TV antenna dish: depth, f/D, or focal distance

Click For Summary
The discussion revolves around determining the depth, focal distance, and f/D ratio of a 20-inch offset parabolic antenna for an elementary physics project. The original poster is struggling to find specific measurements and specifications from the manufacturer's website, which only provides consumer information. Suggestions include using a side-view photo for photogrammetry to estimate the focal length and the importance of measuring the distance from the focus to the closest point on the dish surface. The conversation highlights the challenges of obtaining precise values for educational purposes and emphasizes that typical values can be used for simpler calculations. Ultimately, the focus is on finding practical methods to estimate necessary parameters for the student's project.
  • #31
nomadreid said:
I am afraid that I must have expressed myself badly.
And my reply was far too heavy - sorry.
I strongly approve of linking Science education with life and, personally, I reckon I have always given better value to students when I am fully across what I am telling them about. There is plenty of juice in your particular lemon of choice and I think you can deliver 'a useful package' as long as you steer her into areas where she is likely to ask questions that are in your level of expertise. So, why not just consider the Area that intercepts the microwaves? How would you deal with "Why is f/D important?"? The f number of camera lenses would be an approach but there's no way she'd be aware of that being an issue on her smart phone. Otoh, the actual Area would make sense to anybody. Offset feeding and shading are no more than asides at this level. A lamp and a mirror would show how offsetting a plane reflector works and that's something she may well have come across before. Looking directly behind you in a mirror is not possible; you have to offset it to get your head out of the way.

In many countries, the size of domestic dishes is limited by planning authorities (makes sense to an 11 year old). The power of a satellite transmitter is limited by the size and cost of the satellite and solar panels. (plenty of example specs for TV satellites).
The satellite has a very wide antenna, to focus all its power into the 'footprint'. That power is spread over the whole of the service area so individuals get much less. 20cm diameter 'just works' most of the time. Work out the area of their country in m2, compared with a 20cm dish area, to give the share that each TV gets.

The geometry of a parabolic reflector is great material and there are several ways she could draw a 2D parabola. This link contains a download with a method involving string and a set square / any rectangle. The cleverness of having equal distances on all paths to the focus could (should) be exciting.

Heavy rain will sometimes block reception (she may be aware of that or she can take a look if there is a rainstorm). Those are all issues that an 11 year old can grasp.
 
Engineering news on Phys.org
  • #32
sophiecentaur said:
why not just consider
Thank you, sophiecentaur. Your post contains some excellent suggestions which I will definitely follow up on.
sophiecentaur said:
This link contains
Which link?
 
  • #33
nomadreid said:
Thank you, sophiecentaur. Your post contains some excellent suggestions which I will definitely follow up on.

Which link?
Sorry - senior moment. This was the link (one of many). There are a lot of different methods but the link near the bottom is a document that's obviously aimed at minimal algebra.

It's a development from the ellipse drawing method which uses a length of string between two pins. If you have about three times the length of string (iirc) as the distance between the pins and keep the string taught with a pencil, the pencil will trace out an ellipse. A parabola is the extreme version of an ellipse, with its other focus out at infinity.

Do some searching and you'll find other, harder methods.
 
  • Like
Likes nomadreid
  • #34
Super! :woot:The link looks like it contains lots of useful stuff, thanks.

The two-pin-and-a-string method is of course classic for an ellipse; wandering around the Internet a while back I was amused to find that carpenters (or at least the traditional type, not the ones that mass-produce and design by computer) make their oval tables which very closely approximate an ellipse by conjoining four circular arcs. It's a method that is a little too tricky for most school children, but I tried it (making the curve, not making an actual table), and it works beautifully. However, for the students I collect easier methods.

A great method for conic sections is to get a few Styrofoam/Styropor (closed-cell extruded polystyrene foam, if one needs to avoid brand names) cones, and a big knife, color the cones, and then start hacking! Students like it...the bigger the knife and the faster the slices, the more attention one gets.

Thanks again!
 
  • Like
Likes Tom.G and sophiecentaur
  • #35
nomadreid said:
conjoining four circular arcs
The sandpaper interpolator does a great job. :biggrin: There are worse errors using a jigsaw (for me, at least).
 
  • Haha
Likes nomadreid
  • #36
sophiecentaur said:
sandpaper interpolator

Actually, very little sanding is necessary. 😁Try distinguishing (by eye; furniture buyers don't come with their calculators ready) the conjunction of the four arcs;
centre (1.5, 0), radius 2.5, for x between 3 and 4
centre(-1.5, 0), radius 2.5, for x between -4 and -3
centre (0, 2), radius 5, for y between -3 and -2
centre (0, -2), radius 5, for y between 2 and 3
from:
the ellipse centred at the origin with horizontal width 8 and vertical height 6.
 
  • Like
Likes sophiecentaur
  • #37
I presume the purpose of wanting to know f/D would be typically to assess focussing quality, magnification (gain) and angular resolution?

But for geostationary satellite signals the resolution can be 'poor' because the dish's only real objective is to gather up the radio energy from one spot in the sky and not resolve it with the next satellite along, and the footprint of the satellites can be well-focussed on a given region without requiring extreme antenna gain.

(... in fact I have picked up one or two strong Astra 2 transmission in the UK on a Yagi array!)

Therefore, for that reason I'd suggest f/D for a satellite dish is of no significant regard. 'f' is the length of the arm, because this is the length between a known reflection point (the base of the arm) and the LNB receiver part. The 'f' will be different from other parts of the dish, but it doesn't matter much because angular resolution of the signal is simply not needed.

In fact, quite the opposite!

A satellite dish is often designed to 'lose' angular resolution not have good angular resolution.

Here in the UK, all domestically supplied fixed satellite dishes point at 'Astra 2', but this is not one satellite but 3 satellites.

Nominally, 'Astra 2' is at 28.2deg East, but it consists of 3 satellites, 'Astra 2E' at 28.46E, 'Astra 2F' at 28.17E and 'Astra 2G' at 28.22E.

So, Astra 2 is a cluster of satellites across 0.3 degrees of arc.

If you have a dish resolution better than 5 milliRadians (at 12MHz that means a Rayleigh criterion around 20 cm) then you are not going to catch all the satellites at once. So in fact the geometry can be deliberately a bit poor (in terms of 'optical' resolution quality).

In the UK, domestic dishes are often oblong shaped affairs about 20cm to 30cm across each axis (respectively), with wings and the overall reflection pattern is out of focus, but the signal strong enough that a resolvable signal can be obtained at the LNB for all the satellites spanning those 0.3 degrees of arc.

I found all this out because I installed a 1m motorised dish for myself and found I could not pick up all the Astra 2 channels at once! Yup, it (presumably) had an angular resolution better than 0.3 degrees and didn't want to focus them all in one go. So I had to scan at 28.20E and invent a 'second' satellite in the menu at 28.50E. Such detail is invisible to domestic users.

Most 'satellites' are actually groups of satellites, two in general but AFAIK the most is 'Badr' with 5 satellites in its cluster (although these 5 occupy only 0.22 deg of arc from 25.80E to 26.02E, so not as widespread as Astra). 'Astra 1' is 4 satellites all between 19.18E and 19.20E. Obviously, more than one geostationary satellite cannot occupy precisely the same longitude.

HTH!
 
  • Informative
  • Like
Likes berkeman and nomadreid
  • #38
Gain is pretty strongly (inversely) linked with directivity. Where pointing accuracy has to be poor, the gain must be sacrificed. Also, to control sidelobes, it is often worth losing gain.

This is all rather on the hard side for the original 11 year old customer. bigger the better for most purposes. 20cm has to be enough.
 
Last edited:
  • Like
Likes nomadreid
  • #39
cmb said:
I presume the purpose of wanting to know f/D would be typically to assess focussing quality, magnification (gain) and angular resolution?
sophiecentaur said:
This is all rather on the hard side for the original 11 year old
Thanks for the information, cmb and sophiecentaur. The original purpose of wanting to know f/D was to be able to draw a simplified cross-sectional parabola with an 11-year old, which, given that I had "D", could give me "f", which would allow the pupil to figure out the equation of a parabola and then to draw it on Desmos. I actually started this thread being extremely ignorant about antennas (but now I am only moderately ignorant), and the original purpose can now be carried through, but the information has given me some additional background that I can use to give some rough non-mathematical answers to questions that an observant 11-year old might pose (such as: why is the dish oval, why isn't the focus in the centre, etc.) as well as information that I will not pass on to her but nonetheless find interesting (and, who knows, might actually be necessary in the future). Hence my gratitude grows with each new post.
 
  • #40
nomadreid said:
to draw a simplified cross-sectional parabola with an 11-year old, which, given that I had "D", could give me "f",
If you get her to plot a parabola with focal length f from feed to base of feed arm (that will lay on the focal axis) and then choose half of it - to about 20 cm from the centre of the curve. If you assume that the feed is symmetrical and wants to see a circle when it's looking at the reflector, the centre axis of the beam of the feed should aim half way out to the parabola.
Drawing (and cutting out) both the offset curve and the centre fed parabola will show her that the offset is noticeably flatter than the regular parabola. That will agree with the one on her wall - it's very flat.
 
  • Like
Likes nomadreid
  • #41
sophiecentaur said:
Drawing (and cutting out) both the offset curve and the centre fed parabola will show her that
Thanks for the suggestions, sophiecentaur. I shall try them. Hopefully she will be able to compare the drawing with what she has on her roof, since she can't actually get too close to it.

By the way, where is your signature saying "If God had intended us to use analogies, he would not have released Mathematics for our use" from?
 
  • #42
If your student can’t actually get to the roof then you will be RIGHT. (Result for the teacher.) She will get the shape and the shadowing advantage. The subtlety of the detail may escape her but it brings Physics (Engineering, actually) to her home.
 
  • Like
Likes nomadreid
  • #43
@nomadreid
That was a home made quote of mine. I have a serious problem with badly drawn analogies. 🤨
 
  • Love
Likes nomadreid
  • #44
nomadreid said:
Summary:: What is the approximate depth, f/D, or focal distance of a standard TV 20 inch (58.8 cm) parabolic (offset) antenna?

I am covering parabolic antennas with a school student in elementary physics, and I was looking for the specs of the (offset) antenna on her family's roof in order to calculate things like focal distance. She sent me a photo (she is on another continent), I found the brand specs, which lists it as a 20 inch antenna... but I could find nowhere the depth or focal distance, or f/D ratio listed. Not being an engineer, I am probably looking incorrectly.

I am not sure whether it is allowed to list a particular brand here, so out of caution, I won't. But if someone tells me it is allowed and desirable, I will post the precise model.
Rule of Thumb: a parabolic dish diameter should be 10x the longest wavelength. If you don't have that, you are better off using a different type of antenna.
 
  • #45
jsg2021 said:
Rule of Thumb: a parabolic dish diameter should be 10x the longest wavelength. If you don't have that, you are better off using a different type of antenna.
That rule has to apply to a particular application. The beam width for receiving a source that's not fixed needs to cover the range of directions that the signal will arrive from (so-called geostationary orbits wander about in fact). The 'image magnification' consideration actually applies in many situations - even down to choosing the appropriate magnification for your binoculars. The 'rule of thumb' for marine binoculars used to be to have no more than 7X, so you could stand on deck in a bumpy sea and hold a distant object in view. Life's different nowadays - if you have a spare Grand for Canon stabilised binos.
 
  • Like
Likes nomadreid and Tom.G

Similar threads

Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K