"rickman" wrote in message ...

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.

Yes, that can be a tricky subject. If an isotropic radiator is assumed,
then the surface area of the surrounding sphere will vary with the square of
the radius. That produces less power per square unit as distance increases.
So, path loss from the squared term depends on the radiation spreading out
over distance.

There is no loss due to distance itself, but to the radiation spreading.

Example: for a lossless microwave dish, if all the radiation transmitted
arrives at the receiving dish, then there is no path loss.

"Wayne" wrote in :

There is no loss due to distance itself, but to the radiation spreading.


I'm wondering now if what Gareth is concerned with is the same as divergence
in an asperic lens output with a laser diode. Assuming the diode has an
emitter width of a very few microns (is already usually only one micron in
one axis even in a multimode diode with a single 'stripe' emitter pattern)
then a large enough single asperic lens will make a finely directed but broad
beam, but if you want it very narrow as well, it diverges more widely and
various optic methods will tame it a bit, but there's no real substitute for
a single mode diode if possible to use one for the wanted power.

Assuming it is NOT possible, the multimode diode needed will demand a bigger
lens to match its power efficiently into a well directed, 'collimated' beam.
It seems to me that this is more than just an analogy, but maybe fundmentally
similar to the difficulties with energy density, accuracy of form, low loss
of materials used, aperture size for emission, and maybe several other things
I've seen mentioned recently about this subject of small antennas. Including
the fact that eben if the laser beam IS highly divergent, the small aspheric
lens is just as efficnt at prjecting the power as the larger one, so long as
all light from the diode gets coupled through it without spill or reflection.

I hope that's not too off-topic, but it seems to be that I might get some
learning from responses to this one...

rickman wrote:
On 10/14/2014 1:21 AM, wrote:
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.

I have no idea what you are going on about. Ok, 9.87 is a unitless
number. So is 33.043. Now what?

Exactly.

Since it is a unitless number, it can not be power as claimed.

Nor can it be a rule of thumb for power versus antenna length as a full
wave antenna does not radiate 4 (39.48/9.87) more power than a 1/2
wave antenna.

It is just nonsense.


--
Jim Pennino

Lostgallifreyan wrote:
wrote in :

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


Context is everything. There are cases where this reciprocated form gets
used to good effect. Maybe convention demands 1/20 gal/mile just to indicate
reciprocation of a normal convention.

True but it is the answer to a different question.


--
Jim Pennino

"Lostgallifreyan" wrote in message
. ..

"Wayne" wrote in :

There is no loss due to distance itself, but to the radiation spreading.


I'm wondering now if what Gareth is concerned with is the same as divergence
in an asperic lens output with a laser diode. Assuming the diode has an
emitter width of a very few microns (is already usually only one micron in
one axis even in a multimode diode with a single 'stripe' emitter pattern)
then a large enough single asperic lens will make a finely directed but
broad
beam, but if you want it very narrow as well, it diverges more widely and
various optic methods will tame it a bit, but there's no real substitute for
a single mode diode if possible to use one for the wanted power.

Assuming it is NOT possible, the multimode diode needed will demand a bigger
lens to match its power efficiently into a well directed, 'collimated' beam.
It seems to me that this is more than just an analogy, but maybe
fundmentally
similar to the difficulties with energy density, accuracy of form, low loss
of materials used, aperture size for emission, and maybe several other
things
I've seen mentioned recently about this subject of small antennas. Including
the fact that eben if the laser beam IS highly divergent, the small aspheric
lens is just as efficnt at prjecting the power as the larger one, so long as
all light from the diode gets coupled through it without spill or
reflection.

I hope that's not too off-topic, but it seems to be that I might get some
learning from responses to this one...
############

Well, it may be slightly off topic and certainly out of my field of
experience, but I find it more interesting than Gareth's misused equations

gareth wrote:
"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...


Ramo, Whinnery and Van Duzer gives the same derivation, including
the derivation of radiation resistance.

What you have posted is a unitless number with no relationship to anything.



--
Jim Pennino

On 10/14/2014 12:58 PM, wrote:
rickman wrote:
On 10/14/2014 1:21 AM, wrote:
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.

I have no idea what you are going on about. Ok, 9.87 is a unitless
number. So is 33.043. Now what?

Exactly.

Since it is a unitless number, it can not be power as claimed.

Nor can it be a rule of thumb for power versus antenna length as a full
wave antenna does not radiate 4 (39.48/9.87) more power than a 1/2
wave antenna.

It is just nonsense.

Did you read the OP? He says:

"Has in the equation for radiated power the term"

He is just giving us a portion of the equation to show the dependence on
wavelength vs antenna length. But without the full equation we can't
know if there are mitigating factors.

Nothing that you have posted makes any sense in this context.

--

Rick

wrote in :

True but it is the answer to a different question.


Ok. True enough.

"Wayne" wrote in :

Well, it may be slightly off topic and certainly out of my field of
experience, but I find it more interesting than Gareth's misused equations



Thanks. I do try. I figure even if I am imprecise, I can either try to
entertain or at least come at it from an angle that might be useful, perhaps
not just to me.

rickman wrote:
On 10/14/2014 12:58 PM, wrote:
rickman wrote:
On 10/14/2014 1:21 AM, wrote:
rickman wrote:
On 10/13/2014 11:45 PM, wrote:

snip

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.

I have no idea what you are going on about. Ok, 9.87 is a unitless
number. So is 33.043. Now what?

Exactly.

Since it is a unitless number, it can not be power as claimed.

Nor can it be a rule of thumb for power versus antenna length as a full
wave antenna does not radiate 4 (39.48/9.87) more power than a 1/2
wave antenna.

It is just nonsense.

Did you read the OP? He says:

"Has in the equation for radiated power the term"

He is just giving us a portion of the equation to show the dependence on
wavelength vs antenna length. But without the full equation we can't
know if there are mitigating factors.

And it is utter nonsense.

There is NOTHING about a 1 wavelength antenna that is 4 times that of
a 1/2 wave antenna, or 16 times than of a 1/4 wave antenna, which is
what the expression evaluates to.

The part L/LAMBDA is the fractional size of the antenna, and the rest
just numbers and I assume you have a calculator of some kind.

It has already been shown by others that, neglecting loss, all power is
radiated by an antenna no matter what the size.


--
Jim Pennino