

ASK HN: Can any of you solve this? - eam
http://zombal.com/zomb/scientific-calculation/calculate-percent-of-visible-sky-from-flat-surface
I was driving the other day, having lots of time to think, I wonder what percentage of the entire sky was I looking at.
======
svrocks
Assumptions: 1) Earth is a sphere with radius r 2) "Sky" is a hollow spherical
shell with radius R

The surface area of the entire spherical sky is: 4\piR^2

This can be represented by a spherical integral that I'm not sure I can write
cleanly here.

We just need to change the bounds of that integral to find the area of the
observable part of that shell. The Intersecting Chord Theorem along with some
trigonometry can be used to find these bounds.

The answer I get is:

(1 - cos(x)) / 2 where x = Arctan(sqrt(r(R-r)) / r)

This seems to have the correct asymptotic behavior (as r approaches 0, cos(x)
approaches cos(pi/2) = 0, and the answer approaches 50%

EDIT: My previous answer assumed the shapes were cones instead of spheres.
Sorry about the confusion.

~~~
ColinWright
That can't be right - it doesn't have the right behavior as R->oo. As R->oo
you need to get an answer of 0.5, but this gives an answer of 0.25.

~~~
svrocks
Ah you're right, I missed a factor of 2 in the top integral. Will edit
shortly.

------
davidhollander
[edit, Made a mistake, added sanity checks]

sin theta = r_earth / (r_earth + altitude)

theta = arcsin( r_earth/(r_earth + altitude))

p_visible = (2pi - 2*theta)/(2pi)

p_visible = 1 - arcsin ( r_earth / (r_earth + altitude))/pi

.

.

Sanity checks:

r_earth = 6378.1 X 1000 m

altitude = 1m

p_visible = .50006

.

altitude = 10^7 m

p_visible = .66766

This fits intuitively: the further away you are from the Earth's surface, the
more of the sky you can see without the horizon interfering.

------
mdpm
most of that is simple geometry (heh). The interesting parts creep in with the
'non-ideal' conditions.

You're not just referring to the percentage of the surface area of a sphere
(which earth isn't), atmospheric distribution isn't uniform even if we go by
volume (or should we be going by density?), then there's the curvature of
light in the atmosphere to take into consideration, and the arbitrary
descisions as to what height above sea level our observer is standing at, the
variable nature of the tropopause ...

an interesting problem, but likely more interesting as a mental exercise than
in actual execution.

------
ColinWright
The question is ill defined - it depends.

Define "The Sky" as an enclosing sphere. When it has the same size as the
Earth, the percentage of it you can see is 0. As it gets larger, so the
percentage you can see goes to 50%.

~~~
phamilton
That was my reaction too. Without a better definition of what the enclosing
sphere is, there is no answer.

What portion of the night sky (constellations) can we see? About 50%

------
eam
I was driving the other day, having lots of time to think, I wondered what
percentage of the entire sky was I looking at.

~~~
dkersten
I don't know the answer and don't care to try, but if I did, I would use
Frink[1] for this task. Seems the perfect tool for these kinds of problems.
Hell, I bet if you posted on the mailing list, Alan Eliasen would happily
solve this problem for you - seems like the kind of problem he likes to solve
on the mailing list.

[1] <http://futureboy.us/frinkdocs/>

------
hfinney
What % of the sky is visible from a point on a sphere.

50%, duh.

~~~
sorbus
This is one of those things where it would be much, much easier to explain
with images, but bear with me.

Draw a circle. Within that circle, with the same center as the first circle,
draw a smaller circle. Choose a point on that circle, and draw a tangent line
from that point. Now, consider the fact that the portion of the exterior
circle between the two intersections the tangent line has to it is not half of
the circle.

The same principle applies when expanding to three dimensions: a plain can
only divide a sphere in half if it passes through the center of the sphere.
When that plain must be tangent to a smaller sphere with the same center, it
is only possible to divide the larger sphere in half when the radius of the
smaller sphere is zero.

As such, the portion of sky visible from any point on a sphere is less than
50%.

~~~
eam
I agree, I don't think it's exactly 50%, should be less. Here's a simple
drawing: <http://zombal-zomb.s3.amazonaws.com/70/earth.png>

