I'm working on an on-line app to show the Equation of Time for arbitrary eccentricity and obliquity of the Earth's orbit.

It looks good for small eccentricities and obliquities:

Equation of time with small eccentricity and obliquity

but for bigger ones the components of the Equation (the so called Equation of the Center and Projection to the Equator) loose all the "sine-like" look:

Equation of time with big eccentricity and obliquity

I don't know if that is the expected behaviour; if it is, I would like to undestand why. :)

Thank you very much!

Edit (in response of the excellent analysis of d_e):

I compute the two curves independently; in other words, ignoring the effect of one on the other. After reading d_e's answer, I see now that there's a problem with it. But first, I make explicit the methods I used:

Equation of Center

For the Equation of Center I compute the difference between the mean anomaly ($M$) and the true anomaly ($\nu$). The mean anomaly is the angle between the perihelion and the position of a fictitious Earth, moving in a circle -whose radius is the semi-major axis of the Earth's orbit- with constant velocity, with vertex in the center of that circle. The true anomaly is the angle between the perihelion and the real Earth, with the Sun at its vertex.

The mean anomaly is easy; for the true anomaly I compute first the eccentric anomaly ($E$) using Kepler's equation:

$E=M+e\sin E$, where $e$ is the eccentricity. I use a fixed-point iteration.

Then I get the true anomaly with


Reduction to the Equator

For the Reduction to the Equator I compute the difference between a fictitiuos Earth moving with constant velocity along the plane of the Equator ($\lambda$) and the projection to the Equator of another fictitious mean Earth, moving with constant velocity in a circle inclined an angle $i$ with respect to the Equator ($\alpha$).

For $\lambda$, I compute the mean anomaly, but counting from the vernal equinox (*: see below). For $\alpha$, I use the spherical trigonometry relation:

$\tan \alpha = \frac{\sin \lambda \cos i}{\cos \lambda}$

(I don't use $\tan \alpha = \tan \lambda \cos i$ because I employ the function atan(a,b) for getting the correct quadrant).

(*) After d_e's analysis, now I think that this is a mistake: I should use the true anomaly as a value for $\lambda$, and then project that to get $\alpha$... am I right?

This is a screenshot of the case (as is) of the app when $e$ is low but $i$ is high:

Equation of time with small eccentricity and big obliquity

By the way, the app is up an running; if you want to change values yourself and see the results, it's here: on-line app of Equation of Time (as is)

Second Edit:

I changed the way of computing the Equation of Center according to the analysis of d_e. Now I think it's visually clear that the time between equinoxes is not half a year apart when $e$ is big. :)

Eccentricity affects the interval between equinoxes

I updated the on-line version of the app.

I think that now it correctly represents the effects of arbitrary inclination and eccentricity in the Equation of Time... ¡Thank you very much! I could not have done it alone.

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    $\begingroup$ Thanks for the more information you provided. I added few lines to my answer. This is very cool app! $\endgroup$
    – d_e
    Nov 4, 2021 at 15:32
  • $\begingroup$ @d_e: I updated the original post. Now I think the app works correctly. :) $\endgroup$ Nov 6, 2021 at 0:48
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    $\begingroup$ This page does something similar, but with the analemma: mtirado.com/blog/demystifying-the-analemma $\endgroup$ Dec 10, 2022 at 22:55
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    $\begingroup$ Related: astronomy.stackexchange.com/q/40529/16685 $\endgroup$
    – PM 2Ring
    Dec 11, 2022 at 3:33
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    $\begingroup$ BTW, it's easy to invert Kepler's equation via Newton's method. Live Python demo $\endgroup$
    – PM 2Ring
    Dec 11, 2022 at 4:41

1 Answer 1


Let's first considers what happens to the eccentricity part of the equation (equation of the center - green line), when we change eccentricity:

This component is due to the different speeds the Sun has in its elliptical orb. According to Kepler's law, the pace is faster as we go closer to the Sun. (1) what happens when the eccentricity is 0 (i.e., perfect circular orbit)? the speed of the planet is always the same, hence the green line should be flat at 0. (2) what happens when when we some eccentricity? The speed is not equal - we are running faster at the pehilion and it makes the days longer, since the Sun's eastward motion (on the Right Ascension line - let's assume no obliquely) is the fastest, hence after 24 hours the Sun will be east of the Meridian - Earth will need to rotate little more than that in order to put Sun's at the Meridian. Now, longer days, as it seems, implies lower equation - so we should see the green line go down around the point of the pehilion (4th Jan) with the maximum derivative at 4th Jan. (3) Now, the Sun gains speed before the pehilion and still has some high speed after the pehilion: the green line reflects this accumulated effect in that it reaching minimum value after the pehilion (but again the pace of going down must be higherst at the pehilion). The question now arises in what span of time (starting before the pehilion and ending after the pehilion) the Sun's speed is higher than the avg. which would effect the green line going down? The answer to this question is a function of the eccentricity: larger the eccentricity so is smaller the number of days the Sun's speed is higher than average; and so is the time the green line should go down, will be smaller. But why is that? shouldn't we expect equal time of being below and above the average speed? No. this is, again, due to the fact that the planet spends more time when far from the Sun (when the speed is slow and hardly change), and short amount of time close to the Sun (when the speed is fast but accelerate fast) - this result in kind of long tail distribution and hence the average speed is higher than the median speed - and we have fewer days where the speed is higher than average.

This concludes the green graph: the green graph when the eccentricity is high behaves as expected: very short time of going down and the stiffest pace of going down is at 4th Jan. As for losing the "sine shape" I wouldn't be that concerned it simply due the fact the the graph is shrined and extended so the sine is hardly visible. But in this answer I only provide heuristic to judge if the shape makes sense, not exact numbers.

Now, Let's move on to the blue graph of obliquity.

This component is due to the fact that the time of Sun to be seen on the Merdian depends not upon the elliptical longitude of the Sun, but rather on it's Right Ascension (RA). Hence the speed (i.e., the pace of moving longitudinally on the ecliptic) is not enough to determine the length of days. If, for example the Sun moves 30deg eastward of its 360deg of the orbit, we cannot expect it to move also 30deg (or 2 hours) on RA. this is because the plane of the ecliptic has angle with the plane of the equator. if the angle was 0, then we would expect the blue line to be flat at zero (since at that case we have no problem in converting ecliptic longtitude to RA - they are equal).

Now, how exactly this angle effect us? If we divide the great circle of the ecliptic into equal parts (equal arcs) - say even 365 parts -, each arc is of the same length, but this length is divided into 2 components: RA and declination. The problem is that this inner split between the two components is not the same in all the parts. (1) the greater the RA component is, it means the Sun moves faster eastward, hence the days are longer. (2) So our question is that: When (on what days) the RA component is greater? Well, obviously when the declination (dec) component is smaller. So when the dec component is smaller? Let's take a look on this picture from Wikipedia:

enter image description here

The dec is basically the height a point in the circle of the ecliptic above/or below the the green circle of the equator. We can easily see that when the Sun reaches the top point (max declination) or bottom point (min declination) - at those point the declination does not move and the declination component of the daily change is 0 (consider also the derivative at maxima/minima is 0); Hence they call those points solstices (=Sun stop); But for our concerns here the Sun does not stop, only the declination component is 0. So not only the Sun does not stop, rather it moves fastest in the RA. So at the solstices the RA component is high which means longer days. (3) Now, what happens at the nodes (equinoxes), where the dec is 0? it is easy to understand that on those points the declination coponent is the highest, the RA changes the least on those days making the Sun slow and the days shorter. (4) accordingly, we should expect to see rising blue graph at the equinoxes (rising because the days are shorted and shorted days means higher equation as we saw earlier); and also at the equinoxes the pace of rising is the highest: i.e., the derivative is of max positive value there. (5) by the same logic at the solstices we should expect the blue graph going down (longer days) and at the stiffest. (6) all this is well seen in the upper (of Earth) blue graph at those point of 21Mar 21Jun, 22Sep, 22Dec. (7) it follows indeed from (5) and (6) that somewhere in the middle between a solstice and equinox we should find the maxima (or minima). (8) Now, what happens when we increase the angle between the equator and ecliptic? Basically it changes nothing in our analysis before, but only the quantities and the amplitude of the equation. In other words, if you could provide a third graph where the eccentricity is kept and only the angle is changed - I believe we should see the exact same blue graph just in a different scale. (9) If you survived this far, here comes the most confusing (IMO) part of this answer: the location (in time) of the solstices and the other equinox do not stay the same if we change the eccentricity; for changing the eccentricity changes the time it takes for the Sun to move between node (=equinox) and solstices and then another node. Note, that even in our Earth (with his very small eccentricity) the time between March equinox and Sep equinox is not half a year exactly. if we change the eccentric and/or the node location with respect to the aphelion things are going to change. Therefore I find the analysis of how it would effect the blue line little more difficult.

Edit (some second thought):

  • With respect to point (8), it is right the analysis would stay the same, but I'm not sure if we should see the "exact same" graph as I said - maybe indeed we should find the minima and maxima in different places. I'll try to think of it later on.

  • On second look on the blue graph of second picture: it does seem to behave like our analysis in the 4 points: maximum positive derivative on 21Mar and 22Sep. and maximum negative derivative on the solstices. But as I said: as a result of changing the eccentricity those point should move in time (I'm quite convinced in that); So it would be interesting to see if your analysis included them moving? for it doesn't look it did.

After question Edit:

about point (8): indeed, in case of extreme inclination angle the RA component hardly changes when we are near the equinox, but rather very mostly at the area of the solstices where most of the change in the RA takes place there - so we should see a sharp short-time decrease as we are seeing in the blue graph around the solstices. To understand intuitively why this is so, I think it is quite analogous to the azimuth of the Sun at latitude 0 (Earth equator) on the days very near the equinox: say 24 Mar. The sun path in this day is almost perpendicular to the horizon (like in our example the ecliptic is almost perpendicular to the equator plane), so when the Sun rises from East it hardly changes its azimuth - but eventually near the Zenith almost the entire action of Azimuth change takes place: when the Sun moves through North, to the near the West. So in this sense the blue graph also behaves well and as expected

As for the question if we should take the Mean or True anomaly. My understanding is that we have to take the True anomaly. For $\lambda$ should represent the degrees different on the ecliptic between the current true location and the node (equinox). There are always 180 deg between the equinoxes, but this not necessary mapped to 1/2 of the period (not even in Earth). Taking the Mean anomaly for this calculation would imply the second equinox is 1/2 of the period apart.

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    $\begingroup$ Thank you very much for your answer; you trully convinced me with your arguments. I'm changing the app as soon as I can, and edit the original answer to reflect that. :) $\endgroup$ Nov 4, 2021 at 16:19
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    $\begingroup$ more accurate example is not 24Mar at the equator, but rather latitude of 1 (or very low) at the equinox. for 2 reasons: (1) at the equinox the Sun on dec=0 hence this is great circle like the ecliptic; (2) actually at latitude=0 , not matter what the declination it almost always the same behavior and perpendicular even in the solstices. $\endgroup$
    – d_e
    Nov 4, 2021 at 17:59

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