Qs re aspects of the Holmdel Horn Antenna used to find the CMB

[Buzz Bloom]

Are you talking of a secondary reflector inside the horn?

Hi Sophie:

Yes. Based on the previous discussion in the thread, the internal parabolic antenna has a gain of 43.3 dBi. It is the directionality of this component that makes it possible for the apparatus as a whole to avoid the thermal noise from the 290 K radiation from the aluminum of the horn. This gain corresponds to that provided by a 1 m parabolic reflector with a simple antenna at its focus.

Solving this equation for for D gives 1 m, with G = 43.3, k = 76%, and
λ_2.7K = c/ν = 1.889 cm, where
ν = 158.7 GHz corresponding to the peak frequency at 2.73 K.

Regards,
Buzz

 

[davenn]

I did find that the reflector diameter was about 1 m.

Don't know where you got that from ??
It is very much bigger than that and it doesn't have a diameter as such ... it is basically rectangular with a slightly curved surface 6.1m at it's longest length

The antenna is 50 feet (15 m) in length with a radiating aperture of 20 by 20 feet (6 by 6 m) and is constructed of aluminium.

all this info was in the links in your original post

Hi Sophie:

Yes. Based on the previous discussion in the thread, the internal parabolic antenna has a gain of 43.3 dBi. It is the directionality of this component that makes it possible for the apparatus as a whole to avoid the thermal noise from the 290 K radiation from the aluminium of the horn. This gain corresponds to that provided by a 1 m parabolic reflector with a simple antenna at its focus.
View attachment 221888
Solving this equation for for D gives 1 m, with G = 43.3, k = 76%, and
λ_2.7K = c/ν = 1.889 cm, where
ν = 158.7 GHz corresponding to the peak frequency at 2.73 K.

Regards,
Buzz

As noted by Sophi and myself , that is pretty much all incorrect

It is the directionality of this component that makes it possible for the apparatus as a whole to avoid the thermal noise from the 290 K radiation from the aluminum of the horn

please read carefully ... getting frustrated having to constantly repeat myself
Thermal radiation from the horn material is so low that it is basically irrelevant .. I have given you links to that several times
The metal sides of the horn SHIELD THE ANTENNA from the thermal radiation from the groundDave

 

[Buzz Bloom]

Hi @davenn and @sophiecentaur:

I am sorry that I misunderstand what I have read.

https://www.revolvy.com/main/index.php?s=Holmdel Horn Antenna&item_type=topic

The reflector is a segment of a parabolic reflector, so the antenna is really a parabolic antenna which is fed off-axis. A Hogg horn combines several characteristics useful for radio astronomy. It is extremely broad-band, has calculable aperture efficiency, and the walls of the horn shield it from radiation coming from angles outside the main beam axis. The back and side lobes are therefore so minimal that scarcely any thermal energy is received from the ground. Consequently, it is an ideal radio telescope for accurate measurements of low levels of weak background radiation. The antenna has a gain of about 43.3 dBi and a beamwidth of about 1.5° at 2.39 GHz and an aperture efficiency of 76%.​

The underlining is mine.

From this text I decided that I could determine the radius of the "parabolic antenna". In my post #36 I showed the calculation of a parabolic telescope diameter (1 m) from the other data I had. I also used another formula for determining the beamwidth.

Using the value ψ=1.5^o I calculated D = 88 cm.

I am completely baffled about why this is wrong.

Regards,
Buzz

 

[davenn]

I am completely baffled about why this is wrong.

because you used 1 metre instead of 6 metres

The reflector is a segment of a parabolic reflector, so the antenna is really a parabolic antenna which is fed off-axis. A Hogg horn combines several characteristics useful for radio astronomy. It is extremely broad-band, has calculable aperture efficiency, and the walls of the horn shield it from radiation coming from angles outside the main beam axis. The back and side lobes are therefore so minimal that scarcely any thermal energy is received from the ground. Consequently, it is an ideal radio telescope for accurate measurements of low levels of weak background radiation. The antenna has a gain of about 43.3 dBi and a beamwidth of about 1.5° at 2.39 GHz and an aperture efficiency of 76%.

The underlining is mine.

and you didn't underline the important sentence that tells you what I have been trying to impress upon you several times
read the bolded bit ... it gives you the result of the effectiveness of the horn as commented on in the previous sentence that you did underline

D

 

[Buzz Bloom]

because you used 1 metre instead of 6 metres

Hi Dave:

ADDED
I have just again looked over my spreadsheet I used to do calculations and found another error. I suggest we postpone any further discussion until I figure out how to fix the error.

PREVIOUS POST VERSION

Two more things I don't understand.
(1) Why do you say I "used" 1 meter?
The 1 m was a calculation based on the other data:
G = 43.3, k = 76%, and λ_2.7K = c/ν = 1.889 cm.
(2 Why doesn't

a radiating aperture of 20 by 20 feet (6 by 6 m)​

mean an opening in the horn, rather than a parabolic reflector.
In yout post #36, you indicate that the "antenna" is in the "shed" attached to the small left end of the horn. Perhaps the shed also houses the 1 m diameter parabolic reflector. If not, what is the means by which the CMB signal reaches the "antenna" if there is a 6 m by 6 m reflector?

https://en.wikipedia.org/wiki/Antenna_aperture

In electromagnetics and antenna theory, antenna aperture, effective area, or receiving cross section, is a measure of how effective an antenna is at receiving the power of electromagnetic radiation (such as radio waves). The aperture is defined as the area, oriented perpendicular to the direction of an incoming electromagnetic wave, which would intercept the same amount of power from that wave as is produced by the antenna receiving it. At any point, a beam of electromagnetic radiation has an irradiance or power flux density (PFD) which is the amount of energy passing through a unit area of one square meter. If an antenna delivers Po watts to the load connected to its output terminals (e.g. the receiver) when irradiated by a uniform field of power density PFD watts per square meter, the antenna's aperture Aeff in square meters is given by:

A_eff = P_o/PFD.​

This concept seems to me to mean something other than πD²/4 where D is the diameter of the open circular boundary of a parabolic reflector. A 6 m by 6 m square area doesn't seem to me to have any relationship to the parabolic reflector described in the quote in my post #38.

One more reference
https://en.mimi.hu/astronomy/aperture.html
Aperture

The aperture of a telescope is the diameter of the light collecting region, assuming that the light collecting region has a circular geometry .
...the size of the opening through which light passes in an optical instrument such as a camera or telescope. A higher number represents a smaller opening while a lower number represents a larger opening.​

In addition to the 2 confusions I mentioned above, how do you explain in the absence of a parabolic reflector the avoidance of the 290 K noise radiation emitted by the walls of the horn, and which then is received by the ultimate receiver antenna within the horn, and which (both entirely and also within the frequencies of interest) is much greater than that of the CMB radiation entering through the 36 m² aperture?
It is much greater for two reasons:
(a) the 290 K radiation is much greater per m² than the CMB radiation;
(b) the area of the 290 K radiation is much greater than the (aperture) 36 m² area.

Regards,
Buzz

 

[sophiecentaur]

290 K radiation from the aluminum of the horn

What do you mean by that? If the aluminium of the rest of the horn is radiating the full black body radiation then how does a standard paraboloid antenna mange to look at weak signals? What is different about the aluminium face of the 'reflector?

I am completely baffled about why this is wrong.

If you look at the patterns of a range of antennae of the same basic aperture, you will get a wide variation. This is because the basic aperture is affixed by the 'illumination' of the reflector by the feed antenna. You can choose a narrower beam with poor side lobe performance and that will give you maximum gain but wouldn't;t be suitable for low signal reception. You can also choose a wider beam with good side lobe performance and that will have lower gain but be much more immune to well-off-axis sources. The formulae that you are using are based on rule of thumb and 'typical' antennae.
Yet again, I am surprised that you are surprised that you keep coming across things that don't make sense to you. You just don't know enough about the topic. That's not a problem but the only way round it is to learn a lot more and then revisit the topic. It will make more sense then. This is yet another example of not asking the 'right' question because you are not coming to the topic from the right direction.

If I had found textbooks as a source and tried to learn from them, I am pretty sure it would have taken me months to get to this point.

I just read this again. The fact is that you don't know what point you are 'at'. How do you know that, if you stray away from this particular place in your knowledge space, that you will be in a position to answer another closely related question? This is why people need to sit Degree and Masters Courses etc. before they are considered to be good candidates for Engineering jobs.

 

[Buzz Bloom]

What do you mean by that? If the aluminium of the rest of the horn is radiating the full black body radiation then how does a standard paraboloid antenna mange to look at weak signals? What is different about the aluminium face of the 'reflector?

My conceptual model of the Holmdel, which you have convinced me is wrong, is that there is a parabolic reflector which focuses the target radiation onto the end-point antenna (epa). Almost all of the target radiation which enters the horn through the aperture, gets focused onto this epa. The radiation from the horn only send a small part of its radiation onto the epa. I am not sure how to properly calculate this "small part" of the radiation, but I will give it a try. (See (9) below.) The final signal (filtered 2.73 K radiation) to noise (filtered 290 K radiation) calculation is

SNR = 16.5​

Suppose we assume that the epa has a functional area which "sees" the radiation from the walls of the horn. Given an assumed parabolic reflector of about 1 m diameter (I now have come to understand that 1 m is a wrong calculation), assume the epa is a straight rod of length about 5 cm, and about a 1 mm radius. The area of the rod is then
(1) A_rod = π × 0.001 × 0.05 ~= 0.00015 m².
There are other factors:
(2) The power ratio corresponding to the temperatures:

R_power = (290/2.73)⁴ ~= 1.3 × 10⁸​

(3) The 290 K radiating area

A₂₉₀ ~= (1/2) * (4*6 m) * 16 m =~= 200 m²​

(4) The 2.73 K radiating area

A_2.73 = (1/4) × π × 36 m² ~= 12 m²​

Assume all of the (4) area is focused onto the epa.
(5) The fraction of the 290 K wall area radiation that hits the epa
Assume an average effective distance from a random wall point to the epa is 2 m.
Each point on the wall radiates on the average onto a hemisphere of

2 π 4 m².​

The fraction of the radiation that hits the epa is

R₂₉₀ ~= 0.00015 m² / 2 π 4 m² ~= 0.000006​

(5) The ratio of 290 K power hitting the epa to the 2.73 K power hitting the epa is

R_power = 1.3 × 10⁸ × 0.000006 × 200 m² / 12 m² ~= 13,000​

(6) A useful black body radiation integral

f(x) = x³/(e^x-1)
F(x) = ∫₀^x f(x) dx
F(∞) = π⁴ / 15 ~= 6.494​

(7) The fraction of 2.73 K radiation between 90% and 110% of the peak 2.73 K frequency.

F_2.73 = (F(β - F(α))/F(∞) ~= 0.24
α = 2.257
β = 3.386​

(8) The fraction of 290 K radiation between 90% and 110% of the peak 2.73 K frequency.

F₂₉₀ = (F(β* - F(α*))/F(∞) ~= 1.12 × 10^-6.
α* = (2.73/290) × α = 0.0210
β* = (2.73/290) × β = 0.0315​

(9) The final calculation! The signal to noise ratio (SNR) = ratio of the 2.73 K to 290 K power received by the epa and after filtering with a flat frequency range between 90% and 110% of the peak 2.73 K frequency.

SNR = F_2.73 / (F₂₉₀ × R_power)

~= 0.24 / (13,000 × 1.12 × 10^-6)~= 16.5​

BTW Sophie, I like the bit you have at the end of your posts.

Q. Can you play the piano?
A. Dunno, I have never tried.​

I am not sure what you think it means, but I think it means that the answerer is naive in thinking that if he had previously tried, he would know. I have also decided that the answerer is probably a male.

Regards,
Buzz

 

[sophiecentaur]

I am not sure what you think it means, but I think it means that the answerer is naive in thinking that if he had previously tried, he would know. I have also decided that the answerer is probably a male.

Male and totally self confident. (Or is that tautology?)
I actually don't understand whether or not your analysis makes sense. I'm afraid.
Why not try following through the process with a simple circular paraboloid with a simple Horn feed? The feed can give a range of illumination percentages and a range of side and back lobe values. You can pick and mix in a very few steps. I still don't know why you are considering the 290K aluminium when what counts is the Noise Temperature of the reflecting surface. The hot bit that counts is the part of the ground etc that the feed sees directly (feed horns have some side sensitivity and are not normally buried down inside the paraboloid.

 

[Buzz Bloom]

The feed can give a range of illumination percentages and a range of side and back lobe values. You can pick and mix in a very few steps.

Hi Sophie:

Can you recommend a reference that explains how to work with "illumination percentages" and "side and back lobe values"? These are unfamiliar concepts to me, although there is a sort of hint in Post #7.

I still don't know why you are considering the 290K aluminium when what counts is the Noise Temperature of the reflecting surface. The hot bit that counts is the part of the ground etc that the feed sees directly (feed horns have some side sensitivity and are not normally buried down inside the paraboloid.

Generally, the temperature of the horn will be in equilibrium with the air temperature. The radiation from the ground, and from any other solid areas outside the horn, will not ever directly reach the epa, except for the possibility that the source is on a direct (or reflected) line of sight to the epa. This exception will not happen if the aperture points towards the sky. Instead this radiation from outside solids will hit the horn wall and then will raise the wall's temperature a small amount before the wall re-radiates enough to reestablish equilibrium with the air temperature.

The "Noise Temperature of the reflecting surface" can be considered as an additional 290 K area nearer the epa than the wall. (I assume this is the parabolic surface.) My intuition is that this is a smaller amount of radiation hitting the epa than the wall radiation. To take this into account, I would need to know (or guess) (a) the area of the parabolic reflector, and (b) the average distance (actually the average inverse distance squared) from a random reflector point to the epa at the focus point.

How do you know that, if you stray away from this particular place in your knowledge space, that you will be in a position to answer another closely related question? This is why people need to sit Degree and Masters Courses etc. before they are considered to be good candidates for Engineering jobs.

I am an elderly retired person. My participation in the PFs is a hobby, not related to any job, not am I seeking any job. This thread was started because I was curious about how the Holmdel was able to do what it did that made it famous. In post #1 I asked 4 questions. The first 3 were quickly answered, but the 4th has continues to be a confusing conversation for me, in part due to my careless errors in calculations.
The most confusing part of the conversation I think is what I perceive to be the ambiguous usage of he term "antenna".
(a) The entire big Holmdel hon is called an antenna.
(b) A parabolic reflector with a limited directionality antenna at the focus is an antenna.
(c) The limited directionality antenna at the focus of a parabolic reflector is an antenna.is an antenna.
I have tried in my questions to be specific about which of these three things I am talking about, but in both responses and reference quotes, there is some ambiguity.

One aspect of my question (4) for post #1 that seems still unresolved in my mind is whether or not the Holmdel apparatus contains inside the pyramidal horn exterior a parabolic reflector component. In post #1 I quoted from Wikipedia:

This type of antenna ... consists of a flaring metal horn with a curved reflecting surface mounted in its mouth, at a 45° angle to the long axis of the horn. The reflector is a segment of a parabolic reflector, so the antenna is really a parabolic antenna which is fed off-axis.

I interpreted the Wikipedia tesxt as follows: the text "a segment of" means "a partial", and "mounted in its mouth" means "inside the pyramidal horn exterior".

Male and totally self confident. (Or is that tautology?)

Just probably a tautology.

Regards,
Buzz

 

[davenn]

Two more things I don't understand.
(1) Why do you say I "used" 1 meter?
The 1 m was a calculation based on the other data:
G = 43.3, k = 76%, and λ2.7K = c/ν = 1.889 cm.

which is obviously a wrong result since you can see the reflector is substantially larger than 1m and the text indicates that as well

(2 Why doesn't
a radiating aperture of 20 by 20 feet (6 by 6 m)mean an opening in the horn, rather than a parabolic reflector.

in this case, it is the same thing

In yout post #36, you indicate that the "antenna" is in the "shed" attached to the small left end of the horn. Perhaps the shed also houses the 1 m diameter parabolic reflector. If not, what is the means by which the CMB signal reaches the "antenna" if there is a 6 m by 6 m reflector?

absolutely and utterly definitely NOT

it reflects off the reflector that you can see at the opening and is funnelled down the horn to the antenna element

it really is as simple as that ... I showed you a basic horn antenna way earlier on in the thread
stick a reflector at around 45 deg angle at the open end of the horn

This concept seems to me to mean something other than πD2/4 where D is the diameter of the open circular boundary of a parabolic reflector. A 6 m by 6 m square area doesn't seem to me to have any relationship to the parabolic reflector described in the quote in my post #38.

what don't you understand about it ?

One more reference
https://en.mimi.hu/astronomy/aperture.html
Aperture
The aperture of a telescope is the diameter of the light collecting region, assuming that the light collecting region has a circular geometry .
...the size of the opening through which light passes in an optical instrument such as a camera or telescope. A higher number represents a smaller opening while a lower number represents a larger opening.

that's all true and that is specifically referring to an optical situation telescope or camera lens ... no problems there

Aperture is just the size of the opening regardless of its shape ...
and in the case of the of the antenna being discussed, it is roughly 6m x 6m according to supplied info

In addition to the 2 confusions I mentioned above, how do you explain in the absence of a parabolic reflector the avoidance of the 290 K noise radiation emitted by the walls of the horn, and which then is received by the ultimate receiver antenna within the horn, and which (both entirely and also within the frequencies of interest) is much greater than that of the CMB radiation entering through the 36 m2 aperture
It is much greater for two reasons:
(a) the 290 K radiation is much greater per m2 than the CMB radiation;
(b) the area of the 290 K radiation is much greater than the (aperture) 36 m2 area.

As I have previously stated, neither of which are relevant

Do I really have to repeat previous answers for a 4th time ??
I'm begging you ... PLEASE reread them and let it sink in

As Sophi said many posts ago
because you haven't done any basic study on antenna systems and theory. You continue to ask all these Q's in a haphazard way
when with a bit of good background study, you would already have the answers Dave

 

[Buzz Bloom]

you can see the reflector is substantially larger than 1m and the text indicates that as well

Hi Dave:

Can you highlight where in the Holmdel picture I can see a greater than 1 M reflector?
Where does the text say that there is a greater than 1 M reflector?

in this case, it is the same thing

Please explain to me how an opening can be the same thing as a reflector.

Your post #45 has more in it that convinces me that I still do not understand your concept of what is the reflector in the Holmdel apparatus. Please explain the concept.
(a) Is it parabolic?
(b) Is it the same thing as the opening in the horn?
(c) Does the red arrow on the right end of the edited photo in your post #32 point to it? Is it a part of the exterior wall of the horn? If not, please describe in detail exactly where it is.

I showed you a basic horn antenna way earlier on in the thread

Do you mean the diagram in your post #9? If so, this diagram has nothing that looks like a parabolic reflector. Please describe what part of this diagram you intend to represent the reflector.Regards,
Buzz

 

[Buzz Bloom]

THIS POST WAS AN ACCIDENT

 

[davenn]

Can you highlight where in the Holmdel picture I can see a greater than 1 M reflector?

BUZZ ... seriously !

LOOK at my post of the antenna with the red arrows on it. the blunt end of the right arrow ends on the reflector

Where does the text say that there is a greater than 1 M reflector?

you even quoted the correct text in a post a couple of posts ago
it approximates both measurements ...

Buzz Bloom said: ↑
(2 Why doesn't
a radiating aperture of 20 by 20 feet (6 by 6 m)mean an opening in the horn, rather than a parabolic reflector.

(b) A parabolic reflector with a limited directionality antenna at the focus is an antenna.

this also from another of your posts is wrong ... the reflector is only part of the antenna system ... it ISNT the actual antenna

Please explain to me how an opening can be the same thing as a reflector.

Your post #45 has more in it that convinces me that I still do not understand your concept of what is the reflector in the Holmdel apparatus. Please explain the concept.
(a) Is it parabolic?

the opening size is the APERTURE

The reflector is according to the text parabolic

(b) Is it the same thing as the opening in the horn?

the reflector is the reflector
the opening is the aperture
the measurement of both is the same

(c) Does the red arrow on the right end of the edited photo in your post #32 point to it? Is it a part of the exterior wall of the horn? If not, please describe in detail exactly where it is.

read first comment

I will do a cross section side view in my next post

Dave

 

[davenn]

basic cross section drawing of the Holmdel antenna ( not to scale)

the actual horn and reflector showing the reflector surface ... lines are dotted where the surface is hidden by the front part of the horn panelling
Look at the shaded/outlined area compared to the size of the two guys ... the reflector area is HUGE ... it ISNT 1m x 1m

Dave

 

[Buzz Bloom]

I will do a cross section side view in my next post

H i Dave:

I apologize for my being dense and not previously understanding your concept of the parabolic reflector. I think I got it now.

The reflector is part of one of the four walls of the Holmdel horn which has a truncated pyramid shape with the exception of one of the reflector wall which is described later. The particular reflector wall is either (1) attached to the side of the truncated pyramid opposite the side with the large aperture/opening, or (2) physically a part of that wall. The reflector is geometrically a curved shape positioned at a 45^o angle to the 6m by 6m base of the pyramid. At this angle incoming radiation will be reflected to pass in the direction of the central axis of the pyramid toward the narrow end. The radiation will also be focused by the curved shape onto the "antenna" at the end of the apparatus near or in the shed.

If this is correct, then I thank you for being patient with me, and I am please that I finally understand your concept. I will now have to do some thinking to calculate an approximation for the implied SNR.

A few minor questions?
(a) Is the diameter of the parabolic reflector 6m or something different?
(b) Is the reflector part of the wall or attached to it? (Your diagram in post #49, which I saw after stating this post, seems to be saying "part of the wall")
(c) Do you have an estimate for the geometry and size of the antenna at the left of the apparatus?

Regards,
Buzz

 

[davenn]

The reflector is part of one of the four walls of the Holmdel horn which has a truncated pyramid shape.

no ... the reflector is the curved bit at the opening ... what I have shaded in red
the rest is the horn, a funnelling tunnel to get the signal to the antenna

The particular reflector wall is either (1) attached to the side of the truncated pyramid opposite the side with the large aperture/opening,

it IS that wall you can see shaded in red

The reflector is geometrically a parabolic shape positioned at a 45o angle to the 6m by 6m base of the pyramid. At this angle incoming radiation will be reflected to pass in the direction of the central axis of the pyramid toward the narrow end.

it probably isn't 45deg ... it may be ... I just used 45 deg earlier as an example
No, NOT at that angle ... because it is a parabola shape signal at ANY ANGLE within the field of view of the parabola reflector will be focussed dowqn the horn to the focal point where the antenna is located

A few minor questions?
(a) Is the diameter of the parabolic reflector 6m or something smaller?

already answered

(b) Is the reflector part of the wall or attached to it?

which wall ... see the edited photo

(c) Do you have an estimate for the geometry and size of the antenna at the left of the apparatus?

it will be an electrical 1/4 or 1/2 wavelength long inside the waveguide at the end of the horn... again ...see my drawing of a 1/4 wave antenna

The definition of a horn is a flared waveguideDave

 

[Buzz Bloom]

it IS that wall you can see shaded in red

Hi Dave:

Thanks for your answers. The shaded picture is helpful. Unfortunately I have terrible visualization skills. It would be very helpful if you could sketch a top view of the horn. The sketch you have gives me the general idea, but I can't visualize the 3D shape.

I am curious about the distance from the focus to the shaped surface point where horn the axis intersects the curved shape. Would this be the same distance if the target and the focus were along the same axis, rather than at an angle?

Regards,
Buzz

 

[davenn]

It would be very helpful if you could sketch a top view of the horn.

top view ? ... not sure what you mean ?

I am curious about the distance from the focus to the shaped surface point where horn the axis intersects the curved shape. Would this be the same distance if the target and the focus were along the same axis, rather than at an angle?

trying to decipher that question

I will ponder on it

 

[Buzz Bloom]

top view ? ... not sure what you mean ?

I am thinking of the view in you sketch in post #49. It is not actually the geometry one would see looking down on the Holmdel apparatus for two reasons.
(a) It does not show the general footprint which should include the trapezoidal shape with a curve for the curved wall.
(b) It would also help if it showed as a dotted line the top of a part of the side where there is an opening below it.

trying to decipher that question

If you draw the sketch for me, I will edit it to explain my question about the distance to the focus.

Regards,
Buzz

 

[davenn]

(a) It does not show the general footprint which should include the trapezoidal shape with a curve for the curved wall.

I can't draw and you wouldn't see much of a curve when looking straight down on it, it will look relatively flat.
You need to look at it on an angle as in the photo to clearly see the curve

The photo shows all you need to know, I cannot do better than that ... I can't even equal that quality

Dave

 

[davenn]

I am curious about the distance from the focus to the shaped surface point where horn the axis intersects the curved shape. Would this be the same distance if the target and the focus were along the same axis, rather than at an angle?

still trying to decipher what you mean ??

I am curious about the distance from the focus to the shaped surface point where horn the axis intersects the curved shape

comparing the length to the two guys guestimating approx 10 - 12m from the centre of the parabola reflector to the antenna

is that what you are asking ?

Would this be the same distance if the target and the focus were along the same axis, rather than at an angle?

if it was a direct path, then there would be no parabola and as a result there would be no focus

D

 

[sophiecentaur]

These are unfamiliar concepts to me, although there is a sort of hint in Post #7.

They are unfamiliar to you because of your linear path straight to this rather unusual form af antenna. As I commented already, you seem to be refusing to take the necessary time to get into this subject so you will always be coming across terms that mean nothing to you. If you are really keen on finding out about this stuff then why not take the time to do it properly? Doing it your way, you will never be able to rely on any conclusion you reach about it. You have already spent (wasted) many hours down blind alleys and that time could have given a moderately useful area of sound knowledge. That's the way Engineering and Physics has to work.

 

[Buzz Bloom]

is that what you are asking ?

Hi Dave:

On the figure below the red line is the axis of the parabola. The top end of the red line is the focus of the parabola with respect to rays moving parallel to the red axis. My question is a mathematical one about a property of parabolas. Is the distance (a) from the focus at the end of the red line to the center of the parabola, the same distance as (b) the estimated 10-12 m of the offset focus for rays moving vertically downward in the diagram?

Another question: The photograph in post #22 shows the monopole antenna in the shed rather than inside the horn. Is this correct? Or does the horn continue into the shed, and the monopole is inside the end of the horn?

Regards,
Buzz

 

[Buzz Bloom]

You have already spent (wasted) many hours down blind alleys and that time could have given a moderately useful area of sound knowledge. That's the way Engineering and Physics has to work.

Hi Sophie:

You may not believe this, but in my long technical career before retiring, I actually did come to know what works for me, and what is a waste of my time. I also came to know colleagues who had different roads to success. A principle I feel is generally true is that there is more than one right answer to just about everything, and especially about the right way to do things.

Regards,
Buzz

 

[Vanadium 50]

I actually did come to know what works for me, and what is a waste of my time.

I think the case can be made that since this thread is already five dozen messages long, sophiecentaur is right. It's not working for you. I think his advice is likely to help you.

 

[sophiecentaur]

Hi Sophie:

You may not believe this, but in my long technical career before retiring, I actually did come to know what works for me, and what is a waste of my time. I also came to know colleagues who had different roads to success. A principle I feel is generally true is that there is more than one right answer to just about everything, and especially about the right way to do things.

Regards,
Buzz

Yet again you have proved me right about your lack of the essential basics. That diagram of yours is total nonsense. It is an offset paraboloid. (Look it up!) What good would a focus outside the horn do? The term Offset Paraboloid was mentioned earlier (also in that article iirc). The focus (of course) is at the feed antenna. Where else would it be?
Perhaps, by the end of your "long technical career" you have managed to assemble some useful stuff but I would guess it took you a long time to assemble what you (think you) know.
I will stop answering this thread because you clearly haven't taken much on board and you seem to be determined to do things without help.

 

[davenn]

On the figure below the red line is the axis of the parabola. The top end of the red line is the focus of the parabola with respect to rays moving parallel to the red axis. My question is a mathematical one about a property of parabolas. Is the distance (a) from the focus at the end of the red line to the center of the parabola, the same distance as (b) the estimated 10-12 m of the offset focus for rays moving vertically downward in the diagram?

No this is incorrect
It would be incorrect IF it was a normal full dish antenna

Go read up on offset fed dish antennas ... there's lots of ray diagrams and info on the net
You really start need to read up on this stuff.

Another question: The photograph in post #22 shows the monopole antenna in the shed rather than inside the horn. Is this correct? Or does the horn continue into the shed, and the monopole is inside the end of the horn?

it's inside the horn, right at the end of the horn as I drew in my diagram (which is inside the shed)

 

[Buzz Bloom]

Go read up on offset fed dish antennas

Hi Dave:

Thank you for mentioning "offset fed dish antennas". I found several articles that I was unable to find without knowing the proper vocabulary.

https://en.wikipedia.org/wiki/Offset_dish_antenna
http://www.qsl.net/n1bwt/chap5.pdf
http://www.john-legon.co.uk/offsetdish.htm​

The third one had a particularly useful picture.

The term "offset paraboloid" in the last post from @sophiecentaur was also helpful, although he was mistaken that the term had been in a previous post. (I could not find it by searching through all the thread pages. I also could not find the reference he referred to as "iirc".)

The reason I asked you about the equality of focal lengths from different angles was because my intuition told me that they should be different, and also that the cross sections of the offset dish would be elliptical rather than circular. I tried to confirm this by doing the math, but it was to difficult for me. I was able to deduce that the focal length F for a circular parabolic dish was

F = 1/(4a),​

where the parabola is defined by

y = a x².​

Unfortunately that was as far as I could take the math.

Thank you again for all your help and for your patience.

ADDED
Here is a better photo of the Holmdel. It is much clearer to see its geometry.

Regards,
Buzz

 

[sophiecentaur]

lthough he was mistaken that the term had been in a previous post.

The reflector is a segment of a parabolic reflector, so the antenna is really a parabolic antenna which is fed off-axis.

It was quoted in a post of your own - quoting from that paper with the photo and other info. (I assume that off-axis and offset are near enough for the meaning to have got across).
Edit: "Off Axis" is not a good term for it as the focus is always on the axis of the parabola. It's just that only part of the parabola is actually there. Off axis distorts the image.

 

[Buzz Bloom]

"Off Axis" is not a good term for it as the focus is always on the axis of the parabola.

Hi Sophie:

As I have come to understand this, the axis of the reflector is perpendicular to that point, P, on the surface of the reflector that has the greatest curvature. Since the cross-section is not a circle but a ellipse, the curvature refers to the maximum curvature in both the major and minor semi-axis directions. If you imagine a flat mirror at this point tangent to the surface, the parallel incoming rays that hit the mirror at an angle relative to the axis, θ, are all reflected in parallel towards the focus, F. The line between F and P is also at an angle θ relative to the axis, on the opposite side of the axis than the incoming ray that hits P.

This explanation is well illustrated in the "Figure 5-1" diagram in post #62.

It was quoted in a post of your own

I confess that I tend to read technical things in a formal way, and that may well cause me to miss the proper use of vocabulary. I failed at the time to interpret

"a parabolic antenna which is fed off-axis"​

as being the proper phrase from which to choose the words to use for an Internet search.

Regards,
Buzz

 

[sophiecentaur]

I confess that I tend to read technical things in a formal way, and that may well cause me to miss the proper use of vocabulary. I failed at the time to interpret
"a parabolic antenna which is fed off-axis"as being the proper phrase from which to choose the words to use for an Internet search.

Yet again you are proving that your approach is not inherently successful. Why not do this thing properly?

 

[Buzz Bloom]

Yet again you are proving that your approach is not inherently successful. Why not do this thing properly?

Hi Sophie:

I do not think it will be fruitful for us to continue to debate "what is the proper way" for me to learn new things as a hobby. Although I have a large number of friends, some of whom have technical backgrounds, none are interested in my particular technical hobbies. I also do not have a good technical library easily available. I have found the PFs very helpful regarding most of the topics about which I have sought help. A lot of the help has been citing specific references, and/or suggesting particular key words to use when searching the Internet.

When I have, for various topics, made an effort to get my local librarian to find a copy of a textbook and arrange for an inter-library loan, it generally has not been particularly helpful. The reason, as I see it, is that the purpose of the textbook is to teach a broad subject in an organized fashion. The main difficulty I have had in this thread, aside from the errors in my calculations, is the use of vocabulary which I found confusing. When I use a textbook about a broad subject, I make an effort to generally read the few chapters that directly relate to the problem that interests me. Frequently there are terms I don't know. Text books generally do not seem to believe that a glossary is useful. I use the index, and after some searching, I find the definition of the term I did not understand, Unfortunately, the definition I found uses more terms I do not know, and the chase continues for more levels than I can manage. I agree, that to understand a non-elementary topic in a textbook, I would need to carefully read and absorb the entire book, at least up to the chapters that contain the specific narrow topic of my interest. I find it takes months to do this, and my skills deteriorate as I age.

If I had found much sooner the diagram and photograph posted in post #63, I would have been able to understand Dave's and your discussions much sooner. However, the 66 post process still took much less time than reading a textbook would have taken, assuming I had the where-with-all to finish it.

Regards,
Buzz

 

[sophiecentaur]

I do not think it will be fruitful for us to continue to debate "what is the proper way" for me to learn new things as a hobby.

Perhaps not. It is clear that your method leaves so many holes in your knowledge that my comments are demonstrably valid.

Unfortunately, the definition I found uses more terms I do not know, and the chase continues for more levels than I can manage.

Which is just my point. There are many aspects of Science and Engineering that are interesting and entertaining and appreciation at a superficial level is a good thing. But you can't dip in, here and there and be surprised when you come to a problem you cannot solve. One doesn't attempt a concerto before being able to do perfect scales and arpeggios.
BTW, I am not a girl. I am a boy - see my profile.

 

[Buzz Bloom]

BTW, I am not a girl. I am a boy - see my profile.

Hi Sophie:

I did look that up and made an effort to use masculine pronouns. I apologize for my failure to correct all of my gender errors.

But you can't dip in, here and there and be surprised when you come to a problem you cannot solve.

I confess there have been some problems I posted questions about that I failed to ever understand. (For example, I have completely given up regarding quantum mechanics.) However, these have been relatively few. Most of the questions I have explored I have learned what I wanted to learn.

Regards,
Buzz

 

[sophiecentaur]

I apologize for my failure to correct all of my gender errors.

No problem at all. I just thought you should know a bit about who you are talking to.