<p>If I understand your question, you're interested in the effect of dip on the visible part of the sky. Dip is the angle between the apparent horizon and a plane tangent to the Earth's surface at the location of the observer. It depends on the distance of the observer's eyes above the surface of the earth. An observer in ocean water submerged up to his eyes would see 0 dip.</p>
<p>Variations in surface curvature have no significant effect. For small angles of dip, refraction dominates, dip is negligible.  Having said that, if the Earth had no atmosphere - no refraction - the effect of observer height would be as follows.</p>
<p>Dip (D) can be calculated by solving a right triangle with vertices at the center of the Earth, at the apparent horizon, and at the observer's eye level. The known sides of the triangle are the Earth's radius (R), and the Earth's radius plus the observer's eye level above the surface of the Earth (R+h), the hypotenuse. The angle at the apparent horizon is the right angle. Then the formula for dip in radians is <span class="math-container">$D = \pi - \arcsin(R/(R+h))$</span>.</p>
<p>The effect of dip is to make the visible sky in steradians, using Archimedes' formula for a spherical cap is <span class="math-container">$\Omega = 2\pi (1+\cos(\pi/2 + D))$</span>.</p>
<p>and the answer to your original question is <span class="math-container">$\frac{1}{2}+\cos(\pi/2+D)$</span>.
Which is somewhat bigger than 1/2.</p>

CC BY-SA 4.0
2023-08-07T14:58:31.897Z
2023-10-16T00:26:22.020Z
2023-10-16T00:26:22.020Z
stretch
<p>Yeah I know, 'Half'.  But if you're in the middle of the pacific Pacific, where the surface curvature is greatest(?), with the sun and moon are at opposition, the nadir, on a cloudless night, even sitting in the bottom of a row(ing) boat, you're looking down a tangent cone to infinity, which is looking below the plane of the equator projected on the celestial sphere.  How much?</p>

CC BY-SA 4.0
2023-08-07T09:23:46.133Z
Can you clarify. You mean at any instant of time? Why do you think that is different to 50%?
ProfRob
CC BY-SA 4.0
2023-08-07T10:03:41.083Z
Aye, at an instant in time, what proportion would be visible.  Not that the eye can take more than half in.  It must be different because of the cone.
Martin
CC BY-SA 4.0
2023-08-07T10:25:51.983Z
I realise the need for an instant: https://astronomy.stackexchange.com/questions/8267/can-an-observer-on-earth-only-see-half-of-the-sky?rq=1
Martin
CC BY-SA 4.0
2023-08-07T11:14:05.833Z
I don't understand: How is your question different from the one you link in the comment? What cone (of visibility?) do you mean? Do you mean the field of view of the human eye?
planetmaker
<observational-astronomy>
How much of the celestial sphere can a naked eye observer see?
Martin
# How much of the celestial sphere can a naked eye observer see?

(Asked by: username_0 on 2023-08-07)

Yeah I know, 'Half'. But if you're in the middle of the pacific Pacific, where the surface curvature is greatest(?), with the sun and moon are at opposition, the nadir, on a cloudless night, even sitting in the bottom of a row(ing) boat, you're looking down a tangent cone to infinity, which is looking below the plane of the equator projected on the celestial sphere. How much?

**Tags:** observational-astronomy

> **Comments:**
> **username_1**: Can you clarify. You mean at any instant of time? Why do you think that is different to 50%?  
> **username_0**: Aye, at an instant in time, what proportion would be visible.  Not that the eye can take more than half in.  It must be different because of the cone.  
> **username_0**: I realise the need for an instant: https://astronomy.stackexchange.com/questions/8267/can-an-observer-on-earth-only-see-half-of-the-sky?rq=1  
> **username_2**: I don't understand: How is your question different from the one you link in the comment? What cone (of visibility?) do you mean? Do you mean the field of view of the human eye?  

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## 1 answers

### Answer by username_3 on 2023-08-07. Score: 4 | Accepted answer

If I understand your question, you're interested in the effect of dip on the visible part of the sky. Dip is the angle between the apparent horizon and a plane tangent to the Earth's surface at the location of the observer. It depends on the distance of the observer's eyes above the surface of the earth. An observer in ocean water submerged up to his eyes would see 0 dip.

Variations in surface curvature have no significant effect. For small angles of dip, refraction dominates, dip is negligible. Having said that, if the Earth had no atmosphere - no refraction - the effect of observer height would be as follows.

Dip (D) can be calculated by solving a right triangle with vertices at the center of the Earth, at the apparent horizon, and at the observer's eye level. The known sides of the triangle are the Earth's radius (R), and the Earth's radius plus the observer's eye level above the surface of the Earth (R+h), the hypotenuse. The angle at the apparent horizon is the right angle. Then the formula for dip in radians is $D = \pi - \arcsin(R/(R+h))$.

The effect of dip is to make the visible sky in steradians, using Archimedes' formula for a spherical cap is $\Omega = 2\pi (1+\cos(\pi/2 + D))$.

and the answer to your original question is $\frac{1}{2}+\cos(\pi/2+D)$.
Which is somewhat bigger than 1/2.


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