Sundial: Most probably you have seen one in a science museum or in an observatory. This very simple and amazing apparatus (do you call a stick buried in the ground an apparatus!) can be considered the man’s second important invention after the wheel. With no gears or moving wheels it tells us, with sufficient accuracy, the time of the day. It basically consists of a rod that sticks out of a graduated base. The shadow of the rod reflected by the sun on the base indicates the time of the day. It comes in many forms and shapes. The simplest is the equatorial sundial whose face is parallel to the equatorial plane and the gnomon (a protruding pole from the dial center) is to it. Its dial perimeter is divided into equal 24 parts and configured so that its shadow points to 12 o’clock on Sun culmination. The Sun culmination (noon) is the epoch at which the Sun attains its highest altitude, and when the shadow of an upright pole would be the shortest of the day. It is the time when the Sun, pole and its shadow mark the local meridian (North-South) plane. With this setup, the Sun shadow indicates the hour of the day. History of sundials goes deep in the ancient Egyptian, Babylonian and ancient Greek times. In this site [1] we may find a wealth of the history of this amazing tool. You may also be able to construct your own sundial. To convert to local time you will need to know the longitude of sundial position, as well as of course, the Equation of Time.

Of course the sundial has drawbacks, but—considering the times in which it was used—was very effective method for measuring the time during the day and indicating the start of a new day (days used to start at noon). And by a stretch of imagination we could also determine the day or the season of the year. The time elapsed between two culminations is called solar day. The time system based on this unit is called the apparent solar time (based on the apparent Sun, which is the celestial body that we observe daily revolving about the Earth). Its position in the sky is determined by its spherical angle with the local meridian and it is this angle that is measured by the Sundial.

An annoying issue: Since antiquity astronomers have observed that the period of the solar day varies throughout the year. That is with a “uniform” running watch; we find that the noon intervals vary throughout the course of the year. If this uniform watch is so mechanized that its days are the average of all the solar days throughout the year we arrive at our modern watch. As will be seen later, the variation in the solar day, defined earlier as the period between consecutive culminations, is not constant, but a result of the orbit eccentricity and obliquity. Now we have two time scales the apparent time i.e. sundial time (which we have no means for measuring it in modern times) and the mean solar time (which we can measure accurately by our clocks). The Equation of Time (EoT), is the difference between the apparent solar time and the mean solar time:

$EoT=\text{apparent solar time}\left(\text{dial}\right)-\text{mean solar time}\left(\text{clock}\right)$ (1.1)

It varies roughly between ±15 minutes. The word “equation” is meant to be the balance tip between the two times.

Despite its simple appearance, deriving the mathematical formula of the EoT requires a bit of math, analysis and a bit of imagination. Its two terms appear to be completely unrelated. It is because they miss the many angles that relate them. We aim to describe the relevant angles in both of the ecliptic and equatorial and tie them to a common computational frame. We start with the ecliptic plane.

Herein, we describe the motion of Earth in the ecliptic plane. This motion is controlled by the Kepler laws. The first law states that the orbit of the Earth is an ellipse with Sun resides at its focus.

As shown in Figure 1 , a and b are the semi major and semi minor axes of the ellipse, and its eccentricity is $e=\sqrt{1-b^2/a^2}$. The end points of the major axis are called the perihelion, the closest point to the Sun, and the aphelion its farthest point.

Figure 2 shows the angle,ν, called the true anomaly that the Earth-Sun radius vector makes with the semi-major axis. In the figure, c is the center of the ellipse. It also shows the eccentric anomaly E the angle between radius cD and the semi-major axis. D is the intersection of the orthogonal of the Earth to the semi-major axis and the circle formed by the semi-major axis. Ellipse geometry shows that the true anomaly ν is related to the eccentric anomaly angle E, by

$\tan \frac{\nu }{2}=\frac{1}{\cos \alpha }\tan \frac{E}{2}$(1.2)

where $\cos \alpha =\sqrt{\frac{1-e}{1+e}}$(1.3)

As the Earth leaves the perihelion, its Earth-Sun radius vector would sweep an elliptic area (bounded by the semi-major axis, the radius vector and the elliptical arc that joins them). Kepler’s 2^nd law states that the Earth-Sun vector sweeps equal areas in equal times; i.e. the rate of swept area is constant. Thus the period T takes the Earth to sweep the entire ellipse is constant. If Mean motion is $n=1/T$ rev/day; then swept area/ellipse area $=t/T=nt$. Thus Kepler’s second law can be expressed by the more familiar equations

$E-e\sin E=M$ (1.4)

$M=nt$ (1.5)

where M is the Sun mean anomaly.

From now on, we will adopt the geocentric view for Sun-Earth motion, in which the Earth is at the center of the universe and the Sun describes an ellipse in the ecliptic, as shown in Figure 3 . In it, we will be able to depict all the angles needed to continue with the mathematical derivations as shown in Figure 4 . This figure—which is a variant of Figure 3 —depicts a key player “the vernal equinox line”. This line is the intersection of the ecliptic and the equator planes and to which all angles—called ecliptic longitudes—of the celestial bodies are referenced. It points to an almost a fixed position in the sky, which changes very slowly over time. The ecliptic is inclined to the equator by an angle, ε.

The linearity of M with time suggests the motion of an imaginary Sun m that orbits the ecliptic at a uniform rate. Astronomers [2] - [5] have invented a hypothetical Sun (called the dynamic Sun)—runs in the ecliptic—and orbits the Earth at constant angular rate. Its rate is selected such that it completes a 360 circle around the Earth in exactly the same time as the true Sun. This Sun is nothing but the imaginary Sun m described earlier.

Figure 4 depicts l and l_p the ecliptic longitudes of the Apparent Sun and the perihelion respectively. It also shows l_M, defined to be the longitude of the mean Sun. Moreover, ν is the true anomaly, M is the mean Sun anomaly and P is the angle between the perihelion and the winter solstice. From Figure 4 we get,

$l_M=l_p+M$ (1.6)

$l=l_p+\nu$ (1.7)

Even though the dynamic Sun has overcome the uniformity issue, yet we still face another problem. This dynamic Sun runs in the ecliptic that is inclined to the equator, hence its time on a sundial will not be uniform. Because the dynamic Sun runs in the ecliptic, its observed time will not appear to be uniform. Thus it can not be used as a time keeper. This is averted by inventing another hypothetical Sun [2] , called the mean Sun that runs in the equator and has exactly the same angular rate as the dynamic Sun. Both will always meet, at the vernal equinox. Time is associated with the equator plane, and it is time, no pun, to discuss it and some of its relevant angles.

Figure 5 depicts the plane and three relevant angles. If we imagine the projection of the Earth on the equator plane, we can see the radii that branch from the equator center are just the lines of global longitudes. These longitudes—similar to the ecliptic longitudes—provide alternative way to determine the position of celestial bodies. It uses two angles; the first is the right ascension (RA) the global longitude but referenced relative to the vernal equinox line, see Figure 5 . The second angle, not shown in the figure, is determined by the angle of declination of the body relative to the equator, (its global latitude so to speak). Even though RA is a geometric angle whose value is determined by degrees, it is typically represented by hours by dividing it by (15 = 360 deg/24 hrs per day). The Greenwich Hour Angle (GHA) is the longitude of the celestial body. GHA, like the RA, is also given in hours. It should be noticed that in astronomy, longitudes are positive westward and negative eastwards. Finally, the angle between the Greenwich line and the vernal equinox is called Greenwich Apparent Sidereal Time (GAST).

With all these angles described in the ecliptic and equatorial planes, we can proceed to derive the EoT formula.

With no loss of generality, we shall use the Greenwich longitude as the line on which observers are situated and to which all angles are referenced as indicated in Figure 6 . If the apparent Sun is on the longitude NA, then the apparent time is determined by how the Sun is far from the Greenwich longitude. But this is exactly given by the central angle that is bounded by AG in the figure. (Do you still remember the sundial? If the gnomon is protruding from the center of the celestial sphere, the Sun shadow would fall along cA; and the angle GcA would indicate the Apparent Sun time.) This angle is the same as the hour angle h_A, or the GHA just described. Likewise, the mean Sun time is indicated by the hour angle h_M. Therefore Equation (1.1) becomes

$EoT=h_A\left(\text{hour angle of Apparent Sun}\right)-h_M\left(\text{hour angle of Mean Sun}\right)$ (1.8)

The geometry of Figure 6 implies

${\alpha }_A\left(\text{right ascension}\right)=GAST-h_A\left(\text{hour angle}\right)$ (1.9)

and ${\alpha }_M\left(\text{right ascension}\right)=GAST-h_M\left(\text{hour angle}\right)$ (1.10)

Substituting from Equations (1.9) and (1.10) in Equation (1.8) gives

$\begin{array}{c}EoT=\left(GAST-RA\text{of Apparent Sun}\right)-\left(GAST-RA\text{of Mean Sun}\right)\\ ={\alpha }_M-{\alpha }_A\end{array}$(1.11)

Since, by choice, $l_M={\alpha }_M$, then the above becomes

$EoT=l_M-{\alpha }_A$ (1.12)

In Figure 6 , γAS is a spherical triangle with right angle A; thus has the properties

$\begin{array}{l}\sin SA=\sin \epsilon \sin \gamma S\\ \tan \gamma A=\cos \epsilon \tan \gamma S\end{array}$

Since AS = δ; γA = α_A; γS = l, where δ is the Sun declination angle, then:

$\begin{array}{l}\sin \delta =\sin \epsilon \sin l\\ \tan {\alpha }_A=\cos \epsilon \tan l\end{array}$ (1.13)

Equation (1.13) implies

${\alpha }_A={\tan }^{-1}\left(\cos \epsilon \tan l\right)$(1.14)

Thus Equation (1.12) becomes:

$EoT=l_M-{\tan }^{-1}\left(\cos \epsilon \tan l\right)$ (1.15)

Equation is similar to Meeus [6] formula albeit in absence of the small terms that account for aberration and nutation.

By virtue of Equations (1.6) and (1.7) we get an alternate, yet identical form of the EoT

$EoT=M+P-{\tan }^{-1}\frac{\tan \left(P+\nu \right)}{\cos \epsilon }$ (1.16)

The first step for solving the EoT is to compute the mean anomaly. For a given Julian day, JD, the mean anomaly M, is given by Meeus [6]

$\begin{array}{l}T=\left(JD-2451545\right)/36525\\ M=357.52911+35999.05029T-0.0001537T^2\left[\mathrm{deg}\right]\end{array}$ (1.17)

The mean longitude, the first term of Equation (1.15), is given by

$l_M=280.46646+36000.76983T+0.0003032T^2\left[\mathrm{deg}\right]$ (1.18)

From Equations (1.4) and (1.2), the true anomaly, ν is computed. The celestial longitude l is then computed from,

$l=l_M-M+\nu$ (1.19)

Alternative to the mean longitude is the argument to perihelion, P from which

$l_M=M+P+3\pi /2$ (1.20)

Now, we have all the terms for computing the EoT. We have many methods for executing the computations. Common to all the methods are the executions of Equation (1.17) and Equation (1.18) or Equation (1.20). We start with the analytical method.

We shall see that the EoT is caused by the Earth orbit’s ellipticity and its obliquity.

The paper explained the EoT and its relevance to astronomical measurements. It has presented the common algorithms for computing it. It should be noticed that all share the same root. They differ only on the approximation methods used and the available parameters needed for computing it.

The author accredits the referenced authors for many of the ideas presented herein. He is also grateful to Dr. A. Bekir for his careful editing of the paper and suggesting many improvements.

This equation implies:

$y={\tan }^{-1}\left(\frac{1}{\cos \alpha }\tan x\right)$ (A.1)

The derivative of this equation gives

$\frac{\text{d}y}{\text{d}x}=\frac{1}{\cos \alpha }\frac{{\sec }^{2}x}{\frac{{\tan }^{2}x}{{\cos }^2\alpha }+1}=\frac{\cos \alpha {\sec }^{2}x}{{\sec }^{2}x-1+{\cos }^2\alpha }=\frac{\cos \alpha }{1-{\sin }^2\alpha {\cos }^{2}x}\Rightarrow$

$\frac{\text{d}y}{\text{d}x}=\frac{2\cos \alpha }{2-{\sin }^2\alpha \left(1+\cos 2x\right)}=\frac{2\cos \alpha }{1+{\cos }^2\alpha -{\sin }^2\alpha \cos 2x}\Rightarrow$

$\frac{\text{d}y}{\text{d}x}=\frac{4\cos \alpha }{2\left(1+{\cos }^2\alpha \right)-{\sin }^2\alpha \left({\text{e}}^{i2x}+{\text{e}}^{-i2x}\right)}=\frac{-4\cos \alpha /{\sin }^2\alpha }{\left({\text{e}}^{i2x}+{\text{e}}^{-i2x}\right)-\frac{2\left(1+{\cos }^2\alpha \right)}{{\sin }^2\alpha }}\Rightarrow$

$\frac{\text{d}y}{\text{d}x}=\frac{-4\cos \alpha /{\sin }^2\alpha {\text{e}}^{i2x}}{{\text{e}}^{i4x}-\frac{2\left(1+{\cos }^2\alpha \right)}{{\sin }^2\alpha }{\text{e}}^{i2x}+1}=\frac{-4\cos \alpha /{\sin }^2\alpha {\text{e}}^{i2x}}{\left({\text{e}}^{i2x}-\beta \right)\left({\text{e}}^{i2x}-{\beta }^{-1}\right)}$ (A.2)

In the above, the roots of the denominator $\beta$ and ${\beta }^{-1}$ are determined as follows

$\begin{array}{c}\beta =\frac{1+{\cos }^2\alpha }{{\sin }^2\alpha }-\sqrt{{\left(\frac{1+{\cos }^2\alpha }{{\sin }^2\alpha }\right)}^2-1}\\ =\frac{\left(1+{\cos }^2\alpha \right)-\sqrt{\left(\left(1+{\cos }^2\alpha \right)-{\sin }^2\alpha \right)\left(\left(1+{\cos }^2\alpha \right)+{\sin }^2\alpha \right)}}{{\sin }^2\alpha }\\ =\frac{\left(1+{\cos }^2\alpha \right)-\sqrt{2{\cos }^2\alpha \cdot 2}}{{\sin }^2\alpha }=\frac{\left(1+{\cos }^2\alpha \right)-2\cos \alpha }{1-{\cos }^2\alpha }\\ =\frac{{\left(1-\cos \alpha \right)}^2}{\left(1-\cos \alpha \right)\left(1+\cos \alpha \right)}=\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{2{\sin }^2\frac{\alpha }{2}}{2{\cos }^2\frac{\alpha }{2}}={\tan }^2\frac{\alpha }{2}\end{array}$ (A.3)

Applying the partial fractions to (A.2) gives

$\begin{array}{c}\frac{\text{d}y}{\text{d}x}=\frac{-4\cos \alpha }{{\sin }^2\alpha \left(\beta -{\beta }^{-1}\right)}\left(\frac{{\text{e}}^{i2x}}{{\text{e}}^{i2x}-\beta }-\frac{{\text{e}}^{i2x}}{{\text{e}}^{i2x}-{\beta }^{-1}}\right)\\ =\frac{-4\cos \alpha }{{\sin }^2\alpha \left(\beta -{\beta }^{-1}\right)}\left(\frac{1}{1-\beta {\text{e}}^{-i2x}}+\frac{\beta {\text{e}}^{i2x}}{1-\beta {\text{e}}^{i2x}}\right)\end{array}$ (A.4)

Equation (A.3) gives

$\beta -{\beta }^{-1}=\frac{1-\cos \alpha }{1+\cos \alpha }-\frac{1+\cos \alpha }{1-\cos \alpha }=\frac{{\left(1-\cos \alpha \right)}^2-{\left(1+\cos \alpha \right)}^2}{1-{\cos }^2\alpha }=\frac{-4\cos \alpha }{{\sin }^2\alpha }$ (A.5)

Supposing that $\beta <1$ and using (A.5), Equation (A.4) can be expanded into

$\begin{array}{c}\frac{\text{d}y}{\text{d}x}={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\beta }^n{\text{e}}^{-i2nx}}+\beta {\text{e}}^{i2x}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{\beta }^n{\text{e}}^{i2nx}}=1+{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}{\beta }^n{\text{e}}^{-i2nx}}+{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}{\beta }^n{\text{e}}^{i2nx}}\\ =1+{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}{\beta }^n\left({\text{e}}^{i2nx}+{\text{e}}^{-i2nx}\right)}\end{array}$

$\frac{\text{d}y}{\text{d}x}=1+{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}2{\beta }^n\cos \left(2nx\right)}$ (A.6)

Integrating (A.6) gives

$y=x+{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{{\beta }^n}{n}\sin \left(2nx\right)}+const$

From (A.1), y = 0 at x = 0, thus the integration constant in the above is 0, therefore

$y=x+{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{1}{n}{\tan }^{2n}\frac{\alpha }{2}\sin \left(2nx\right)}\Rightarrow$

$y=x+{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\frac{1}{n}{\left(\frac{1-\cos \alpha }{1+\cos \alpha }\right)}^n\sin \left(2nx\right)}$ (A.7)

Special Case: $\cos \alpha =\sqrt{\frac{1-e}{1+e}}$

The coefficients of (A.7) can be simplified as follows:

$\begin{array}{c}\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{{\left(1-\cos \alpha \right)}^2}{1-{\cos }^2\alpha }=\frac{1+{\cos }^2\alpha -2\cos \alpha }{1-{\cos }^2\alpha }=\frac{1+\frac{1-e}{1+e}-2\sqrt{\frac{1-e}{1+e}}}{1-\frac{1-e}{1+e}}\\ =\frac{1+e+1-e-2\sqrt{1-e^2}}{1+e-1+e}=\frac{1-\sqrt{1-e^2}}{e}\end{array}$

When e is small, then to an accuracy of e³ the above expression is approximated to

$\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{1}{e}\left(1-1+\frac{1}{2}{e}^2+\frac{1}{8}{e}^4\right)=\frac{1}{2}e+\frac{1}{8}{e}^3$

Hence

${\left(\frac{1-\cos \alpha }{1+\cos \alpha }\right)}^2=\frac{1}{4}{e}^2$ and ${\left(\frac{1-\cos \alpha }{1+\cos \alpha }\right)}^3=\frac{1}{8}{e}^3$

Consequently (A.7) is simplified to

$y=x+\left(\frac{1}{2}e+\frac{1}{8}{e}^3\right)\sin 2x+\frac{1}{8}{e}^2\sin 4x+\frac{1}{24}{e}^3\sin 6x$ (A.8)

The Lagrange reversion theorem, Battin [10] , Deakin [11] , states that, if

$z=y+xf\left(z\right)$ (B.1)

x, y and z are arbitrary variables, then g(z) can be expanded into a series of x and y

$\begin{array}{c}g\left(z\right)=g\left(y\right)+x{g}^{\prime }\left(y\right)f\left(y\right)+\frac{x^2}{2!}\frac{\partial }{\partial y}\left\{g^{\prime }\left(y\right){\left[f\left(y\right)\right]}^2\right\}\\ +\frac{x^3}{3!}\frac{{\partial }^2}{\partial y^2}\left\{g^{\prime }\left(y\right){\left[f\left(y\right)\right]}^3\right\}+\cdots \end{array}$(B.2)

Applying to the Kepler equation yields,

$E=M+e\sin E$ (B.3)

Comparing with (B.1) shows that expansion is obtained when:

$z=Ey=Mx=ef\left(z\right)=\sin \left(z\right)g\left(z\right)=z$(B.4)

Substituting in the Lagrange equation yields

$E=M+e\sin M+\frac{e^2}{2!}\frac{\partial }{\partial M}{\sin }^{2}M+\frac{e^3}{3!}\frac{{\partial }^2}{\partial M^2}{\sin }^{3}M+\cdots$

$E=M+e\sin M+\frac{e^2}{2}\sin 2M+\frac{e^3}{8}\left(3\sin 3M-\sin M\right)+\frac{e^4}{6}\left(2\sin 4M-\sin 2M\right)+\cdots$

Rearranging terms yields the expansion of the Kepler equation

$E=M+\left(e-\frac{1}{8}{e}^3\right)\sin M+\left(\frac{1}{2}{e}^2-\frac{1}{6}{e}^4\right)\sin 2M+\frac{3}{8}{e}^3\sin 3M+\frac{1}{3}{e}^4\sin 4M+\cdots$(B.5)

Substituting for

$\begin{array}{l}{\alpha }_A=90-{\alpha }_c\\ l=90-l_c\end{array}$

in the above equation yields $\tan \left(90-{\alpha }_c\right)=\cos \epsilon \tan \left(90-l_c\right)$

From the properties of the tangent function, the above becomes

$\begin{array}{l}1/\tan \left({\alpha }_c\right)=\cos \epsilon /\tan \left(l_c\right)\Rightarrow \\ \tan \left({\alpha }_c\right)=\left(1/\cos \epsilon \right)\tan \left(l_c\right)\end{array}$

for which the series expansion is given by

${\alpha }_c=l_c+y\sin \left(2l_c\right)+1/2y^2\sin \left(4l_c\right)$

where $y={\tan }^2\epsilon /2$

Substituting back for α_A and l results in

$90-{\alpha }_A=90-l+y\sin \left(180-2l\right)+1/2y^2\sin \left(360-4l\right)$

which yields,

${\alpha }_A=l-y\sin \left(2l\right)+1/2y^2\sin \left(4l\right)$