wrote:

Of course.

Everyone knows the gain of a parabola is directly proportional to
the size in wavelengths, or:

G=10*log k(pi*D/L)^2

Where G= gain in DB over an isotropic, k ~ .55 for most real parabolas,
D is the diameter, and L is the wavelength (wavelength and diameter
in the same units.

So a 2,000 foot parabola on 20m would have just about 58db gain.


Hm. I get 47.

Roy Lewallen, W7EL

Roy Lewallen wrote:
wrote:

Of course.

Everyone knows the gain of a parabola is directly proportional to
the size in wavelengths, or:

G=10*log k(pi*D/L)^2

Where G= gain in DB over an isotropic, k ~ .55 for most real parabolas,
D is the diameter, and L is the wavelength (wavelength and diameter
in the same units.

So a 2,000 foot parabola on 20m would have just about 58db gain.


Hm. I get 47.

Roy Lewallen, W7EL

Hmm, when I use 14 Mhz and 6 decimal places I get 37; must have fat
fingered it the first time.

Working backward from 47 I get a wavelength of 21 feet.


--
Jim Pennino

Remove .spam.sux to reply.

writes:

Roy Lewallen wrote:
wrote:

Of course.

Everyone knows the gain of a parabola is directly proportional to
the size in wavelengths, or:

G=10*log k(pi*D/L)^2

Where G= gain in DB over an isotropic, k ~ .55 for most real parabolas,
D is the diameter, and L is the wavelength (wavelength and diameter
in the same units.

So a 2,000 foot parabola on 20m would have just about 58db gain.


Hm. I get 47.

Roy Lewallen, W7EL

Hmm, when I use 14 Mhz and 6 decimal places I get 37; must have fat
fingered it the first time.

Going a different way, I also get about 37.

Aperture of a dish is the area, pi*r^2. r is about 14.2 wl, so area
is about 635 sq. wl.

Aperture of a dipole is 1/4 * 1/2 wl = 1/8 sq. wl.

That makes gain 635/(1/8) = 635*8, i.e about 5100 or just over 37
dBd. This assumes 100 % illumination of the dish, which we won't
achieve. So make it 35 dBd or so, i.e. 37 dBi. Using the o.p.'s
formula, I get 36.5 dBi.

It's odd that pi is squared in the formula. The squared part must be
to account for the area of the dish, which is pi*r^2. Obviously, this
can has been compensated for by the choice of 'k'.

On Sep 11, 2:53*am, Jon Kåre Hellan wrote:
writes:
Roy Lewallen wrote:
wrote:

Of course.

Everyone knows the gain of a parabola is directly proportional to
the size in wavelengths, or:

G=10*log k(pi*D/L)^2

Where G= gain in DB over an isotropic, k ~ .55 for most real parabolas,
D is the diameter, and L is the wavelength (wavelength and diameter
in the same units.

So a 2,000 foot parabola on 20m would have just about 58db gain.

Hm. I get 47.

Roy Lewallen, W7EL

Hmm, when I use 14 Mhz and 6 decimal places I get 37; must have fat
fingered it the first time.

Going a different way, I also get about 37.

Aperture of a dish is the area, pi*r^2. r is about 14.2 wl, so area
is about 635 sq. wl.

Aperture of a dipole is 1/4 * 1/2 wl = 1/8 sq. wl.

That makes gain 635/(1/8) = 635*8, i.e about 5100 or just over 37
dBd. This assumes 100 % illumination of the dish, which we won't
achieve. So make it 35 dBd or so, i.e. 37 dBi. Using the o.p.'s
formula, I get 36.5 dBi. *

It's odd that pi is squared in the formula. The squared part must be
to account for the area of the dish, which is pi*r^2. Obviously, this
can has been compensated for by the choice of 'k'.

Whoaaa guys............!

Let us think a bit more regarding the basics presented instead of
parrotting
dish's as used in the present state of the art.
Isn't a dish built around phase change of a half wave dipole in inter
magnetic coupling?
If I have a flash light that is focussed does this wavelength aproach
still apply?
I thought it would be a question of action and reaction. Trow a ball
against the wall and it bounces back
in a reflective manner to the angle of velocity.
A dish as presently used changes the phase of a given signal to
reverse it's direction.
In physics we can also talk about mechanical force that rebound and
rebound has nothing to do
with wavelength! If we consider radiation as being the projection
of particles instead of wavelike oscillation
then surely the size of the reflector is solely based on what can be
collected from the
emmitter such that it rebounds to a point or a focussed form ?
I ask the question as I know nothing about the reflective phenomina
of dish's tho I have visited
the one in P.R. where the dish is formed with the knoweledge that the
radiation spreads out
according to the emmiter used and thus when it reaches the reflector
the unit strength is weaker which the
dish attempts to reverse by refocussing. But then I could be totally
in error thus the question to the experts
Best regards
Art Unwin KB9MZ
..

On Thu, 11 Sep 2008 06:18:14 -0700 (PDT), Art Unwin
wrote:

If I have a flash light that is focussed does this wavelength aproach
still apply?

The reflector (or magnifier lens, take your pick) is on order of at
least 1 centimeter. The light wavelength is on order of 500
nanometers.

Ratio = 20,000:1

Beam is generally no narrower than 15 degrees. At a distance of, say,
6 feet, that beam would cover a diameter of 18 inches. Nothing like a
Lazer (sic) if that is the goal.

73's
Richard Clark, KB7QHC

On Sep 11, 10:37*am, Richard Clark wrote:
On Thu, 11 Sep 2008 06:18:14 -0700 (PDT), Art Unwin

wrote:
If I have a flash light that is focussed does this wavelength aproach
still apply?

The reflector (or magnifier lens, take your pick) is on order of at
least 1 centimeter. *The light wavelength is on order of 500
nanometers.

Ratio = 20,000:1

Beam is generally no narrower than 15 degrees. *At a distance of, say,
6 feet, that beam would cover a diameter of 18 inches. *Nothing like a
Lazer (sic) if that is the goal.

73's
Richard Clark, KB7QHC

I see no basis for the inclusion of wavelengths when one is not using
a straight radiator
A straight radiator requires one type of reflector an array that is
condensed to a smaller volume
requires a reflector that is based on the propagation from that
radiator. If propagation flares out
then you can calculate dish size via WL. If propagation is of a
different form then
the dish must be designed accordingly.The important factor as I see
it is the mode of propagation
and what area is required at a distance to account for tha propagation
mode. If one starts with a
lazer then the reflecting surface need not be larger than the
initiating beam area assuming zero scattering.
Your thinking is based solely on the state of the art via reading
matter. You need to go back in physics
to the four forces of the standard model to analyse this question on
the basis of the unification
theory which is all conclusive where one can determine relative
ejection paths from the radiator.
The latter may well gyrate to WL I suppose

On Thu, 11 Sep 2008 09:38:46 -0700 (PDT), Art Unwin
wrote:

On Sep 11, 10:37*am, Richard Clark wrote:
On Thu, 11 Sep 2008 06:18:14 -0700 (PDT), Art Unwin
wrote:
If I have a flash light that is focussed does this wavelength aproach
still apply?

The reflector (or magnifier lens, take your pick) is on order of at
least 1 centimeter. *The light wavelength is on order of 500
nanometers.

Ratio = 20,000:1

Beam is generally no narrower than 15 degrees. *At a distance of, say,
6 feet, that beam would cover a diameter of 18 inches. *Nothing like a
Lazer (sic) if that is the goal.

I see no basis for the inclusion of wavelengths when one is not using
a straight radiator

Read your own question. There is no such thing as a "straight
radiator" of light. There is everything to do with wavelength or you
could never see light.

73's
Richard Clark, KB7QHC

On Sep 11, 9:18*am, Art Unwin wrote:
On Sep 11, 2:53*am, Jon Kåre Hellan wrote:





writes:
Roy Lewallen wrote:
wrote:

Of course.

Everyone knows the gain of a parabola is directly proportional to
the size in wavelengths, or:

G=10*log k(pi*D/L)^2

Where G= gain in DB over an isotropic, k ~ .55 for most real parabolas,
D is the diameter, and L is the wavelength (wavelength and diameter
in the same units.

So a 2,000 foot parabola on 20m would have just about 58db gain.

Hm. I get 47.

Roy Lewallen, W7EL

Hmm, when I use 14 Mhz and 6 decimal places I get 37; must have fat
fingered it the first time.

Going a different way, I also get about 37.

Aperture of a dish is the area, pi*r^2. r is about 14.2 wl, so area
is about 635 sq. wl.

Aperture of a dipole is 1/4 * 1/2 wl = 1/8 sq. wl.

That makes gain 635/(1/8) = 635*8, i.e about 5100 or just over 37
dBd. This assumes 100 % illumination of the dish, which we won't
achieve. So make it 35 dBd or so, i.e. 37 dBi. Using the o.p.'s
formula, I get 36.5 dBi. *

It's odd that pi is squared in the formula. The squared part must be
to account for the area of the dish, which is pi*r^2. Obviously, this
can has been compensated for by the choice of 'k'.

Whoaaa guys............!

*Let us think a bit more regarding the basics presented instead of
parrotting
dish's as used in the present state of the art.
Isn't a dish built around phase change of a half wave dipole in inter
magnetic coupling?
If I have a flash light that is focussed does this wavelength aproach
still apply?
*I thought it would be a question of action and reaction. Trow a ball
against the wall and it bounces back
in a reflective manner to the angle of velocity.
A dish as presently used changes the phase of a given signal to
reverse it's direction.
In physics we can also talk about mechanical force that rebound and
rebound has nothing to do
with wavelength! * * *If we consider radiation as being the projection
of particles instead of wavelike oscillation
then surely the size of the reflector is solely based on what can be
collected from the
*emmitter such that it rebounds to a point or a focussed form ?
*I ask the question as I know nothing about the reflective phenomina
of dish's tho I have visited
*the one in P.R. where the dish is formed with the knoweledge that the
radiation spreads out
according to the emmiter used and thus when it reaches the reflector
the unit strength is weaker which the
dish attempts to reverse by refocussing. But then I could be totally
in error thus the question to the experts
Best regards
Art Unwin KB9MZ
.- Hide quoted text -

- Show quoted text -

I think you cannot use the particle analogy with HF when dish size is
not greater than wavelength. For a small dish at HF, the waves will
simply bend around the dish and act as if it wasn't there. At much
higher frequencies, particle concepts become more accurate.

Jon K??re Hellan wrote:
writes:

Roy Lewallen wrote:
wrote:

Of course.

Everyone knows the gain of a parabola is directly proportional to
the size in wavelengths, or:

G=10*log k(pi*D/L)^2

Where G= gain in DB over an isotropic, k ~ .55 for most real parabolas,
D is the diameter, and L is the wavelength (wavelength and diameter
in the same units.

So a 2,000 foot parabola on 20m would have just about 58db gain.


Hm. I get 47.

Roy Lewallen, W7EL

Hmm, when I use 14 Mhz and 6 decimal places I get 37; must have fat
fingered it the first time.

Going a different way, I also get about 37.

Aperture of a dish is the area, pi*r^2. r is about 14.2 wl, so area
is about 635 sq. wl.

Aperture of a dipole is 1/4 * 1/2 wl = 1/8 sq. wl.

That makes gain 635/(1/8) = 635*8, i.e about 5100 or just over 37
dBd. This assumes 100 % illumination of the dish, which we won't
achieve. So make it 35 dBd or so, i.e. 37 dBi. Using the o.p.'s
formula, I get 36.5 dBi.

It's odd that pi is squared in the formula. The squared part must be
to account for the area of the dish, which is pi*r^2. Obviously, this
can has been compensated for by the choice of 'k'.

The k is generally called the "efficiency factor" which is supposed to
account for diversions from the theoretical optimum.

From what I've read, it appears most real, well constructed and fed
parabolas wind up with a k of around .55, which is why I used that number.


--
Jim Pennino

Remove .spam.sux to reply.

[Slaps self upside the head] 47 dB for a 2000 meter dish, 37 dB for a
2000 foot dish. And that's why I didn't choose bridge design for a
career. . .

Roy Lewallen, W7EL

wrote:
Roy Lewallen wrote:
wrote:
Of course.

Everyone knows the gain of a parabola is directly proportional to
the size in wavelengths, or:

G=10*log k(pi*D/L)^2

Where G= gain in DB over an isotropic, k ~ .55 for most real parabolas,
D is the diameter, and L is the wavelength (wavelength and diameter
in the same units.

So a 2,000 foot parabola on 20m would have just about 58db gain.

Hm. I get 47.

Roy Lewallen, W7EL

Hmm, when I use 14 Mhz and 6 decimal places I get 37; must have fat
fingered it the first time.

Working backward from 47 I get a wavelength of 21 feet.