,
the calculation again becomes much more complicated because the quality
of the ground becomes a factor. If you're interested in the numerical
methods used, locate the NEC-2 manual (available on the web), which
describes it.

If you're satisfied with approximate results, the work by S.I.
Shelkunoff provides formulas for free-space input impedance of antennas
with finite diameter wire which can be solved with a programmable
calculator or computer. They're detailed in "Theory of Antennas of
Arbitrary Size and Shape", in Sept. 1941 Proceedings of the I.R.E. The
formulas for R and X contain many terms involving sine and cosine
integrals, which can be approximated with numerical series. You'll find
additional information in his book _Advanced Antenna Theory_. For
approximate calculations of mutual impedance of thin linear antennas,
see "Coupled Antennas" by C.T. Tai, in April 1948 Proceedings of the
I.R.E. Those also involve multiple terms of sine and cosine integrals.

Before numerical calculations became possible, many very good
mathematicians and engineers devised a number of approximation methods
of varying complexity and accuracy. You'll find their works in various
journals primarily in the 1940s - 1960s.

The complexity and difficulty of the problem is why virtually all
antenna calculations are done today with computers, using numerical
methods such as the moment method.

In summary, here are your choices:

1. You can calculate the approximate radiation resistance but not
reactance of a thin, free-space antenna by assuming a sinusoidal current
distribution and using the method Reg described. To include the effect
of ground, you have to calculate or look up from a table the mutual
impedance between the antenna and its "image", and modify the feedpoint
impedance accordingly by applying the mesh equations for two coupled
antennas. This method of including the effect of ground becomes
inaccurate below around 0.2 wavelength, if the antenna is over typical
earth.

2. You can use various approximation methods to calculate reactance, and
resistance with better accuracy. But for the effect of ground, you're
still limited to being greater than about 0.2 wavelength high.

3. To accurately include the effect of real ground with low antennas,
and/or to get resistance and reactance values with arbitrarily good
accuracy requires numerical methods. A computer program is the only
practical way to do this. A very good basic description of the moment
method can be found in the second and later editions of Kraus' _Antennas_.
[*] Cin(x) = ln(gamma * x) - Ci(x), where gamma = Euler's constant,
0.577. . . Ci(x) = the integral from -infinity to x of [cos(v)/v dv] =
ln(gamma * x) - (x^2)/(2!2) + (x^4)/(4!4) - (x^6)/(6!6). . .

Roy Lewallen, W7EL

Harry wrote:
Hi Tim and Reg,

Thank you for your valuable information. Is there any website or
textbook that actually shows the step-by-step calculation of this magic
number which has been quoted so often in the cable industry?

You know most video cables and connectors have characteristic
impedance, 75 Ohms.

I am not afraid of math. I just like to understand the details of its
derivation.

-- Harry

"Cecil Moore" wrote
Prad is found by integrating the Poynting Vector over a
certain radius.

====================================

Cec, what's the Poynting Vector?
---
Reg.

Reg Edwards wrote:

There's no such value as 'exact'.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Nonsense. I have exactly two pencils in my pencil holder. :-)

73, Bill W6WRT

Reg Edwards wrote:

"Cecil Moore" wrote

Prad is found by integrating the Poynting Vector over a
certain radius.

Cec, what's the Poynting Vector?

Same thing as a power flow vector. The dimensions are
energy per second per unit area. If the power flow vector
is integrated over the entire surface of a sphere of radius
r, the result is total radiated power. I thought maybe you
and Mr. Poynting might have worked together at some time in
the 20th century. :-)

Poynting, John Henry (1852-1914)
English physicist, mathematician, and inventor. He devised
an equation by which the rate of flow of electromagnetic energy
(now called the Poynting vector) can be determined.

http://www.cartage.org.lb/en/themes/...oynting/1.html
--
73, Cecil http://www.qsl.net/w5dxp


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"Reg Edwards" wrote
It is equally correct to say that the radiation resistance of a
dipole
is 2*73 = 146 ohms, since this is the value when the radiation
resistance is assumed to be uniformly distributed along its length
as
is the conductor resistance.

For calculating purposes the radiation and conductor resistances can
simply be added together.
=====================================

Hints and tips -

By the same arithmetical reasoning -

The centre feedpoint input resistance of a 1/2-wave dipole is 73 ohms
plus HALF of the conductor end-to-end HF resistance, which to be
precise sometimes matters. This is exact insofar as 73 is exact - not
an approximation.

The HF resistance of a copper wire at 20 degrees C is -

Rhf = Sqrt( F ) / 12 / d ohms per metre,

Where F is the frequency in MHz and d is the wire diameter in
millimetres.

For example, the end-to-end loss resistance of a 1/2-wave dipole at
1.9 MHz using 16-gauge copper wire is 7.07 ohms, which increases the
feedpoint resistance to 73 + 3.53 = 76.53 ohms, to give a radiating
efficiency of 95.4 percent. Some people would consider that's enough
to lose a contest!

Why not start a little notebook to record useful, simple, little
formulae such as above which don't appear in Terman et al? Or if they
do appear then you can never find the right page in the right volume.
----
Reg.

Hi Cecil,

From "Antenna Theory" by Balanis:

Rr = 2*Prad/|Io^2| = 73 ohms (4-93)

I got Balanis' book and found the above formula.
Thanks a lot! (Tom also mentioned the same book.)

4-94 (I have the 1982 edition) is from formula 4-70.
The math is OK to me. I happen to have a PH.D. degree in math.

-- Harry

Hi Roy,

Thanks for all the valuable information. Are you a professor?

You may want to write a book titled "Story of the 73 Ohms".

-- Harry

Harry wrote:
Hi Roy,

Thanks for all the valuable information. Are you a professor?

Not by a long shot. I'm way too poor a student to ever be considered for
a job as a teacher.

You may want to write a book titled "Story of the 73 Ohms".

It's already been written, by many people who know a great deal more than I.

Roy Lewallen, W7EL