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

[Buzz Bloom]

Q and A has its limits and can be pretty inefficient as a way of learning about a wide topic like this one.

Hi Sophie:

Since you have so very helpful to me, I feel it to awkward for me to disagree with you. However, I have learned what I was seeking to learn here on the PF in about 10 days. 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.

Thank you a gain for your help.

Regards,
Buzz

 

[davenn]

Yes, but I do not understand the geometry of the Holmdel apparatus.The diagram in Dave's post #9 doesn't seem to match the picture of the Holmdel.

It's just a different shaped horn ... it's just a convenience thing for ease of pointing at the sky whilst being able to rotate it in 2 directions...
don't get hung up on it ... it's irrelevent

regardless of if it is the simple horn I posted in #9 or the one the topic is about. If the horn is pointing at the sky, then there is VERY LOW (close to zero) level of ground (290K) noise entering the opening of the horn. This is what makes dish and horn antennas so very effective. Their directionality is VERY SHARP.

Do you know in this picture where the antenna is, and what direction it's directionality is pointing? Is the horn supposed to reflect the CMB radiation to the antenna? Do you know what material the horn is made of? Is this material highly reflective at the 159 GHz peak frequency for 2.7 K?

swing the opening arrow towards you a little
The antenna is right at the apex of the horn right in front of the receiver
I'm sure that article you originally posted a link to commented on the structure
it is aluminium ... and you can observe the inside surface is made of unpolished panels

Dave

 

[sophiecentaur]

It would help if I could see an actual construction diagram of the Holmdel apparatus.

I guess the search for that would be up to you, I'm afraid. That article tells us that the horn has a section of parabolic form for its reflector. Your response rather goes to show that you really need to approach this along a tried and tested path of learning, which has facts and ideas presented in a logical order, not leaving out the important bits on the way. Your attempt to get anywhere by just asking questions has left you without the tools to understand the whole system - rather proving my point.
How would you possibly know the right questions to ask if you do it your way? It would be Brownian Motion, which never gets anywhere fast.
Remember, there is nothing fundamentally better about a Holmdel horn, any more than the receiving equipment was better than we use these days. It's an interesting historical bit of kit of a design which is seldom used these days. Better to look into the sort of set ups that are used in conventional comms reflectors and radio telescopes. There is far more written about them and that will help you far more than one fuzzy photo and a very limited paper.
The CMBR is observed every day by modern equipment.

 

[Buzz Bloom]

Remember, there is nothing fundamentally better about a Holmdel horn, any more than the receiving equipment was better than we use these days.

Hi Sophie:

BTW: Although the material I read did not describe the specific characteristics of the Holmdel parabolic reflector, I was able to make a calculation from the data I did find that the reflector diameter was about 1 m.

The reason I chose to learn about the Holmdel rather than current technology was mostly a historical-technical interest.

Regards,
Buzz

 

[sophiecentaur]

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

This doesn't seem right. The Photograph shows a rectangular shaped aperture which is four or five human heights in each axis. Are you talking of a secondary reflector inside the horn? How is this relevant to the basic gain of the antenna?
D'you know, I think we have reached the end of the line with this thread. You keep introducing ideas and questions that are only serving to impede your learning about this topic. There is no alternative but for you to go through the topic of microwave receiving antennae from square one.

 

[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.