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

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?


3 Answers 3


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 projection centered on the equator, the equator (brown) is a straight line, and the ecliptic (blue) is approximately sinusoidal.

equator-centered cylindrical projection

If the same projection is centered on the ecliptic instead, the ecliptic is a straight line and the equator is approximately sinusoidal. As uhoh's answer illustrates, neither curve is exactly a sinusoid.

ecliptic-centered cylindrical projection

Lucey 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

He also provides a stereographic projection centered on the midday zenith at 55°N. Here both equator (red) and ecliptic (green) are mapped as circular arcs.

zenith-centered stereographic by Lucey


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 that crosses the celestial equator at the equinox points and which is tilted to the celestial equator by the same amount as the Earth's axial tilt (currently 23°26′12.0″).

Here's a diagram from Wikipedia that might be helpful:

Celestial sphere

  • $\begingroup$ Would the path still be sinusoidal if Earth wasn’t tilted? Or would it then be a straight line - similar to this animation? community.dur.ac.uk/john.lucey/users/sun_ecliptic.gif $\endgroup$
    – keynes
    Feb 18, 2020 at 13:11
  • $\begingroup$ @keynes Yes, if the Earth's axis wasn't tilted, then the ecliptic would coincide with the equator. $\endgroup$
    – PM 2Ring
    Feb 18, 2020 at 13:22

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}

enter image description here

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))

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)
  • 1
    $\begingroup$ Interesting. Maybe they'd be true sinusoids in some Mercator-like projection? $\endgroup$
    – Mike G
    Oct 13, 2020 at 18:00
  • $\begingroup$ @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. $\endgroup$
    – uhoh
    Oct 13, 2020 at 18:05
  • $\begingroup$ @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. $\endgroup$
    – uhoh
    Oct 13, 2020 at 18:07
  • $\begingroup$ @MikeG Maybe not, since the Mercator represents paths of constant bearing (loxodromes) as straight lines instead of as spirals. But I could be wrong... :) $\endgroup$
    – PM 2Ring
    Oct 13, 2020 at 18:08
  • 2
    $\begingroup$ GIS SE has a related question. Stereographic would plot great circles as circular arcs. Central cylindrical looks promising. $\endgroup$
    – Mike G
    Oct 13, 2020 at 18:26

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