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Old 2009-03-16, 13:13   #1
Mini-Geek
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"Tim Sorbera"
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Default area by percent of sun in sky

I'm wondering how much of the sky is taken up by the sun, as seen from earth. From earth, the sun appears approximately 0.5 degrees across. This means the question is equivalent to asking what is the fraction of a disk that is 0.5 degrees on the surface of a sphere, to the surface area of the sphere.
I think it might be 1/720th, or .13888...%, since a half degree line on a circle would be 1/720th of the circumference of the circle, and you could equate a sphere to a series of circles such that each would have 1/720th taken up by this line. Is that right? It seems too low, but then so does that the sun is only half a degree, and that's right.

Last fiddled with by Mini-Geek on 2009-03-16 at 13:14
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Old 2009-03-16, 13:43   #2
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Surface area goes as R^2, so I make it to be 1/518400 or about 1.9e-4 %
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Old 2009-03-16, 14:28   #3
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The surface area of a sphere is 4*pi*r^2, so the unit half-sphere has surface area 2pi. Approximating the sun as a circle, it has area 1/720^2 * pi/4. The ratio is thus 1/(720^2*8) = 1/4147200 ~= 0.0000241%.
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Old 2009-03-16, 16:40   #4
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The number of square degrees on a sphere's surface is 4 * pi * (180 / pi)^2 = 129600 / pi = ~ 41253. ([URL]http://en.wikipedia.org/wiki/Square_degree[/URL]) The Sun's disc is an area of about pi * (1/4 degree)^2, which is about (pi * 1/16 square degree) / (129600 / pi) = pi^2 / 2073600 = ~ 4.76e-6 of the celestial sphere. Now, if [I]sky[/I] is taken to mean only half of the celestial sphere, the the Sun occupies ~9.52e-6 of the sky.

Lessee ... are we all in agreement? Apparently not, since we have four different answers. Back to the calculator ...

Let's try a different method. The Sun is ~865,000 miles wide. Its distance is ~93 million miles. A circle with diameter of 865,000 miles has surface area pi * (865000 / 2)^2 = pi * 187056250000 square miles. A sphere with radius of 93 million miles has surface area 4 * pi * (93,000,000)^2 = pi * 34596000000000000 square miles. The ratio is 187056250000 / 34596000000000000 = 5.4e-6.

Hmmm... My two answers are 4.76e-6 and 5.4e-6. Aha!! The first one assumed a solar diameter of 1/2 degree; it's actually more like 32 arc-minutes = 8/15 of a degree. Correcting the first figure by multiplying by (8/7.5)^2 gives ~5.4e-6, which agrees with the second figure (= ~1.08e-5 of the half-sphere sky).

Last fiddled with by cheesehead on 2009-03-16 at 17:27
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Old 2009-03-16, 17:15   #5
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Quote:
Originally Posted by CRGreathouse View Post
The surface area of a sphere is 4*pi*r^2, so the unit half-sphere has surface area 2pi. Approximating the sun as a circle, it has area 1/720^2 * pi/4. The ratio is thus 1/(720^2*8) = 1/4147200 ~= 0.0000241%.
1/720^2 * pi/4? I can't figure out where you get this number. I get that the image of the sun on the unit half-sphere is approximated by a circle with radius pi/720, so the area of the approximating circle is pi^3/(720^2), so the ratio is pi^2/(2*720^2) ~ 0.0000095

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Old 2009-03-16, 17:16   #6
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Cheesehead: Your initial analysis and mine agree
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Old 2009-03-16, 17:25   #7
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Quote:
Originally Posted by Mini-Geek View Post
I think it might be 1/720th, or .13888...%, since a half degree line on a circle would be 1/720th of the circumference of the circle, and you could equate a sphere to a series of circles such that each would have 1/720th taken up by this line. Is that right? It seems too low, but then so does that the sun is only half a degree, and that's right.
I'm not sure what the answer is yet, but everyone's giving different answers that are in the same general area of being several orders of magnitude smaller than my answer. I'm wondering where the mistake in my logic is. Can not a sphere be equated to a series of circles? Will not each of those circles, if taken as I described, contain 1/720th sun and 719/720ths open sky? Will not that mean the the sphere is 1/720th sun?
I'm thinking that, possibly, equating a sphere to an infinite number of 0 thickness lines with the same radius as the sphere does not result in consistent arithmetic. I think if wanting to equate a sphere to circles with the same radius as the sphere, we would have to instead give them each a finite, however small, thickness based on degrees, which gives us many arcs that are probably harder to calculate for than a sphere, and so are useless.

Last fiddled with by Mini-Geek on 2009-03-16 at 17:31
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Old 2009-03-16, 17:37   #8
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I was going to post, but Orgasmic Troll beat me to it - I agree with him.
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Old 2009-03-16, 17:40   #9
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Quote:
Originally Posted by Mini-Geek View Post
I'm wondering where the mistake in my logic is. Can not a sphere be equated to a series of circles? Will not each of those circles, if taken as I described, contain 1/720th sun and 719/720ths open sky? Will not that mean the the sphere is 1/720th sun?
Here's the mistake: While it's true that the circle that bisects the Sun and goes all the way around will have 1/720th sun and 719/720ths open sky, the other parallel circles above and below that one will have a smaller and smaller length of Sun to cross, until reaching zero for all circles that are more than 1/4 degree above or below the circle that bisects the Sun, so the vast majority of circles contain [I]only[/I] open sky.

And, yes, you have to use only parallel circles (which will have smaller and smaller radii as you go north or south, until reaching zero at 90 degrees above or below the Sun), or you'd be double-counting the (individually infinitesimal, but significant in total) areas where they cross and overlap.

Last fiddled with by cheesehead on 2009-03-16 at 17:51
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Old 2009-03-16, 18:25   #10
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Quote:
Originally Posted by Mini-Geek View Post
I'm wondering where the mistake in my logic is. Can not a sphere be equated to a series of circles?
Consider two concentric (planar) circles radius r and 2r, The area of the larger circle is four times that of the smaller, but your reasoning, applied to radii would imply that it is just two times the size. So what's gone wrong?

The problem is that you can't legitimately reason from length to area in this way. Instead, consider an infinitesimally thin wedge bounded by two radii an infinitesimal angle apart. Just as before, it intersects with the smaller circle over about half its length, but because it's the thin end, that's only a quarter of its area.
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Old 2009-03-16, 22:33   #11
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Quote:
Originally Posted by cheesehead View Post
Here's the mistake: While it's true that the circle that bisects the Sun and goes all the way around will have 1/720th sun and 719/720ths open sky, the other parallel circles above and below that one will have a smaller and smaller length of Sun to cross, until reaching zero for all circles that are more than 1/4 degree above or below the circle that bisects the Sun, so the vast majority of circles contain [I]only[/I] open sky.

And, yes, you have to use only parallel circles (which will have smaller and smaller radii as you go north or south, until reaching zero at 90 degrees above or below the Sun), or you'd be double-counting the (individually infinitesimal, but significant in total) areas where they cross and overlap.
(back to the rotated circles instead of the proper parallel circles) If each has a width of 0, then must they each also have an area, and therefore potential overlap size, of 0 and not some infinitesimal amount? If they did overlap, even completely, wouldn't that mean that 0+0+0+... is not 0 but instead some infinitesimal number?
I know that my original statement was wrong, and I think I get why (thanks for the example Mr. P-1), but it's a bit of a brain twister.
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