5 ms·

ASK HN: Can any of you solve this?

14 points by eam 15y ago

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

eam 15y ago
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 15y ago
  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/ http://futureboy.us/frinkdocs/
hfinney 15y ago
What % of the sky is visible from a point on a sphere. 50%, duh.
- sorbus 15y ago
  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 15y ago
    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 http://zombal-zomb.s3.amazonaws.com/70/earth.png
- deleted 15y ago
  [deleted]
ColinWright 15y ago
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 15y ago
  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%
svrocks 15y ago
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 15y ago
  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 15y ago
    Ah you're right, I missed a factor of 2 in the top integral. Will edit shortly.
- YuriNiyazov 15y ago
  The way the question is posed, it seems to me that the radius R of hollow spherical shell should not figure in the calculation. Assume there is no Earth, and you are hanging out in the center of a large hollow sphere. In that case, you can see 100% of that sphere, no matter how large it is. Now, let's say that earth magically appears, and now r' varies from 0 to r. The question asks to find percentage of the sky visible as a function of r'
- deleted 15y ago
  [deleted]
- tolo 15y ago
  I'm not sure if I'm missing something, but I get a much simpler answer. Surface area of a spherical cap is 2 pi R h, with h = R - r. Divided by the total surface area (4 pi R^2), gives (R-r)/(2 R)
mdpm 15y ago
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.
deleted 15y ago
[deleted]
davidhollander 15y ago
[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.

14 points by eam 15y ago

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

eam 15y ago

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 15y ago
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/ http://futureboy.us/frinkdocs/

hfinney 15y ago

What % of the sky is visible from a point on a sphere. 50%, duh.

sorbus 15y ago
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 15y ago
  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 http://zombal-zomb.s3.amazonaws.com/70/earth.png
deleted 15y ago
[deleted]

ColinWright 15y ago

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 15y ago
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%

svrocks 15y ago

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 15y ago
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 15y ago
  Ah you're right, I missed a factor of 2 in the top integral. Will edit shortly.
YuriNiyazov 15y ago
The way the question is posed, it seems to me that the radius R of hollow spherical shell should not figure in the calculation. Assume there is no Earth, and you are hanging out in the center of a large hollow sphere. In that case, you can see 100% of that sphere, no matter how large it is. Now, let's say that earth magically appears, and now r' varies from 0 to r. The question asks to find percentage of the sky visible as a function of r'
deleted 15y ago
[deleted]
tolo 15y ago
I'm not sure if I'm missing something, but I get a much simpler answer. Surface area of a spherical cap is 2 pi R h, with h = R - r. Divided by the total surface area (4 pi R^2), gives (R-r)/(2 R)

mdpm 15y ago

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.

deleted 15y ago

[deleted]

davidhollander 15y ago

[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.