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?
Best Answer
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%.