<p>From <a href="https://mathworld.wolfram.com/Circle-CircleIntersection.html" rel="nofollow noreferrer">BowlOfRed's Wolfrom link</a>, the area of the asymmetric lense labeled <span class="math-container">$A_1$</span> in the O.P's post is
<span class="math-container">$$A_1 = r^2\cos^{-1}\bigg(\frac{d^2+r^2-R^2}{2dr}\bigg) +R^2\cos^{-1}\bigg(\frac{d^2+R^2-r^2}{2dR}\bigg)- \frac{1}{2}\sqrt{(-d+r+R)(d+r-R)(d-r+R)(d+r+R)}$$</span></p>
<p>Here, <span class="math-container">$d$</span> is the distance between the Sun center and Moon center, <span class="math-container">$R$</span> is the apparent radius of the Sun, and <span class="math-container">$r$</span> is the apparent radius of the Moon.</p>
<p>Then the visible area of the Sun is simply:</p>
<p><span class="math-container">$A_2 = \pi R^2-A_1 $</span></p>
<p>I suspect it's not possible to calculate this simply as a ratio of solid angles.</p>

CC BY-SA 4.0
2023-06-05T21:04:16.743Z
It might be worht including that $r\approx R$ , and this approximation simplifies the formulae somewhat.
James K
CC BY-SA 4.0
2023-06-05T20:56:52.340Z
2023-06-05T20:56:52.340Z
Connor Garcia
<p>Assume a perfectly spherical Sun and Moon, and ignore atmospheric effects for simplicity. I'm an observer in the origin of my coordinate system, and I know the positions of the centers of the Sun and the Moon. I also know their radii.</p>
<p>At some point during an eclipse, my view of the Moon partially overlaps that of the Sun. Without the eclipse occuring, I could see 100% of the Sun in terms of solid angles. But since the view of Moon now partially blocks the Sun, I can see less. How do I calculate how much of the Sun is still visible to me at that time, as a ratio of solid angles, using the position and radii information?</p>
<p>This is not an observation-based question, since if I was actually observing the event, I could simply shoot a few photos and count the number of pixels and save myself the headache. This is more of a pure geometry question in that sense, to which I haven't found a simple and satisfying answer until now - meaning an (or a set of) equation(s) into which I can plug in the positions and the radii and get the result. To me it looks like that much information is just enough to get the answer, I just can't wrap my head around deriving the maths of it.</p>
<p><a href="https://i.sstatic.net/CDhZ5.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CDhZ5.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-06-04T22:43:45.803Z
Might be useful: https://mathworld.wolfram.com/Circle-CircleIntersection.html
BowlOfRed
CC BY-SA 4.0
2023-06-05T00:01:49.963Z
Some of the geometry you need can be found in this answer of mine: https://astronomy.stackexchange.com/a/47731/16685 That tells you how to project  the spheres to circles. BowlOfRed's link tells you how to get the overlap area.
PM 2Ring
CC BY-SA 4.0
2023-06-05T10:59:25.380Z
Do you mean the projected area (2D) or the surface area (3D)?
planetmaker
CC BY-SA 4.0
2023-06-05T17:28:04.623Z
@planetmaker My question is about the ratio of solid angles, so I'm asking the projected area (2D).
arda güler
CC BY-SA 4.0
2023-06-05T22:42:43.333Z
Ptolemy addressed the issue. You might want to read the Almagest, Book 6.
Pierre Paquette
<solar-eclipse><mathematics><occultation><space-geometry>
By analytical geometry, how to calculate the fraction of the visible area of the Sun at a point during an eclipse?
arda güler
# By analytical geometry, how to calculate the fraction of the visible area of the Sun at a point during an eclipse?

(Asked by: username_0 on 2023-06-04)

Assume a perfectly spherical Sun and Moon, and ignore atmospheric effects for simplicity. I'm an observer in the origin of my coordinate system, and I know the positions of the centers of the Sun and the Moon. I also know their radii.

At some point during an eclipse, my view of the Moon partially overlaps that of the Sun. Without the eclipse occuring, I could see 100% of the Sun in terms of solid angles. But since the view of Moon now partially blocks the Sun, I can see less. How do I calculate how much of the Sun is still visible to me at that time, as a ratio of solid angles, using the position and radii information?

This is not an observation-based question, since if I was actually observing the event, I could simply shoot a few photos and count the number of pixels and save myself the headache. This is more of a pure geometry question in that sense, to which I haven't found a simple and satisfying answer until now - meaning an (or a set of) equation(s) into which I can plug in the positions and the radii and get the result. To me it looks like that much information is just enough to get the answer, I just can't wrap my head around deriving the maths of it.

[![enter image description here](https://i.sstatic.net/CDhZ5.png)](https://i.sstatic.net/CDhZ5.png)

**Tags:** solar-eclipse, mathematics, occultation, space-geometry

> **Comments:**
> **username_1**: Might be useful: https://mathworld.wolfram.com/Circle-CircleIntersection.html  
> **username_2**: Some of the geometry you need can be found in this answer of mine: https://astronomy.stackexchange.com/a/47731/16685 That tells you how to project  the spheres to circles. username_1's link tells you how to get the overlap area.  
> **username_3**: Do you mean the projected area (2D) or the surface area (3D)?  
> **username_0**: @username_3 My question is about the ratio of solid angles, so I'm asking the projected area (2D).  
> **username_4**: Ptolemy addressed the issue. You might want to read the Almagest, Book 6.  

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## 1 answers

### Answer by username_5 on 2023-06-05. Score: 2

From [username_1's Wolfrom link](https://mathworld.wolfram.com/Circle-CircleIntersection.html), the area of the asymmetric lense labeled $A\_1$ in the O.P's post is
$$A\_1 = r^2\cos^{-1}\bigg(\frac{d^2+r^2-R^2}{2dr}\bigg) +R^2\cos^{-1}\bigg(\frac{d^2+R^2-r^2}{2dR}\bigg)- \frac{1}{2}\sqrt{(-d+r+R)(d+r-R)(d-r+R)(d+r+R)}$$

Here, $d$ is the distance between the Sun center and Moon center, $R$ is the apparent radius of the Sun, and $r$ is the apparent radius of the Moon.

Then the visible area of the Sun is simply:

$A\_2 = \pi R^2-A\_1 $

I suspect it's not possible to calculate this simply as a ratio of solid angles.

> **Comments:**
> **username_6**: It might be worht including that $r\approx R$ , and this approximation simplifies the formulae somewhat.  

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