Talk:Parabolic reflector: Difference between revisions

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This article seems more like a stub to me, but the links were extremely useful. Maybye add a description of how this works (the angle of incidence = the angle of reflection)

Maybe more practical uses and images should be added. Such as how Archimedes used mirrors in a parabolic path to catch invading Roman ships on fire, solar ovens, etc.

I was looking for more information on how parabolic reflectors create a virtual image as seen in the wok pan illusion. Any explanation or diagrams of this?

This is interesting appropedia:Aleiha's parabolic solar cooker. Bawolff 00:52, 7 February 2007 (UTC)[reply]

I'm sorry to say, but the figure Image:Parabolic reflection 1.svg is just terrible. While it expresses the general idea of the parabolic reflector, the position of the focal point is such that the law of reflection is painfully disobeyed, with the reflected ray on the same side of the normal as the incoming ray. A similar problem appears in Image:Waves reflecting from a curved mirror.PNG (which now I notice was drawn by the same user). — AdiJapan ☎ 11:45, 13 June 2008 (UTC)[reply]

I agree. I have proposed deletion of both images. Inaccurate illustrations of technical subjects are worse than useless.--Srleffler (talk) 04:54, 11 November 2008 (UTC)[reply]

A possibly useful reference for this article:

• Lord Oxmantown (1840). "An account of experiments on the reflecting telescope" (pdf). Philosophical Transactions of the Royal Society of London. 130. The Royal Society,: 503–27, esp. p. 522. doi:10.1098/rstl.1840.0023.{{cite journal}}: CS1 maint: extra punctuation (link)

This appears to be an early effort at the manufacture of a large non-spherical reflector. --Srleffler (talk) 04:57, 11 November 2008 (UTC)[reply]

Their size and lack of electricity means that solar cookers and solar furnaces have problems tracking the sun. (Solar cookers are frequently adjusted by hand every few minutes.) By using a much deeper bowl and a more complex shape the sun can be with in 30^o or 45^o of the centre of the mirror. The Compound Parabolic shape is made by merging two parabolas with the same focal point. Compound Parabolic Mirror video

Andrew Swallow (talk) 07:15, 1 June 2009 (UTC)[reply]

This might be a good reference link to real-world applications of the paraboloid algorithm, but I leave it to more learned minds to decide: http://www.freeantennas.com/projects/template/ --Roland (talk) 18:58, 21 November 2011 (UTC)[reply]

Probably worth adding something about off-axis parabolic reflectors [1]. The focal point of a full paraboloidal mirror lies in the path of the incident collimated beam, meaning that a receiver/detector placed at this point will block part of the incident energy. The figure in the article illustrates how an off-axis paraboloid (i.e., a section of a full parabolic reflector) focuses radiation to a point outside the incident beam, meaning that the detector does not cast a shadow.

Separate article perhaps... or is that overkill? Papa November (talk) 10:45, 17 May 2013 (UTC)[reply]

It's mentioned in the Rotating furnace article, about making parabolic reflectors by rotating molten glass. Off-axis reflectors are made by rotating the container of glass about an axis that does not pass through it. `DOwenWilliams (talk) 21:30, 17 May 2013 (UTC)[reply]

Ok. I've added a few lines about off-axis reflectors. DOwenWilliams (talk) 01:39, 18 May 2013 (UTC)[reply]

Great! Thanks :) Papa November (talk) 16:21, 20 May 2013 (UTC)[reply]

The article mentions the volume of a parabolic dish, but not the surface area. I don't know if this would be useful, but suppose I wanted to paint the surface of a dish, how would I work out how many m² of paint to buy? I tried to derive it but it doesn't seem correct, I got: ${\displaystyle A={\frac {4}{3}}\pi rh}$ does that sound right? I doesn't to me... (h is height of dish, from vertex to centre of opening, r is radius at opening) --Jamietwells (talk) 12:23, 25 June 2015 (UTC)[reply]

There's a good reason why this calculation is not given in the article. It's difficult to do. Probably the least difficult way would be to program a computer to grind out a numerical value by summing the areas of a lot of "hoops" around the dish. Or you could just buy a lot of paint. DOwenWilliams (talk) 15:00, 25 June 2015 (UTC)[reply]

You may like to look at this:

http://www.appropedia.org/Focus-balanced_paraboloid

It's something I wrote years go that does a similar calculation. You should be able to modify it to do what you want.

DOwenWilliams (talk) 20:24, 25 June 2015 (UTC)[reply]

"program a computer to grind out a numerical value by summing the areas of a lot of "hoops" around the dish." Could we not just find an expression for the area of the hoops in terms of the distance from the vertex and then integrate? Or is that not possible? I'm not sure I trust the source you linked to. There are no references and I don't understand QBasic. Also his maths is a bit vague (undefined terms etc) Do you understand it? Could you explain it? I came to this article to find the equation for the area of a parabolic dish, but it wasn't mentioned and it also didn't say why it wasn't here. Perhaps even just a note to say that the area formula is very complicated but an approximate result exists? (With a reference to that same information from a trusted source)

--Jamietwells (talk) 08:01, 26 June 2015 (UTC)[reply]

As I said above, the reference I gave is to something that *I* wrote, years ago. It wouldn't be acceptable in the main body of Wikipedia, but on talk pages, the rules are more relaxed. As I said in the article, my result was confirmed, to an accuracy of ten significant digits, by a university mathematician, who used a different method, so I don't think there's any doubt that the result is correct.

If you read through the explanation I wrote, before the program itself, you'll find an expression for the area of a single "hoop". If you can integrate that to find an expression for the area of the dish, then that's all you have to do to solve your problem. But integrating that expression is not something that can be done by any method of which I am aware. That's why I chose to do a numerical summation, using a computer. You could do the same.

You asked a question, and I tried to help you to find the answer. Now you are criticizing what I wrote and demanding that I do more for you. No. It's your problem. You solve it.

DOwenWilliams (talk) 15:09, 26 June 2015 (UTC)[reply]

That was certainly my mistake, I had forgotten (after reading the page you linked to) that you had said you wrote it. I suppose to then say that I don't trust the author probably seems quite offensive now. I didn't mean to offend you, but I still do not trust that your calculations are correct without understanding them. Or rather, I don't want an answer, I want an understanding of how to find the answer, so even if I trusted you had done it correctly I still wouldn't be satisfied. I don't need the answer, I'm just curious about the expression for the area of a parabolic dish as a mathematical idea.

You say I'm criticising what you wrote, and I get why you would say that, and I suppose I am, but what I'm really criticising is the article. sigh this is quite difficult to explain when written...

I'm trying to say, when I went looking for the Wikipedia article I was looking for one fact in particular, specifically the expression for the area of a parabolic dish. When I didn't find it I asked on the talk page, and you very kindly tried to help, but unfortunately you've not been able to because I wasn't able to follow or understand the page you linked to (which I'd forgotten you said you wrote so asked if you understood it and was probably more blunt than I should have been). This is my failing, I'm fully aware of that, and if you can't explain your calculation, or don't want to, then that is totally fine, I can't force you to give up your time to help me. When I talked about adding "a note to say that the area formula is very complicated" I was talking about improving this page, and it was just a suggestion that adding why we haven't included the area calculation in the article might help future visitors who come to the article. I'm not brave enough to edit pages myself in case I ruin someone's work, so I usually suggest edits on the talk pages. That's all I was doing.

I'm really not trying to demand you do work for me, honestly. I think this has been a bit of a misunderstanding (mostly on my part) and I'm really sorry if I have offended you in any way.

--Jamietwells (talk) 14:40, 27 June 2015 (UTC)[reply]

Ok. Apology accepted. Let's put the misunderstandings behind us.

But I don't really know what more I can say to explain the calculation, other than what is already in that article I linked you to. The first part of it goes through the calculation step by step, explaining each one It's just in math notation. The only computerese in it is the use of "^" to indicate exponentiation, so for example x^2 means x², and the use of SQR to indicate square-root. so SQR(expression) means the square-root of the expression. I used periods to indicate multiplication. There are some spaces that can be ignored. Other than that, it's all plain and simple, I think. Let me know if there's any specific thing you don't understand.

Of course, I was interested in the specific problem of calculating the dimensions of a "focus balanced" dish. The later parts of the calculation have to do with force moments and the like. But the earlier parts have more general application, calculating the area of the dish.

When I did this, I was also a bit uncertain about its accuracy, so I sent out a public appeal for anyone to confirm or deny it. Robert Israel, of the mathematics department of the University of British Columbia, volunteered to help, and re-did the calculation by some other method (which, frankly, I don't fully understand), and came up with answers that were exactly the same as mine, to ten significant digits. So then I posted it on the web, and several people actually used my result to design dishes, which turned out to balance exactly as I had calculated. So I am now highly confident that my calculation is correct.

Do, please, try to puzzle your way through the explanation in the article I pointed you to.

DOwenWilliams (talk) 15:52, 27 June 2015 (UTC)[reply]

Thanks. I've asked a question on stack exchange if you're interested. My puzzle is over. The formula for the surface area of a parabolic dish is given by

${\displaystyle A={\frac {\pi d}{6h^{2}}}(4h^{2}+d^{2})^{\frac {3}{2}}}$

--Jamietwells (talk) 20:38, 27 June 2015 (UTC)[reply]

Let me see if I've got this straight. d is the diameter of the opening of the dish and h is its height, from the vertex to the plane of the opening. Correct? So if the dish is a flat sheet, h=0 and the area is infinite. Hmmm.... DOwenWilliams (talk) 21:44, 27 June 2015 (UTC)[reply]

I was really using d as the radius, but yes. I'll have a think about it tomorrow. It's time for bed here really. --Jamietwells (talk) 22:13, 27 June 2015 (UTC)[reply]