How wide is the observable horizon, at sea level?

[Lonious]

I'm quite aware of how to compute how FAR you are from the horizon, but my question is, how WIDE is the observable horizon at sea level (like, from left to right, how many kilometers is this):

http://www.jeicentral.com/wp-content/uploads/2014/10/far_sunset_in_te_ocean_horizon-wide.jpg

Thanks!

 

[Simon Bridge]

Question makes little sense - the horizon is all around you unless blocked by a nearby rise.
So the width would be 360deg (max)

The length could be considered the circumference of the circle whose radius is the distance to the horizon.
The left-right length in the photo? It's the angle subtended by the view times the radius.

 

[Doug Huffman]

The height of eye must be specified. The distance to the visible horizon is an important calculation in dead reckoning navigation.
https://en.wikipedia.org/wiki/Horizon#Distance_to_the_horizon

 

[Simon Bridge]

The height of eye must be specified. The distance to the visible horizon is an important calculation in dead reckoning navigation.

... this is correct, except the question asks for the width of the horizon, not the distance to it.

 

[Doug Huffman]

Better a good error than a bad question. Maybe we can help OPie.

 

[Simon Bridge]

Maybe so - see post #2. But what is your take on the question?

 

[montoyas7940]

42

 

[David Morgan]

If you know the distance to the horizon, and we're assuming you're on a spherical surface with uniform sea all around you...

The distance to the horizon would be your radius r.
Circumference C of your 360 degree view would be 2πr
Factor in your field of view... for example a FOV of 108 degrees = 108/360 = 0.3
Then multiply your C value by your FOV ratio

...and derp... somehow missed Simon's #2 post which says the same thing. Ignore me.

 

[Lonious]

Thanks for the replies!

Question makes little sense - the horizon is all around you unless blocked by a nearby rise.
So the width would be 360deg (max)

The length could be considered the circumference of the circle whose radius is the distance to the horizon.
The left-right length in the photo? It's the angle subtended by the view times the radius.

Thank you very much, Simon! I have used your post in a debate in another forum: http://forums.spacebattles.com/threads/iron-giant-vs-supcom-aeon-gc.332686/page-2.

I appreciate your help.

Regards,
Lonious

 

[Nugatory]

The distance to the horizon would be your radius r.
Circumference C of your 360 degree view would be 2πr

You will need a small additional correction because on the curved surface of the earth, the circumference of a circle is not $2\pi{r}$, it's a bit less. The plane that the circle lies in cuts through the earth, so the observer on the surface of the Earth at the center of the circle is standing above that plane, not on it, and the radius, the distance from him to the circumference of the circle, is a bit longer than it would be on a flat surface.

This is a very small correction for a human-sized observer standing on an earth-sized object, but it's there.

 

[Simon Bridge]

The usual "distance to horizon" calculation is for line of sight... so the correction is for that line and the horizon marking out a cone with apex at eye height above the ground. However, considering the distance in question is less than 3mi, that correction will be much smaller than the ones for refraction and attenuation in the air which are usually, also, neglected. I imagine the discussion referenced has debate along these lines with nobody crunching the numbers... it's a good exercize.