The length of the visible horizon

[alitanha]

How much is the length of horizon that human's eye can see at sea level? How much is the length of horizon that human's eye can see at higher level? any formula?

I am not sure i should use wide or length. i mean the rightest point of horizon to the leftest point of it for human' eye angles.

 

[Baluncore]

Welcome to PF.
Without moving your head you can see about 180° of the horizon.

You can estimate it by looking at the horizon and moving your hands to the point where you can just see them.

Then stand with your back to a wall. Without turning your head, can you see both ends of the wall in your peripheral vision?

 

[willem2]

The distance of the horizon is d ≈ 3.57 sqrt(h), assuming flat terrain (sea), and if h is very small compared to the radius of the earth.
h = your height in meters, d = the distance in km.
The length of the horizon is the distance times the horizontal viewing angle in radians.
According to https://en.wikipedia.org/wiki/Human_eye#Field_of_view this is 200 degrees using both eyes, so 200 * π/180 = 3.49 radians.
So the length will be 12.45 * sqrt(h) = 16.7 km if h is 1.8m
If your viewing point is higher, you can just add your altitude to h. If the horizon is higher, it will be very hard to compute, because the distance to the horizon will likely not be the same everywhere.

 

[alitanha]

According to https://en.wikipedia.org/wiki/Human_eye#Field_of_view this is 200 degrees using both eyes, so 200 * π/180 = 3.49 radians.
So the length will be 12.45 * sqrt(h) = 16.7 km if h is 1.8m

Where does the 12.45*sqrt(h) come from ?

 

[DrGreg]

Where does the 12.45*sqrt(h) come from ?

$$3.49 \times 3.57 \sqrt{h} $$

 

[alitanha]

$$3.49 \times 3.57 \sqrt{h} $$

Why? Why you used a circle rather than a triangle? Willem2 supposed Earth is round then he calculated distance of horizon and length of it.

I don't want to suppose anything i see horizon is a straight line so let's calculate for a plane: i know Earth is round but i want to calculate for another possibility.

1. how far away horizon creates at 1.8 m height level for a plane without atmosphere,refraction or anything else?
2. how far away horizon creates at 1.8 km height level for a plane without atmosphere,refraction or anything else?
3. how much the length of horizon that human's eye can see at sea level for a plane without atmosphere,refraction or anything else?

I am talking about a pure geometry in physical reality.

 

[willem2]

The d = 3.57√h is without atmosphere or refraction.
All the points of the horizon will be at the same distance for a flat terrain, so they form a circle.

 

[alitanha]

The d = 3.57√h is without atmosphere or refraction.

Horizon has a dip in a planet but it hasn't any dip in a plane. so they have a totally different mathematics. the d = 3.57√h is a approximating equation for a planet not a plane . If I made a mistake, correct me.
After all, I actually want to know is there any formula for calculating distance horizon in a plane without atmosphere or refraction?

All the points of the horizon will be at the same distance for a flat terrain, so they form a circle.

Interesting. i didn't know that. Thank you.
To ensure i ask again.Are you saying the rightest point and the leftest point and the middle point of horizon line have same distance to human's eye even in a plane?

 

[nasu]

What do you mean by horizon in a plane?

 

[DrGreg]

On a flat plane you can, in theory, see to infinity.

 

[alitanha]

On a flat plane you can, in theory, see to infinity.

So i was right and Willem2 was wrong?

After all a new problem is coming in my head. Willem2 told all point of horizon will be at the same distance for a flat terrain, so they form a circle. what about a planet? if it is same for a planet, another problem rises up here:
Horizon has two curves. one curve to human eyes same as Willem2 said. one curve to another coordinate system for example sphere's center. So his equation for calculating length of horizon does not work here.

Thank you

 

[jbriggs444]

So i was right and Willem2 was wrong?

After all a new problem is coming in my head. Willem2 told all point of horizon will be at the same distance for a flat terrain,

Willem was working the problem assuming a flat spherical earth. The ##3.57\sqrt{h}## applies on the Earth's surface and yields distance to the horizon in kilometers for a viewing height above the surface of h meters. He made those assumptions fairly clear. The 3.57 figure can be arrived at using basic trigonometry and the radius of the earth.

If you want to make up a different problem, the answer will, of course, be different.

Horizon has two curves. one curve to human eyes same as Willem2 said. one curve to another coordinate system for example sphere's center.

Before you can talk about the curve of the horizon relative to a coordinate system or about a coordinate system based on a sphere's center, you have to define those terms. As it stands, it is nonsense.

 

[Baluncore]

For a perfect sphere, without atmosphere, the horizon forms a perfect circle centred on the observer.

The radial distance from the observer to the horizon is a function of the sphere radius and the height of the observer above the surface. That is where the approximate equation d(km) = 3.57 √ h(m) comes from.

The circumferential length of horizon observed is the radius to that horizon circle multiplied by the observers horizontal angular field of view measured in radians.

The observers horizontal angular field of view will depend on the bone structure of their skull. If the observer keeps their eyes forward they will see a wider view in their peripheral vision. That is because when they look sideways the eyeball rotates and the pupil moves deeper into the eye socket.

The lines from the observer to the horizon form a cone, with the 'horizon' usually being below the observer's local 'horizontal' reference plane.

The Earth is a flattened ellipsoid, with slight hollows and bumps in places. So the horizon will not be a true circle. It will depend on latitude and the local bumps.

The Earth's atmosphere has a vertical pressure gradient, the resultant refraction will make the rays curve with the curvature of the Earth's surface, which will extend the range to the observed horizon.

Temperature variations in the atmosphere near the surface may cause widely variable diffraction. That may create extreme mirages or obscure an identifiable horizon.

The OP question is insufficiently defined to have anyone answer.

 

[alitanha]

Willem was working the problem assuming a flat spherical earth. The ##3.57\sqrt{h}## applies on the Earth's surface and yields distance to the horizon in kilometers for a viewing height above the surface of h meters. He made those assumptions fairly clear. The 3.57 figure can be arrived at using basic trigonometry and the radius of the earth.

If you want to make up a different problem, the answer will, of course, be different.Before you can talk about the curve of the horizon relative to a coordinate system or about a coordinate system based on a sphere's center, you have to define those terms. As it stands, it is nonsense.

I need to put a image for delivering my point. See image below:
https://i.cubeupload.com/z8ybaR.jpeg

According to Welleim2 and Buluncore, horizon forms a perfect circle centered on the observer for a perfect sphere. let's assume Earth is a perfect sphere.

So horizon must have a curve to x,z coordinate and one curve due to curvature of Earth at x,y coordinate. which one is true? if both how his equation for length of horizon works here?

For a perfect sphere, without atmosphere, the horizon forms a perfect circle centred on the observer.

The radial distance from the observer to the horizon is a function of the sphere radius and the height of the observer above the surface. That is where the approximate equation d(km) = 3.57 √ h(m) comes from.

The circumferential length of horizon observed is the radius to that horizon circle multiplied by the observers horizontal angular field of view measured in radians.

The observers horizontal angular field of view will depend on the bone structure of their skull. If the observer keeps their eyes forward they will see a wider view in their peripheral vision. That is because when they look sideways the eyeball rotates and the pupil moves deeper into the eye socket.

The lines from the observer to the horizon form a cone, with the 'horizon' usually being below the observer's local 'horizontal' reference plane.

Assume i stand exactly at north pole and Earth is a perfected sphere.see image below:
https://i.cubeupload.com/gSnvRN.jpg

The horizon forms a perfect circle centered to my standing point but a little below respect to my eye level. so you are right about "The lines from the observer to the horizon form a cone".

But horizon has another curve because of curvature of sphere. the problem is here. horizon is a line that required two curve at same time.
so that equation don't work here. In another word this statement is wrong "The circumferential length of horizon observed is the radius to that horizon circle multiplied by the observers horizontal angular field of view measured in radians."

 

[Baluncore]

Ideally, if you set up a theodolite where the observers head would be, you could measure the azimuth (= direction) to a point on the horizon. There will always be 2π radians (= 360°) of possible azimuth. The ideal elevation (= height) of the horizon will always be the same small angle below the observers horizontal reference plane.

But horizon has another curve because of curvature of sphere. the problem is here. horizon is a line that required two curve at same time. so that equation don't work here. In another word this statement is wrong "The circumferential length of horizon observed is the radius to that horizon circle multiplied by the observers horizontal angular field of view measured in radians."

There needs to be a conversion from range in spherical coordinates (centred on the observers head) to radius in cylindrical coordinates based on the vertical axis through the centre of the Earth and observer.
Cylindrical Radius = Range * Cos( depression angle). The effect is very small. Although a 1° depression angle gives a vertical drop of about 1 in 60, Cos(–1°) is 0.99985 which gives only a 0.015% difference between radius and range.

The second curvature you hypothesise comes into existence only when you try to project the observers 3D view of the horizon onto a 2D map as viewed by the observer. There is no problem if you plot it as azimuth and elevation, or as a plan view.

Lines to the horizon are tangent to the surface of the sphere. Geometrically the horizon will be a plane circle defined by the tangent contact of the sphere and the cone. In reality, grains on the surface will make the horizon line a fractal of infinite length with a discontinuous range and elevation. That is because you are looking at the side profile of the grains of sand on the surface in the region where the observation cone is tangent to the Earth's surface.

Any answer will really depend on the meaning chosen for the term “length of the horizon”.