<p>I'd like to expand more on <a href="https://astronomy.stackexchange.com/a/19696/42349">Arvind H's answer</a>. The Earth inclination is 23°, this means that the sun can have a maximum elevation above and below the celestial equator of <span class="math-container">$\alpha=23°$</span>, as shown in the first figure.
<a href="https://i.sstatic.net/rc264.jpg" rel="noreferrer"><img src="https://i.sstatic.net/rc264.jpg" alt="the portion of the celestial sphere covered by the Sun" /></a></p>
<p>Let's start by calculating the solid angle of the band on the celestial sphere between <span class="math-container">$\alpha$</span> and <span class="math-container">$-\alpha$</span> on the equator. In spherical coordinates where <span class="math-container">$\phi$</span> is the longitude and <span class="math-container">$\theta$</span> is the latitude:</p>
<p><span class="math-container">$$\Omega = \int_0^{2\pi} \int_{-\alpha}^{+\alpha} \cos(\theta)d\theta d\phi = 4\pi \sin(\alpha)$$</span></p>
<p>Let's rapidly check that this is correct. If <span class="math-container">$\alpha=0$</span> this gives 0, and if <span class="math-container">$\alpha = \pi/2$</span> this gives <span class="math-container">$4\pi$</span>, which is the whole sphere, as expected.</p>
<p>Now let's tackle the OP question. An observer in a particular location can only see half of the celestial sphere, that has non-trivial intersections with the area we have just calculated, as shown in the second figure.</p>
<p><a href="https://i.sstatic.net/VOFcb.jpg" rel="noreferrer"><img src="https://i.sstatic.net/VOFcb.jpg" alt="intersection with the observer plane" /></a></p>
<p>Yet, the intersected area highlighted in the second figure is always exactly half of the band, i.e. <span class="math-container">$2 \pi \sin(\alpha)$</span>. This is clearly true for the three cases I have drawn, but the symmetry of the problem implies that it is true in every case. A nice way to see it is by looking at the top and bottom pole cups where the Sun will never pass. The observer plane cuts them in a symmetrical way, so that if you combine the visible part of the top cup with the visible part of the bottom cup, you will always get a whole cup, as shown in the third figure.</p>
<p><a href="https://i.sstatic.net/Pe5gc.jpg" rel="noreferrer"><img src="https://i.sstatic.net/Pe5gc.jpg" alt="enter image description here" /></a></p>
<p>This means that the percentage of the sky covered by the Sun at any given location is <span class="math-container">${2\pi \sin\alpha \over 2\pi} = \sin\alpha \approx 40\%$</span></p>

CC BY-SA 4.0
2022-05-26T11:12:59.043Z
2022-05-26T11:12:59.043Z
Prallax
<p>I was surprised to see that the 40% confirmed in other answers does not depend on latitude (but certainly does depend on Earth's axial inclination!) and curious what the patterns looked like and needed to satisfy my &quot;one script per day&quot; minimum to keep my coding skills from getting rusty, so I wrote the script below.</p>
<p>I considered 0, 15, 30, 45, 60, 75, and 90 degrees latitudes and sampled one million equally spaced time points in the year 2022.</p>
<p>Sure enough it's about 40% for the Sun and in 2022<sup>1</sup> about 46% for the Moon.</p>
<hr />
<p><sup>1</sup>The <a href="https://en.wikipedia.org/wiki/Orbit_of_the_Moon" rel="nofollow noreferrer">Moon's orbit slowly varies</a> over timescales longer than one year.</p>
<hr />
<p><a href="https://i.sstatic.net/ZpqYQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ZpqYQ.png" alt="enter image description here" /></a></p>
<p><strong>below:</strong> boolean registry of &quot;hits&quot; for each degree elevation (vertical from -90 to +90 wrt horizon) and azimuth (horizontal) for one million equally spaced time samples in 2022 at 90, 75, 60, 45, 30, 15 and 0 degrees latitude on the prime meridian. <em>There's still a few missing spots due to finite sampling</em> but I won't rerun with ten million timepoints unless I figure out over how many years I'd need to sample to get a good understanding of the Moon's behavior.</p>
<p><a href="https://i.sstatic.net/8P2z2.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/8P2z2.png" alt="enter image description here" /></a></p>
<hr />
<p>Script is set to 100,000 time points so that it does't take to long to run, feel free to increase</p>
<pre><code>from skyfield.api import Loader, Topos
import numpy as np
import matplotlib.pyplot as plt
from skyfield import searchlib

load = Loader('~/Documents/fishing/SkyData') # avoid multiple copies of large files
ts = load.timescale()
eph = load('de421.bsp')

earth, sun, moon = [eph[x] for x in ('earth', 'sun', 'moon')]

nlats = 7
latitudes = np.linspace(0, 90, nlats) # degrees

np.save('latitudes.npy', latitudes)

t0, t1 = ts.utc([2022, 2023], 1, 1)
n_points = 100000
times =  ts.linspace(t0, t1, n_points)

sun_data, moon_data = [], []

for lat in latitudes:
    location = Topos(latitude_degrees = lat,
                     longitude_degrees = 0.0)
    moon_astrometric = (earth + location).at(times).observe(moon)
    moon_alts, moon_azs, d = moon_astrometric.apparent().altaz()

    sun_astrometric = (earth + location).at(times).observe(sun)
    sun_alts, sun_azs, d = sun_astrometric.apparent().altaz()

    sun_data.append(np.vstack([sun_alts.degrees, sun_azs.degrees]))
    moon_data.append(np.vstack([moon_alts.degrees, moon_azs.degrees]))

nalt, naz = 180, 361  # nalt is even so there's no bin cetered on horizon
sun_data = np.rint(sun_data).astype(int)
moon_data = np.rint(moon_data).astype(int)

np.save('sun_data.npy', sun_data)
np.save('moon_data.npy', moon_data)

sun_plots, moon_plots = np.zeros((2, nlats, nalt+1, naz), dtype=bool)

for i, (sd, md) in enumerate(zip(sun_data, moon_data)):
    sun_plots[i, sd[0]+(nalt&gt;&gt;1), sd[1]] = True
    moon_plots[i, md[0]+(nalt&gt;&gt;1), md[1]] = True

halfpi, pi, twopi, fourpi = [f * np.pi for f in (0.5, 1, 2, 4)] 
altitude, azimuth = np.mgrid[-halfpi:halfpi:(nalt+1)*1j, 0:twopi:naz*1j] 

dalt, daz = np.radians(1), np.radians(1) ### implicit from .degrees() and .int()

d_Omega = np.cos(altitude) * dalt * daz
in_the_sky = altitude &gt; 0

solid_angle_of_sky = d_Omega[in_the_sky].sum()
print('solid_angle_of_sky / fourpi: ', solid_angle_of_sky / fourpi)

sun_percents = 100 * (sun_plots * d_Omega * in_the_sky).sum(axis=(1, 2)) / twopi
moon_percents = 100 * (moon_plots * d_Omega * in_the_sky).sum(axis=(1, 2)) / twopi
np.save('sun_percents.npy', sun_percents)
np.save('moon_percents.npy', moon_percents)

big_sun = sun_plots.reshape(-1, sun_plots.shape[-1])
big_moon = moon_plots.reshape(-1, moon_plots.shape[-1])

np.save('big_sun.npy', big_sun)
np.save('big_moon.npy', big_moon)

if True:
    if False:
        import numpy as np
        import matplotlib.pyplot as plt
        big_sun = np.load('big_sun.npy')        
        big_moon = np.load('big_moon.npy')        
    fig, (ax_sun, ax_moon) = plt.subplots(1, 2, figsize=[12, 7.5])
    ax_sun.imshow(big_sun)
    ax_sun.set_title('Sun', fontsize=16)
    ax_moon.imshow(big_moon)
    ax_moon.set_title('Moon', fontsize=16)
    plt.show()

if True:
    if False:
        import numpy as np
        import matplotlib.pyplot as plt
        sun_percents = np.load('sun_percents.npy')
        moon_percents = np.load('moon_percents.npy')
        latitudes = np.load('latitudes.npy')
        nlats = 7
        latitudes = np.linspace(0, 90, nlats) # degrees
    fig, ax = plt.subplots(1, 1)
    ax.plot(latitudes, sun_percents, '-r', label='Sun')
    ax.plot(latitudes, sun_percents, 'ok')
    ax.plot(latitudes, moon_percents, '--g', label='Moon')
    ax.plot(latitudes, moon_percents, 'ok')
    ax.set_ylim(0, 100)
    ax.set_xlabel('latitude')
    ax.set_ylabel('approx percent of sky')
    plt.legend()
    plt.show()
</code></pre>

CC BY-SA 4.0
2022-06-10T07:45:12.770Z
Hi, nice work! But, from what I understand by reading the code and the description, you sample elevation from -90 to 90 and azimuth from 0 to 360, so you are always plotting the whole celestial sphere, not just the visible hemisphere from a given latitude, right? If it is the case, I think it is trivial to say that the percentage of the sky covered by the Sun doesn't vary with latitude, because you are always showing the same sphere, just changing the coordinate system. In the context of this question, it may be more useful to sample the elevation only from 0 to 90.
Prallax
CC BY-SA 4.0
2022-06-10T11:19:14.517Z
@Prallax I did! See how the boolean array `in_the_sky = altitude > 0` is created and then used to select only data visible at the given latitude via multiplication. I did plot the full celestial sphere with the local zenith at the top for each latitude because it looks cool. I ran out of steam otherwise I would have drawn a line across the middle and perhaps used a different colormap for the lower half.
uhoh
CC BY-SA 4.0
2022-06-11T12:15:56.113Z
Thank you for the clarification!
Prallax
CC BY-SA 4.0
2022-05-31T01:41:02.657Z
2022-05-31T01:41:02.657Z
uhoh
<p>Yes, you're right. The current inclination of the earth's axis of rotation to the plane of orbit around sun is 23.43708° degrees. 
The sun can effectively cover the area from latitudes +23.43708° to -23.43708° on a sphere. Ignoring the night side of the celestial sphere gives the required area as Sin(23.43708°) = 39.774% ~40%.</p>

CC BY-SA 3.0
2017-01-10T10:33:24.027Z
Do you mean it covers from the Tropic of Cancer to the Tropic of Capricorn? ;)
Dean
CC BY-SA 3.0
2017-01-10T14:50:49.773Z
No, I mean the percentage of the sky as viewed from ANY ONE PLACE. The view is of a HEMIsphere, not of the notional SPHERE encompassing the entire earth.
My current thinking is, 73% (47°) everywhere between the x-arctic circles. WITHIN the x-arctic circles, the percentage falls gradually to 40% (23.5°) at the poles.
I BELIEVE the 47° figure holds good throughout the tropics, as well as the temperate latitudes.
N. Joseph Potts
CC BY-SA 4.0
2020-01-15T12:54:03.963Z
I don't understand your last step 'sin(23.4)`.
usernumber
CC BY-SA 4.0
2022-05-26T10:11:31.983Z
Your result is correct, but you should show how you got it
Prallax
CC BY-SA 4.0
2022-05-27T15:14:45.753Z
Well, it's been 5 years at this point, and my memory is ever worse, but: for non-arctic areas, the percentage is given by sin(2 *23.5), or 73%. From the x-arctic circles proceeding toward the poles, the percentage declines with latitude from 73% to 40% at the poles, calculated (there) as sin(23.5). I have not developed the formula for calculating percentage as a function of latitude (>67.5), but I'm sure it's simple (and cold).
N. Joseph Potts
CC BY-SA 4.0
2022-06-01T02:33:22.920Z
I've been "working" (thinking) about the patterns the ecliptic traces over skies over the course of one year, and I've so far come up with this: at the poles, the sun traverses a band along the horizon through the (constant) height of 23.5 degrees above the horizon over one year. At the equator, the sun traverses a band arching directly overhead from west to east that is 47 degrees across from side to side. In between, it does ... what? Pictures would be nice, but precise descriptions might be just as good, or maybe even better.
N. Joseph Potts
CC BY-SA 3.0
2017-01-10T05:58:45.530Z
2017-01-10T05:58:45.530Z
Arvind H
<p>For every location on earth, there are parts of the sky that the sun can never be in (e.g., directly overhead anywhere outside the tropics).</p>

<p>I think the sun can be in about 40% of the sky (a hemisphere).</p>

<p>I'll withhold my reasoning/calculations for now, because I'd like the opinion(s) of others more-expert (most are) than me in this subject.</p>

<p>Please weigh in. I believe this "simple" geometric question will have been answered centuries ago, but I have not found it as an answer, nor even as a subject.</p>

CC BY-SA 3.0
2017-01-10T00:56:10.153Z
You shouldn't withhold your calculations on this.  Either your math is correct or it isn't.  When asking a question, it is important that you give as much information as you can.
Phiteros
CC BY-SA 3.0
2017-01-10T00:59:45.580Z
OK - I used sin(23.5) (degrees, the inclination of the axis of the rotation of the earth vis-a-vis the plane of its orbit around the sun). The sun's angle at any given time of day varies by twice that much across the year, but half of that angle is occluded by the earth itself (at night).
N. Joseph Potts
CC BY-SA 3.0
2017-01-10T12:24:48.113Z
My first thought is "all of it". Consider an observer at the South Pole. Over the course of a six month long summer, the Sun will have painted every bit of the sky at an elevation below 23.5 degrees. That's 40% of the sky right there. Add in strings of observers along the Arctic and Antarctic Circles and the Sun will have painted every bit of the sky at an elevation below 47 degrees. That's 73 percent of the sky. A couple of more strings of observers along the Tropics should get that remaining 27 percent.
David Hammen
CC BY-SA 4.0
2022-05-26T09:07:40.350Z
I dont have an answer…only questions… i cant figure out how to ask the proper question but i know the answers im reading arent it. I want to know what percentage of the sky(as seen from earth) is taken up by the sun? It seems to me it should be something like 1% not 43% or 73%. If the sun took up 43% or 73% of the sky(as seen from earth) earth would either be WAY closer to the sun or the sun would be WAY bigger and earth would be a molten ball of lava.
user46282
CC BY-SA 4.0
2022-05-26T09:45:12.050Z
I think the question is not referring to the percentage of the sky the sun occupies _in a particular moment_. In that case you would be right to say it is very small (pi*(0.5°)^2)/(2pi) = 0.004%. I think the question refers to the total area that can be covered by the sun in one year. Imagine the sun left behind a permanent trail everywhere it went. After one year, what percentage of the sky would be covered by that trail?
Prallax
CC BY-SA 4.0
2022-05-26T16:12:21.417Z
A more visual approach. The link below shows the path of the Sun through the sky on any given day.  Just select the summer and winter solstice, and the sun will always lie between those arcs. https://celestialprogramming.com/Sky_Path/
Greg Miller
CC BY-SA 4.0
2022-05-26T21:35:06.607Z
Referred to the fixed stars, the sun covers only a very narrow (0,5 deg wide) strip on the sky.
Thomas
<the-sun>
What percentage of the sky could the sun be in?
N. Joseph Potts
# What percentage of the sky could the sun be in?

(Asked by: username_0 on 2017-01-10)

For every location on earth, there are parts of the sky that the sun can never be in (e.g., directly overhead anywhere outside the tropics).

I think the sun can be in about 40% of the sky (a hemisphere).

I'll withhold my reasoning/calculations for now, because I'd like the opinion(s) of others more-expert (most are) than me in this subject.

Please weigh in. I believe this "simple" geometric question will have been answered centuries ago, but I have not found it as an answer, nor even as a subject.

**Tags:** the-sun

> **Comments:**
> **username_1**: You shouldn't withhold your calculations on this.  Either your math is correct or it isn't.  When asking a question, it is important that you give as much information as you can.  
> **username_0**: OK - I used sin(23.5) (degrees, the inclination of the axis of the rotation of the earth vis-a-vis the plane of its orbit around the sun). The sun's angle at any given time of day varies by twice that much across the year, but half of that angle is occluded by the earth itself (at night).  
> **username_2**: My first thought is "all of it". Consider an observer at the South Pole. Over the course of a six month long summer, the Sun will have painted every bit of the sky at an elevation below 23.5 degrees. That's 40% of the sky right there. Add in strings of observers along the Arctic and Antarctic Circles and the Sun will have painted every bit of the sky at an elevation below 47 degrees. That's 73 percent of the sky. A couple of more strings of observers along the Tropics should get that remaining 27 percent.  
> **username_3**: I dont have an answer…only questions… i cant figure out how to ask the proper question but i know the answers im reading arent it. I want to know what percentage of the sky(as seen from earth) is taken up by the sun? It seems to me it should be something like 1% not 43% or 73%. If the sun took up 43% or 73% of the sky(as seen from earth) earth would either be WAY closer to the sun or the sun would be WAY bigger and earth would be a molten ball of lava.  
> **username_4**: I think the question is not referring to the percentage of the sky the sun occupies _in a particular moment_. In that case you would be right to say it is very small (pi*(0.5°)^2)/(2pi) = 0.004%. I think the question refers to the total area that can be covered by the sun in one year. Imagine the sun left behind a permanent trail everywhere it went. After one year, what percentage of the sky would be covered by that trail?  
> **username_5**: A more visual approach. The link below shows the path of the Sun through the sky on any given day.  Just select the summer and winter solstice, and the sun will always lie between those arcs. https://celestialprogramming.com/Sky_Path/  
> **username_6**: Referred to the fixed stars, the sun covers only a very narrow (0,5 deg wide) strip on the sky.  

---

## 3 answers

### Answer by username_4 on 2022-05-26. Score: 10

I'd like to expand more on [username_8's answer](https://astronomy.stackexchange.com/a/19696/42349). The Earth inclination is 23°, this means that the sun can have a maximum elevation above and below the celestial equator of $\alpha=23°$, as shown in the first figure.
[![the portion of the celestial sphere covered by the Sun](https://i.sstatic.net/rc264.jpg)](https://i.sstatic.net/rc264.jpg)

Let's start by calculating the solid angle of the band on the celestial sphere between $\alpha$ and $-\alpha$ on the equator. In spherical coordinates where $\phi$ is the longitude and $\theta$ is the latitude:

$$\Omega = \int\_0^{2\pi} \int\_{-\alpha}^{+\alpha} \cos(\theta)d\theta d\phi = 4\pi \sin(\alpha)$$

Let's rapidly check that this is correct. If $\alpha=0$ this gives 0, and if $\alpha = \pi/2$ this gives $4\pi$, which is the whole sphere, as expected.

Now let's tackle the OP question. An observer in a particular location can only see half of the celestial sphere, that has non-trivial intersections with the area we have just calculated, as shown in the second figure.

[![intersection with the observer plane](https://i.sstatic.net/VOFcb.jpg)](https://i.sstatic.net/VOFcb.jpg)

Yet, the intersected area highlighted in the second figure is always exactly half of the band, i.e. $2 \pi \sin(\alpha)$. This is clearly true for the three cases I have drawn, but the symmetry of the problem implies that it is true in every case. A nice way to see it is by looking at the top and bottom pole cups where the Sun will never pass. The observer plane cuts them in a symmetrical way, so that if you combine the visible part of the top cup with the visible part of the bottom cup, you will always get a whole cup, as shown in the third figure.

[![enter image description here](https://i.sstatic.net/Pe5gc.jpg)](https://i.sstatic.net/Pe5gc.jpg)

This means that the percentage of the sky covered by the Sun at any given location is ${2\pi \sin\alpha \over 2\pi} = \sin\alpha \approx 40\%$


---

### Answer by username_7 on 2022-05-31. Score: 3

I was surprised to see that the 40% confirmed in other answers does not depend on latitude (but certainly does depend on Earth's axial inclination!) and curious what the patterns looked like and needed to satisfy my "one script per day" minimum to keep my coding skills from getting rusty, so I wrote the script below.
I considered 0, 15, 30, 45, 60, 75, and 90 degrees latitudes and sampled one million equally spaced time points in the year 2022.
Sure enough it's about 40% for the Sun and in 20221 about 46% for the Moon.

1The Moon's orbit slowly varies over timescales longer than one year.


below: boolean registry of "hits" for each degree elevation (vertical from -90 to +90 wrt horizon) and azimuth (horizontal) for one million equally spaced time samples in 2022 at 90, 75, 60, 45, 30, 15 and 0 degrees latitude on the prime meridian. There's still a few missing spots due to finite sampling but I won't rerun with ten million timepoints unless I figure out over how many years I'd need to sample to get a good understanding of the Moon's behavior.


Script is set to 100,000 time points so that it does't take to long to run, feel free to increase
from skyfield.api import Loader, Topos
import numpy as np
import matplotlib.pyplot as plt
from skyfield import searchlib

load = Loader('~/Documents/fishing/SkyData') # avoid multiple copies of large files
ts = load.timescale()
eph = load('de421.bsp')

earth, sun, moon = [eph[x] for x in ('earth', 'sun', 'moon')]

nlats = 7
latitudes = np.linspace(0, 90, nlats) # degrees

np.save('latitudes.npy', latitudes)

t0, t1 = ts.utc([2022, 2023], 1, 1)
n_points = 100000
times =  ts.linspace(t0, t1, n_points)

sun_data, moon_data = [], []

for lat in latitudes:
    location = Topos(latitude_degrees = lat,
                     longitude_degrees = 0.0)
    moon_astrometric = (earth + location).at(times).observe(moon)
    moon_alts, moon_azs, d = moon_astrometric.apparent().altaz()

    sun_astrometric = (earth + location).at(times).observe(sun)
    sun_alts, sun_azs, d = sun_astrometric.apparent().altaz()

    sun_data.append(np.vstack([sun_alts.degrees, sun_azs.degrees]))
    moon_data.append(np.vstack([moon_alts.degrees, moon_azs.degrees]))

nalt, naz = 180, 361  # nalt is even so there's no bin cetered on horizon
sun_data = np.rint(sun_data).astype(int)
moon_data = np.rint(moon_data).astype(int)

np.save('sun_data.npy', sun_data)
np.save('moon_data.npy', moon_data)

sun_plots, moon_plots = np.zeros((2, nlats, nalt+1, naz), dtype=bool)

for i, (sd, md) in enumerate(zip(sun_data, moon_data)):
    sun_plots[i, sd[0]+(nalt>>1), sd[1]] = True
    moon_plots[i, md[0]+(nalt>>1), md[1]] = True

halfpi, pi, twopi, fourpi = [f * np.pi for f in (0.5, 1, 2, 4)] 
altitude, azimuth = np.mgrid[-halfpi:halfpi:(nalt+1)*1j, 0:twopi:naz*1j] 

dalt, daz = np.radians(1), np.radians(1) ### implicit from .degrees() and .int()

d_Omega = np.cos(altitude) * dalt * daz
in_the_sky = altitude > 0

solid_angle_of_sky = d_Omega[in_the_sky].sum()
print('solid_angle_of_sky / fourpi: ', solid_angle_of_sky / fourpi)

sun_percents = 100 * (sun_plots * d_Omega * in_the_sky).sum(axis=(1, 2)) / twopi
moon_percents = 100 * (moon_plots * d_Omega * in_the_sky).sum(axis=(1, 2)) / twopi
np.save('sun_percents.npy', sun_percents)
np.save('moon_percents.npy', moon_percents)

big_sun = sun_plots.reshape(-1, sun_plots.shape[-1])
big_moon = moon_plots.reshape(-1, moon_plots.shape[-1])

np.save('big_sun.npy', big_sun)
np.save('big_moon.npy', big_moon)

if True:
    if False:
        import numpy as np
        import matplotlib.pyplot as plt
        big_sun = np.load('big_sun.npy')        
        big_moon = np.load('big_moon.npy')        
    fig, (ax_sun, ax_moon) = plt.subplots(1, 2, figsize=[12, 7.5])
    ax_sun.imshow(big_sun)
    ax_sun.set_title('Sun', fontsize=16)
    ax_moon.imshow(big_moon)
    ax_moon.set_title('Moon', fontsize=16)
    plt.show()

if True:
    if False:
        import numpy as np
        import matplotlib.pyplot as plt
        sun_percents = np.load('sun_percents.npy')
        moon_percents = np.load('moon_percents.npy')
        latitudes = np.load('latitudes.npy')
        nlats = 7
        latitudes = np.linspace(0, 90, nlats) # degrees
    fig, ax = plt.subplots(1, 1)
    ax.plot(latitudes, sun_percents, '-r', label='Sun')
    ax.plot(latitudes, sun_percents, 'ok')
    ax.plot(latitudes, moon_percents, '--g', label='Moon')
    ax.plot(latitudes, moon_percents, 'ok')
    ax.set_ylim(0, 100)
    ax.set_xlabel('latitude')
    ax.set_ylabel('approx percent of sky')
    plt.legend()
    plt.show()

> **Comments:**
> **username_4**: Hi, nice work! But, from what I understand by reading the code and the description, you sample elevation from -90 to 90 and azimuth from 0 to 360, so you are always plotting the whole celestial sphere, not just the visible hemisphere from a given latitude, right? If it is the case, I think it is trivial to say that the percentage of the sky covered by the Sun doesn't vary with latitude, because you are always showing the same sphere, just changing the coordinate system. In the context of this question, it may be more useful to sample the elevation only from 0 to 90.  
> **username_7**: @username_4 I did! See how the boolean array `in_the_sky = altitude > 0` is created and then used to select only data visible at the given latitude via multiplication. I did plot the full celestial sphere with the local zenith at the top for each latitude because it looks cool. I ran out of steam otherwise I would have drawn a line across the middle and perhaps used a different colormap for the lower half.  
> **username_4**: Thank you for the clarification!  

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### Answer by username_8 on 2017-01-10. Score: 2

Yes, you're right. The current inclination of the earth's axis of rotation to the plane of orbit around sun is 23.43708° degrees.
The sun can effectively cover the area from latitudes +23.43708° to -23.43708° on a sphere. Ignoring the night side of the celestial sphere gives the required area as Sin(23.43708°) = 39.774% ~40%.

> **Comments:**
> **username_9**: Do you mean it covers from the Tropic of Cancer to the Tropic of Capricorn? ;)  
> **username_0**: No, I mean the percentage of the sky as viewed from ANY ONE PLACE. The view is of a HEMIsphere, not of the notional SPHERE encompassing the entire earth. My current thinking is, 73% (47°) everywhere between the x-arctic circles. WITHIN the x-arctic circles, the percentage falls gradually to 40% (23.5°) at the poles. I BELIEVE the 47° figure holds good throughout the tropics, as well as the temperate latitudes.  
> **username_10**: I don't understand your last step 'sin(23.4)`.  
> **username_4**: Your result is correct, but you should show how you got it  
> **username_0**: Well, it's been 5 years at this point, and my memory is ever worse, but: for non-arctic areas, the percentage is given by sin(2 *23.5), or 73%. From the x-arctic circles proceeding toward the poles, the percentage declines with latitude from 73% to 40% at the poles, calculated (there) as sin(23.5). I have not developed the formula for calculating percentage as a function of latitude (>67.5), but I'm sure it's simple (and cold).  
> **username_0**: I've been "working" (thinking) about the patterns the ecliptic traces over skies over the course of one year, and I've so far come up with this: at the poles, the sun traverses a band along the horizon through the (constant) height of 23.5 degrees above the horizon over one year. At the equator, the sun traverses a band arching directly overhead from west to east that is 47 degrees across from side to side. In between, it does ... what? Pictures would be nice, but precise descriptions might be just as good, or maybe even better.  

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