What % of the sky can i see?

What % of the sky can i see?

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Discussion

dkatwa

Original Poster:

572 posts

252 months

Thursday 8th October 2015
quotequote all
Can't find answer on Google...
I was wondering, as an average 5'8" man, if I was in the middle of the sea (calm conditions, so I can see the horizon in all directions), what % of the sky can I see? OK, ignore the fact that I can't stand in the middle of the sea, but you get the picture?
My brain does not allow me to even attempt a useful answer...hence, why the brains on this site can surely answer this?
cheers

Einion Yrth

19,575 posts

251 months

Thursday 8th October 2015
quotequote all
5' 8" isn't average, it's short. HTH.


mike_knott

343 posts

231 months

Thursday 8th October 2015
quotequote all
This website has both charts AND equations, so it must be true:

http://physics.stackexchange.com/questions/129317/...

V8FGO

1,651 posts

212 months

Thursday 8th October 2015
quotequote all
The area you can see is approx 17352 km2
Assuming a local horizon distance of 4.7km. Just need to work out the hemisphere area of the earth for its radius and take % of that.
I think the above is correct, if not I'm sure someone will be at ng to tell me

MartG

21,250 posts

211 months

Friday 9th October 2015
quotequote all
I think a few people have misread the question....

Anyway, assuming no atmospheric haze, you'd be able to see 50% of the sky - your height compared to the size of the earth makes pretty much zero difference

m1dg3

128 posts

161 months

Friday 9th October 2015
quotequote all
MartG said:
I think a few people have misread the question....

Anyway, assuming no atmospheric haze, you'd be able to see 50% of the sky - your height compared to the size of the earth makes pretty much zero difference
That's exactly what I thought otherwise the day would be really short...

...however, maybe the OP means the earth's atmosphere when saying the sky in which case it is simple Pythagoras:

a^2 + b^2 = c^2

where a is the distance to the edge of the atmosphere if you look towards the horizon, b is the radius of the earth and c is the radius of the earth plus atmosphere. Google says the earth's radius is 6,371km and the atmosphere is 12km thick on average. If you run the numbers it comes out at about 400km.

I think the percentage by volume works out to around 0.15% but maths was never my strongest subject.

Simpo Two

87,113 posts

272 months

Friday 9th October 2015
quotequote all
MartG said:
I think a few people have misread the question....
Indeed, percentage not distance!

MartG said:
Anyway, assuming no atmospheric haze, you'd be able to see 50% of the sky - your height compared to the size of the earth makes pretty much zero difference
I'd say slightly less, because the Earth is blocking some.

MartG

21,250 posts

211 months

Friday 9th October 2015
quotequote all
Simpo Two said:
MartG said:
Anyway, assuming no atmospheric haze, you'd be able to see 50% of the sky - your height compared to the size of the earth makes pretty much zero difference
I'd say slightly less, because the Earth is blocking some.
The Earth is only blocking the lower half of the sky, so you can see 50% - anything above the horizon is visible

Einion Yrth

19,575 posts

251 months

Friday 9th October 2015
quotequote all
MartG said:
Simpo Two said:
MartG said:
Anyway, assuming no atmospheric haze, you'd be able to see 50% of the sky - your height compared to the size of the earth makes pretty much zero difference
I'd say slightly less, because the Earth is blocking some.
The Earth is only blocking the lower half of the sky, so you can see 50% - anything above the horizon is visible
Unless you are (a calculable number that I can't be arsed to calculate right now)(units of choice) tall, then the horizon will not be a great circle, hence as height tends towards infinity, percentage of "sky" visible will tend towards 50. At 5' 8" it'll fall waaaaaaay short.
ETA
By the time you're high enough to see 50% of the sky, it's the floor.

Edited by Einion Yrth on Friday 9th October 22:10

MartG

21,250 posts

211 months

Friday 9th October 2015
quotequote all
Einion Yrth said:
MartG said:
Simpo Two said:
MartG said:
Anyway, assuming no atmospheric haze, you'd be able to see 50% of the sky - your height compared to the size of the earth makes pretty much zero difference
I'd say slightly less, because the Earth is blocking some.
The Earth is only blocking the lower half of the sky, so you can see 50% - anything above the horizon is visible
Unless you are (a calculable number that I can't be arsed to calculate right now)(units of choice) tall, then the horizon will not be a great circle, hence as height tends towards infinity, percentage of "sky" visible will tend towards 50. At 5' 8" it'll fall waaaaaaay short.
As height tends to infinity, visible sky tends to 100%

On the surface, the horizon is at 0 degrees elevation all around you - you can see all parts of the sky with a +ve elevation for your location, but can't see any of the sky with a -ve elevation - so you can see 50% of the sky

DocJock

8,487 posts

247 months

Friday 9th October 2015
quotequote all
But space is infinite.

You cannot assign a numerical value to a fraction of infinity.

wink

MartG

21,250 posts

211 months

Friday 9th October 2015
quotequote all
DocJock said:
But space is infinite.

You cannot assign a numerical value to a fraction of infinity.

wink
The question doesn't refer to space, just 'the sky'

DocJock

8,487 posts

247 months

Friday 9th October 2015
quotequote all
https://en.wikipedia.org/wiki/Sky

The sky (or celestial dome) is everything that lies above the surface of the Earth, including the atmosphere and outer space.[/snip] tongue out

Simpo Two

87,113 posts

272 months

Saturday 10th October 2015
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DocJock said:
But space is infinite.

You cannot assign a numerical value to a fraction of infinity.

wink
We could do it in angles and degrees rather than percent...

Eric Mc

122,861 posts

272 months

Saturday 10th October 2015
quotequote all
DocJock said:
https://en.wikipedia.org/wiki/Sky

The sky (or celestial dome) is everything that lies above the surface of the Earth, including the atmosphere and outer space.[/snip] tongue out
What if it's cloudy?

DocJock

8,487 posts

247 months

Saturday 10th October 2015
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Take an umbrella.

Eric Mc

122,861 posts

272 months

Saturday 10th October 2015
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What percentage of the sky can you see if you put your umbrella up?

davepoth

29,395 posts

206 months

Sunday 11th October 2015
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Serious answer then.

Working on the assumption that the earth is round (it's ovoid in reality, with a fat bottom) we can do this with geometry.



Should be familiar. We're just doing a bit of geometry. The hypotenuse is the radius plus the height of the man, and one of the other sides of the triangle is the radius.

The radius of the earth is 6371km, the height of a man is 1.72m (ish) so the other number is 6371.00172km. We also know one of the angles. That's enough for sines, cosines, and tangents, which of course everyone knows how to do from school, don't they?

I had to google it. biggrin

Anyway, the upshot is that at 0m above the surface the field of view is 180 degrees - half the sky. At 1.72m above the surface that increases to 180.084 degrees. At 10m, 180.2 degrees. At 100m, 180.64 degrees. At 1km, 182 degrees. And so on, and so on.

180.084/360 is .5002333, so you can see 50.023% of the sky.


Edited by davepoth on Sunday 11th October 00:49

zollburgers

1,278 posts

190 months

Sunday 11th October 2015
quotequote all
Assuming you are in the sea and can see no hills and things then yes 180 degrees.

If on a hill then you can see more, if around other things that are tall then less.

I wouldn't say it's 50% of the total sky though.



Dr Jekyll

23,820 posts

268 months

Sunday 11th October 2015
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anonymous said:
[redacted]
If the earth was covered by clouds then the percentage of the cloud area you could see would depend on how high the clouds were. So I took the question to mean how much sky could you see as a proportion of the 360deg all round view of space you would have if the earth wasn't in the way.