Quoting from Electromagnetism
By F.N.H.Robinson
in the Oxford Physics Series
1973 edition
ISBN 0 19 8518913
Chapter 11, Radiation,
page 102
Formula 11.11

Has in the equation for radiated power the term

(2*PI*L/LAMBDA)**2

where L is the antenna length and LAMBDA the wavelength,
thereby showing that the radiated power decreases when the
antenna length decreases.

I will read up further and report further...

"gareth" wrote in message
...
Quoting from Electromagnetism
By F.N.H.Robinson
in the Oxford Physics Series
1973 edition
ISBN 0 19 8518913
Chapter 11, Radiation,
page 102
Formula 11.11

Has in the equation for radiated power the term

(2*PI*L/LAMBDA)**2

where L is the antenna length and LAMBDA the wavelength,
thereby showing that the radiated power decreases when the
antenna length decreases.

I will read up further and report further...


Damned typo!

ISBN 0 19 851801 3

"Jeff" wrote in message ...

Gareth, please have a look around the web and find a copy of Kraus to
download; then read, in particular, chapters 3 and 5.

In particular note the following in relation to short dipoles:

"Assuming no losses it [the power radiated] is also equal to the power
delivered to the [short] dipole"

"The maximum effective aperture of the 1/2 wavelength antenna is about 10%
greater than that of the short dipole"

The gains of a short and 1/2 wave dipole is also quoted as 1.76 and 2.14dBi
respectively.

So can we now put this to bed, the short dipole radiates well it is the
practicabilities that make it a poor antenna.

Jeff

And along the same lines, antennas are often described in terms of isotropic
(point) antennas. With radiation being related to length, isotropic
antennas would not radiate.

Also with effective aperture, the 10% greater you mention is a result of
orientation of the aperture with respect to the maximum part of the
individual antenna pattern. Considering the entire pattern of both
antennas, reciprocity is maintained.

On 10/13/2014 11:00 AM, Wayne wrote:


"Jeff" wrote in message ...

Gareth, please have a look around the web and find a copy of Kraus to
download; then read, in particular, chapters 3 and 5.

In particular note the following in relation to short dipoles:

"Assuming no losses it [the power radiated] is also equal to the power
delivered to the [short] dipole"

"The maximum effective aperture of the 1/2 wavelength antenna is about
10% greater than that of the short dipole"

The gains of a short and 1/2 wave dipole is also quoted as 1.76 and
2.14dBi respectively.

So can we now put this to bed, the short dipole radiates well it is
the practicabilities that make it a poor antenna.

Jeff

And along the same lines, antennas are often described in terms of
isotropic (point) antennas. With radiation being related to length,
isotropic antennas would not radiate.

Also with effective aperture, the 10% greater you mention is a result of
orientation of the aperture with respect to the maximum part of the
individual antenna pattern. Considering the entire pattern of both
antennas, reciprocity is maintained.

Yes, but an isotropic source is an imaginary tool used for comparisons.
It obviously cannot exist in the real world, but it's spherical
radiation pattern can be used as a standard for comparisons.

Similar to an inductor or capacitor with no resistance - only reactance.
They don't exist in the real world, but are used to simplify
calculations. Once you get an answer, you can tweak the results for the
resistance.

--
==================
Remove the "x" from my email address
Jerry, AI0K

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

"gareth" wrote in news:m1g1n8$39o$1@dont-
email.me:

(2*PI*L/LAMBDA)**2

where L is the antenna length and LAMBDA the wavelength,
thereby showing that the radiated power decreases when the
antenna length decreases.


Ok, but again, doesn't this just mean the system, as in taking into account
feeding it? I'm not up to the maths of it, I'm just imaging a kind of logical
extreme where you have a tiddly bit of wire stub out of the end of a coax
instead of a 9m tall vertical whip. It seems obvious to me that to get the
same efficiency, same power, you have a vastly increased energy density, so
even without the maths I have no problem seeing the relevance of comments
like Jim's (Jeff's?) allusion to room temperature superconductors and such.

In other words, any actual reduction is based on practical limits, not theory
itself. It's not so different with laser diodes, in that a diffraction
limited spot may be obtained easily with a simple aspheric lens from any size
apeture so long as it's a signle lattidutinal mode emitter, but try actually
MAKING an emitter that size. Theory says sure, no problem, energy density and
nature of materials says otherwise.

gareth wrote:
Quoting from Electromagnetism
By F.N.H.Robinson
in the Oxford Physics Series
1973 edition
ISBN 0 19 8518913
Chapter 11, Radiation,
page 102
Formula 11.11

Has in the equation for radiated power the term

(2*PI*L/LAMBDA)**2

where L is the antenna length and LAMBDA the wavelength,
thereby showing that the radiated power decreases when the
antenna length decreases.

I will read up further and report further...

You do that and while you are at it take note of the fact that the
expression you give is unitless and can not be power.

You will also find that the total power radiated by an antenna is the
surface integral of the average Poynting vector over a surface enclosing
the antenna. The surface usually chosen is a sphere in the far field to
keep the equations "simple".



--
Jim Pennino

On 10/13/2014 1:36 PM, wrote:
gareth wrote:
Quoting from Electromagnetism
By F.N.H.Robinson
in the Oxford Physics Series
1973 edition
ISBN 0 19 8518913
Chapter 11, Radiation,
page 102
Formula 11.11

Has in the equation for radiated power the term

(2*PI*L/LAMBDA)**2

where L is the antenna length and LAMBDA the wavelength,
thereby showing that the radiated power decreases when the
antenna length decreases.

I will read up further and report further...

You do that and while you are at it take note of the fact that the
expression you give is unitless and can not be power.

You will also find that the total power radiated by an antenna is the
surface integral of the average Poynting vector over a surface enclosing
the antenna. The surface usually chosen is a sphere in the far field to
keep the equations "simple".

He is taking a portion of the equation and presenting it out of context
assuming that this is a valid way to consider what he wishes to show. I
would like to see the full equation. The devil is in the details.

I remember once when I was looking at a link budget and an equation I
was presented with contained a relationship with the distance which was
not a square. I questioned the source of the equation meaning how it
was derived. The person who gave it to me brought me the book and said
it was by one of the authorities in the field. lol I'm sure the guy
was an expert, but I wanted to know why the power didn't drop off with
the distance. I expect this was an equation that was empirical as the
context was over ground distance including likely obstructions and many
factors changed the formula from the free space model.

--

Rick

rickman wrote:
On 10/13/2014 1:36 PM, wrote:
gareth wrote:
Quoting from Electromagnetism
By F.N.H.Robinson
in the Oxford Physics Series
1973 edition
ISBN 0 19 8518913
Chapter 11, Radiation,
page 102
Formula 11.11

Has in the equation for radiated power the term

(2*PI*L/LAMBDA)**2

where L is the antenna length and LAMBDA the wavelength,
thereby showing that the radiated power decreases when the
antenna length decreases.

I will read up further and report further...

You do that and while you are at it take note of the fact that the
expression you give is unitless and can not be power.

You will also find that the total power radiated by an antenna is the
surface integral of the average Poynting vector over a surface enclosing
the antenna. The surface usually chosen is a sphere in the far field to
keep the equations "simple".

He is taking a portion of the equation and presenting it out of context
assuming that this is a valid way to consider what he wishes to show. I
would like to see the full equation. The devil is in the details.

Actually, there is no "the" equation for the power radiated by an
antenna other than the surface integral of the average Poynting vector
over a surface enclosing the antenna.

There are some approximate rules for specific cases and limiting
conditions, but this isn't one of them.

What he presented is for a 1/2 wavelegth antenna 9.87 and a full wave
antenna 39.48.

WTF is that??

--
Jim Pennino

On 10/13/2014 11:45 PM, wrote:
rickman wrote:
On 10/13/2014 1:36 PM, wrote:
gareth wrote:
Quoting from Electromagnetism
By F.N.H.Robinson
in the Oxford Physics Series
1973 edition
ISBN 0 19 8518913
Chapter 11, Radiation,
page 102
Formula 11.11

Has in the equation for radiated power the term

(2*PI*L/LAMBDA)**2

where L is the antenna length and LAMBDA the wavelength,
thereby showing that the radiated power decreases when the
antenna length decreases.

I will read up further and report further...

You do that and while you are at it take note of the fact that the
expression you give is unitless and can not be power.

You will also find that the total power radiated by an antenna is the
surface integral of the average Poynting vector over a surface enclosing
the antenna. The surface usually chosen is a sphere in the far field to
keep the equations "simple".

He is taking a portion of the equation and presenting it out of context
assuming that this is a valid way to consider what he wishes to show. I
would like to see the full equation. The devil is in the details.

Actually, there is no "the" equation for the power radiated by an
antenna other than the surface integral of the average Poynting vector
over a surface enclosing the antenna.

There are some approximate rules for specific cases and limiting
conditions, but this isn't one of them.

What he presented is for a 1/2 wavelegth antenna 9.87 and a full wave
antenna 39.48.

WTF is that??

I have no idea what you are talking about. Where did you get these
numbers? 9.87 what?

--

Rick

rickman wrote:
On 10/13/2014 11:45 PM, wrote:
rickman wrote:
On 10/13/2014 1:36 PM, wrote:
gareth wrote:
Quoting from Electromagnetism
By F.N.H.Robinson
in the Oxford Physics Series
1973 edition
ISBN 0 19 8518913
Chapter 11, Radiation,
page 102
Formula 11.11

Has in the equation for radiated power the term

(2*PI*L/LAMBDA)**2

where L is the antenna length and LAMBDA the wavelength,
thereby showing that the radiated power decreases when the
antenna length decreases.

I will read up further and report further...

You do that and while you are at it take note of the fact that the
expression you give is unitless and can not be power.

You will also find that the total power radiated by an antenna is the
surface integral of the average Poynting vector over a surface enclosing
the antenna. The surface usually chosen is a sphere in the far field to
keep the equations "simple".

He is taking a portion of the equation and presenting it out of context
assuming that this is a valid way to consider what he wishes to show. I
would like to see the full equation. The devil is in the details.

Actually, there is no "the" equation for the power radiated by an
antenna other than the surface integral of the average Poynting vector
over a surface enclosing the antenna.

There are some approximate rules for specific cases and limiting
conditions, but this isn't one of them.

What he presented is for a 1/2 wavelegth antenna 9.87 and a full wave
antenna 39.48.

WTF is that??

I have no idea what you are talking about. Where did you get these
numbers? 9.87 what?

(2 * 3.14 * 5 meters / 10 meters) ^ 2

9.87 nothing; the expression is unitless, i.e. a pure number without
units.

In the expression you have a length divided by a length, which cancels
into a unitles number.

As my old physics professor used to say, always check the units of your
answer; the arithmatic may be correct but it is meaningless unless the
units are correct.

Sample problem:

You drove 100 miles and used 5 gallons of gas. What was your mileage?

5 gallons * 100 miles = 500 gallon-miles --- wrong units.

5 gallons / 100 miles = .05 gal/mile --- wrong units.

100 miles / 5 gallons = 20 miles/gal --- correct units and correct answer.


--
Jim Pennino