How much of sky can be seen on a typical night from a mid-latitude location?

I'm at about 40deg north so, assuming a clear southern horizon, I can't see things below about -30 or so (I actually don't know how far south). I also have a large portion that is circumpolar so it's always visible. I assume there's an equal size area south that is never visible.

On an equinox, the average night, I have less than 12 hours of darkness, but as the Earth rotates some stars will set and other rise, so I'm guessing something approximating 70% of the sky will be visible west to east.

So, how much of the sky is actually visible on a typical night?

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You pretty much answered it yourself. –  Florin Andrei Oct 8 '11 at 1:20
It also depends on where you are. Like in a valley or on top of a mountain. –  xmjx Oct 8 '11 at 9:17
You could install some planetarium software and try for yourself. –  xmjx Oct 8 '11 at 9:19
Terrain has some effect on total amount of visible sky, but only because it may slice a wedge off the far east or west edges. The middle will still be visible for some time. Its biggest effect is to decrease the length of time any one slice is visible. –  Andrew Oct 8 '11 at 11:16
I have tried this with Stellarium and Starry Nights. It's fun to see but it doesn't do the math. I am really curious to get an actual number. –  forestplay Oct 8 '11 at 17:06

Following Larry's response, but with approximate numbers:

Assume an "average" night means September 21 / March 21 (nights about that length are more common than others).

From http://www.sunrisesunset.com/calendar.asp, I get that the night lasts 11:45 in Boulder, CO (which is at 40N).

I'd assume that everything about 10 degrees above the horizon is visible, and everything below is not - that's true in many locations either because of trees, atmosphere opacity, mountains, or city lights. If you're using a telescope, I wouldn't ever go below 20 degrees.

A general formula using a surface integral on the surface of a sphere:

$a = \text{angle above horizon something is considered visible}$

$b = \text{latitude}$

$c = \text{number of degrees in the night = number of hours in the night / 24 * 360}$

$d = 180 - 2*a + c$

$e = (90-b)-a$

$\text{visible fraction} = \left( \int_0^d \int_{90-e}^{180} \sin(\varphi) d\varphi d\theta \right) / 4 \pi$ $= -(\cos(180) - \cos(90-e)) * d / (4\pi)$ $= d (\sin(e) + 1) / (4\pi)$

(d and e must be converted to radians)

For an 11h45m night, that comes out to 38.0%, 29.9%, 22.0% for $a=$10, 20, and 30 degrees respectively. If you consider that $\cos(90-b) / 2$ of the sky is never visible (because it's always below the horizon), these become 61.6%, 48.5%, 35.7% of the sky that you could ever see.

These calculations were somewhat hasty... I expect to be brutally corrected. The real answer is much more complicated - you need to do an integral over a sphere after rotating the pole, which gets into Euler parameters and quaternions. Still, I think my first guess is probably correct to within about 5-10%.

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A precise answer is not simple, as the night's length depends upon the season and your latitude. To make things even more tricky, you could add in the length of twilight, which also varies by season and latitude. To make things even more tricky, you could account for the sun's relative motion during the course of 24 hours. To make things even more tricky, you could account for the Earth's wobbling around its axis (nutation).

Having said all that, this code is precise enough for casual use:

http://www.astro.ucla.edu/~mperrin/IDL/sources/suntimes.pro

OK, so to put it together:

Putting aside the curvature of the Earth and refraction and terrain and so forth, we can say that at any moment, half the sky is available to view.

Over the course of 24 hours, the percentage of the entire celestial sphere that's available to view varies from 50% (at the poles) to 100% (at the equator). For a given latitude, this value is 50% + cos(|lat|) * 50%.

But of course the sun blocks the sky during the course of a day. So let's consider "dark" hours to be between twilights -- you can calculate that duration using the algorithm embodied in the code above.

Since you can see all the way to the E and W horizons at any time, you can subtract 6H RA to the value at sunset twilight and add 6H to the RA to the value at dawn twilight (roughly). So now you have some number of hours of RA that are visible over the course of the night. Normalize that to 24H if necessary (i.e., you can't have more than 100%!). Divide your RA visibility by 24 to get the percentage of the "potential sky" that is available to you during the dark.

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How does computing the approximate sunrise/set time tell me how much of the sky I can see on a typical night? Also, how can I run that program? I'm guessing a file that ends with .pro is a prolog program. Is that really something to which a casual reader would have access? –  forestplay Oct 11 '11 at 21:39
It's actually in IDL, a language commonly used in astronomy, and runnable under GNU Data Language, which is freely available. –  Larry OBrien Oct 11 '11 at 23:19
Yes, it's "runnable" if you get around the unending hurdle of dependencies. To post IDL code is not a good idea, it's not for everybody in audience:) –  Tigran Khanzadyan Oct 12 '11 at 20:18
I 100% agree that IDL is not as desirable as, say, Python or what-have-you, but it is very common in astronomy and there are tons and tons of IDL functions available on the Web. To suggest "no IDL" on an astronomy site is going to lead to a lot of unanswered questions. –  Larry OBrien Oct 13 '11 at 2:36
Mind you, I'm not suggesting that IDL should not be used. I use it almost everyday. The point is some institutions are dropping their support in favour to Python since IDL license is quite pricey. –  Tigran Khanzadyan Oct 13 '11 at 20:47