# How much of the sky is visible from a particular location?

From a particular point how much of the sky can be observed. For simplicity sake let us assume the particular point is the head of a 6 foot tall man floating in the middle of the ocean with no visible barriers between him and the horizon. If he stays in exactly the same location being able to turn 360 degrees how much of the sky is visible?

Would we calculate the circumference of the circle with the distance to the horizon as the radius? The surface area of a portion of a sphere?

Taking it further what percentage of the total sky can be seen at any given time. The answer above divided by the total surface area of earths atmosphere?

• When I float in the ocean, my head's not six feet above the surface … – rob Aug 3 '14 at 17:32
• rob, OK, so consider this friend of yours: google.cz/… – Luboš Motl Aug 3 '14 at 17:33
• Assuming your head is not as dense as rob's and you don't sink the hypothetical boat, or you are able to float like our friend the giraffe how then would we best calculate this? – blissorblight Aug 3 '14 at 18:22
• Perhaps you can clarify what you mean by "percentage of the total sky" - it seems you are not liking the answers given so far much (please comment to let the people who bothered to write an answer know what is still missing... there seems to be a gap between what you are looking for, and what they have written in explanation). – Floris Aug 4 '14 at 14:15
• @Floris I agree. I tried to answer what you describe in your last paragraph but apparently it was not popular. – Enredanrestos Aug 4 '14 at 21:15

Ignoring refraction (which will be important at the scales we discuss below, so ought not be ignored in actuality, see Floris' answer ), this is a geometry problem. Whenever you are trying to calculate a result, it is always good to start by reasoning out some limiting cases. In one limit, we can imagine being infinitely far away from the surface, where surely we will be able to see the whole sky. In the other limit, we can imagine being infinitesimally close to the surface, at which point our planet will appear to be an infinite wall and we will see precisely half of the surface. We expect the answer to be monotonic in between. Let's try to work it out.

At any height above the surface, our planet, assuming its a sphere, will block out a circle of possible sky from our view. If we assume that the edge of the circle, forms and angle of $\theta$ with respect to its center, by our view, the solid angle blocked by our planet is $$\int_{0}^{\theta} d\theta\, \sin\theta \int_0^{2\pi} d\phi = 2\pi ( 1 - \cos \theta)$$

and the full sky is $4\pi$ steradians, so the fraction of the sky we can see is

$$S(\theta) = \frac 12 \left( 1 + \cos \theta \right)$$

Let's check to make sure this agrees with our intuition. When we are on the surface, $\theta$, the angle to the horizon would be $\pi/2$ and we get $S(\pi/2) = \frac 12$ as expected, and if we are infinitely far away, $\theta \to 0$ and we get $S(\infty) \to 1$ as we expect. Perfect. It remains to determine $\theta$ as a function of height. For that, we rely on a diagram:

We know that where our gaze meets the planet, the line should be tangent to the circle, meaning our triangle is a right triangle and it is easy for us to determine:

$$\cos \theta = \frac{ \sqrt{ (r+h)^2 - r^2 } }{ r + h }$$

Giving us for our fraction of the sky visible:

$$S(\theta) = \frac 12 \left( 1 + \frac{ \sqrt{ 2 r h + h^2 }}{ r + h } \right)$$

Plotting this function over large distances, we see that it behaves as we expect:

It starts at 50% and goes to 100% as we get to distances a few earth radii out.

But you were interested in very short distances:

Here we can see the effect of small height changes. In particular, for your 6 foot scenario, you can see

$$S(6 \text{ feet}) = 50.04 \%$$ of the sky from purely geometric considerations. Hardly an improvement over the $50\%$ visible if you were against a wall, and we can't expect this to contribute anything appreciable in light of optical effects like refraction or the effect of a nonuniform horizon. But, given the nature of the square root dependence, this sky viewing function grows rather rapidly over modest height differences:

Here I've plotted the semilog plot against the $x$ axis. For instance, at the cruising altitude of a commercial plane (~ 38,000 ft), you can see 53% of the sky, a whole 3% more than on the ground from purely geometric considerations. This we could hope to notice over optical effects.

• Given the precision of your calculation, "ignoring refraction" is a rather large omission... – Floris Aug 4 '14 at 3:46
• @Floris I know, but you talked about refraction and I thought the geometry was interesting in its own right. The precision was not meant to be taken literally, but to give a sense of the scale of growth in a purely geometric fashion. I'll edit a bit to emphasize. – alemi Aug 4 '14 at 3:50
• OK now your answer makes sense. I wonder what the person who is doing all the downvoting is thinking?... – Floris Aug 4 '14 at 14:13

In principle from the surface of a flat ocean you can see $2\pi$ of the $4\pi$ that is the full celestial sphere. However there is some refraction of the atmosphere (you can see that from the shape of the setting sun...) - this actually means you can see a little bit more than half the entire sky. At the level of the horizon the refraction is just over half a degree. From http://en.m.wikipedia.org/wiki/Atmospheric_refraction :

... only 5.3′ at 10° altitude; it quickly increases as altitude decreases, reaching 9.9′ at 5° altitude, 18.4′ at 2° altitude, and 35.4′ at the horizon

Of course if you then wait for the earth to rotate you will see more, depending on your latitude. And some of the sky may not be visible until you wait a whole year. And if you float long enough, the precession of the earth will give you and even more complete picture of the night sky. After twenty thousand years your skin may have gotten a bit crinkly, floating in the water...

From a point on the sphere exactly one half of space is visible; it is divided in half by the plane that touches the sphere at this given point. This is the average horizon in practice. So we can use Pythagoras' theorem to calculate the distance to an object at a certain height, when it rises above the horizon: $$(\textrm{Earth’s Radius})^2 + (\textrm{ground Distance})^2 = (\textrm{Earth’s Radius} + \textrm{clouds Height})^2$$ Radius of Earth is about $$6400\;\mathrm{km}$$. If rainy clouds lie at $$2\;\mathrm{km}$$ and we can see them at the horizon, then its raining in about $$160\;\mathrm{km}$$ from our location. For double directions ($$320\;\mathrm{km}$$) this is $$0.8$$ percents of the length of Earth's circle ($$40000\;\mathrm{km}$$). The corresponding area is about $$0.02$$ percents of Earth's sphere.

Notice, the most of area is observed close to the horizon. And, for example, if a small cloud is visible at $$45$$ degrees above the horizon, then it floats right above the place (ground) at same distance from us as the altitude of this cloud.

I am assuming you mean the fraction of the "astronomical" sky. If you mean the fraction of the atmosphere (?), then that is a different question.

Well, the answer for the fraction of the total sky can be seen at any given time is $$\begin{cases} \frac{1}{2}\left(1+\frac{R}{R+h}\cos(\ell)+\sqrt{1-\left(\frac{R}{R+h}\right)^2}\sin(|\ell|)\right) &\text{for \cos(\ell)\le\frac{R}{R+h}}\\ 1 &\text{otherwise}, \end{cases},$$ where $R$ is the radius of the Earth (assumed spherical) and $h$ is your height over the surface. Note that for $h\approx0$, if you are located in the Equator you can see the entire sky.

• The formula has to be wrong. It should be independent of latitude by symmetry and in the limit $h \to \infty$ the answer should be 1 – alemi Aug 4 '14 at 1:33
• @alemi There, it needed correction. As the earth rotates, you see more sky during a 24 hr period. – Enredanrestos Aug 4 '14 at 2:38

Visible objects in low-Earth orbit (such as the International Space Station) take about 90 minutes—5,400 seconds—to complete a single orbit. Unequivocally far beyond the mesosphere at 240-250 miles above Earth, such objects might provide a rough proxy for measuring the portion of blue "sky" that is visible.

Now when viewed from their backyard, most people will estimate that an object in low-Earth orbit passes directly over a fixed location in about five to ten seconds. By this measure, from a single point on Earth you can usefully “see” 0.185% (about 2 tenths of one percent) of the local sky.

The nice thing about this method is that you can fine-tune the approximation yourself to tailor it to your specific location, situation, visual acuity, or other circumstance(s).