<p>Both the equator and the ecliptic are great circles on the celestial sphere.
The appearance of each on a map depends on the map projection.</p>
<p>In an <a href="https://en.wikipedia.org/wiki/Equirectangular_projection" rel="nofollow noreferrer">equirectangular</a> projection centered on the equator, the equator (brown) is a straight line, and the ecliptic (blue) is approximately sinusoidal.</p>
<p><img src="https://i.sstatic.net/93uT0.png" alt="equator-centered cylindrical projection" /></p>
<p>If the same projection is centered on the ecliptic instead, the <em>ecliptic</em> is a straight line and the <em>equator</em> is approximately sinusoidal.
As <a href="/a/39356">uhoh's answer</a> illustrates, neither curve is exactly a sinusoid.</p>
<p><img src="https://i.sstatic.net/MAHYJ.png" alt="ecliptic-centered cylindrical projection" /></p>
<p><a href="https://johnlucey.webspace.durham.ac.uk/users/solar-year/" rel="nofollow noreferrer">Lucey</a> provides both views too, but his ecliptic-centered map doesn't show the equator.
The axes are labeled in ecliptic rather than equatorial coordinates.</p>
<p><img src="https://i.sstatic.net/i3WQJ.png" alt="ecliptic-centered cylindrical by Lucey" /></p>
<p>He also provides a <a href="https://en.wikipedia.org/wiki/Stereographic_projection" rel="nofollow noreferrer">stereographic</a> projection centered on the <a href="https://community.dur.ac.uk/john.lucey/users/solar_year_durham.html" rel="nofollow noreferrer">midday zenith at 55°N</a>.
Here both equator (red) and ecliptic (green) are mapped as circular arcs.</p>
<p><img src="https://i.sstatic.net/tGAYY.png" alt="zenith-centered stereographic by Lucey" /></p>

CC BY-SA 4.0
2020-02-19T03:56:28.313Z
2022-12-20T17:41:46.133Z
2022-12-20T17:41:46.133Z
Mike G
<p>The ecliptic is a plane. It looks like a sine wave because that diagram maps the sky onto a flat plane. On the celestial sphere, the ecliptic is a <a href="https://en.wikipedia.org/wiki/Great_circle" rel="nofollow noreferrer">great circle</a> that crosses the celestial equator at the <a href="https://en.wikipedia.org/wiki/First_Point_of_Aries" rel="nofollow noreferrer">equinox points</a> and which is tilted to the celestial equator by the same amount as the Earth's axial tilt (currently <a href="https://en.wikipedia.org/wiki/Axial_tilt" rel="nofollow noreferrer">23°26′12.0″</a>).</p>

<p>Here's a diagram from <a href="https://en.m.wikipedia.org/wiki/Equatorial_coordinate_system" rel="nofollow noreferrer">Wikipedia</a> that might be helpful:</p>

<p><img src="https://upload.wikimedia.org/wikipedia/commons/6/66/Ra_and_dec_demo_animation_small.gif" alt="Celestial sphere"></p>

CC BY-SA 4.0
2020-02-18T13:11:33.410Z
Would the path still be sinusoidal if Earth wasn’t tilted? Or would it then be a straight line - similar to this animation? https://community.dur.ac.uk/john.lucey/users/sun_ecliptic.gif
keynes
CC BY-SA 4.0
2020-02-18T13:22:56.013Z
@keynes Yes, if the Earth's axis wasn't tilted, then the ecliptic would coincide with the equator.
PM 2Ring
CC BY-SA 4.0
2020-02-18T09:44:28.183Z
2020-02-18T09:51:19.797Z
2020-02-18T09:51:19.797Z
PM 2Ring
<p>This answer supplements the other, better, clearer answers here already.</p>
<blockquote>
<p>Why does the Sun track out a seemingly sinusoidal path on the celestial sphere?</p>
</blockquote>
<p>It seems to be sinusoidal because for low inclinations the shape is roughly to sinusoidal (straight when crossing zero, has gently curved and symmetric extrema) and so we don't stop and ask what shape it is.</p>
<p><a href="https://en.wikipedia.org/wiki/Equirectangular_projection" rel="nofollow noreferrer">Equirectangular projection</a> maps spherical coordinates <span class="math-container">$\varphi, \theta$</span> or lon, lat or RA, Dec (but with zero at the equator) on to cartesian <span class="math-container">$X, Y$</span> axes with the mind-numbingly simple transform:</p>
<p><span class="math-container">\begin{align}
X &amp; = \varphi \\
Y &amp; = \pi/2 - \theta, \\
\end{align}</span></p>
<p>but when you do that an inclined plane intersecting a unit (or celestial) sphere sphere doesn't really give you a sinusoidal wave in spherical coordinates.</p>
<p>Lifted from <a href="https://space.stackexchange.com/a/43416">this answer</a> to <em>Analytical expression for the ground track of the International Space Station:</em></p>
<p>For an inclination <span class="math-container">$i$</span> and intersection along the <span class="math-container">$x$</span> axis the intersection can be described parametrically as:</p>
<p><span class="math-container">\begin{align}
x &amp; = \cos t \\
y &amp; = \sin t \ \cos i\\
z &amp; = \sin t \ \sin i\\
\end{align}</span></p>
<p>where <span class="math-container">$t$</span> is the distance traveled around the circle from 0 to <span class="math-container">$2 \pi$</span>, which you can think of as one orbit or one year, and</p>
<p><span class="math-container">\begin{align}
\varphi &amp; = \arctan2(y, x)\\
\theta &amp; = \arcsin(z).\\
\end{align}</span></p>
<p><a href="https://i.sstatic.net/UtHNf.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UtHNf.png" alt="enter image description here" /></a></p>
<pre><code>import numpy as np
import matplotlib.pyplot as plt

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
to_degs, to_rads = 180/pi, pi/180

incs = to_rads * np.arange(0, 90, 11)
t = to_rads * np.arange(-179, 180) # left out endpoints to avoid wraparound in plot

ct, st = np.cos(t), np.sin(t)

curves = []

for inc in incs:
    
    cinc, sinc = np.cos(inc), np.sin(inc)
    x, y, z = ct, st * cinc, st * sinc
    phi = np.arctan2(y, x)
    # phi = np.mod(phi + pi, twopi) - pi
    theta = np.arcsin(z)
    curves.append((inc, theta, phi))

plt.figure()
m, n = 9, 10
for i, (inc, theta, phi) in enumerate(curves):
    plt.plot(to_degs * phi, to_degs * theta)
    plt.plot(to_degs * phi[m::n], to_degs * theta[m::n], '.k')
plt.xlim(-180, 180)
plt.ylim(-90, 90)
plt.title('inclinations: 0, 11, 22, 33, 44, 55, 66, 77, 88 degrees')
plt.xlabel('RA', fontsize=12)
plt.ylabel('Dec', fontsize=12)
plt.gca().set_aspect('equal')
plt.show()
</code></pre>

CC BY-SA 4.0
2020-10-13T18:00:30.703Z
Interesting. Maybe they'd be true sinusoids in some Mercator-like projection?
Mike G
CC BY-SA 4.0
2020-10-13T18:05:56.307Z
@MikeG ya that's exactly what I was hoping when I saw your answer and that you used the fancy word "equirectangular". I was sad when I found out what it was and there wasn't really anything "equi" about it. I think your question is an excellent candidate for Math SE. *Is there some transform from an unwrapped sphere to a rectangle that would map great circles on to sinusoids?* You could relax it so equally spaced points on the circle can be substantially unequally spaced (as they are above). If you frame it cleanly and just let it sit, I'll bet it will get a definitive answer.
uhoh
CC BY-SA 4.0
2020-10-13T18:07:32.470Z
@MikeG What I like about Math SE is that if the answer is No then they will explain why or possibly even *prove it* definitively.
uhoh
CC BY-SA 4.0
2020-10-13T18:08:33.853Z
@MikeG Maybe not, since the Mercator represents paths of constant bearing ([loxodromes](https://en.wikipedia.org/wiki/Rhumb_line)) as straight lines instead of as spirals. But I could be wrong... :)
PM 2Ring
CC BY-SA 4.0
2020-10-13T18:10:21.580Z
@PM2Ring If all planes were limited to contain the x-axis (as drawn here) I wonder if that would make it sufficiently easier? That might be a work around for the loxodrome thing. It would be a silly projection probably, but silly never scares Math SE: [1](https://math.stackexchange.com/q/3097867), [2](https://math.stackexchange.com/q/2156273) :-) (yes closed, but well-received and multi-answered)
uhoh
CC BY-SA 4.0
2020-10-13T18:26:53.890Z
GIS SE has a [related question](https://gis.stackexchange.com/q/52949). Stereographic would plot great circles as circular arcs. [Central cylindrical](https://en.wikipedia.org/wiki/Central_cylindrical_projection) looks promising.
Mike G
CC BY-SA 4.0
2020-10-13T16:45:08.470Z
2020-10-13T18:54:23.903Z
2020-10-13T18:54:23.903Z
uhoh
<p>The question arose because I wanted to understand the path of the ecliptic after reading about it here: <a href="https://johnlucey.webspace.durham.ac.uk/users/solar-year/" rel="nofollow noreferrer">https://johnlucey.webspace.durham.ac.uk/users/solar-year/</a></p>
<p><a href="https://i.sstatic.net/dyR1c.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dyR1c.gif" alt="Sun's path on ecliptic" /></a></p>
<p>Is the sinusoidal path due to Earth’s axial tilt? Or is it related to mapping the path on a 2D pane? Is there a way to visualise this to aid understanding?</p>

<celestial-mechanics><ecliptic>
Why does the Sun track out a seemingly sinusoidal path on the celestial sphere?
keynes
# Why does the Sun track out a seemingly sinusoidal path on the celestial sphere?

(Asked by: username_0 on 2020-02-18)

The question arose because I wanted to understand the path of the ecliptic after reading about it here: <https://johnlucey.webspace.durham.ac.uk/users/solar-year/>

[![Sun's path on ecliptic](https://i.sstatic.net/dyR1c.gif)](https://i.sstatic.net/dyR1c.gif)

Is the sinusoidal path due to Earth’s axial tilt? Or is it related to mapping the path on a 2D pane? Is there a way to visualise this to aid understanding?

**Tags:** celestial-mechanics, ecliptic


---

## 3 answers

### Answer by username_1 on 2020-02-19. Score: 4

Both the equator and the ecliptic are great circles on the celestial sphere.
The appearance of each on a map depends on the map projection.

In an [equirectangular](https://en.wikipedia.org/wiki/Equirectangular_projection) projection centered on the equator, the equator (brown) is a straight line, and the ecliptic (blue) is approximately sinusoidal.

![equator-centered cylindrical projection](https://i.sstatic.net/93uT0.png)

If the same projection is centered on the ecliptic instead, the *ecliptic* is a straight line and the *equator* is approximately sinusoidal.
As [username_3's answer](/a/39356) illustrates, neither curve is exactly a sinusoid.

![ecliptic-centered cylindrical projection](https://i.sstatic.net/MAHYJ.png)

[Lucey](https://johnlucey.webspace.durham.ac.uk/users/solar-year/) provides both views too, but his ecliptic-centered map doesn't show the equator.
The axes are labeled in ecliptic rather than equatorial coordinates.

![ecliptic-centered cylindrical by Lucey](https://i.sstatic.net/i3WQJ.png)

He also provides a [stereographic](https://en.wikipedia.org/wiki/Stereographic_projection) projection centered on the [midday zenith at 55°N](https://community.dur.ac.uk/john.lucey/users/solar_year_durham.html).
Here both equator (red) and ecliptic (green) are mapped as circular arcs.

![zenith-centered stereographic by Lucey](https://i.sstatic.net/tGAYY.png)


---

### Answer by username_2 on 2020-02-18. Score: 2

The ecliptic is a plane. It looks like a sine wave because that diagram maps the sky onto a flat plane. On the celestial sphere, the ecliptic is a [great circle](https://en.wikipedia.org/wiki/Great_circle) that crosses the celestial equator at the [equinox points](https://en.wikipedia.org/wiki/First_Point_of_Aries) and which is tilted to the celestial equator by the same amount as the Earth's axial tilt (currently [23°26′12.0″](https://en.wikipedia.org/wiki/Axial_tilt)).

Here's a diagram from [Wikipedia](https://en.m.wikipedia.org/wiki/Equatorial_coordinate_system) that might be helpful:

![Celestial sphere](https://upload.wikimedia.org/wikipedia/commons/6/66/Ra_and_dec_demo_animation_small.gif)

> **Comments:**
> **username_0**: Would the path still be sinusoidal if Earth wasn’t tilted? Or would it then be a straight line - similar to this animation? https://community.dur.ac.uk/john.lucey/users/sun_ecliptic.gif  
> **username_2**: @username_0 Yes, if the Earth's axis wasn't tilted, then the ecliptic would coincide with the equator.  

---

### Answer by username_3 on 2020-10-13. Score: 2

This answer supplements the other, better, clearer answers here already.

Why does the Sun track out a seemingly sinusoidal path on the celestial sphere?

It seems to be sinusoidal because for low inclinations the shape is roughly to sinusoidal (straight when crossing zero, has gently curved and symmetric extrema) and so we don't stop and ask what shape it is.
Equirectangular projection maps spherical coordinates $\varphi, \theta$ or lon, lat or RA, Dec (but with zero at the equator) on to cartesian $X, Y$ axes with the mind-numbingly simple transform:
\begin{align}
X & = \varphi \\
Y & = \pi/2 - \theta, \\
\end{align}
but when you do that an inclined plane intersecting a unit (or celestial) sphere sphere doesn't really give you a sinusoidal wave in spherical coordinates.
Lifted from this answer to Analytical expression for the ground track of the International Space Station:
For an inclination $i$ and intersection along the $x$ axis the intersection can be described parametrically as:
\begin{align}
x & = \cos t \\
y & = \sin t \ \cos i\\
z & = \sin t \ \sin i\\
\end{align}
where $t$ is the distance traveled around the circle from 0 to $2 \pi$, which you can think of as one orbit or one year, and
\begin{align}
\varphi & = \arctan2(y, x)\\
\theta & = \arcsin(z).\\
\end{align}

import numpy as np
import matplotlib.pyplot as plt

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
to_degs, to_rads = 180/pi, pi/180

incs = to_rads * np.arange(0, 90, 11)
t = to_rads * np.arange(-179, 180) # left out endpoints to avoid wraparound in plot

ct, st = np.cos(t), np.sin(t)

curves = []

for inc in incs:
    
    cinc, sinc = np.cos(inc), np.sin(inc)
    x, y, z = ct, st * cinc, st * sinc
    phi = np.arctan2(y, x)
    # phi = np.mod(phi + pi, twopi) - pi
    theta = np.arcsin(z)
    curves.append((inc, theta, phi))

plt.figure()
m, n = 9, 10
for i, (inc, theta, phi) in enumerate(curves):
    plt.plot(to_degs * phi, to_degs * theta)
    plt.plot(to_degs * phi[m::n], to_degs * theta[m::n], '.k')
plt.xlim(-180, 180)
plt.ylim(-90, 90)
plt.title('inclinations: 0, 11, 22, 33, 44, 55, 66, 77, 88 degrees')
plt.xlabel('RA', fontsize=12)
plt.ylabel('Dec', fontsize=12)
plt.gca().set_aspect('equal')
plt.show()

> **Comments:**
> **username_1**: Interesting. Maybe they'd be true sinusoids in some Mercator-like projection?  
> **username_3**: @MikeG ya that's exactly what I was hoping when I saw your answer and that you used the fancy word "equirectangular". I was sad when I found out what it was and there wasn't really anything "equi" about it. I think your question is an excellent candidate for Math SE. *Is there some transform from an unwrapped sphere to a rectangle that would map great circles on to sinusoids?* You could relax it so equally spaced points on the circle can be substantially unequally spaced (as they are above). If you frame it cleanly and just let it sit, I'll bet it will get a definitive answer.  
> **username_3**: @MikeG What I like about Math SE is that if the answer is No then they will explain why or possibly even *prove it* definitively.  
> **username_2**: @MikeG Maybe not, since the Mercator represents paths of constant bearing ([loxodromes](https://en.wikipedia.org/wiki/Rhumb_line)) as straight lines instead of as spirals. But I could be wrong... :)  
> **username_3**: @PM2Ring If all planes were limited to contain the x-axis (as drawn here) I wonder if that would make it sufficiently easier? That might be a work around for the loxodrome thing. It would be a silly projection probably, but silly never scares Math SE: [1](https://math.stackexchange.com/q/3097867), [2](https://math.stackexchange.com/q/2156273) :-) (yes closed, but well-received and multi-answered)  
> **username_1**: GIS SE has a [related question](https://gis.stackexchange.com/q/52949). Stereographic would plot great circles as circular arcs. [Central cylindrical](https://en.wikipedia.org/wiki/Central_cylindrical_projection) looks promising.  

---

