Talk:Horizon/Archive 1

Archive 1

Unheadered stuff

so why is it purple then?

I came to this page to see if it said anything about why the sky (during a clear day) gets grayer as it gets closer to the horizon. If anyone has a good treatment or link, that would be swell. Jake 00:02, 19 Aug 2004 (UTC)

Maybe this would be better treated on sky since that is where they talk a little about the color? Jake

See airmass or extinction (astronomy) for an explanation of this phenomenon. --Lasunncty 00:18, 29 January 2006 (UTC)

Added geometrical horizon

I added the part about perspective drawing and the "geometrical horizon". In a way, it is similar to the description of "astronomical horizon". However, astronomical horizon was described as a plane, while I believe the horizon is alway a line. Note that it compares the astronomical horizon to the true horizon (one being lower than the other), even though one is described as a line and one as a plane. It seems like this could use some clean up to clarify this distinction.

Robbrown 21:06, 29 January 2006 (UTC)

Mental Calculation

What if you do not have a calculator? Naval officers often use a thumb rule calculation that you can do in your head, that is:

The distance to the horizon in miles is equal to the square root of the height of eye in feet, multiplied by 1.1.

It usually suffices to choose a height of eye that is a square of a round number, like 4, 9, 16, 25, 36, etc. If in a small boat or looking through a periscope, assume a height of eye of 4 feet, the square root of 4 is 2, multiply by 1.1 to get the distance to the horizon of 2.2 miles. If you are on a larger boat, use 9 feet [3.3 miles], a bigger boat, 16 feet [4.4 miles], and so on. If your height if eye is mid-way between 9 and 16 feet, the distance would be half way between 4.4 and 5.5 miles, say 5 miles.

The distance provided by this thumb rule is roughly 10 percent shorter than the exact distance. That accuracy is usually quite sufficient for an eye-ball approximation of a ship or lighthouse that far away.

24.15.3.112 18:22, 24 February 2007 (UTC)tvbanfield

Etymology

The Zeitgeist Movie says that Horizon is a contraction of Horus-is-risen, in the sense that the horizon is from where the god of the sky rises. Is there anything in this or are they talking crap?

Straussian 14:38, 2 July 2007 (UTC)

The OED indicates that horizon comes from the Old French orizonte. The relation between horus and the word horizon is wishful coincidence. Horizon is related to the word meaning limit in greek. This word was accepted into Latin to designate a limit of time "hora" and space "horison." The word horison was in latin long before english existed. Horison cannot come from an english contraction because it was a word already. The sentance "Horus is risin" in latin would be "Horus oriras" or "Horus orirbas" depending on tense of the statement (latin may be off). So a contraction of the statement would sound nothing like Horison. The same film makes the assertion that we get our word sunset from the egyptian god set. This is just as bogus as set comes from the germanic languages and more specifically the old English sitten, derived from setten. —Preceding unsigned comment added by 131.107.0.73 (talk) 21:16, 6 September 2007 (UTC)

That's what I thought. The film does indeed seem to contain a lot of wishful etymologies. Thank you for clearing that up. Straussian 11:54, 13 November 2007 (UTC)

References

confused by the approximate formula and the exact formula because of the units

The approximate formula says d=sqrt(13*h), but the exact formula says d=sqrt(h^2+2 R h). If I assume h<<R, then I could ignore the h^2 part in the exact formula, leaving d=sqrt(2 R h). Here's where I got confused, if I compare this to the approximate formula, it looks like 13= 2 R. How can the radius of the Earth be only 6.5? After alot of thinking, I realized that in the approximate formula, you are supposed to use h in meters and it gives you an answer in km, but in the exact formula, all of the symbols are supposed to have the same units of measurement. So, if I take my approximation to the exact formula and h is given to me in meters, then d = sqrt(2 R h/1000). When this is compared to the approximate formula, I saw that 13 = 2 R /1000; in other words, R = 6,500 km. In fact, the true radius is about 6378 km (no idea how this defined for an imperfect sphere). I will try to edit the article to make it less confusing to other people. —Preceding unsigned comment added by Johnlv12 (talk • contribs) 21:14, 19 November 2008 (UTC)

Theoretical model

I am having a little trouble with my diagram. I will come back to it later. Martinvl (talk) 11:42, 10 May 2010 (UTC)

Calculating the theoretical radius of vision

$\pi r{\cfrac {\cos ^{-1}{\cfrac {r}{r+h}}}{360}}$

Is that the correct formula? If it is, shouldn't it be added in? 114.94.240.29 (talk) 07:14, 18 September 2010 (UTC)

I don't think that it is correct - for start it contains both π and 360, implying a mixture of radians and degrees. One should either find a reference for it, or give a one-liner to show how it was generated. Martinvl (talk) 21:10, 19 September 2010 (UTC)

Headline text

Is this article referring only to the horizon as observed from earth? 69.106.224.48 (talk) 05:42, 12 October 2010 (UTC)

Dawn and dusk under See also

I question the relevance to this article of the links to Dawn and Dusk, and disagree with the appended descriptions. The description for Dawn is too imprecise, and that for Dusk is simply wrong. Twilight is the period during which the Sun is below the horizon but still provides some light in the sky, so dusk occurs sometime during that period—it is not followed by twilight. The linked articles describe dawn and dusk as equivalent to the beginning and end of civil twilight, but lack support from authoritative astronomical sources. Given the questionable relevance, I think the best approach here would be to simply remove the links. JeffConrad (talk) 09:36, 21 October 2010 (UTC)

Reference format

The current reference format, with the titles first, is unusual. If no one objects, I’d like to change it to have the author’s names first, following Chicago format. JeffConrad (talk) 07:39, 18 April 2011 (UTC)

I suggest that the standard wikipedia templates "<ref>{{cite ....}}</ref>" be used. They might well produce Chicago-style referencing (I am not sure) - at any rate they produce references that have a format which is consistent with the rest of Wikipedia. Martinvl (talk) 07:52, 18 April 2011 (UTC)

I oppose using templates; they’re much harder to enter and read, and to me, solve a nonexistent problem. They’re also far from “standard” in Wikipedia, and are neither encouraged nor discouraged by WP:CITECONSENSUS. I normally use author-date, but that’s not the established style here. The article is loosely described as “footnote”, but I’m not aware of any major style guides that puts the article title first. JeffConrad (talk) 08:38, 18 April 2011 (UTC)

I actually like templates - it means that references are consistenly formatted. Almost all the references in the article Metric system (an article that I am have recently overhauled) use templates. If you go into Edit mode in that article, you can see how I lay the templates out. In real life, my career has been in writing computer programs, hence the reason that I lay them out in that manner. Martinvl (talk) 11:40, 18 April 2011 (UTC)

There obviously are many others who like citation templates, but there also are many who don’t; I happen to be one of the latter. I use templates when editing an article that already uses them, but avoid them otherwise. So we have different preferences, which Wikipedia allows, but definitely don’t have consensus for changing to templates.

I think there is something to be said for a vertical format when declaring variables in a program, if for no other reason than it’s easy to include an identifying comment after each. But in this context, I generally prefer the more compact horizontal format (if every parameter needed a comment, the vertical format would probably be better).

The question then remains whether there’s an objection to putting the authors’ names first, as is called for by every style manual and every example I can find in WP:CITE. I’m honestly not quite sure how to format the citations to Liddell and Scott (with or without templates), because I’ve never seen any quite like them—they really aren’t chapters of a book, and I’m hesitant even to treat them as Web pages (though that may prove the only option). JeffConrad (talk) 12:46, 18 April 2011 (UTC)

Jeff, by all means go ahead - I agree that the norm is to put the author's names first, how it is done is of secondary importance. Martinvl (talk) 12:59, 18 April 2011 (UTC)

I’ve cleaned up the references, leaving the format of the dictionary references largely intact. Chicago and Oxford would use something like

A Greek-English Lexicon, Henry George Liddell and Robert Scott, s.v. “ὁρίζων”. On Perseus Digital Library. Accessed 19 April 2011.

In most cases, they would omit the authors’ names when citing widely used references such as dictionaries or encyclopedias, although if the works are commonly known by the authors’ names, those names can appear first. I’m in no position to make such an assessment here, so I’ve opted for a bit of a compromise; if indeed the work is more commonly cited as “Liddell and Scott”, by all means put the authors first. Use of “s.v.” seems a bit hifalutin for Wikipedia; MLA and APA use a format similar to what we have, so I think we’re OK.

In any event, see if this works. JeffConrad (talk) 02:33, 20 April 2011 (UTC)

Distance to the horizon

Given the space to the derivation of the theoretical model, even though the treatment neglects refraction, it seems to me that we should give at least some mention to models that consider refraction. The question, then, is how much to present.

Rigorous method—Sweer
As far as I know, the classic article is John Sweer's “The Path of a Ray of Light Tangent to the Surface of the Earth” in the September 1938 issue of the Journal of the Optical Society of America. The result, briefly, is that the distance to the horizon d is given by

$d={{R}_{\text{E}}}\left(\psi +\delta \right)\,,$

where ψ is the dip of the horizon and δ is the refraction of the horizon. The dip is determined fairly simply from

$\cos \psi ={\frac {{{R}_{\text{E}}}{{\mu }_{0}}}{\left({{R}_{\text{E}}}+y_{\text{obs}}\right)\mu }}\,,$

where R_E is the radius of the Earth, y_obs is the height above the Earth, μ is the index of refraction of air at the observer's height, and μ₀ is the index of refraction of air at Earth's surface. For many purposes, the index of refraction is given to sufficient accuracy by the Dale-Gladstone relation

${\frac {\mu -1}{{{\mu }_{0}}-1}}={\frac {\rho }{{\rho }_{0}}}\,,$

where ρ is the atmospheric density at the observer's height and ρ₀ is the atmospheric density at Earth's surface. The variation of density with height depends on the atmospheric model; for an isothermal atmosphere, it is given by

${\frac {\rho }{{\rho }_{0}}}={{e}^{-\left(y-{{y}_{0}}\right)/H}}\,,$

where H is the pressure scale height of the atmosphere. The isothermal model suffices for many purposes; a more realistic model is a combination of a polytropic troposphere and an isothermal stratosphere.

The refraction must be found by integration of

$\delta =-\int _{{\mu }_{0}}^{u_{\text{obs}}}{\tan \phi {\frac {{\text{d}}\mu }{\mu }}}=-\int _{{y}_{0}}^{y_{\text{obs}}}{\tan \phi {\frac {{\text{d}}\mu }{\mu }}}\,,$

where $\phi \,\!$ is the angle between the ray and a line through the center of the Earth. The angles ψ and $\phi \,\!$ are related by

$\phi =90{}^{\circ }-\psi \,,$

so that

$\sin \phi ={\frac {{{R}_{\text{E}}}{{\mu }_{0}}}{\left({{R}_{\text{E}}}+y\right)\mu }}\,.$

The integral is the same as that used to determine astronomical refraction, except that the upper limit of integration is the observer's height rather than infinity. Strictly, distinction should be made between the observer's elevation above sea level and the observer's height above the horizon (which could be considerably above sea level); the distinction would be important for an east-facing observer on Pikes Peak.

Simple method—Young
A much simpler approach is described by Andrew T. Young in Distance to the Horizon; it uses the geometrical model but uses a radius R′ = 7/6 R_E. When y ≪ R_E, this can be otherwise stated as

$d={\sqrt {\frac {2{{R}_{\text{E}}}\,y_{\text{obs}}}{1-k}}}\,;$

k is the refraction constant, with typically used values ranging from about 0.13 to 0.15.

Results from Young's method are quite close to those from Sweer's method, and are sufficiently accurate for many purposes. JeffConrad (talk) 05:58, 26 October 2010 (UTC)

I've added links to Young's pages, but I think at least some of the material above should be included in the article. Adding this might make the article too technical for some readers; an alternative would be to create a separate article Distance to the horizon. JeffConrad (talk) 05:51, 26 October 2010 (UTC)

Additional cleanup

I still think this section needs a bit of work:

• Do we really need two separate derivations of the geometrical formulas? I think we should eliminate one to address the comments about excessive math. My preference would be to keep the derivation from the Pythagorean Theorem because the conversions from the segment lengths can be eliminated. This probably would require a new figure, but the current figure really isn’t correct anyway with the expression “OA.OB” rather than “OA×OB” or “OA·OB”.
• There’s some redundancy (e.g., the second reference to the Pythagorean Theorem) that should be eliminated to reduce the length of the section.
• I see no need to mention Earth’s diameter; the commonly tabulated values are for the radius, and I think multiplication by 2 in a couple of cases is simple enough. By eliminating the symbol D, we’d eliminate potential confusion with the symbol D_BL used for the boat and lighthouse. It would be better to use d throughout, but we′d probably need to change the figure of the boat and lighthouse to avoid confusion between the equation and the figure.
• The formulas for arc length given in Exact geometrical formula seem like they belong in the subsection that gives the derivations.
• The figure “Three types of horizon” refers to a true horizon that is more properly termed a geometrical horizon that is distinct from the apparent horizon that includes the effect of refraction; for a 1.7 m observer height, the difference is minimal, but it becomes significant for an elevated observer. I’ll concede that this probably isn’t our most pressing issue, however.
• The examples use the geometrical formulas; if one of the simplified forms is used, it’s no more difficult to use the formulas that include the effects of refraction. Again, should we move the examples to the beginning of the section and use the more accurate formulas? We might need to leave the example of the boat and lighthouse where it is, however.

If there’s no objection, I’ll try to address some of these issues. JeffConrad (talk) 13:17, 18 April 2011 (UTC)

I’ve tried to address most of these issues. Unless there’s an express consensus for eliminating the geometrical derivation based on the secant tangent theorem, I’m not going to touch it. I really don’t have a problem with it; perhaps we can consider removal if too many protest “Too much math!”

I’d still like to move the examples to the beginning of the section in order to

1. Make them more accessible to those who desire the time rather than the means of building a watch, and
2. Use the more accurate formulas that consider refraction.

Any objections? JeffConrad (talk) 03:14, 20 April 2011 (UTC)

I’ve moved the examples to the beginning of the section and recalculated the values to account for atmospheric refraction; restore the previous examples if this doesn’t work. I increased the precision of the formulas to give the distance from the summit of Aconcagua as 200 mi rather than the 199.495 mi that results from rounding to three significant figures. JeffConrad (talk) 03:25, 21 April 2011 (UTC)

Units, please

This article is confused somewhat (a great deal?) by not being clear on units of measure.

Miles are mentioned. In the most practical uses you are looking to find horizon distance "at sea". (On land is much more difficult problem -- i.e. define horizon on land...) As such, Nautical Miles would be the more expected _assumption_!

This needs to be cleared up in this article. The distinction between Statute (Land) Miles and Nautical Miles needs to be addressed explicitly... That, or you need to drop working in miles altogether and work in feet or yards and let the user convert to Statue or Nautical Miles, as they prefer.

(And this last is often fudged, as Naut. Miles is often rounded to 6000 ft (2000 yds) -- as compared to 6076 ft (or such) in reality. And the quick approximation of 1 n mi = 1.15 st mi)

It's already pretty well mentioned... but Emphasize these are Approximations.

I've seen several approximations for this that do not agree (and it's harder to tell if they do w/o knowing if units are st. mi. or n. mi) when units conversion (st. mi. vs. n. mi. vs. metric) are made.

Surely, some of this is due to some approximations taking into account atmospheric effects and some not.

And -- atmospheric effects are only an approximation in themselves. The vary with temperature, humidity, viewing conditions, etc.

Perhaps the titled of the article needs to have "APPROXIMATIONS" screaming out...

75.136.98.204 (talk) 14:34, 17 May 2011 (UTC)

For a sailor, the nautical mile is probably the default unit of measure; for others, who far outnumber sailors, the more natural unit is undoubtedly the “land” mile. Except for English-speaking countries, few people use the mile, and unsurprisingly, there is no SI definition. Those who do use the mile define it simply as the “mile” rather than the “statute mile”, whereas the nautical mile is specifically defined as such. In Special Publication 811, NIST mention a “statute mile” that is equal to the U.S. survey mile, which differs by about 2 ppm from the the “international” mile of 5280 “international” feet (1609.344 meters). So technically, a reference to a “statute mile” would be improper as well as slightly ambiguous. (The article Mile indicates that a statute mile is equal to 5280 “international” feet, but is contradicted by the source it cites). Perhaps we could add a note explaining that the “mile” is a “land” mile.

Note added—see if it works. JeffConrad (talk) 01:42, 20 May 2011 (UTC)

Many surveyors still use feet (both “international” and U.S. survey), and geodesists work almost exclusively in meters; however, neither group is typically concerned with distance to the horizon. The general public, who presumably are the primary audience for this article, usually think of distances in miles or kilometers, so I think the distance units used in this article.

I’m not quite sure what’s meant by “I’ve seen several approximations for this that do not agree”, so I cannot comment.

There probably are more different derivations in this article than needed, but there doesn’t seem to be much consensus for removing them. The formulas that include the effect of atmospheric refraction clearly indicate that they are only approximations, because of the variability of refraction, and in the case of Young′s method, use of the approximate geometrical formula. Perhaps it could be mentioned that phenomena such as temperature inversions can result in very large deviations from what is predicted by any of the formulas, though I wonder if this would be getting further into atmospheric phenomena than is indicated. An approximation not mentioned is that Earth’s flattening is ignored. That said, the treatment here is consistent with that in most other sources with which I am familiar. JeffConrad (talk) 19:54, 17 May 2011 (UTC)

Inconsistent approximation formulae

The simple formula for approximating the horizon distance is given several times here, but they don't seem to be consistent. Working in metric units all the time, we have:

$d\approx 3.570{\sqrt {h}}$

(from Young, including the effect of refraction)

$d\approx 3.57{\sqrt {h}}$

(approximate geometrical formula, derived from Euclidean principles)

$d\approx 3.86{\sqrt {h}}$:

(from Young, including the effect of refraction)

I think the first one of these should be changed so that either it uses 3.86, or the explanatory text says that it doesn't include the effects of refection. I'm not 100% confident that I'm right so I haven't made the change myself, and I'm not close enough to the development of this article to have an opinion as to whether the most appropriate figure to show at this point would be with or without refraction.

Interestingly, the non-metric approximations seem to be consistent: the multipliers used in the three formulae are 1.323, 1.22 and 1.32 respectively.

Stewart Robertson (talk) 14:15, 15 February 2012 (UTC)

Corrected (partly, at least). Tim Zukas (talk) 19:13, 29 March 2012 (UTC)

ashutosh kumar mishra — Preceding unsigned comment added by 117.227.39.130 (talk) 02:03, 17 June 2012 (UTC)

Projective Geometry section

The last section seems too much like a book review or summary of Stillwell's book, and I question if it really should be included in the article. It would be more appropriate if there were a separate article on the book, and this section was moved there. (That section was written in recent edits by Rgdboer.) --209.2.238.10 (talk) 15:04, 18 June 2012 (UTC)

The section is about Horizon as the term is developed in a single chapter of one of Stillwell's books. It is placed last in this article as it treats profound implications of the horizon experience in vision and understanding. Above anon contributor is from New York City; "more appropriate" suggestion has little merit.Rgdboer (talk) 19:36, 18 June 2012 (UTC)

Today the section was modified to include an exerpt from Girard Desargues work Perspective from 1636. He uses the horizon as a central line in graphical perspective.Rgdboer (talk) 20:13, 22 October 2012 (UTC)

Today the section title was changed to "Vanishing points" as that is a feature of the horizon which suggests some projective geometry. The section was rewritten, starting from graphical perspective. Improvements are due to access to works by Kirsti Andersen. The heavy reliance on Stillwell's chapter "Horizon" has been reduced.Rgdboer (talk) 21:36, 19 April 2013 (UTC)

Question about links at bottom of page.

Looking for more information, I happened to click on the first external link, a personal website. Does this have any pertinence to the discussion? Not that I can tell.

Ed8r (talk) 22:23, 24 May 2013 (UTC)

Yes. I guess the guy had some stuff about the horizon on his website, and has since deleted it. So I've deleted the link to it. The other link is ok. DOwenWilliams (talk) 00:17, 25 May 2013 (UTC)

curvature of the horizon

I used a ruler to check the image that allegedly shows the curvature of the horizon. In fact, it is exactly straight.

Usually, images that appear to show the curvature are affected by optical defects in the camera.

DOwenWilliams (talk) 19:35, 29 November 2012 (UTC)\

• Hello D0wen, Actually the caption makes no claims as to showing the curvature of the Earth. It just states a fact that where the Ocean meets the sky is the best place to see it. The illusion created in that photo was accomplished by rotating the camera slightly to the left. The Human eye tends to use the foreground as its balancing point, thus the Earth looks bent under the horizon. Thanks....Pocketthis (talk) 19:57, 29 November 2012 (UTC)

Hmmm... Suppose you're looking at a wall which has two parallel horizontal lines marked on it. One line is at the same height as your eyes. The other is a bit lower. If you look straight ahead, perpendicular to the wall, you see the lines a certain angular distance apart. If you look to one side, at the lines where they are further from you, they are, in angular terms, closer together. You know that they are both horizontal, so you see them as horizontal. Even if the upper line is removed, you still see the lower line as horizontal, although, in angular terms, it is a bit higher out to the sides than in front of you.

Now suppose that the line is replaced by the real horizon. To the sides, it is lowwr than the painted line, so it appears to be curved downward, even though, really, it isn't.

Of course it's true that, seen from space, the Earth is obviously round. But seen from close to ground level, any curvature of the horizon is more an optical illusion than anything else.

DOwenWilliams (talk) 03:08, 30 November 2012 (UTC)

Are you saying that when we go to the beach on a clear day and we have a look at the horizon that no wide angle camera lens can duplicate, we aren't seeing a slight curvature of the Earth? Lets look at it in another perspective: When we look straight out to sea at a large Boat traveling directly away from us, the boat doesn't fade into nothingness; it lowers itself as we see only the top of the boat last as it drops below the horizon due to the curvature. However, if I am incorrect about being able to see the curvature from a beach on a crisp clear day, I will be happy to re-write the caption. I read that you were a PHD student of chemistry. I'll take your word for it. I can only rely on what I learned from my professors and my eyes; living on a beach in Florida for many years. Also, I am a licensed Pilot. When I am flying above 7 or 8 thousand feet, if it isn't the curvature of the Earth I'm seeing on a clear day over water, please notify the F.A.A. to revoke my license, because I'm halucinating. :) Many mountains are well over 7,000 feet, and some are close enough to the Ocean to have as good a look at the horizon, as I am in my aircraft. Interesting conversation. Thanks, Pocketthis (talk) 15:58, 30 November 2012 (UTC)

I live in Toronto, Canada, a couple of hundred metres (or yards) from the north shore of Lake Ontario, which is about 50 km wide. On the south shore, there's a narrow strip of low-lying land, behind which is a range of hills, the Niagara Escarpment, which is about 70 metres high. (Originally, Niagara Falls fell over it, but erosion has now moved the falls a long way upstream.)

Simple geometry says that I shouldn't be able to see the escarpment if I stand on the beach near my house, and indeed that is usually true. But not always. Sometimes, usually when the water is a lot colder than the air above it, as often happens in spring and summer, atmospheric refraction bends light downward so the escarpment is clearly visible. Occasionally, at night, the lights of vehicles moving along the highway which is on the coastal strip of ground can also be seen from here, with binoculars.

Looking in the opposite direction, the tall buildings in Toronto can be seen from the south shore. The CN Tower, for example, which is about 550 metres high, can always be seen, unless the atmosphere is foggy. But the amount of it that is visible varies substantially. Sometimes, the pod that contains a restaurant, etc., which is about 350 m above the ground, appears to be low on the tower. Sometimes it looks much higher.

What I'm getting around to saying is that the stuff that is often written about ships on the horizon, etc., is largely poppycock. Atmospheric refraction often has a far greater effect than the curvature of the earth's surface, for an observer who is close to the ground. For someone who is flying at considerable altitude, the earth's curvature dominates.

Yes. This is interesting, and largely unknown to people who live far from any coast.

DOwenWilliams (talk) 16:36, 30 November 2012 (UTC)

• Hmmmm.... lol. I was with you until you used the words: Poppycock. :) If you read the Horizon page under the section: "Objects above the horizon", the authors explain it quite clearly. Funny, I put up that photo and caption without ever reading the article in detail. I knew I had a photo that was much more suited for that spot than the one of some guy diving off a pier; so I replaced it. Since you and I have been going back and forth, I started doing a little research; and I didn't have to go any further than our own article. been fun as always DOwen.... thanksPocketthis (talk) 16:54, 30 November 2012 (UTC)

Thanks for saying it's clear. I wrote that section. At least, I edited it so heavily that what's there now is almost all my work.

But most of the section makes the common assumption that light travels in straight lines which, under many conditions is simply not true. The little sub-section at the end tries to explain the effects of refraction. Unfortunately, in many other publications, this is not mentioned. The rectilinear assumption is unquestioned, which is, I maintain, poppycock.

Lake Ontario gives me opportunities to see things more clearly than can be done by folks by an ocean. I can look at something from my local beach today, maybe take a picture of it, then go back to the exact same place months later, when conditions are quite different, and compare what I see then with what I saw previously. At sea, it's difficult to be sure that a ship is in the same place now as it was previously. I suspect that people often see effects that are actually caused by refraction and mistakenly think they are caused by the earth's curvature.

I read someplace that on Venus, with its very dense atmosphere, there is theoretically no horizon. Light gets refracted right around the planet, so you can look in the distance and see the back of your own head. On the real Venus, the atmosphere is too hazy to allow that, not to mention other problems...

Polynesian navigators have/had many techniques for finding their way from island to island in the Pacific Ocean. One of them was to look for a patch of green on the horizon. This would be the highly distorted image of the destination island, refracted for hundreds of kilometres through the atmosphere.

Fun stuff....

DOwenWilliams (talk) 19:49, 30 November 2012 (UTC)

It is fun stuff, and getting 'funner' as we go along. :) Have you ever been on a beach on a crystal clear day, with the water completely flat? I have many times. I don't live in Florida anymore, but back in the day, I would sit on the Beach every single day. I didn't have to be to work till 4:PM in the afternoon. In Ft. Lauderdale, there is a cargo lane for a major commercial port. Giant Freighters, literally with buildings on them 12 stories high, would come and go out of Port Everglades like traffic on a Freeway. I had a very strong pair of binoculars (16 power), and I would love to watch these monsters go out to sea for destinations unknown. On a day such as I described above, I would watch these giant ships head across the South Atlantic directly toward the Horizon; and take special note on how the last thing I could see through my binoculars was the extreme top of these vessels until they seemed to drop into a hole. If what I was seeing on days with those perfect conditions wasn't the curvature of the Earth swallowing those ships, please change the caption under the photo to what ever you think works, and thanks for the education, and great conversation. Pocketthis (talk) 00:53, 1 December 2012 (UTC)

Lake Ontario is often very flat in summertime. We do get ocean-going ships sailing past. It's part of the St Lawrence Seaway system, which lets ships travel between the Atlantic Ocean and the western end of Lake Superior. So, yes, I do know what ships look like as they sail over the horizon.

Some years ago, I stayed for a couple of weeks in Playa del Carmen, Mexico, on the Yucatan coast, south of Cancun. There's an island called Cozumel about 20 km offshore, which is a favourite tourist destination. Huge cruise ships sail in and out of there every day. As they do so, seen from the mainland, they are often unrecognizable as ships. They look like big rectangular blocks moving on the horizon. The buildings on Cozumel also look like blocks, but not moving. The block shapes are the result of distortion by refraction. It's impossible to tell, even with binoculars, what part of the ship corresponds to what part of the block. Presumably, the top of the superstructure is the last part to be visible as a ship sails away, but this is not often obvious.

I'm not denying that ships do sometimes look just like textbooks say they look as they go over the horizon. I've seen that too. But it happens only when conditions are close to ideal. Maybe in Florida that is common, but it isn't in many other places.

Refraction distorts the horizon itself, as well as ships. Sometime, usually in summertime, things looking like tsunamis seem to move along the horizon on Lake Ontario. Moving packets of warm or cold air make the horizon appear to move up or down.

I'm not sure what caption I would put on that picture. I don't feel that it shows anything interesting...

DOwenWilliams (talk) 03:53, 1 December 2012 (UTC)

Ahhh...now you've hit on the main problem with associating photos with this article. "Interesting Horizon Shots". I agree. In fact, one of the most boring photos on any Article page is the featured photo here; Photo number one. When I first saw it I said to myself: "Blue on Blue nothing". At least the one in question has some shrubs in it..lol. I'll find an interesting horizon shot buried in my photo pile, scan it and put it up with a caption that doesn't infer curvature. Perhaps a land horizon shot would work there. I've just got to get my lazy butt off this chair and dig through thousands of photos. In the meantime I'll change the caption under the existing photo back to a Travelogue. Thanks DOwen. Pocketthis (talk) 16:25, 1 December 2012 (UTC)

I found an image in Commons of a horizon photographed from a space shuttle. It is obviously curved. So I've put this image in the article to illustrate the paragraph about curvature. Ok? DOwenWilliams (talk) 20:23, 1 December 2012 (UTC)

Thanks for being so considerate, however, you certainly don't have to ask me if it's OK. That section is your baby. If you like that shot....consider the issue closed. I really enjoyed our conversation, and I learned from it as well. That's what it's all about my friend. All the best.....Pocketthis (talk) 01:20, 2 December 2012 (UTC)

I've put up a note requesting some references and more clarity about the statements, to highlight the issues with this section as currently written and I hope encouraging others to respond (and more still not to rely on what is stated)Robma (talk) 11:09, 1 June 2013 (UTC)

Too much math

I think this article is too math heavy, most people that want to know "distance to horizon" are interested in an actual distance answer, and a little theory of how far a you can see. Right now the page is like a physics book. —Preceding unsigned comment added by 83.76.117.159 (talk) 13:45, 16 April 2011 (UTC)

Any better?

As I previously mentioned above, I think the article is a bit heavy on the math, especially in the space devoted to purely geometrical considerations that neglect atmospheric refraction. The problem with the edit of 16 April 2011 is that the simplified formulas given at the top of the section also neglect refraction

I’ve revised the new introduction slightly, using formulas that include the effect of refraction; Young has been widely published on this topic, in books as well as peer-reviewed journals, so his personal web page should be citable under WP:RS. I removed the unsupported qualifier in the interest of simplicity (and of avoiding a tag); Young, Woolard and Clemence, and Yallop and Hohenkerk (in the Explanatory Supplement to the Astronomical Almanac) omit the qualifier, so doing this seems consistent with accepted practice.

Perhaps we should move the examples to the beginning of the section and use the more accurate formulas. We may also need to make it more clear that the formulas in the following subsections do not include the effect of refraction.

I’ve added a link to Sweer’s paper, which appears to be the basis for Woolard and Clemence as well as the Explanatory Supplement. I think this paper might merit mention in the article, perhaps with only the most minimal treatment of the math that I previously discussed above. JeffConrad (talk) 01:56, 17 April 2011 (UTC)

I’ve added a much simplified discussion of Sweer and Young; see if this works. JeffConrad (talk) 07:29, 18 April 2011 (UTC)

The ideal article should cover several levels of math from no or little to intermediate (easily accesible to those who have studied science at high school or college level) and, if necessary advanced (for matematicians, theoretical physicist and people needing to know as much as possible). I think the article has a rather good balance perhaps with too much repetition and non-metric units. — Preceding unsigned comment added by 150.227.15.253 (talk) 09:15, 19 June 2013 (UTC)

Approximate formula

I could not reproduce the (better) approximation d= Sqrt(2Rh+h^2). Expanding Cos or Arccos I find d=Sqrt(2Rh-2h^2). Maybe this is correct? — Preceding unsigned comment added by 144.85.153.27 (talk) 11:41, 9 August 2013 (UTC)

Refraction

I added brief mention at the end of the fact that the actual visual horizon is a little further than the calculation in the article would suggest, because of refraction.

Actually, I would like to have changed the nomenclature completely here, but I didn't feel like taking the time to redraw the picture. The word "visual horizon" ought to include the effect of refraction, while "geometric horizon" is the horizon for a spherical planet, but without the effect of atmosphere. I don't know of a phrase meaning "where the horizon would be if the Earth were a plane," but "geometrical horizon" isn't it.

I think we need to be more clear that the current “theoretical model” doesn't include the effects of refraction; perhaps even better would be to give both geometrical and refracted models and make the distinction clear. At least two approaches are possible for the refracted model:

• The simple “rule of thumb” method described by Andrew T. Young.
• A more exact method, such as described by John Sweer in the September 1938 issue of the Journal of the Optical Society of America.

The best approach would probably be to give both, but if giving only one, I'd go with the former.

I have a problem with “visual” horizon as it's currently indicated—from the observer to an elevated object; I don't think I've ever seen it described this way. I think we should show

• The astronomical (or sensible) horizon (altitude 0°)
• The geometrical horizon
• The “visual” (or “visible”) horizon, which includes the effect of refraction.

JeffConrad (talk) 09:23, 21 October 2010 (UTC)

Section on radar added: Outside the visual wavelength range the refraction will be different. For radar (e.g. for wavelengths 30 to 300 mm i.e. frequencies between 10 and 1 GHz) the radius of the Earth may be multiplied by 4/3 to obtain an effective radius giving a factor of 4.12 in the metric formula i.e. the radar horizon will be 15% beyond the geometrical horizon or 7% beyond the visual. The 4/3 factor is not exact, as in the visual case the refraction depends on atmospheric conditions.

The 4/3 approximation will be found in many radar books e.g. Radar Principles, Nadav Levanon, John Wiley & Sons 1988, ISBN 0-471-85881-1. There seem to be a number of different ways of adding references to Wikipedia. Previously one could often see how other references were inserted and learn by example, but now they are often hidden so I don't know how to include it but give the information here in the talk section.

"I don't know of a phrase meaning "where the horizon would be if the Earth were a plane," but "geometrical horizon" isn't it." But if the Earth were a plane i.e. flat then the horizon would be infinity excepting for any mirage effects. As for geometrical horizon my dictionary has geometrical defined as 'of or relating to or determined by geometry' which seems to suggest to me that the phrase is appropriate. 31.96.241.116 (talk) 19:28, 13 December 2015 (UTC)

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About the "curvature of the horizon" section

Failing to find sources for the formula given in this section, despite thorough researches, I wrote my own derivation and found the result to be true (and, most importantly, why and under which assumptions it is so). For a visitor not to take my word for it, I had left it in the talk page for others to double-check it, which would not be very complicated IMHO as it only implies high-school mathematics. Unfortunately some admin judged that, because of WP:NOR, this derivation doesn't even deserve to be published in this talk page. For those interested, you can retrieve it in the view history (timestamp : 19:17, 13 October 2015‎). My reading of the WP:NOR rule is that it doesn't apply to talk pages so you should be able to show it here.82.25.22.100 (talk) 17:45, 8 December 2015 (UTC)

The point in signaling this is not to argue with the administrators' decisions, but just how can some unverified result be left in the main page (along with nonsensical, ill-formatted gibberish), while at the same time refusing any attempt of clarification ? I understand the reason underlying WP:NOR, but when it comes to mathematics - at least basic ones - a truth is a truth ! This is completely different from an unquoted affirmation in the fields of, say, history or litterature ; or even science, when it comes to some most recent theories and one has to trust a reputed third-part source.

Fact : the page needs to be fixed. Fact : I propose a fix. Personal thought : this fix is relevant. That's because this thought is personal and I know it that I posted it in the talk page. I already made contributions under other pseudonyms, mainly on french pages because english is not my native language, and mainly (if not only) in maths and physics ; never once have I been opposed the WP:NOR rule when proposing a derivation that was deemed true by posters with a knowledge background on the subject, and able to do third-party verifications.

So I kindly ask to any math-qualified person to review this derivation, and maybe if enough people give approval/improvements/troubleshooting on it, it can make it to the main page. Or, after all, let's blindly follow the rules and leave that flawed article as is. Readers knowledge won't be improved but the holy rules will be saved (sorry for the bitter talk but this had to be told). NoahhaoN (talk) 20:30, 13 October 2015 (UTC)

Edit : WP:IAR NoahhaoN (talk) 20:30, 13 October 2015 (UTC)

If the equation is going to be in the article, it seems worth having access to the derivation here in the talk page, if only so that it can be checked or compared to an external source at some point.

Derivation

I have found a derivation of the formula giving the geometrical curvature of the horizon (I had to do it myself because I couldn't find something similar anywhere on the Internet). I submit it here, feel free to put a hide/show paragraph (e.g.) in the right section (if it turns out I made no mistakes...). So, assume the Earth is spherical ; let C be its center, O the position of an observer, looking towards a point A on the horizon line. Define a plane P, normal to (OA) and intersecting [OA) at a distance from O taken for unity. This plane acts like a virtual screen on which we will project the view (at this point, it would be good to draw some scheme). It is then only natural to attach to the observer a vector basis $(e_{x},e_{y},e_{z})$ such that $e_{z}$ would be the unit vector alongside [OA) and $e_{y}$ the unit vector along [AC) -- because [AC) being by construction orthogonal to [OA) is parallel to the projection plane. We would then put $e_{x}=e_{y}\times e_{z}$ to complete this into a direct orthonormal basis (note that in this construction, the positive y-axis points towards the ground for the observer). From now on, we'll express coordinates in the $(O,e_{x},e_{y},e_{z})$ system. Ok so first, notice that every point M in space (apart from O ...) projects itself on a point M' of P, such that if ${\overrightarrow {OM}}=Xe_{x}+Ye_{y}+Ze_{z}=(X,Y,Z)$ then ${\overrightarrow {OM'}}={\frac {1}{Z}}{\overrightarrow {OM}}=(X/Z,Y/Z,1)$. This is not difficult to see, it's kind of a Thales-like relation (though we can derive it more formally by putting OM'=kOM where k is unknown but would be found through O'M'.ez = 0 where O' is the point where (OA) intersects P). The next step is to derive an equation of the line of horizon in our coordinates system. If L be this line, then we have $M\in {\mathcal {L}}\iff (OM)\perp (CM){\text{ and }}CM=R$ -- this translates both the tangency of the sight line to the earth sphere and the fact that M is a point on Earth, and that's all we need to define the horizon. Because, then, OM²+CM²=OM²+R²=OC²=(R+h)² where R is the Earth's radius and h the observer's altitude, the distance OM is the same for every point of L. For convenience we put this distance = $\ell ={\sqrt {(R+h)^{2}-R^{2}}}={\sqrt {2hR+h^{2}}}$. So, L can also be viewed as an intersection between two spheres : ${\mathcal {L}}={\mathcal {S}}(O,\ell )\cap {\mathcal {S}}(C,R)$ (the latter obviously being the Earth itself). So, this translates to : $M\in {\mathcal {L}}\iff OM^{2}=\ell ^{2}{\text{ and }}CM^{2}=R^{2}$ In our coordinates system, where OM²=X²+Y²+Z², the former raises no issue ; the latter is a bit trickier : we decompose ${\overrightarrow {CM}}={\overrightarrow {CA}}+{\overrightarrow {AO}}+{\overrightarrow {OM}}$, then plug in ${\overrightarrow {CA}}=-Re_{y}$ and ${\overrightarrow {AO}}=-\ell e_{z}$ (again, a drawing would help). Developing this yields, after simplification : $\ell ^{2}-RY-\ell Z=0$. Hence the following system characterizes the horizon line : $\ell ^{2}=X^{2}+Y^{2}+Z^{2}=RY+\ell Z$ Now we have to project this on the plane P. Recall that for M=(X,Y,Z), its image M'=(X/Z,Y/Z,1). So if we want an equation for the projected horizon line, all we've got to do is rewrite the above equations in terms of x=X/Z and y=Y/Z. This is done by dividing the first by Z² and the other by Z : $\left\lbrace {\begin{matrix}x^{2}+y^{2}+1={\frac {\ell ^{2}}{Z^{2}}}\\Ry+\ell ={\frac {\ell ^{2}}{Z}}\end{matrix}}\right.$ It is then straightforward that $(Ry+\ell )^{2}=\ell ^{4}/Z^{2}=\ell ^{2}(x^{2}+y^{2}+1)$. Thus, by eliminating Z we obtained the equation in (x,y) which represents the projection of the horizon line on the plane P, which we assumed to be the exact shape of the horizon as viewed by the observer. Since we squared Ry+l, a parasite curve appears, which will be dealt with later. Keeping this in mind, the equation can be rewritten : $\ell ^{2}x^{2}+y^{2}(\ell ^{2}-R^{2})-2\ell Ry=0$ Introducing the factor k=l/R, a few simple manipulations yield : ${\bigl (}y(k-k^{-1})-1{\bigr )}^{2}+x^{2}(k^{2}-1)=1$ To compute the curvature, things will be made easier if we express directly y as a function of x : $y={\frac {k}{k^{2}-1}}{\bigl (}1-{\sqrt {1-x^{2}(k^{2}-1)}}{\bigr )}$ We chose the minus sign in front of the square root, because this is the only choice that yields y=0 when x=0, thus eliminating the parasite branch we mentioned above (now, remember that y is counted positively towards the ground, so if you display the graph in your favorite math software, the curvature you will see will actually point towards the top of the screen). Now, using the formula for the curvature of an y=f(x) curve (which is, indeed, the reciprocal of the radius of the circle that best fits the curve at a given point) : $\gamma (x)={\frac {y''}{(1+y'^{2})^{3/2}}}$ We find that : $\gamma (x)={\frac {k}{(1+x^{2})^{3/2}}}$ So the curvature at the point the observer stares at is $\gamma (0)=k$, and this is indeed the result quoted in the article, because : $k={\frac {\ell }{R}}={\sqrt {\frac {(R+h)^{2}-R^{2}}{R^{2}}}}={\sqrt {\left(1+{\frac {h}{R}}\right)^{2}-1}}$ QED ! NoahhaoN (talk) 19:17, 13 October 2015 (UTC)

There. --tronvillain (talk) 18:20, 12 August 2016 (UTC)

Curvature of the horizon

I am bit a confused about the formula for the curvature of the horizon. Are you sure this section is correct? I am a complete novice to such calculations, but I am interested in space flight so I wanted to do some quick calculations and encountered some problems...

I agree on the part, that in 10 km altitude, the curvature is 0.056. Since the curvature is the reciprocal of the angular radius in radians, this results in 1/0.056 rad = 17,84 rad = 17,84 * 180/PI degrees = 1022 degrees as angular radius.

In 2640 km altitude the curvature is 1. The angular radius then is 1/1 rad = 1 * 180/PI degrees = 57,28 degrees as angular radius. But as far as I understand the text in the section says it should be 45 degrees!?!?

Most probably I am getting something wrong, but it would be great if someone could comment on that.

I am also bit clueless about how to handle "angular radius" since I have not found a clear definition here in Wikipedia. Is there a figure somewhere that explains this concept? Maybe that would make clear where my fault is.

Thanks! Chris

---

Let me add that I also found this section confusing. I think it needs:

• a reference
• hyperlinks to 'curvature' and 'angular diameter' and anything else technical that is referred to here.
• reiteration of the quantities: h is the height above ground, R is the radius of the earth
• An improvement to the 'plain english' explanation. The first sentence is a good start, but it should be clarified that this calculation is the apparent curvature of the circle of the horizon, and not the appearance of the earth's actual curvature (if indeed this is what it is)

thanks tombleyboo

It seems that the equation for the curvature of the horizon is actually just the ratio between the horizon distance and the earth's radius. This is the same as tan(θ) = √(2rh + h²) / R, which is not a measure of curvature. That would be 1/tan⁻¹(√(2rh + h²)/R). An equivalent and simpler expression is 1/sin⁻¹(R/(R + h)).

However, the actual curvature seen is not represented accurately with either expression. Instead, you have to also consider the curvature of the earth's limb and field of view. Such an expression would be:

sin⁻¹(r/(r+h)) − tan⁻¹[rsin(cos⁻¹(r/(r+h)))cos(f/2)/(h+r-r²/(r+h))] where f = your field of view

To calculate the tan⁻¹(...), it would be best to use the atan2 function.

Jmckaskle (talk) 06:55, 28 March 2017 (UTC)

That formula for the apparent curvature of the horizon is based on the altitude of the observer (h) and the radius of the Earth (R). --tronvillain (talk) 14:03, 28 March 2017 (UTC)

Horizon at eye level

Can the horizon be classed as eye level from the observer, it makes more sense since we are talking about the perspective from earth. As believe it or not I have heard flat-earthers, trying to argue absurd ideas. They say they argue about the "North" star. They don't get the idea that the earth wobbles on its axis UserHerName (talk) 20:25, 4 May 2018 (UTC)

I don't think it says that the horizon is "eye level" anywhere in the article, but at sea level if your eyes are at a height of two meters, the difference between the horizon and a line tangential to Earth is going to be a fraction of a degree, which would be indistinguishable. Was there something in the article that needed to be changed? --tronvillain (talk) 21:30, 4 May 2018 (UTC)

My point was can we not include that the horizon is always at eye level from the perspective of the observer? Or would that be inaccurate? UserHerName (talk) 18:20, 5 May 2018 (UTC)

It's approximately at eye level from the perspective of the observer at ground level, but that becomes progressively less accurate with elevation.--tronvillain (talk) 18:26, 5 May 2018 (UTC)

Can we include that in the article? Becuase flat-earthers are using this article to claim that the horizon changes from eye-level at the ground, which isn't true. If we can include that in the article what you just said, it would clear up some of the confusion. Tell me what you think. UserHerName (talk) 19:36, 5 May 2018 (UTC)

I made a edit. Is it good?

So, i made a edit to this page today. I added the actual info first (people should know that the horizon is not just a straight line.), and then explained that viewed from Earth, it divides the sky and the ground. --2601:249:8181:3980:6033:CB67:52CB:36C9 (talk) 19:23, 25 May 2021 (UTC)