The Azimuth of Ecliptic Ascension Point

Consider the azimuth of the point on the ecliptic circle which is ascending at the eastern horizon. According to Eq. (52), the azimuth of any point on the horizon (i.e., $a=0^\circ$) satisfies $\cos A = {\bf r}\cdot {\bf n}$. It follows from Eqs. (33) and (49) that

$$\cos A= -\cos\lambda\,\sin L\,\sin\alpha + \sin\lambda\,\... ...\sin \epsilon + \sin\lambda\,\sin L\,\cos\epsilon\,\cos\alpha.$$ (67)

Here, we have made use of the fact that the point in question also lies on the ecliptic (i.e., $\beta = 0$), as well as the fact that $\alpha_0=\alpha-90^\circ$, where $\alpha$ is the right ascension of the simultaneously rising point on the celestial equator. Here, $\lambda$ is the ecliptic longitude of the point in question, and $L$ the terrestrial latitude of the observation site. Now, $\lambda$ and $\alpha$ satisfy Eq. (64), as well as the above equation. Thus, eliminating $\alpha$ between these two equations, we obtain

$$\cos A = \frac{\sin\lambda\,\sin\epsilon}{\cos L}.$$ (68)

This expression gives the azimuth, $A$, of the ascending point of the ecliptic as a function of its ecliptic longitude, $\lambda$, and the latitude, $L$, of the observation site.

For instance, suppose that we wish to find the azimuth of the point at which the sun rises on the eastern horizon at an observation site of terrestrial latitude $+60^\circ$, on a day on which the sun's ecliptic longitude is 08PI00. It follows from Eq. (68) that $A= \cos^{-1}[\sin (338^\circ)\,\sin (23^\circ 26')/\cos (60^\circ)] = 107^\circ 20'$. We conclude that the sun rises $17^\circ 20'$ to the south of the east compass point on the day in question.