<p>The curve you are seeking is just the intersection of a plane (the back of the camera) and a right circular cone. This is not really a question about the earth, or views of planets from space; it's just plain simple 3D coordinate geometry. To find a reference, I'd recommend searching for "intersection of a plane and a cone" or "plane section of a cone", or "plane section of quadric", something like that.</p>

<p>I'd expect that you can find the relevant formulae (and derivations) in any standard text on 3D coordinate geometry. Some likely places are:</p>

<ul>
<li>Salmon -- A Treatise on the Analytic Geometry of Three Dimensions</li>
<li>Somerville -- Analytical Geometry of Three Dimensions</li>
<li>Snyder and Sisam -- Analytic Geometry of Space</li>
</ul>

<p>These are all pretty old books, and you may have trouble finding them. </p>

<p>You might also try asking on Math.StackExchange.</p>

<p>Calling the derivation "original research" seems absurd, to me. It's a high-school homework problem in analytic geometry.</p>

CC BY-SA 3.0
2018-03-25T18:47:14.927Z
Thank you for the answer! I will try to locate these sources.

As far as your statements that this is not really original research or even about the Earth: to see why *as far as Wikipedia goes* both of these are disputable, see [here](https://en.wikipedia.org/wiki/Wikipedia:No_original_research#Synthesis_of_published_material) and also e.g. [here](https://en.wikipedia.org/wiki/Wikipedia:No_original_research#Routine_calculations). Many Wikipedia editors will ultimately agree with you, but some will create problems. The easiest way to deal with the latter ones is to show an appropriate source.
linguisticturn
CC BY-SA 3.0
2017-07-11T13:24:03.003Z
2017-07-11T14:30:23.097Z
2017-07-11T14:30:23.097Z
bubba
<p><strong>What I'm asking for</strong></p>

<p>I stress that I am <strong>not asking for the formula</strong>---I know the formula, and how to derive it. Several different versions of it are reproduced near the end of the post. In fact, someone else has not only derived it as well, but also nicely presented one of the derivations <a href="https://goo.gl/FJjkHQ" rel="nofollow noreferrer">here</a>. </p>

<p>What I need is a <strong><em>reputable source</em></strong> for the formula so that, for example, one could put it on Wikipedia without violating its <a href="https://en.wikipedia.org/wiki/Wikipedia:No_original_research" rel="nofollow noreferrer">ban on reporting original research.</a> [People have <a href="https://en.wikipedia.org/wiki/Talk:Horizon#About_the_.22curvature_of_the_horizon.22_section" rel="nofollow noreferrer">actually tried</a>... But the <a href="https://en.wikipedia.org/wiki/Horizon" rel="nofollow noreferrer">relevant article</a> has some very conscientious editor who deleted the section on the grounds of it being original research... and, well, unfortunately, the editor is correct, so there's not much point in trying to fight it.]</p>

<p><strong>The reason I'm posting in Computer Graphics stackexchange</strong></p>

<p>Since someone here might have modeled the way the Earth looks form orbit, perhaps he or she might know if this formula (or, more likely, some generalization of it) is published in some book, or journal, or conference proceedings, or class notes, etc.</p>

<p><strong>I've done the "due googling"</strong></p>

<p>Please understand that I'm not asking anyone to go searching for the answer on my behalf. I've done <em>lots</em> of googling already, and am only posting here as a last resort. My (far-fetched) hope is that someone here will simply <em>know</em> a reference right off the bat; if not... well, I hope at least you enjoyed the pretty picture below (if I do say so myself, with full awareness I'm talking to people interested in <em>computer graphics</em> of all things) before you moved on to bigger and better things.</p>

<p><strong>Two sources that come close</strong></p>

<ol>
<li><p>D. K. Lynch, "Visually discerning the curvature of the Earth," Applied Optics vol. 47, H39 (2008). It is freely available <a href="http://thulescientific.com/Lynch%20Curvature%202008.pdf" rel="nofollow noreferrer">here</a>. Unfortunately, instead of doing it the right way (which is not that hard), the author opted for a hack, which (a) I don't completely understand, and (b) which doesn't agree with what I know to be the correct formula.</p></li>
<li><p>R. Hartley and A. Zisserman, <a href="https://rads.stackoverflow.com/amzn/click/0521540518" rel="nofollow noreferrer"><em>Multiple View Geometry in Computer Vision,</em></a> 2nd ed. (Cambridge University Press, Cambridge UK, 2004). In Sec. 8.3, "Action of a projective camera on quadrics," we <a href="https://goo.gl/IHvX2d" rel="nofollow noreferrer">read</a>:</p></li>
</ol>

<blockquote>
  <p>Suppose the quadric is a sphere, then the cone of rays between the
  camera centre and quadric is right-circular, i.e. the contour
  generator is a circle, with the plane of the circle orthogonal to the
  line joining the camera and sphere centres. This can be seen from the
  rotational symmetry of the geometry about this line. The image of the
  sphere is obtained by intersecting the cone with the image plane. It
  is clear that this is a classical conic section, so that the apparent
  contour of a sphere is a conic.</p>
</blockquote>

<p>In principle, this would be exactly what is needed, if only just a bit more information were included---at least an expression for the eccentricity of the conic as a function of the distance to the sphere and the sphere radius (in the case when the image plane is perpendicular to a <a href="https://en.wikipedia.org/wiki/Generatrix" rel="nofollow noreferrer">generatrix</a> of the cone, as is the case when the pinhole camera is directed at a point on the horizon).</p>

<p><strong>Details about the formula for which I need a scholarly reference</strong></p>

<p>We assume a perfectly spherical, perfectly smooth Earth with no atmosphere. We point an idealized pinhole camera at the horizon, and, using straightforward central projection, compute the shape of the image of the horizon on the back of the camera (i.e. the shape it will have on film---the "film plane"). Here's a graphic (made in <a href="http://asymptote.sourceforge.net/" rel="nofollow noreferrer">Asymptote</a>, for those interested) that should make this clearer:</p>

<p><a href="https://i.sstatic.net/Shaud.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Shaud.png" alt="enter image description here"></a></p>

<p>As we saw above, the image of the horizon is a portion of a conic section. Let $\varepsilon$ be the eccentricity of the conic; the derivation <a href="https://goo.gl/FJjkHQ" rel="nofollow noreferrer">I mentioned above</a> instead uses a parameter $k$, which is just the inverse eccentricity: $k=1/\varepsilon$. The eccentricity itself is given as $\varepsilon=1/\sqrt{\epsilon(2+\epsilon)}$, where $\epsilon=h/R$ is the ratio of the altitude $h$ of the pinhole above the surface of the Earth and the Earth radius $R$. [Instead of using $\epsilon$, which is the ratio  of the <em>altitude</em> to $R$, it may be useful to use $\eta$, the ratio of <em>the pinhole's distance to the Earth's center</em>, $h+R$, to the Earth's radius: $\eta=(R+h)/R=1+\epsilon$. In terms of $\eta$, we have $\varepsilon=1/\sqrt{\eta^{2}-1}$.]</p>

<p>The distance from the pinhole (point $P$ in the graphic) to the film plane is taken to be one unit length.</p>

<p>The $y$-axis in the film plane is chosen to be parallel to the line joining the center of the Earth $C$ (not shown in the image) and the point on the horizon (labeled $V$ in the image) at which the camera is trained. This choice is well-defined because the line $CV$ must be parallel to the film plane. The reason for this is that both $CV$ and the film plane are perpendicular to the line of sight $PV$ (the line joining $P$ and $V$). And <em>that</em> is because 1. the line $PV$ is tangent to the Earth at $V$, thus perpendicular to $CV$, and 2. $PV$ is perpendicular to the film plane because the camera is trained at $V$. The $x$ axis is of course perpendicular to the $y$ axis and lies in the film plane, and the origin is chosen as the projection of the point $V$.</p>

<p>With these definitions out of the way, we are ready to write down a representation of the conic section that is the image of the Earth's horizon. This can be written in many ways, some of which are given below. <strong><em>What I need is a reputable reference for any one of these formulas, or for a formula equivalent to them.</em></strong></p>

<p><strong>1. The explicit formula given in the derivation mentioned above</strong></p>

<p>The derivation <a href="https://goo.gl/FJjkHQ" rel="nofollow noreferrer">I mentioned above</a> gives this as the final version:</p>

<p>$[\,y\,(1/\varepsilon-\varepsilon)-1\,]^{2}+x^{2}(1/\varepsilon^{2}-1)=1.$</p>

<p>Let's represent this in a couple of additional ways.</p>

<p><strong>2. Expression in terms of the canonical equation of a conic section</strong></p>

<p>In this case, the equation takes <a href="https://goo.gl/wIPjWs" rel="nofollow noreferrer">the following form</a>:</p>

<p>$x^2=2\mu y-(1-\varepsilon^{2}) y^2$,</p>

<p>where, in our case, $\mu=\varepsilon$. </p>

<p>The advantage of the canonical form is that it can deal with all conics on an equal footing, including in particular the case of the parabola, $\varepsilon=1$. In the ``standard'' formulation (see below), the case of the parabola can only be dealt with by taking the limit $\varepsilon\to 1$.</p>

<p>Details: the above formula holds in the case of a right circular cone, whose sides subtend an angle of $2\theta$, being intersected---at a distance $d$ from the vertex of the cone---by a plane at an angle $\omega$ relative to the cone axis. (To clarify: $d$ is the distance from the cone vertex to the point on the ellipse that is the closest to the cone vertex; that point is always one of the ends of the major axis of the ellipse). In this general case, the eccentricity is given as $\varepsilon=\cos \omega/\cos \theta$, while $\mu=d(\varepsilon-\cos|\omega+\theta|)$.</p>

<p>In terms of the above graphic: $d$ is the distance from $P$ to the film plane (i.e., the distance along the dotted red line); $\theta$ is the angle between the dotted red line and the axis of the cone (which is the line joining $P$ and the center of the Earth---the extension of the black line labeled $h$ in the graphic); the angle $\omega$ is the angle between the axis of the cone and the film plane. </p>

<p>Given that the film plane is perpendicular to the dotted red line, we have $\omega+\theta=\pi/2$; in addition, we take $d=1$, which then together give that $\mu=\varepsilon$. </p>

<p><strong>3. Expression in terms of the ``standard form' of a conic section</strong></p>

<p>This form is perhaps the most familiar:</p>

<p>$\frac{(x-x_{0})^{2}}{p^2}+\frac{(y-y_{0})^{2}}{q^2}=1$.</p>

<p>It is related to the parameters entering the canonical equation (see 2., above) as follows:</p>

<p>$x_{0}=0$;</p>

<p>$y_{0}=q=\frac{\mu}{1-\varepsilon^{2}}$ (which is $\frac{\varepsilon}{1-\varepsilon^{2}}$ in our case&mdash;note that $y_{0}=q$ follows from the fact that the ellipse passes through the origin); and</p>

<p>$p=q\smash[b]{\sqrt{|1-\varepsilon^{2}|}}=\frac{\mu}{\smash[b]{\sqrt{|1-\varepsilon^{2}|}}}$ (which is $\frac{\varepsilon}{\smash[b]{\sqrt{|1-\varepsilon^{2}|}}}$ in our case).</p>

<p>It is obvious that the parabolic case, $\varepsilon=1$, will create problems; as mentioned above, that case must be dealt with through taking the limit $\varepsilon\to 1$.</p>

<p><strong>4. Expression in terms of a parametric curve</strong></p>

<p>$x=\frac{(\epsilon+1) \cos (\alpha )}{\sin (\alpha )+\epsilon(\epsilon+2)}$</p>

<p>$y=\frac{\sqrt{\epsilon (\epsilon+2)} (\sin (\alpha )-1)}{\sin (\alpha )+\epsilon (\epsilon+2)},$</p>

<p>where $\alpha$ is the longitude of a point on the horizon, defined so that $\alpha=\pi/2$ corresponds to the point $V$ in the image above (i.e. to the point at which the pinhole camera is trained).</p>

<p>For how one might use these formulas, see <a href="https://i.sstatic.net/FMWcM.jpg" rel="nofollow noreferrer">this</a>.</p>

<p><strong>In conclusion...</strong></p>

<p>Has anyone seen the formulas above in some reputable source, possibly in the context of modeling how the Earth looks from space? If so, could you let me know what this source was?</p>

<p>Thanks!</p>

CC BY-SA 3.0
2018-03-25T18:18:21.953Z
@Pikalek Sorry for it took me almost a year to reply. Remember what is at issue: I want to put this formula into the Wikipedia article about the shape of the Earth's horizon. But there are editors who won't let me do this until I find a *reputable source* saying that this is indeed the correct formula for the horizon. It is not about whether the formula is *correct*. What you post on Wikipedia should of course be correct, but that is not enough. It should also be 'attributable to a reliable, published source'. Otherwise it is 'original research', which cannot be posted on Wikipedia.
linguisticturn
CC BY-SA 3.0
2017-07-11T18:24:20.763Z
@linguisticturn Have you tried contacting Hartley &/or Zisserman for the additional details you require? A quick search brought up contact info for both.
Pikalek
CC BY-SA 3.0
2016-03-13T20:29:40.783Z
@trichoplax 1. Why now, after all this time? 2. Moreover, the response to the [meta-question](http://meta.computergraphics.stackexchange.com/questions/211/are-requests-for-reputable-sources-on-topic/212#212), while not firm, was leaning towards allowing the question. 3. Finally, as I have explained there, the argument about "opinionated answers and spam" is completely inapplicable in the case of requests for reputable sources for a specific fact. Either there is such a source or not.
linguisticturn
CC BY-SA 3.0
2016-03-14T00:10:10.030Z
If anyone disagrees that off site resource requests should be off topic, they can have their say in [Are questions asking for off site resources on topic?](http://meta.computergraphics.stackexchange.com/questions/146/are-questions-asking-for-off-site-resources-on-topic) If anyone thinks requests for reputable sources should be an exception to the off site resources rule, they can have their say in [Are requests for reputable sources on topic?](http://meta.computergraphics.stackexchange.com/questions/211/are-requests-for-reputable-sources-on-topic)
trichoplax is on Codidact now
CC BY-SA 3.0
2016-03-20T14:23:28.137Z
@trichoplax You yourself were uncertain whether this question is really a request for off-site resources, and you say that uncertainty remains. Is it a standard practice on this stackexchange that, when there is uncertainty about whether a question is off-topic, to err on the side of it being off-topic?
linguisticturn
CC BY-SA 3.0
2016-03-23T15:34:44.407Z
You make a good point, which I've taken some time to think about. In the absence of any community support for excluding such questions, I have reopened this one.
trichoplax is on Codidact now
CC BY-SA 3.0
2016-03-24T05:29:00.690Z
This is allmost certainly never going to be answered. Besides what does it have to do with 3d graphics.
joojaa
CC BY-SA 3.0
2016-03-25T02:07:53.613Z
@joojaa 1. you are probably right that if it hadn't been answered up to this point, it's has low probability of ever being answered. But (a) "low" is not zero, and (b) there was no way to estimate this probability before posting the question.
linguisticturn
CC BY-SA 3.0
2016-03-25T02:09:05.650Z
@joojaa 2. I have explained my rationale for submitting this question to this stackexchange in the post itself, section **The reason I'm posting in Computer Graphics stackexchange**. Moreover, a book (_Multiple View Geometry in Computer Vision_), which isn't quite computer graphics but is surely quite related to it, does come pretty close to giving the answer, I would suggest that this is a reasonably strong argument that the question was indeed appropriate (see section **Two sources that come close**).
linguisticturn
CC BY-SA 3.0
2016-03-25T15:25:50.633Z
Have you tried posting this in some approriate journal. This should immediately bring oout the trolls that know it has been addressed.
joojaa
CC BY-SA 3.0
2016-03-26T19:52:22.037Z
@joojaa No, but I would be thankful indeed if your can think of some journal which might be appropriate---I'm certainly willing to try...  By the way, I'm not familiar with journal websites offering this sort of discussion as an option... do you happen to know of examples?
linguisticturn
CC BY-SA 3.0
2016-03-31T21:08:16.023Z
Not sure if this helps, but the book [3D Engine Design for Virtual Globes](http://www.virtualglobebook.com/) includes a chapter on GPU ray-casting the globe, which involves computing the horizon in the fragment shader.  Disclaimer, I know the authors of this book, after they wrote it they went on to be the founders of [Cesium](http://cesiumjs.org).
emackey
CC BY-SA 3.0
2016-11-29T21:57:53.133Z
Q&As like these are always difficult for moderators. Apart from the asking for sources - which is kind of begging the question - this is a gem. I wouldn't mind seeing the 'question' protected', until there's something like a S.O.'s documentation implemented. Maybe that's not a bad place for it - under some 3D tags?
Brett Hale
<projections><simulation>
Need a reputable source for the formula for the shape of Earth's horizon
linguisticturn
# Need a reputable source for the formula for the shape of Earth's horizon

(Asked by: username_0 on 2016-01-29)

**What I'm asking for**

I stress that I am **not asking for the formula**---I know the formula, and how to derive it. Several different versions of it are reproduced near the end of the post. In fact, someone else has not only derived it as well, but also nicely presented one of the derivations [here](https://goo.gl/FJjkHQ).

What I need is a ***reputable source*** for the formula so that, for example, one could put it on Wikipedia without violating its [ban on reporting original research.](https://en.wikipedia.org/wiki/Wikipedia:No_original_research) [People have [actually tried](https://en.wikipedia.org/wiki/Talk:Horizon#About_the_.22curvature_of_the_horizon.22_section)... But the [relevant article](https://en.wikipedia.org/wiki/Horizon) has some very conscientious editor who deleted the section on the grounds of it being original research... and, well, unfortunately, the editor is correct, so there's not much point in trying to fight it.]

**The reason I'm posting in Computer Graphics stackexchange**

Since someone here might have modeled the way the Earth looks form orbit, perhaps he or she might know if this formula (or, more likely, some generalization of it) is published in some book, or journal, or conference proceedings, or class notes, etc.

**I've done the "due googling"**

Please understand that I'm not asking anyone to go searching for the answer on my behalf. I've done *lots* of googling already, and am only posting here as a last resort. My (far-fetched) hope is that someone here will simply *know* a reference right off the bat; if not... well, I hope at least you enjoyed the pretty picture below (if I do say so myself, with full awareness I'm talking to people interested in *computer graphics* of all things) before you moved on to bigger and better things.

**Two sources that come close**

1. D. K. Lynch, "Visually discerning the curvature of the Earth," Applied Optics vol. 47, H39 (2008). It is freely available [here](http://thulescientific.com/Lynch%20Curvature%202008.pdf). Unfortunately, instead of doing it the right way (which is not that hard), the author opted for a hack, which (a) I don't completely understand, and (b) which doesn't agree with what I know to be the correct formula.
2. R. Hartley and A. Zisserman, [*Multiple View Geometry in Computer Vision,*](https://rads.stackoverflow.com/amzn/click/0521540518) 2nd ed. (Cambridge University Press, Cambridge UK, 2004). In Sec. 8.3, "Action of a projective camera on quadrics," we [read](https://goo.gl/IHvX2d):

> Suppose the quadric is a sphere, then the cone of rays between the
> camera centre and quadric is right-circular, i.e. the contour
> generator is a circle, with the plane of the circle orthogonal to the
> line joining the camera and sphere centres. This can be seen from the
> rotational symmetry of the geometry about this line. The image of the
> sphere is obtained by intersecting the cone with the image plane. It
> is clear that this is a classical conic section, so that the apparent
> contour of a sphere is a conic.

In principle, this would be exactly what is needed, if only just a bit more information were included---at least an expression for the eccentricity of the conic as a function of the distance to the sphere and the sphere radius (in the case when the image plane is perpendicular to a [generatrix](https://en.wikipedia.org/wiki/Generatrix) of the cone, as is the case when the pinhole camera is directed at a point on the horizon).

**Details about the formula for which I need a scholarly reference**

We assume a perfectly spherical, perfectly smooth Earth with no atmosphere. We point an idealized pinhole camera at the horizon, and, using straightforward central projection, compute the shape of the image of the horizon on the back of the camera (i.e. the shape it will have on film---the "film plane"). Here's a graphic (made in [Asymptote](http://asymptote.sourceforge.net/), for those interested) that should make this clearer:

[![enter image description here](https://i.sstatic.net/Shaud.png)](https://i.sstatic.net/Shaud.png)

As we saw above, the image of the horizon is a portion of a conic section. Let $\varepsilon$ be the eccentricity of the conic; the derivation [I mentioned above](https://goo.gl/FJjkHQ) instead uses a parameter $k$, which is just the inverse eccentricity: $k=1/\varepsilon$. The eccentricity itself is given as $\varepsilon=1/\sqrt{\epsilon(2+\epsilon)}$, where $\epsilon=h/R$ is the ratio of the altitude $h$ of the pinhole above the surface of the Earth and the Earth radius $R$. [Instead of using $\epsilon$, which is the ratio of the *altitude* to $R$, it may be useful to use $\eta$, the ratio of *the pinhole's distance to the Earth's center*, $h+R$, to the Earth's radius: $\eta=(R+h)/R=1+\epsilon$. In terms of $\eta$, we have $\varepsilon=1/\sqrt{\eta^{2}-1}$.]

The distance from the pinhole (point $P$ in the graphic) to the film plane is taken to be one unit length.

The $y$-axis in the film plane is chosen to be parallel to the line joining the center of the Earth $C$ (not shown in the image) and the point on the horizon (labeled $V$ in the image) at which the camera is trained. This choice is well-defined because the line $CV$ must be parallel to the film plane. The reason for this is that both $CV$ and the film plane are perpendicular to the line of sight $PV$ (the line joining $P$ and $V$). And *that* is because 1. the line $PV$ is tangent to the Earth at $V$, thus perpendicular to $CV$, and 2. $PV$ is perpendicular to the film plane because the camera is trained at $V$. The $x$ axis is of course perpendicular to the $y$ axis and lies in the film plane, and the origin is chosen as the projection of the point $V$.

With these definitions out of the way, we are ready to write down a representation of the conic section that is the image of the Earth's horizon. This can be written in many ways, some of which are given below. ***What I need is a reputable reference for any one of these formulas, or for a formula equivalent to them.***

**1. The explicit formula given in the derivation mentioned above**

The derivation [I mentioned above](https://goo.gl/FJjkHQ) gives this as the final version:

$[\,y\,(1/\varepsilon-\varepsilon)-1\,]^{2}+x^{2}(1/\varepsilon^{2}-1)=1.$

Let's represent this in a couple of additional ways.

**2. Expression in terms of the canonical equation of a conic section**

In this case, the equation takes [the following form](https://goo.gl/wIPjWs):

$x^2=2\mu y-(1-\varepsilon^{2}) y^2$,

where, in our case, $\mu=\varepsilon$.

The advantage of the canonical form is that it can deal with all conics on an equal footing, including in particular the case of the parabola, $\varepsilon=1$. In the ``standard'' formulation (see below), the case of the parabola can only be dealt with by taking the limit $\varepsilon\to 1$.

Details: the above formula holds in the case of a right circular cone, whose sides subtend an angle of $2\theta$, being intersected---at a distance $d$ from the vertex of the cone---by a plane at an angle $\omega$ relative to the cone axis. (To clarify: $d$ is the distance from the cone vertex to the point on the ellipse that is the closest to the cone vertex; that point is always one of the ends of the major axis of the ellipse). In this general case, the eccentricity is given as $\varepsilon=\cos \omega/\cos \theta$, while $\mu=d(\varepsilon-\cos|\omega+\theta|)$.

In terms of the above graphic: $d$ is the distance from $P$ to the film plane (i.e., the distance along the dotted red line); $\theta$ is the angle between the dotted red line and the axis of the cone (which is the line joining $P$ and the center of the Earth---the extension of the black line labeled $h$ in the graphic); the angle $\omega$ is the angle between the axis of the cone and the film plane.

Given that the film plane is perpendicular to the dotted red line, we have $\omega+\theta=\pi/2$; in addition, we take $d=1$, which then together give that $\mu=\varepsilon$.

**3. Expression in terms of the ``standard form' of a conic section**

This form is perhaps the most familiar:

$\frac{(x-x\_{0})^{2}}{p^2}+\frac{(y-y\_{0})^{2}}{q^2}=1$.

It is related to the parameters entering the canonical equation (see 2., above) as follows:

$x\_{0}=0$;

$y\_{0}=q=\frac{\mu}{1-\varepsilon^{2}}$ (which is $\frac{\varepsilon}{1-\varepsilon^{2}}$ in our case—note that $y\_{0}=q$ follows from the fact that the ellipse passes through the origin); and

$p=q\smash[b]{\sqrt{|1-\varepsilon^{2}|}}=\frac{\mu}{\smash[b]{\sqrt{|1-\varepsilon^{2}|}}}$ (which is $\frac{\varepsilon}{\smash[b]{\sqrt{|1-\varepsilon^{2}|}}}$ in our case).

It is obvious that the parabolic case, $\varepsilon=1$, will create problems; as mentioned above, that case must be dealt with through taking the limit $\varepsilon\to 1$.

**4. Expression in terms of a parametric curve**

$x=\frac{(\epsilon+1) \cos (\alpha )}{\sin (\alpha )+\epsilon(\epsilon+2)}$

$y=\frac{\sqrt{\epsilon (\epsilon+2)} (\sin (\alpha )-1)}{\sin (\alpha )+\epsilon (\epsilon+2)},$

where $\alpha$ is the longitude of a point on the horizon, defined so that $\alpha=\pi/2$ corresponds to the point $V$ in the image above (i.e. to the point at which the pinhole camera is trained).

For how one might use these formulas, see [this](https://i.sstatic.net/FMWcM.jpg).

**In conclusion...**

Has anyone seen the formulas above in some reputable source, possibly in the context of modeling how the Earth looks from space? If so, could you let me know what this source was?

Thanks!

**Tags:** projections, simulation

> **Comments:**
> **username_0**: @username_2 1. Why now, after all this time? 2. Moreover, the response to the [meta-question](http://meta.computergraphics.stackexchange.com/questions/211/are-requests-for-reputable-sources-on-topic/212#212), while not firm, was leaning towards allowing the question. 3. Finally, as I have explained there, the argument about "opinionated answers and spam" is completely inapplicable in the case of requests for reputable sources for a specific fact. Either there is such a source or not.  
> **username_2**: If anyone disagrees that off site resource requests should be off topic, they can have their say in [Are questions asking for off site resources on topic?](http://meta.computergraphics.stackexchange.com/questions/146/are-questions-asking-for-off-site-resources-on-topic) If anyone thinks requests for reputable sources should be an exception to the off site resources rule, they can have their say in [Are requests for reputable sources on topic?](http://meta.computergraphics.stackexchange.com/questions/211/are-requests-for-reputable-sources-on-topic)  
> **username_0**: @username_2 You yourself were uncertain whether this question is really a request for off-site resources, and you say that uncertainty remains. Is it a standard practice on this stackexchange that, when there is uncertainty about whether a question is off-topic, to err on the side of it being off-topic?  
> **username_2**: You make a good point, which I've taken some time to think about. In the absence of any community support for excluding such questions, I have reopened this one.  
> **username_3**: This is allmost certainly never going to be answered. Besides what does it have to do with 3d graphics.  
> **username_0**: @username_3 1. you are probably right that if it hadn't been answered up to this point, it's has low probability of ever being answered. But (a) "low" is not zero, and (b) there was no way to estimate this probability before posting the question.  
> **username_0**: @username_3 2. I have explained my rationale for submitting this question to this stackexchange in the post itself, section **The reason I'm posting in Computer Graphics stackexchange**. Moreover, a book (_Multiple View Geometry in Computer Vision_), which isn't quite computer graphics but is surely quite related to it, does come pretty close to giving the answer, I would suggest that this is a reasonably strong argument that the question was indeed appropriate (see section **Two sources that come close**).  
> **username_3**: Have you tried posting this in some approriate journal. This should immediately bring oout the trolls that know it has been addressed.  
> **username_0**: @username_3 No, but I would be thankful indeed if your can think of some journal which might be appropriate---I'm certainly willing to try...  By the way, I'm not familiar with journal websites offering this sort of discussion as an option... do you happen to know of examples?  
> **username_4**: Not sure if this helps, but the book [3D Engine Design for Virtual Globes](http://www.virtualglobebook.com/) includes a chapter on GPU ray-casting the globe, which involves computing the horizon in the fragment shader.  Disclaimer, I know the authors of this book, after they wrote it they went on to be the founders of [Cesium](http://cesiumjs.org).  
> **username_5**: Q&As like these are always difficult for moderators. Apart from the asking for sources - which is kind of begging the question - this is a gem. I wouldn't mind seeing the 'question' protected', until there's something like a S.O.'s documentation implemented. Maybe that's not a bad place for it - under some 3D tags?  
> **username_1**: @username_0 Have you tried contacting Hartley &/or Zisserman for the additional details you require? A quick search brought up contact info for both.  
> **username_0**: @username_1 Sorry for it took me almost a year to reply. Remember what is at issue: I want to put this formula into the Wikipedia article about the shape of the Earth's horizon. But there are editors who won't let me do this until I find a *reputable source* saying that this is indeed the correct formula for the horizon. It is not about whether the formula is *correct*. What you post on Wikipedia should of course be correct, but that is not enough. It should also be 'attributable to a reliable, published source'. Otherwise it is 'original research', which cannot be posted on Wikipedia.  

---

## 1 answers

### Answer by username_6 on 2017-07-11. Score: 4

The curve you are seeking is just the intersection of a plane (the back of the camera) and a right circular cone. This is not really a question about the earth, or views of planets from space; it's just plain simple 3D coordinate geometry. To find a reference, I'd recommend searching for "intersection of a plane and a cone" or "plane section of a cone", or "plane section of quadric", something like that.

I'd expect that you can find the relevant formulae (and derivations) in any standard text on 3D coordinate geometry. Some likely places are:

* Salmon -- A Treatise on the Analytic Geometry of Three Dimensions
* Somerville -- Analytical Geometry of Three Dimensions
* Snyder and Sisam -- Analytic Geometry of Space

These are all pretty old books, and you may have trouble finding them.

You might also try asking on Math.StackExchange.

Calling the derivation "original research" seems absurd, to me. It's a high-school homework problem in analytic geometry.

> **Comments:**
> **username_0**: Thank you for the answer! I will try to locate these sources.  As far as your statements that this is not really original research or even about the Earth: to see why *as far as Wikipedia goes* both of these are disputable, see [here](https://en.wikipedia.org/wiki/Wikipedia:No_original_research#Synthesis_of_published_material) and also e.g. [here](https://en.wikipedia.org/wiki/Wikipedia:No_original_research#Routine_calculations). Many Wikipedia editors will ultimately agree with you, but some will create problems. The easiest way to deal with the latter ones is to show an appropriate source.  

---

